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Zeta and q-Zeta Functions and Associated Series and Integrals


H. M. Srivastava Department of Mathematics University of Victoria Victoria Canada

Junesang Choi Department of Mathematics Dongguk University Gyeongju Republic of Korea

AMSTERDAM • BOSTON • HEIDELBERG • LONDON • NEW YORK • OXFORD PARIS • SAN DIEGO • SAN FRANCISCO • SINGAPORE • SYDNEY • TOKYO

Elsevier 32 Jamestown Road, London NW1 7BY 225 Wyman Street, Waltham, MA 02451, USA First edition 2012 c 2012 Elsevier Inc. All rights reserved. Copyright No part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or any information storage and retrieval system, without permission in writing from the publisher. Details on how to seek permission, further information about the Publisher’s permissions policies and our arrangement with organizations such as the Copyright Clearance Center and the Copyright Licensing Agency, can be found at our website: www.elsevier.com/permissions. This book and the individual contributions contained in it are protected under copyright by the Publisher (other than as may be noted herein). Notices Knowledge and best practice in this field are constantly changing. As new research and experience broaden our understanding, changes in research methods, professional practices, or medical treatment may become necessary. Practitioners and researchers must always rely on their own experience and knowledge in evaluating and using any information, methods, compounds, or experiments described herein. In using such information or methods they should be mindful of their own safety and the safety of others, including parties for whom they have a professional responsibility. To the fullest extent of the law, neither the Publisher nor the authors, contributors, or editors, assume any liability for any injury and/or damage to persons or property as a matter of products liability, negligence or otherwise, or from any use or operation of any methods, products, instructions, or ideas contained in the material herein. British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library Library of Congress Cataloging-in-Publication Data A catalog record for this book is available from the Library of Congress ISBN: 978-0-12-385218-2 For information on all Elsevier publications visit our website at elsevierdirect.com This book has been manufactured using Print On Demand technology. Each copy is produced to order and is limited to black ink. The online version of this book will show color figures where appropriate.

Preface

This book is essentially a thoroughly revised, enlarged and updated version of the authors’ work: Series Associated with the Zeta and Related Functions (Kluwer Academic Publishers, Dordrecht, Boston and London, 2001). It aims at presenting a state-of-the-art account of the theories and applications of the various methods and techniques which are used in dealing with many different families of series associated with the Riemann Zeta function and its numerous generalizations and basic (or q-) extensions. Systematic accounts of only some of these methods and techniques, which are widely scattered in journal articles and book chapters, were included in the abovementioned book. In recent years, there has been an increasing interest in problems involving closedform evaluations of (and representations of the Riemann Zeta function at positive integer arguments as) various families of series associated with the Riemann Zeta function ζ (s), the Hurwitz Zeta function ζ (s, a), and their such extensions and generalizations as (for example) Lerch’s transcendent (or the Hurwitz-Lerch Zeta function) 8(z, s, a). Some of these developments have apparently stemmed from an over twocentury-old theorem of Christian Goldbach (1690−1764), which was stated in a letter dated 1729 from Goldbach to Daniel Bernoulli (1700−1782), from recent rediscoveries of a fairly rapidly convergent series representation for ζ (3), which is actually contained in a 1772 paper by Leonhard Euler (1707−1783), and from another known series representation for ζ (3), which was used by Roger Ape´ ry (1916−1994) in 1978 in his celebrated proof of the irrationality of ζ (3). This revised, enlarged and updated version of our 2001 book is motivated essentially by the fact that the theories and applications of the various methods and techniques used in dealing with many different families of series associated with the Riemann Zeta function, its aforementioned relatives and its many different basic (or q-) extensions are to be found so far only in widely scattered journal articles published during the last decade or so. Thus, our systematic (and unified) presentation of these results on the evaluation and representation of the various families of Zeta and q-Zeta functions is expected to fill a conspicuous gap in the existing books dealing exclusively with these Zeta and q-Zeta functions. The main objective of this revised, enlarged and updated version is to provide a systematic collection of various families of series associated with the Riemann and Hurwitz Zeta functions, as well as with many other higher transcendental functions, which are closely related to these functions (including especially the q-Zeta and related functions). It, therefore, aims at presenting a state-of-the-art account of the theory and applications of many different methods (which are available in the rather scattered

xii

Preface

literature on this subject, especially since the publication of our aforementioned 2001 book) for the derivation of the types of results considered here. In our attempt to make this book as self-contained as possible within the obvious constraints, we include in Chapter 1 (Introduction and Preliminaries) a reasonably detailed account of such useful functions as the Gamma and Beta functions, the Polygamma and related functions, multiple Gamma functions, the Gauss hypergeometric function and its familiar generalization, the Stirling numbers of the first and second kind, the Bernoulli, Euler and Genocchi polynomials and numbers, the Apostol-Bernoulli, the Apostol-Euler and the Apostol-Genocchi polynomials and numbers, as well as some interesting inequalities for the Gamma function and the double Gamma function. In Chapter 2 (The Zeta and Related Functions), we present the definitions and various potentially useful properties (and characteristics) of the Riemann, Hurwitz and Hurwitz-Lerch Zeta functions and their generalizations, the Polylogarithm and related functions and the multiple Zeta functions, together with their analytic continuations. In Chapter 3 (Series Involving Zeta Functions), we begin by providing a brief historical introduction to the main subject of this book. We then describe and illustrate some of the most effective methods of evaluating series associated with the Zeta and related functions. Further developments on the evaluations and (rapidly convergent) series representations of ζ (s) when s ∈ N \ {1} are presented in Chapter 4 (Evaluations and Series Representations), which also deals with various computational results on this subject. Chapter 5 (Determinants of the Laplacians) considers the problem involving computations of the determinants of the Laplacians for the n-dimensional sphere Sn (n ∈ N). It is here in this chapter that we show how fruitfully some of the series evaluations (which are presented in the earlier chapters) can be applied in the solution of the aforementioned problem. In a brand new Chapter 6 (q-Extensions of Some Special Functions and Polynomials), we first introduce the concepts of the basic (or q-) numbers, the basic (or q-) series and the basic (or q-) polynomials. We then proceed to apply these concepts and present a reasonably detailed theory of the various basic (or q-) extensions of the Gamma and Beta functions, the derivatives, antiderivatives and integrals, the binomial theorem, the multiple Gamma functions, the Bernoulli numbers and polynomials, the Euler numbers and polynomials, the Apostol-Bernoulli polynomials, the ApostolEuler polynomials and so on. The last chapter (Chapter 7) contains a wide variety of miscellaneous results dealing with (for example) the analysis of several useful mathematical constants, a variety of Log-Sine integrals involving series associated with the Zeta function and Polylogarithms, applications of the Gamma and Polygamma functions involving convolutions of the Rayleigh functions, evaluations of the Bernoulli and Euler polynomials at rational arguments, and the closed-form summation of several classes of trigonometric series. Each chapter in this book begins with a brief outline summarizing the material presented in the chapter and is then divided into a number of sections. Equations in every section are numbered separately. While referring to an equation in another section of

Preface

xiii

the book, we use numbers like 3.2(18) to represent Equation (18) in Section 3.2 (that is, the second section of Chapter 3). At the close of each chapter, we have provided a set of carefully-selected problems, which are based essentially upon the material presented in the chapter. Many of these problems are taken from recent research publications, and (in all such instances) we have chosen to include the precise references for further investigation (if necessary). Another valuable feature of this book is the extensive and up-to-date bibliography on the subject dealt with in the book. Just as its predecessor (that is, the 2001 edition), this book is written primarily as a reference work for various seemingly diverse groups of research workers and other users of series associated with the Zeta and related functions. In particular, teachers, researchers and postgraduate students in the fields of mathematical and applied sciences will find this book especially useful, not only for its detailed and systematic presentations of the theory and applications of the various methods and techniques used in dealing with many different classes of series associated with the Zeta and related functions, or for its stimulating historical accounts of a large number of problems considered here, but also for its well-classified tables of series (and integrals) and its well-motivated presentation of many sets of closely related problems with their precise bibliographical references (if any).

Acknowledgements

Many persons have contributed rather significantly to this thoroughly revised, enlarged and updated version, just as to its predecessor (that is, the 2001 edition), both directly and indirectly. Contribution of subject matter is duly acknowledged throughout the text and in the bibliography. Indeed, we are greatly indebted to the various authors whose works we have freely consulted and who occasionally provided invaluable references and advice serving for the enrichment of the matter presented in this book. The first-named author wishes to express his deep sense of gratitude to his wife and colleague, Professor Rekha Srivastava, for her cooperation and support throughout the preparation of this thoroughly revised, enlarged and updated version of the 2001 book. The collaboration of the authors on the 2001 book project was conceptualized as long ago as August 1995, and the preparation of a preliminary outline was initiated in December 1997, during the first-named author’s visits to Dongguk University at Gyeongju. The first drafts of some of the chapters in this book were written during several subsequent visits of the first-named author to Dongguk University at Gyeongju. The final drafts of most of the chapters in the 2001 book were prepared during the second-named author’s visit to the University of Victoria from August 1999 to August 2000, while he was on Study Leave from Dongguk University at Gyeongju. The preparation of this thoroughly revised, enlarged and updated version was carried out, in most part, during the period from January 2008 to January 2009, during the secondnamed author’s visit to the University of Victoria, while he was on Study Leave from Dongguk University at Gyeongju for the second time. Our sincere thanks are due to the appropriate authorities of each of these universities, to the Korea Research Foundation (Support for Faculty Research Abroad under its Research Fund Program) and to the Natural Sciences and Engineering Research Council of Canada, for providing financial support and other facilities for the completion of each of the projects leading eventually to the 2001 edition and this thoroughly revised, enlarged and updated version. We especially acknowledge and appreciate the financial support that was received under the Basic Science Research Program through the National Research Foundation of the Republic of Korea. We take this opportunity to express our thanks to the editorial (and technical) staff of the Elsevier Science Publishers B.V. (especially the Publisher, Ms. Lisa Tickner, for Serials and Elsevier Insights) for their continued interest in this book and for their proficient (and impeccable) handling of its publication. Springer’s permission to publish this thoroughly revised, enlarged and updated edition of the 2001 book is also greatly appreciated. Finally, we should like to record our indebtedness to the members of our respective families for their understanding, cooperation and support throughout this project.

xvi

Acknowledgements

The second-named author and his family would, especially, like to express their appreciation for the first-named author and his family’s hospitality and every prudent consideration during their stay in Victoria for over one year, first from August 1999 to August 2000 and then again from January 2008 to January 2009, while the secondnamed author was on Study Leave from Dongguk University at Gyeongju. H. M. Srivastava University of Victoria Canada Junesang Choi Dongguk University Republic of Korea February 2011

1 Introduction and Preliminaries In this introductory chapter, we present the definitions and notations (and some of the important properties and characteristics) of the various special functions, polynomials and numbers, which are potentially useful in the remainder of the book. The special functions considered here include (for example) the Gamma, Beta and related functions, the Polygamma functions, the multiple Gamma functions, the Gaussian hypergeometric function and the generalized hypergeometric function. We also consider the Stirling numbers of the first and second kind, the Bernoulli, Euler and Genocchi polynomials and numbers and the various families of the generalized Bernoulli, Euler and Genocchi polynomials and numbers. Relevant connections of some of these functions with other special functions and polynomials, which are not listed above, are also presented here.

1.1 Gamma and Beta Functions The Gamma Function The origin of the Gamma function can be traced back to two letters from Leonhard Euler (1707–1783) to Christian Goldbach (1690–1764), just as a simple desire to extend factorials to values between the integers. The first letter (dated October 13, 1729) dealt with the interpolation problem, whereas the second letter (dated January 8, 1730) dealt with integration and tied the two together. The Gamma function 0(z) developed by Euler is usually defined by 0(z) :=

Z∞

e−t tz−1 dt

( 0).

(1)

0

We also present here several equivalent forms of the Gamma function 0(z), one by Weierstrass: 0(z) =

∞ z −1 z/k e−γ z Y 1+ e z k

(2)

k=1

z ∈ C \ Z− 0;

Z− 0

:= {0, −1, −2, . . .} ,

Zeta and q-Zeta Functions and Associated Series and Integrals. DOI: 10.1016/B978-0-12-385218-2.00001-3 c 2012 Elsevier Inc. All rights reserved.

2


where γ denotes the Euler-Mascheroni constant defined by ! n X 1 γ := lim − log n ∼ = 0.57721 56649 01532 86060 6512 . . . , n→∞ k

(3)

k=1

and the other by Gauss: (n − 1)! nz 0(z) = lim n→∞ z(z + 1) · · · (z + n − 1) n! (n + 1)z = lim n→∞ z(z + 1) · · · (z + n) n! nz = lim n→∞ z(z + 1) · · · (z + n)

(4)

(z ∈ C \ Z− 0 ), since lim

n→∞

n nz = 1 = lim . n→∞ (n + 1)z z+n

In terms of the Pochhammer symbol (λ)n defined (for λ ∈ C) by ( 1 (n = 0) (λ)n := λ(λ + 1) · · · (λ + n − 1) (n ∈ N := {1, 2, 3, . . .}),

(5)

the definition (4) can easily be written in an equivalent form: 0(z) = lim

n→∞

(n − 1)! nz (z)n

(z ∈ C \ Z− 0 ).

(6)

By taking the reciprocal of (2) and applying the definition (3), we have # " n n Y 1 1 1 z −z/k o = z lim exp 1 + + · · · + − log n z lim 1+ e n→∞ n→∞ 0(z) 2 n k k=1 # " Y n n 1 1 z −z/k o = z lim exp 1 + + · · · + − log n z · 1+ e n→∞ 2 n k k=1 ( ) n Y z −z = z lim n 1+ n→∞ k k=1 "(n−1 )# ) (Y n Y 1 −z z 1+ = z lim 1+ n→∞ k k k=1 k=1 ( ) ∞ Y z 1 −z =z 1+ 1+ , k k k=1

Introduction and Preliminaries

3

which yields Euler’s product form of the Gamma function: ∞ 1Y 1 z z −1 0(z) = 1+ 1+ . z k k

(7)

k=1

When t in (1) is replaced by − log t, (1) is also written in an equivalent form: 0(z) =

Z1 1 z−1 dt log t

( 0).

(8)

0

This representation of the Gamma function as well as the symbol 0 are attributed to Legendre. Integration of (1) by parts easily yields the functional relation: 0(z + 1) = z 0(z),

(9)

so that, obviously, 0(z) =

0(z + n) z(z + 1) · · · (z + n − 1)

(n ∈ N0 := N ∪ {0}),

(10)

which enables us to define 0(z) for −n(n ∈ N0 ) as an analytic function except for z = 0, −1, −2, . . . , −n + 1. Thus, 0(z) can be continued analytically to the whole complex z-plane except for simple poles at z ∈ Z− 0. The representation (2) in conjunction with the well-known product formula: ∞ Y z2 sin π z = π z 1− 2 n

(11)

n=1

also yields the following useful relationship between the Gamma and circular functions: 0(z) 0(1 − z) =

π sin πz

(z 6∈ Z := {0, ±1, ±2, . . .}),

(12)

which incidentally provides an immediate analytic continuation of 0(z) from right to the left half of the complex z-plane. Several special values of 0(x), when x is real, are worthy of note. Indeed, from (1) we note that 0(1) =

Z∞ 0

e−t dt = 1,

(13)

4


and that 0(x) > 0 for all x in the open interval (0, ∞). Thus, (12) with z = ately yields √ 1 0 = π, 2

1 2

immedi-

(14)

which, in view of (1), implies Z∞

√ e−t √ dt = π t

(15)

0

or, equivalently, Z∞

√ π exp(−t )dt = . 2 2

(16)

0

By making use of the relation (10), we obtain √ 1 (2n)! π 0 n+ = ; 2 22n n! √ 22n n! 1 0 −n + = (−1)n π ; (n ∈ N0 ). 2 (2n)!

0(n + 1) = n!;

(17)

The last two results in (17) also require the use of (14). The formulas listed under (17) enable us to compute 0(x) when x is a positive integer and when x is half an odd integer, positive or negative. From (2) or (4), it also follows that 0(z) is a meromorphic function on the whole complex z-plane with simple poles at z = −n(n ∈ N0 ) with their respective residues given by Res 0(z) =

z=−n

(−1)n n!

(n ∈ N0 ).

(18)

Since the function 0(1/w) has simple poles at w = −1, − 12 , − 13 , . . . , which implies that w = 0 is an accumulation point of the poles of 0(1/w), the Gamma function 0(z) has an essential singularity at infinity. Furthermore, it follows immediately from (2) that 1/ 0(z) has no poles, and, therefore, 0(z) is never zero.

Pochhammer’s Symbol and the Factorial Function Since (1)n = n!, the Pochhammer symbol (λ)n defined by (5) may be looked upon as a generalization of the elementary factorial; hence, the symbol (λ)n is also referred to as the shifted factorial.


5

In terms of the Gamma function, we have (cf. Definition (5)) (λ)n =

0(λ + n) 0(λ)

λ ∈ C \ Z− 0 ,

(19)

which can easily be verified. Furthermore, the binomial coefficient may now be expressed as λ λ(λ − 1) · · · (λ − n + 1) (−1)n (−λ)n = = (20) n n! n! or, equivalently, as λ 0(λ + 1) . = n! 0(λ − n + 1) n

(21)

It follows from (20) and (21) that 0(λ + 1) = (−1)n (−λ)n , 0(λ − n + 1) which, for λ = α − 1, yields 0(α − n) (−1)n = 0(α) (1 − α)n

(α 6∈ Z).

(22)

Equations (19) and (22) suggest the definition: (λ)−n =

(−1)n (1 − λ)n

(n ∈ N; λ 6∈ Z).

(23)

Equation (19) also yields (λ)m+n = (λ)m (λ + m)n ,

(24)

which, in conjunction with (23), gives (λ)n−k =

(−1)k (λ)n (1 − λ − n)k

(0 5 k 5 n).

(25)

For λ = 1, we have (n − k)! =

(−1)k n! (−n)k

(0 5 k 5 n),

which may alternatively be written in the form:  k   (−1) n! (0 5 k 5 n), (−n)k = (n − k)!  0 (k > n).

(26)

(27)

6


Multiplication Formulas of Legendre and Gauss In view of the definition (5), it is not difficult to show that (λ)2n = 2

2n

1 1 1 λ λ+ 2 n 2 2 n

(n ∈ N0 ),

(28)

which follows also from Legendre’s duplication formula for the Gamma function, viz √

π 0(2z) = 2

2z−1

1 0(z) 0 z + 2

1 3 z 6= 0, − , −1, − , . . . . 2 2

(29)

For every positive integer m, we have (λ)mn = m

mn

m Y λ+j−1 j=1

m

(m ∈ N; n ∈ N0 ),

(30)

n

which reduces to (28) when m = 2. Starting from (30) with λ = mz, it can be proved that 0(mz) = (2π) 2 (1−m) mmz− 2 1

1

m Y j=1

j−1 0 z+ m 1 2 z 6= 0, − , − , . . . ; m ∈ N , m m

(31)

which is known in the literature as Gauss’s multiplication theorem for the Gamma function.

Stirling’s Formula for n! and its Generalizations For a large positive integer n, it naturally becomes tedious to compute n!. An easy way of computing an approximate value of n! for large positive integer n was initiated by Stirling in 1730 and modified subsequently by De Moivre, who showed that n! ∼

n n √ 2πn e

(n → ∞)

(32)

or, more generally, that 0(x + 1) ∼

x x √ 2π x e

(x → ∞; x ∈ R),

where e is the base of the natural logarithm.

(33)


7

For a complex number z, we have the following asymptotic expansion: n X 1 1 B2k −2n−1 + O z log 0(z) = z − log z − z + log(2π) + 2 2 2k(2k − 1) z2k−1 k=1

(|z| → ∞; | arg(z)| 5 π − (0 < < π); n ∈ N0 ), (34) which, upon taking exponentials, yields an asymptotic formula for the Gamma function: r 139 571 2π 1 1 z −z 0(z) = z e − − 1+ + 2 3 z 12 z 288 z 51840 z 2488320 z4 163879 50043869 + + + O z−7 (35) 209018880 z5 75246796800 z6 (|z| → ∞; | arg(z)| 5 π − (0 < < π)). The asymptotic formula (35), in conjunction with the recurrence relation (9), is useful in computing the numerical values of 0(z) for large real values of z. Some useful consequences of (34) or (35) include the asymptotic expansions: 1 1 log 0(z + α) = z + α − log z − z + log(2π) + O z−1 2 2 (36) (|z| → ∞; | arg(z)| 5 π − ; | arg(z + α)| 5 π − ; 0 < < π), and 0(z + α) (α − β)(α + β − 1) α−β =z 1+ + O z−2 0(z + β) 2z (|z| → ∞; | arg(z)| 5 π − ; | arg(z + α)| 5 π − ; 0 < < π),

(37)

where α and β are bounded complex numbers. Yet another interesting consequence of (35) is the following asymptotic expansion of |0(x + iy)|: √ 1 1 |0(x + iy)| ∼ 2π |y|x− 2 e− 2 π |y| (|x| < ∞; |y| → ∞), (38) where x and y take on real values.

The Beta Function The Beta function B(α, β) is a function of two complex variables α and β, defined by B(α, β) :=

Z1 0

tα−1 (1 − t)β−1 dt = B(β, α)

( 0; 0)

(39)

8


or, equivalently, by Zπ/2 B(α, β) = 2 (sin θ)2α−1 (cos θ)2β−1 dθ

( 0; 0),

(40)

0

which follows from (39) on setting t = sin 2 θ. The integrals in (39) and (1) are known as the Eulerian integrals of the first and second kind, respectively. Putting t = u/(1 + u) in (39), we obtain the following representation of B(α, β) as an infinite integral: B(α, β) =

Z∞ 0

uα−1 du ( 0; 0). (1 + u)α+β

(41)

The Beta function is closely related to the Gamma function; in fact, we have B(α, β) =

0(α) 0(β) 0(α + β)

α, β 6∈ Z− 0 ,

(42)

which not only confirms the symmetry property in (39), but also continues the Beta function analytically for all complex values of α and β, except when α, β ∈ Z− 0 . Thus, we may write

B(α, β) =

 Z1     α−1  (1 − t)β−1 dt  t 0    0(α) 0(β)    0(α + β)

( 0; 0) (43)

c; 0; 0). The following functional equations for the Beta function can be deduced easily from (39) and (42): B(α, β + 1) =

β β B(α + 1, β) = B(α, β); α α+β

B(α, β) B(α + β, γ ) = B(β, γ ) B(β + γ , α) = B(γ , α) B(α + γ , β); B(α, β) B(α + β, γ ) B(α + β + γ , δ) =

0(α) 0(β) 0(γ ) 0(δ) , 0(α + β + γ + δ)

(51) (52) (53)

10


or, more generally,   k X 0(α1 ) · · · 0(αn+1 ) (n ∈ N); B αj , αk+1  = 0(α1 + · · · + αn+1 ) j=1 k=1 n+m−1 n+m−1 1 =n (n, m ∈ N). =m B(n, m) n−1 m−1 n Y

(54)

(55)

The Incomplete Gamma Functions The incomplete Gamma function γ (z, α) and its complement 0(z, α) (also known as Prym’s function) are defined by γ (z, α) :=

Zα

tz−1 e−t dt

( 0; | arg(α)| < π),

(56)

tz−1 e−t dt

(| arg(α)| < π),

(57)

0

0(z, α) :=

Z∞ α

so that γ (z, α) + 0(z, α) = 0(z).

(58)

For fixed α, 0(z, α) is an entire (integral) function of z, whereas γ (z, α) is a meromorphic function of z, with simple poles at the points z ∈ Z− 0. The following recursion formulas are worthy of note: γ (z + 1, α) = z γ(z, α) − α z e−α ,

(59)

0(z + 1, α) = z 0(z, α) + α e

(60)

z −α

.

The Incomplete Beta Functions The incomplete Beta function Bx (α, β) is defined by Bx (α, β) :=

Zx

tα−1 (1 − t)β−1 dt

( 0).

(61)

0

For the associated function: Ix (α, β) =

Bx (α, β) , B(α, β)

(62)


11

we note here the following properties that are easily verifiable: Ix (α, β) = 1 − I1−x (β, α), n X n j Ix (k, n − k + 1) = x (1 − x)n−j j

(63) (1 ≤ k ≤ n),

(64)

j=k

Ix (α, β) = x Ix (α − 1, β) + (1 − x) Ix (α, β − 1), (α + β − αx) Ix (α, β) = α(1 − x) Ix (α + 1, β − 1) + βIx (α, β + 1), (α + β) Ix (α, β) = α Ix (α + 1, β) + βIx (α, β + 1).

(65) (66) (67)

The Error Functions The error function erf(z), also known as the probability integral 8(z), is defined for any complex z by 2 erf(z) := √ π

Zz

exp(−t2 ) dt = 8(z),

(68)

0

and its complement by 2 erfc(z) := 1 − erf(z) = √ π

Z∞

exp(−t2 ) dt.

(69)

z

Clearly, we have erf(0) = 0

and

erfc(0) = 1,

(70)

and, in view of the well-known result (16), we also have erf(∞) = 1

and

erfc(∞) = 0.

(71)

The following alternative notations: Erf(z) =

√ π erf(z) and 2

Erfc(z) =

√ π erfc(z) 2

(72)

are sometimes used for the error functions. Many authors use the notations Erf(z) and ˆ Erf(z) for the error functions erf(z) and Erf(z), respectively, defined by (68) and (72), ˆ and the notations Erfc(z) and Erfc(z) for their complements. In terms of the incomplete Gamma functions, it is easily verified that 1 1 2 1 1 2 erf(z) = √ γ ,z and erfc(z) = √ 0 , z . (73) 2 2 π π

12


The Bohr-Mollerup Theorem We have already observed that Euler’s definition (1) and its such consequences as (9) and (13) enable us to compute all the real values of the Gamma function from the knowledge merely of its values in the interval (0, 1), as noted in conjunction with (35). Since the solution to the interpolation problem is not determined uniquely, it makes sense to add more conditions to the problem. After various trials to find those conditions to guarantee the uniqueness of the Gamma function, in 1922, Bohr and Mollerup were able to show the remarkable fact that the Gamma function is the only function that satisfies the recurrence relationship and is logarithmically convex. The original proof was simplified, several years later, by Emil Artin, and the theorem, together with Artin’s method of proof, now constitute the Bohr-Mollerup-Artin theorem: Theorem 1.1 Let f : R+ → R+ satisfy each of the following properties: (a) log f (x) is a convex function; (b) f (x + 1) = x f (x) for all x ∈ R+ ; (c) f (1) = 1.

Then f (x) = 0(x) for all x ∈ R+ . Instead of giving here the proof of Theorem 1.1 (see Conway [339, p. 179] and Artin [72, p. 14]), we simply state the necessary and sufficient condition for the logarithmic convexity of a given function. Theorem 1.2 Let f : [a, b] → R, and suppose that f (x) > 0 for all x ∈ [a, b] and that f has a continuous second derivative f 00 (x) for x ∈ [a, b]. Then f is logarithmically convex, if and only if 2 f 00 (x) f (x) − f 0 (x) = 0 (x ∈ [a, b]). Remmert [973] admires the following Wielandt’s uniqueness theorem for the Gamma function: It is hardly known that there is also an elegant function theoretic characterization of 0(z). This uniqueness theorem was discovered by Helmut Wielandt in 1939. A function theorist ought to be as much fascinated by Wielandt’s complex-analytic characterization as by the Bohr-Mollerup theorem. For further comment and applications for Wielandt’s theorem, see [675, pp. 47–49], [973], and [1065]. Here, without proof, we present Theorem 1.3 (Wielandt’s Theorem) Let F(z) be an analytic function in the right half plane A := {z ∈ C | 0} having the following two properties: (a) F(z + 1) = z F(z) for all z ∈ A; (b) F(z) is bounded in the strip S := {z ∈ C | 1 5 0).

(25)


Z∞

3 γ = +2 2

0

γ=

17

cos x − 1 1 + 2 2(1 + x) x

dx . x

(26)

Zπ/2

h i dx . 1 − sec 2 x cos (tan x) tan x

(27)

0

2 γ = − ln p − π

Z∞

sin (p x) ln x

dx x

(p > 0).

(28)

0

2 γ = 1 − ln(2p) − pπ

Z∞

sin 2 (px) x2

ln x dx

(p > 0).

(29)

0

q

1 q γ = 1+ ln p p−q p

2 + π(p − q)

Z∞

cos (px) − cos (qx) x2

ln x dx

0

(30)

(p > 0; q > 0; p 6= q). Z∞

1 γ = +2 2

0

γ=

Z1 x− 0

γ =−

Z∞

x dx . 2 2π 1+x e x −1

1 1 − log x

dx . x log x Z1

−x

0

(32)

1 dx. log log x

(33)

π x sin π xu − x dudx. 2 sin π u

(34)

log x dx = −

e

(31)

0 1

γ = log 2 − π

Z1 Z 2 tan 0 0

γ=

1 +2 2

Z∞

sin (tan−1 x) dx. √ e2π x − 1 1 + x2

0

γ = log 2 − 2

Z∞ 0

γ = 1+

Z∞ 0

sin (tan−1 x) dx. √ e2π x + 1 1 + x2

sin x cos x − x

log x dx. x

(35)

(36)

(37)

18


γ=

Z∞

dx 1 − cos x . 2 x

(38)

0

1 B2 B4 B2n γ= + + + ··· + − (2n + 1)! 2 2 4 2n

Z∞

Q2n+1 (x) dx, x2n+2

(39)

1

where the functions Qn (x) are defined by  1  (n = 1; 0 < x < 1), x − 2 Qn (x) :=   1 B (x − [x]) (n ∈ N \ {1}; 0 5 x < ∞), n n! Bn := Bn (0) and Bn (x) being the Bernoulli numbers and polynomials, respectively (see [1094, Section 1.6]). As observed by Knopp [676] by an explicit example with n = 3 in (2.61), the approximate value of γ can easily be calculated with much greater accuracy than before (and, theoretically, to any degree of accuracy whatever) by means of the formula (39).

Further Integral Representations for γ Very recently, Choi and Srivastava [302] presented several further integral representations for γ by making use of some formulas in the previous subsection and other known formulas for log 0(z), ψ(z) (Section 1.3) and the Hurwitz (or generalized) Zeta function ζ (s, a) (Section 2.2) or the Riemann Zeta function ζ (s) (Section 2.3) in conjunction with the residue calculus. Here, we choose to record some of them: We begin by recalling an integral formula for log z (see [1225, p. 248]). The following integral formula holds true for log z: Z∞

−t

e

−t z

−e

dt = t

0

Z1 dt = log z tz−1 − 1 log t

0 ,

(40)

0

where the log z is an appropriate branch of the multiple-valued function log z, such as log z = ln |z| + i arg z

|z| > 0; α < arg(z) < α + 2π

for some real α ∈ R with possibly −π 5 α 5 − π2 .


19

If (40) is used in the formulas (25), (28), (29) and (30) and tan x is replaced by x in (27), the following integral formulas for γ are obtained: Each of the following integral representations holds true for γ : γ=

Z∞ 0

2 γ =− π

dx 2 arccot x − e−x , π x

(41)

Z∞ π dx 2 e−x − e−p x + ln x sin (p x) x

(p > 0),

(42)

0

2 γ = 1− pπ

Z∞ h i pπ sin 2 (px) dx −x 2 e − exp −(2p) x + ln x x x

(p > 0),

(43)

0

2 γ = 1+ π(p − q)

Z∞

−x

e

" π # ! qq 2 cos (px) − cos (qx) dx − exp − p x + ln x p x x

0

(p > 0; q > 0)

(44)

and Z∞ h i dx γ= cos 2 (arctan x) − cos x . x

(45)

0

If x is replaced by xp in (10), (12), (21), (23) and (26), the following mildly more general formulas for γ are obtained. Each of the following integral representations holds true for γ : γ =p

Z∞ 0

γ =p

Z∞ 0

γ = 1+p

dx 1 p − exp −x 1 + xp x dx 1 p − cos x 1 + xp x

Z∞ 0

γ = 1+p

Z∞ 0

(p > 0),

(p > 0),

1 exp (−x p ) − 1 + 1 + xp xp 1 sin (x p ) − 1 + xp xp

(46)

dx x

dx x

(47)

(p > 0),

(p > 0)

(48)

(49)

20


and 3 γ = + 2p 2

Z∞ 0

1 cos (x p ) − 1 dx + p 2p 2(1 + x ) x x

(p > 0).

(50)

It is noted that the case p = 2 of (46) would obviously reduce to (13). A class of vanishing integrals is provided just below. The following vanishing integral formula holds true: Z∞ 0

Z∞ q−1 1 dx x − xp−1 1 − = dx = 0 1 + x p 1 + xq x (1 + x p ) (1 + xq )

(51)

0

(p > 0; q > 0). Proof. The integral (51) is separated into two parts as follows:  1 ∞ Z∞ q−1 Z Z q−1 x − xp−1 x − xp−1   dx = + dx, (1 + x p ) (1 + xq ) (1 + x p ) (1 + xq ) 0

0

1

which, upon replacing x by 1/x in the second integral, is seen to vanish to 0.

Applying Eq. (51) to Eqs. (46)–(50), Choi and Srivastava [302] derived much more general integral representations for γ , which are recorded here. γ =p

Z∞

dx 1 p − exp −x 1 + xq x

0

γ =p

Z∞ 0

γ = 1+p

dx 1 p − cos x 1 + xq x

Z∞ 0

γ = 1+p

Z∞ 0

3 γ = + 2p 2

0

γ=

pq q−p

0

(p > 0; q > 0),

1 exp (−x p ) − 1 dx + 1 + xq xp x 1 sin (x p ) dx − 1 + xq xp x

Z∞

Z∞

(p > 0; q > 0),

(p > 0; q > 0),

(p > 0; q > 0),

cos (x p ) − 1 1 dx + q 2p 2(1 + x ) x x

dx exp −xq − exp −x p x

(p > 0; q > 0),

(p > 0; q > 0; p 6= q),

(52)

(53)

(54)

(55)

(56)

(57)


pq γ= q−p

Z∞

21

dx cos xq − cos x p x

(p > 0; q > 0; p 6= q),

(58)

dx x

(59)

0

pq γ = 1+ q−p 3 2pq γ= + 2 q−p

Z∞

sin (xq ) sin (x p ) − xq xp

(p > 0; q > 0; p 6= q),

0

Z∞

cos (x p ) − 1 cos (xq ) − 1 − x2p x2q

dx x

(60)

0

(p > 0; q > 0; p 6= q), γ=

pq q−p

Z∞

cos (x p ) − exp(−x p )

dx x

(p > 0; q > 0; p 6= q),

(61)

0

p pq γ= + p−q p−q

Z∞

exp(−x p ) −

sin (x p ) xp

dx x

(p > 0; q > 0; p 6= q),

0

(62) 3p pq γ= + 2(p − q) p − q

Z∞

exp(−x p ) +

2 [cos (x p ) − 1] x2q

dx x

(63)

0

(p > 0; q > 0; p 6= q), p pq γ= + p−q p−q

Z∞

cos (x p ) −

sin (x p ) xp

dx x

(p > 0; q > 0; p 6= q), (64)

0

3p pq γ= + 2(p − q) p − q

Z∞

2 [cos (x p ) − 1] cos (x ) + x2q p

dx x

0

(65)

(p > 0; q > 0; p 6= q), 3p − 2q pq γ= + 2(p − q) p − q

Z∞

sin (x p ) 2 [cos (x p ) − 1] + xp x2q

dx x

0

(66)

(p > 0; q > 0; p 6= q), p pq γ= + p−q p−q

Z∞

exp (−x p ) − 1 dx p + exp(−x ) p x x

0

(p > 0; q > 0; p 6= q),

(67)

22


pq p + γ= p−q p−q

Z∞

dx exp (−x p ) − 1 p + cos x xp x

(68)

0

(p > 0; q > 0; p 6= q), pq γ = 1+ p−q

Z∞

exp (−x p ) − 1 sin (x p ) + xp xp

dx x

(69)

0

(p > 0; q > 0; p 6= q), pq 2p − 3q + γ= 2(p − q) p − q

Z∞

exp (−x p ) − 1 2 [1 − cos (x p )] + xp x2p

dx x

0

(70)

(p > 0; q > 0; p 6= q). It is noted that the integral formula (57) is recorded in [505, p. 364, Entry 3.476-2] and many (if not all) of the integral formulas in the previous subsection can be seen to be special cases of the corresponding integral formulas asserted in this subsection.

From an Application of the Residue Calculus Consider a function f (z) given by f (z) =

1 z

1 n − ei z 1 + zn

(n ∈ N).

Since lim f (z) =

z→0

(n = 1)

  −1 − i 

(n ∈ N \ {1}),

0

the function f (z) has a removable singularity at z = 0 and simple poles at (2k + 1)πi z = exp n

(k = 0, 1, . . . , n − 1).

We now consider a counterclockwise-oriented simple closed contour: C := Cδ ∪ L1 ∪ CR ∪ L2

(0 < δ < 1 < R),

where Cδ : z = δ eiθ

π θ varies from to 0 , 2n


23

L1 a line segment from δ to R on the positive real axis, π θ varies from 0 to 2n

CR : z = R eiθ and L2 : z = x exp

iπ 2n

(x varies from R to δ) ,

π that is, a line segment on the half-line beginning at the origin with the argument 2n . Since f (z) is analytic throughout the domain interior to and on the closed contour C, it follows from the Cauchy-Goursat theorem that

  Z Z Z Z    + + +  f (z) dz = 0, L1

Cδ

CR

L2

which, upon taking the limits as δ → 0+

and R → ∞

and equating the real and imaginary parts of the last resulting equation, yields the following two interesting integral identities: Z∞ 0

Z∞ dx dx 1 1 n n = − cos x − exp −x 1 + xn x x 1 + x2n

(n ∈ N)

(71)

0

and Z∞ 0

xn−1 dx = 1 + x2n

Z∞

π sin (xn ) dx = x 2n

(n ∈ N).

(72)

0

It is noted that the integral identity (71) is a special case of (52) or (53). Moreover, (72) can be evaluated, as above, by applying the residue calculus to another function f(z) =

exp (izn ) z

(n ∈ N)

and a counterclockwise-oriented simple closed contour C := Cδ ∪ L1 ∪ CR ∪ L2

(0 < δ < R),

24


where Cδ and L1 are the same as above,

CR : z = R eiθ

θ varies from 0 to

π n

and L2 : z = x exp

iπ n

(x varies from R to δ).

We conclude this section by remarking that more integral representations for γ can be obtained by applying the same techniques employed here (see [302]) or other methods (if any) to some other known formulas that have not been used (see [572]).

1.3 Polygamma Functions The Psi (or Digamma) Function The Psi (or Digamma) function ψ(z) defined by d 0 0 (z) ψ(z) := {log 0(z)} = dz 0(z)

or

log 0(z) =

Zz

ψ(t)dt

(1)

1

possesses the following properties: ψ(z) = lim

n→∞

ψ(z) = −γ −

log n −

n X k=0

1 + z

∞ X n=1

= −γ + (z − 1)

! 1 ; z+k

z n(z + n)

∞ X n=0

1 , (n + 1)(z + n)

(2)

(3)

where γ is the Euler-Mascheroni constant defined by 1.1(3) (or 1.2(2)). These results clearly imply that ψ(z) is meromorphic (that is, analytic everywhere in the bounded complex z–plane, except for poles) with simple poles at z = −n(n ∈ N0 ) with its residue −1. Also we have ψ(1) = −γ ,

(4)

which follows at once from (3). It is noted that, very recently, Bagby [83] proved (4) in another way.


25

The following additional properties of ψ(z) can be deduced from known results for 0(z) : log

∞ X 0(z + 1) 1 1 = log z = −γ + − log 1 + ; 0(z) n+1 n+z n=0 ∞ X 1 1 − log 1 + , ψ(z) = log z − n+z n+z

(5)

(6)

n=0

which follows from (3) and (5); ψ(z + n) = ψ(z) +

n X k=1

1 z+k−1

(n ∈ N);

(7)

1 ψ(z) − ψ(−z) = −π cot π z − ; z 1 ψ(1 + z) − ψ(1 − z) = − πcot πz; z ψ(z) − ψ(1 − z) = −π cot π z; 1 1 +z −ψ − z = π tan π z; ψ 2 2 m−1 1 X k ψ(mz) = log m + ψ z+ (m ∈ N). m m

(8) (9) (10) (11) (12)

k=0

Integral Representations for ψ(z) Expanding (1 − t)−1 into a series, integrating term by term and using (3), we get Z1 ψ(z) = −γ + 1 − tz−1 (1 − t)−1 dt

( 0),

(13)

0

which, upon replacing t by e−t , yields ψ(z) = −γ +

Z∞

−1 e−t − e−tz 1 − e−t dt

( 0).

(14)

0

Making use of (10), it follows from (13) and (14) that ψ(z) = −γ − πcot πz +

Z1 0

1 − t−z (1 − t)−1 dt

( 0; x ∈ N0 )

= n! (−α)−n Ln(x−n) (α) x = (−α)−n n! 1 F1 (−n; x − n + 1; α) . n

(73)

In an alternative notation, we have (see Szego¨ [1141, p. 35]) (x − n + 1)n √ 1 F1 (−n; x − n + 1; α) n! 1 √ x = α − 2 n n! 1 F1 (−n; x − n + 1; α) . n 1

pn (x) = α − 2 n

(74)

76


1.6 Stirling Numbers of the First and Second Kind Stirling Numbers of the First Kind The Stirling numbers s(n, k) of the first kind are defined by the generating functions: z(z − 1) · · · (z − n + 1) =

n X

s(n, k) zk

(1)

zn n!

(2)

k=0

and {log(1 + z)}k = k!

∞ X

s(n, k)

n=k

(|z| < 1).

We have the following recurrence relations satisfied by s(n, k): s(n + 1, k) = s(n, k − 1) − n s(n, k)

(n = k = 1);

n−j X k n s(n, k) = s(n − l, j) s(l, k − j) j l

(n = k = j).

(3)

(4)

l=k−j

From the definition (1) of s(n, k), the Pochhammer symbol in 1.1(5) can be written in the form: (z)n = z(z + 1) · · · (z + n − 1) =

n X

(−1)n+k s(n, k) zk ,

(5)

k=0

where (−1)n+k s(n, k) denotes the number of permutations of n symbols, which has exactly k cycles. It is not difficult to see also that ( 1 (n = 0) s(n, n) = 1, s(n, 0) = 0 (n ∈ N), (6) n n+1 s(n, 1) = (−1) (n − 1)!, s(n, n − 1) = − 2 and n X

s(n, k) = 0 (n ∈ N\{1});

k=1 n X j=k

n X k=0

s(n + 1, j + 1) n

j−k

= s(n, k).

(−1)n+k s(n, k) = n!; (7)


77

Yet another recursion formula for s(n, k) is given by (see Shen [1024]): (k − 1) s(n, k) = −

k−1 X

(m)

s(n, k − m) Hn−1 ,

(8)

m=1 (s)

where Hn is the generalized harmonic numbers of order s, defined by Hn(s) :=

n X 1 ks

(n ∈ N; s ∈ C)

(9)

k=1

(1)

and Hn := Hn (n ∈ N) is the harmonic numbers. (0) (m) Here, for (8), we assume H0 := 1 and H0 := 0 (m ∈ N). It readily follows from the recursion formula (8) that s(n, 2) = (−1)n (n − 1)! Hn−1 ; i (n − 1)! h (2) s(n, 3) = (−1)n+1 (Hn−1 )2 − Hn−1 ; 2 h i (n − 1)! (2) (3) s(n, 4) = (−1)n (10) (Hn−1 )3 − 3Hn−1 Hn−1 + 2Hn−1 ; 6 (n − 1)! s(n, 5) = (−1)n+1 24 (3) (2) 2 (4) 4 2 (2) · (Hn−1 ) + 8 Hn−1 Hn−1 − 6 (Hn−1 ) Hn−1 + 3 Hn−1 − 6 Hn−1 . In view of (5), by logarithmically differentiating 1.1(19) and then using 1.2(7), we obtain n

X d {(z)n } = (−1)n+k k s(n, k) zk−1 dz k=1 ! n X 1 = (z)n , z+k−1

(11)

k=1

which, upon employing Leibniz’s rule for differentiation, yields a more general formula: n X dj+1 {(z) } = (−1)n+k+j+1 (−k)j+1 s(n, k) zk−j−1 n dzj+1 k=j+1 ! j n X X j! 1 dj−l l {(z)n } = (−1) (j − l)! (z + k − 1)l+1 dzj−l l=0

(j ∈ N0 ; n ∈ N).

k=1

(12)

78


For j = 1 and j = 2, (12) immediately yields n X

(−1)n+k k(k − 1) s(n, k) zk−2

k=2

 = (z)n 

n X k=1

1 z+k−1

!2 −

n X k=1

(13)

 1  (z + k − 1)2

and n X

(−1)n+k k(k − 1)(k − 2) s(n, k) zk−3

k=3

 = (z)n 

n X k=1

+

1 z+k−1

n X k=1

!3 −3

n X k=1

1 z+k−1

!

n X k=1

1 (z + k − 1)2

! (14)

# 1 . (z + k − 1)3

Stirling Numbers of the Second Kind The Stirling numbers S(n, k) of the second kind are defined by the generating functions: n

z =

n X

S(n, k) z(z − 1) · · · (z − k + 1),

(15)

k=0

(ez − 1)k = k!

∞ X n=k

S(n, k)

zn , n!

(16)

and (1 − z)−1 (1 − 2z)−1 · · · (1 − kz)−1 =

∞ X

S(n, k) zn−k

(|z| < k−1 ),

(17)

n=k

where S(n, k) denotes the number of ways of partitioning a set of n elements into k nonempty subsets. It is not difficult to see also that ( 1 (n = 0) S(n, 0) = 0 (n ∈ N), n S(n, 1) = S(n, n) = 1, and S(n, n − 1) = . (18) 2


79

The recurrence relations for S(n, k) are given by (n = k = 1)

S(n + 1, k) = k S(n, k) + S(n, k − 1)

(19)

and n−j X k n S(n, k) = S(n − i, j) S(i, k − j) j i

(n = k = j).

(20)

i=k−j

The numbers S(n, k) can be expressed in an explicit form:

S(n, k) =

k k n 1 X j . (−1)k−j k! j

(21)

j=0

Some additional properties of S(n, k) are recalled here as follows: n X k=0 n X

(−1)n−k k! S(n, k) = 1;

(22)

S(j − 1, k − 1) kn−j = S(n, k);

(23)

j=k

S(n, k) =

n−k X

(−1)

j

j=0 n X

n−1+j n−k+j

2n − k S(n − k + j, j); n−k−j

S(j, k) S(n, j) = δkn ,

(24)

(25)

j=k

where δmn denotes the Kronecker delta defined by ( δmn =

0 (m 6= n), 1 (m = n).

( and δm =

0 (m 6= 0), 1 (m = 0).

(26)

Relationships Among Stirling Numbers of the First and Second Kind and Bernoulli Numbers Akiyama and Tanigawa [17] presented, to evaluate multiple zeta values at nonpositive integers, the following identities: n X 1 n 1 S(n, `) s(`, k) = Bn−k + δn−k−1 ` n k `=k

(n, k ∈ N; n = k),

(27)

80


where Bk is the Bernoulli number given in 1.6. n X n

n S(n − 1, k − 1) (n, k ∈ N). k `=0 n X S(n, `) s(` + 1, k) Bn+1−k n + 1 = (n, k ∈ N; n = k − 1), `+1 n+1 k `

Bn−` S(`, k) =

(28)

(29)

`=k−1

which, upon setting k = 1 and using (21), yields

Bn =

n X (−1)` `! S(n, `) `+1 `=0

=

n X `=0

` ` n 1 X (−1) j j `+1 j

(n ∈ N0 ) ,

(30)

j=0

or, equivalently, n X

s(n, `) B` =

`=0

(−1)n n! n+1

(n ∈ N0 ) .

(31)

We also recall some known formulas (see, e.g., [982]): max{k, Xj}+1

s(`, j) S(k, `) = δjk .

(32)

s(k, `) S(`, j) = δjk .

(33)

`=0 max{k, Xj}+1 `=0

s(n, i) =

n X k X

s(n, k) s(k, j) S(j, i).

(34)

S(n, k) S(k, j) s(j, i).

(35)

k=i j=0

S(n, i) =

n X k X k=i j=0 n−m X

k+n−1 2n − m S(n, m) = (−1) s(k − m + n, k). k+n−m n−k−m k=0 n−m X 2n − m k k+n−1 s(n, m) = (−1) S(k − m + n, k). k+n−m n−k−m k=0

k

(36)

(37)


81

1.7 Bernoulli, Euler and Genocchi Polynomials and Numbers Bernoulli Polynomials and Numbers The Bernoulli polynomials Bn (x) are defined by the generating function: ∞

X z exz zn = B (x) n ez − 1 n!

(|z| < 2π).

(1)

n=0

The numbers Bn := Bn (0) are called the Bernoulli numbers generated by ∞

X z zn = Bn z e −1 n!

(|z| < 2π).

(2)

n=0

It easily follows from (1) and (2) that Bn (x) =

n X n

k

k=0

Bk xn−k .

(3)

The Bernoulli polynomials Bn (x) satisfy the difference equation: Bn (x + 1) − Bn (x) = n xn−1

(n ∈ N0 ),

(4)

which yields Bn (0) = Bn (1)

(n ∈ N \ {1}).

(5)

Setting x = 1 in (3), in view of (5), we have Bn =

n X n k=0

k

Bk ,

(6)

which gives a recursion formula for computing Bernoulli numbers. The first few of the Bernoulli numbers are already listed with the Euler-Maclaurin summation formula 1.4(68), and (for the sake of completeness) we have the following list:

B0 = 1,

1 B2 = , 6

691 , 2730

7 B14 = , 6

854513 , 138

B24 = −

B12 = − B22 =

1 B1 = − , 2

B4 = −

B16 = −

1 , 30

3617 , 510

236364091 , 2730

B6 = B18 =

B26 =

1 , 42

1 , 30

B10 =

B20 = −

174611 , 330

B8 = −

43867 , 798

8553103 , . . . , B2n+1 = 0 6

5 , 66

(n ∈ N).

(7)

82


The first few of the Bernoulli polynomials are given below: 1 1 B1 (x) = x − , B2 (x) = x2 − x + , 2 6 3 2 1 1 3 4 3 2 B3 (x) = x − x + x, B4 (x) = x − 2x + x − , 2 2 30 5 5 1 B5 (x) = x5 − x4 + x3 − x, 2 3 6 1 1 5 B6 (x) = x6 − 3x5 + x4 − x2 + , 2 2 42 7 6 7 5 7 3 1 7 B7 (x) = x − x + x − x + x, . . . . 2 2 6 6 B0 (x) = 1,

(8)

It is not difficult to derive the following identities for the Bernoulli polynomials: B0n (x) = n Bn−1 (x) (n ∈ N); Bn (1 − x) = (−1)n Bn (x) (n ∈ N0 );

(9) (10)

(−1)n Bn (−x) = Bn (x) + n xn−1

(11)

(n ∈ N0 ).

Multiplication formula:

Bn (mx) = m

n−1

m−1 X

Bn

k=0

k x+ m

(n ∈ N0 , m ∈ N).

(12)

Addition formula: Bn (x + y) =

n X n k=0

k

Bk (x) yn−k

(n ∈ N0 ).

(13)

Integral formulas: Zy x

Bn (t) dt =

Bn+1 (y) − Bn+1 (x) ; n+1

(14)

Zx+1 Bn (t) dt = xn ;

(15)

x

Z1 0

Bn (t) Bm (t) dt = (−1)n−1

m! n! Bm+n (m + n)!

(m, n ∈ N).

(16)


83

It follows from (14) and (15) that the finite sum of powers is expressed as Bernoulli polynomials and numbers: m X

kn =

k=1

Bn+1 (m + 1) − Bn+1 n+1

(m, n ∈ N).

(17)

By writing 2n for n in (3), we can deduce that B2n (x) + n x

2n−1

=

n X 2n

2k

k=0

B2k x2n−2k ,

which, upon integrating from 0 to 12 , yields n X k=0

1 22k B2k = (2k)!(2n − 2k + 1)! (2n)!

(n ∈ N0 ),

(18)

where we have applied (14). It is readily shown that B2n

1 = 21−2n − 1 B2n 2

and B2n+1

1 = 0 (n ∈ N). 2

(19)

The Generalized Bernoulli Polynomials and Numbers (α)

The generalized Bernoulli polynomials Bn (x) of degree n in x are defined by the generating function:

z z e −1

α

exz =

∞ X

B(α) n (x)

n=0

zn n!

|z| < 2π; 1α := 1

(20)

for arbitrary (real or complex) parameter α. Clearly, we have n (α) B(α) n (x) = (−1) Bn (α − x),

(21)

so that n (α) n (α) B(α) n (α) = (−1) Bn (0) =: (−1) Bn ,

(22) (α)

in terms of the generalized Bernoulli numbers Bn defined by the generating function:

z z e −1

α =

∞ X n=0

B(α) n

zn n!

(|z| < 2π ; 1α := 1).

(23)

84


It is easily observed that (1) B(1) n (x) = Bn (x) and Bn = Bn

(n ∈ N0 ).

(24)

From the generating function (20), it is fairly straightforward to deduce the addition theorem: B(α+β) (x + y) = n

n X n

k

k=0

(β)

(α)

Bk (x) Bn−k (y),

(25)

which, for x = β = 0, corresponds to the elegant representation: B(α) n (x) =

n X n k=0

k

(α)

Bk xn−k

(26)

for the generalized Bernoulli polynomials as a finite sum of the generalized Bernoulli numbers. Srivastava et al. [1101, p. 442, Eqs. (4.4) and (4.5)] gave two new classes of addition theorems for the generalized Bernoulli polynomials: ) B(α+λγ (x + γ y) = n

n X γ + n n (α−λk) (λk+λγ ) B (x − ky) Bn−k (ky + γ y) γ +k k k

( 0);

k=0

(27) Bn(α+β+n+1) (x + y + n) =

n X k=0

n (α+k+1) (β+n−k+1) B (x + k) Bn−k (y + n − k). (28) k k

Srivastava and Todorov [1110, p. 510, Eq. (3)] proved the following explicit formula for the generalized Bernoulli polynomials: B(α) n (x) =

k n X n α+k−1 k! X j k 2k (−1) j (x + j)n−k (2k)! j k k j=0

k=0

· 2 F1 [k − n, k − α; 2k + 1; j/(x + j)],

(29)

in terms of the Gaussian hypergeometric function (see Section 1.5). They also applied the representation (29) to derive certain interesting special cases considered earlier by Gould [499] and Todorov [1153]. Indeed, by the Chu-Vandermonde theorem 1.5(9), we have 2 F1 (−N, b; c; 1) =

c − b + N − 1 c + N − 1 −1 N N

(N ∈ N0 ),


85

which, for N = n − k, b = k − α and c = 2k + 1, readily yields 2 F1 (k − n, k − α; 2k + 1; 1) =

α + n (n − k)! (2k)! (n + k)! n−k

(0 ≤ k ≤ n).

(30)

In view of (30), the special case of the Srivastava-Todorov formula (29), when x = 0, gives the following representation for the generalized Bernoulli numbers: B(α) n

=

n X α+n α+k−1 n−k

k=0

k

k X n! j k n+k j (−1) (n + k)! j

(31)

j=0

or, equivalently, B(α) n =

n X

(−1)k

k=0

α+n α+k−1 n! ∆k 0n+k , n−k k (n + k)!

(32)

where, for convenience, ∆ a =∆ x k r

k r

= x=a

k X

(−1)

j=0

k−j

k (a + j)r , j

(33)

∆ being the difference operator defined by (cf. Comtet [337, p. 13 et seq.]) ∆ f (x) = f (x + 1) − f (x),

(34)

so that, in general, ∆ f (x) = k

k X

(−1)

k−j

j=0

k f (x + j). j

(35)

Alternatively, since (Comtet [337, p. 204, Theorem A]; see also Eq. 1.5(20)) S(n, k) =

1 k n ∆ 0 , k!

(36)

where S(n, k) denotes the Stirling number of the second kind defined by 1.6(14), that is, by n

z =

n X z k=0

k

k! S(n, k),

(37)

86


the representation (31) or (32) can be written also as (Todorov [1153, p. 665, Eq. (3)]) B(α) n

=

n X k=0

α + n α + k − 1 n + k −1 (−1) S(n + k, k). n−k k k k

(38)

Formula (38) provides an interesting generalization of the following known result for the Bernoulli numbers Bn : Bn =

n X

(−1)k

k=0

n + 1 n + k −1 ∆k 0n+k , k! k+1 k

(39)

which was considered, for example, by Gould [499, p. 49, Eq. (17)].

Euler Polynomials and Numbers The Euler polynomials En (x) and the Euler numbers En are defined by the following generating functions: ∞

X 2exz zn = E (x) n ez + 1 n!

(|z| < π)

(40)

n=0

and ∞

X 2ez zn = sech z = E n n! e2z + 1 n=0

π |z| < , 2

(41)

respectively. The following formulas are readily derivable from (40) and (41): En (x + 1) + En (x) = 2 xn (n ∈ N0 ); En0 (x) = n En−1 (x) (n ∈ N); En (1 − x) = (−1)n En (x) (n ∈ N0 );

(42) (43) (44)

(−1)n+1 En (−x) = En (x) − 2 xn (n ∈ N0 ); n X n En (x + y) = Ek (x) yn−k (n ∈ N0 ); k k=0 n X n Ek 1 n−k x − En (x) = (n ∈ N0 ), k 2k 2

(45) (46)

(47)

k=0

which, upon taking x = 21 , yields En = 2n En

1 2

(n ∈ N0 );

(48)

Introduction and Preliminaries n X 2n k=0

2k

87

E2k = 0 (n ∈ N).

(49)

Multiplication formulas: En (mx) = m

n

m−1 X

(−1) En k

k=0

k x+ m

(n ∈ N0 ; m = 1, 3, 5, . . .);

m−1 X 2 k n k En (mx) = − (−1) Bn+1 x + m n+1 m

(50)

(n ∈ N0 ; m = 2, 4, 6, . . .).

k=0

(51) Integral formulas: Zy

En (t) dt =

x

Z1

En+1 (y) − En+1 (x) n+1

(m, n ∈ N0 );

En (t) Em (t) dt = (−1)n 4 2m+n+2 − 1

0

(52)

m! n! Bm+n+2 (m + n + 2)!

(m, n ∈ N0 ). (53)

An alternating finite sum of powers can be expressed as the Euler polynomials: m X

(−1)m−k kn =

k=1

1 En (m + 1) + (−1)m En (0) 2

(m, n ∈ N).

(54)

Fourier Series Expansions of Bernoulli and Euler Polynomials By employing suitable contour integrations in complex function theory, we can obtain the following Fourier series expansions of Bernoulli and Euler polynomials: ∞

B2n (x) =

(−1)n−1 2 (2n)! X cos 2kπx (2π)2n k2n

(n ∈ N; 0 5 x 5 1); (55)

k=1

∞

B2n−1 (x) =

(−1)n 2 (2n − 1)! X sin 2kπ x (2π)2n−1 k2n−1 k=1

(n = 1 and 0 < x < 1; n ∈ N \ {1} and 0 5 x 5 1); E2n (x) =

(−1)n 4 (2n)! π 2n+1

∞ X k=0

sin (2k + 1)πx (2k + 1)2n+1

(n = 0 and 0 < x < 1; n ∈ N and 0 5 x 5 1);

(56) (57)

88

Zeta and q-Zeta Functions and Associated Series and Integrals ∞

E2n−1 (x) =

(−1)n 4 (2n − 1)! X cos (2k + 1)πx π 2n (2k + 1)2n

(n ∈ N; 0 5 x 5 1),

(58)

k=0

which, for x = 12 , yields E2n+1 = 0

(n ∈ N0 ),

(59)

where use is made of the relationship (48). The first few of the Euler numbers En are given below: E0 = 1, E2 = −1, E4 = 5, E6 = −61, E8 = 1385, E10 = −50521, . . . .

(60)

Relations Between Bernoulli and Euler Polynomials The following relationships between the Bernoulli and Euler polynomials follow easily from the definitions (1) and (40): x 2n+1 x+1 Bn+1 − Bn+1 n+1 2 2 x o 2 n = Bn+1 (x) − 2n+1 Bn+1 (n ∈ N0 ), n+1 2

En (x) =

(61)

which, in view of (10), can also be written in the form: 2 x+1 n+1 2 Bn+1 − Bn+1 (x) En (1 − x) = (−1) n+1 2 n

(n ∈ N0 ).

(62)

Two additional formulas involving these polynomials are given below: −1 X n−2 n n En−2 (x) = 2 2n−k − 1 Bn−k Bk (x) 2 k k=0 n X n Bn (x) = 2−n Bn−k Ek (2x) (n ∈ N0 ). k

(n ∈ N \ {1});

(63)

(64)

k=0

The Generalized Euler Polynomials and Numbers (α)

(α)

The generalized Euler polynomials En (x) and the generalized Euler numbers En are defined by the generating functions:

2 z e +1

α exp(xz) =

∞ X n=0

En(α) (x)

zn n!

(|z| < π ; 1α := 1)

(65)


89

and

2 ez e2z + 1

α =

∞ X

En(α)

n=0

zn n!

|z|
0 and 0 < s < 1.

108


Bustoz and Ismail [196] established a remarkably more general result. They showed that the two functions 0(x + s) s+1 f1 (x) = exp (1 − s) ψ x + (0 < s < 1) (15) 0(x + 1) 2 and f2 (x) =

s s−1 0(x + 1) x+ 0(x + s) 2

(0 < s < 1)

(16)

are strictly completely monotonic on (0, ∞). Since lim f1 (x) = lim f2 (x) = 1,

x→∞

x→∞

the inequalities (13) and (14) are immediate consequences of the fact that f1 and f2 are strictly decreasing on (0, ∞). Alzer [22] refined one of the Kec˘ kic´ -Vasic´ ’s inequalities in [642] and also gave the following interesting result: For every s ∈ (0, 1), the function x 7−→ fα (x, s) =

0(x + s) (x + 1)x+1/2 0(x + 1) (x + s)x+s−1/2 1 0 0 · exp s − 1 + (α > 0) ψ (x + 1 + α) − ψ (x + s + α) 12

(17)

is strictly completely monotonic on (0, ∞), if and only if α = 12 . Furthermore, for every s ∈ (0, 1), the function x 7−→

1 fβ (x, s)

(β = 0)

is strictly completely monotonic on (0, ∞), if and only if β = 0. Gurland [527] obtained the following inequality [0(λ + α)]2 λ < 0(λ) 0(λ + 2α) α 2 + λ

(α, λ ∈ R; α + λ > 0; λ > 0; α 6= 0; α 6= 1),

(18)

by making a novel use of the Crame´ r-Rao lower bound for the variance of an unbiased estimator (see Crame´ r [343]). Indeed, we consider the density function given by f (x) =

x 1 exp − x λ−1 θ λ 0(λ) θ

From the fact that the expression: 0(λ) xα 0(α + λ)

(x > 0; λ > 0; θ > 0).

(19)


109

is an unbiased estimator of θ α , we have the following Crame´ r-Rao bound for the variance: V

1 0(λ) α x = 2 , 0(α + λ) E ∂θ∂ α {log f (x)}

which yields the inequality (18). A special case of (18) with λ = n/2 and α = 1/2 is seen to be reduced to Gurland’s formula [526]: 4n + 3 (2n + 1)2

(2n)!! (2n − 1)!!

2

1 + . λ [0(λ + α)]2

(23)

We note that the inequality (22) is seen to be stronger than that in (23) for a certain range of α. Indeed, when α 2 (λ − 2) α2 > , that is, λ (λ + α − 1)2 p p − (λ − 1) − λ(λ − 2) < α < −(λ − 1) + λ(λ − 2).

110


By employing the multivariate generalization of (19), that is, the Wishart distribution, Olkin [878] obtained p−1 Y j=0

2 λ2 p2 − 1 + λ p4 [0(λ + α − j/2)]2 ≤ 0(λ − j/2) 0(λ + 2α − j/2) λ2 p2 − 1 2 + λ p4 + α 2 p−1 α, λ ∈ R; α + λ > ; λ > 0; p > 0 . 2

(24)

The special case of (24) with p = 1 reduces to (18). Selliah [1018], by employing the multiparameter version of the Crame´ r-Rao lower bound, using the information matrix for the same problem, obtained the following inequality: p−1 Y j=0

λ [0(λ + α − j/2)]2 ≤ 0(λ − j/2) 0(λ + 2α − j/2) λ + p α 2 p−1 α, λ ∈ R; α + λ > ; λ > 0; p > 0 , 2

(25)

which is seen to be sharper than that given in (24). For x ∈ (−α, ∞), define the function zs,t (x) by   0(x + t) 1/(t−s)   −x   0(x + s) zs,t (x) :=      ψ(x+s) e − x (s = t),

(s 6= t) (26)

+ where s, t ∈ R+ 0 , R0 being the set of nonnegative real numbers, and α = min{s, t}. A monotonicity and convexity of zs,t (x) was proved (see [241, 406, 929, 945]) so that the function zs,t (x) is either convex and decreasing for |t − s| < 1 or concave and increasing for |t − s| > 1. From this fact, the best bounds in the Kershaw’s double inequality (14) could be deduced. Qi [930, 931] further generalized this result. For x ∈ (−ρ, ∞), define the function Ha,b,c (x) by

Ha,b,c (x) := (x + c)b−a

0(x + a) 0(x + b)

(a, b, c ∈ R; ρ = min{a, b, c}).

(27)

Very recently, Qi [932, Theorem 1] proved the following results: Ha,b,c (x) ∈ L[(−ρ, ∞)]

(a, b, c) ∈ D1 (a, b, c)

(28)

and

Ha,b,c (x)

−1

∈ L[(−ρ, ∞)]

(a, b, c) ∈ D2 (a, b, c) ,

(29)


111

where, for convenience, 1 1 ∪ (a, b, c) | a > b = c + D1 (a, b, c) := (a, b, c) | a + b = 1, c 5 b < c + 2 2 ∪ (a, b, c) | 2a + 1 5 a + b 5 1, a < c ∪ (a, b, c) | b − 1 5 a < b 5 c

\ {(a, b, c) | a = c + 1, b = c}

and 1 1 D2 (a, b, c) := (a, b, c) | a + b = 1, c 5 a < c + ∪ (a, b, c) | b > a = c + 2 2 ∪ (a, b, c) | b < a 5 c ∪ (a, b, c) | b + 1 5 a, c 5 a 5 c + 1 ∪ (a, b, c) | b + c + 1 5 a + b 5 1 \ {(a, b, c) | a = c + 1, b = c} \ {(a, b, c) | b = c + 1, a = c} . Qi [932, Theorem 2] made use of (28), (29) and 1.1(37) to prove the following inequalities: (x + c)a−b
0), so ζ (s, a) is an analytic function of s in the half-plane 1. Setting n = 1 in 2.1(6), we have the integral representation: 0(s) ζ (s, a) =

Z∞ 0

Z1 = 0

xs−1 e−ax dx = 1 − e−x xa−1 1−x

log

1 x

Z∞ 0

xs−1 e−(a−1)x dx ex − 1 (2)

s−1 dx

( 1; 0).

Moreover, ζ (s, a) can be continued meromorphically to the whole complex s-plane (except for a simple pole at s = 1 with its residue 1) by means of the contour integral

156


representation (see Theorem 2.5): 0(1 − s) ζ (s, a) = − 2πi

Z C

(−z)s−1 e−az dz, 1 − e−z

(3)

where the contour C is the Hankel loop of Theorem 2.5. The connection between ζ (s, a) and the Bernoulli polynomials Bn (x) is also given in 2.1(17). From the definition (1) of ζ (s, a), it easily follows that ζ (s, a) = ζ (s, n + a) +

n−1 X

(k + a)−s

(n ∈ N);

(4)

k=0

∞ X 1 1 1 ζ s, a − ζ s, a + = 2s (−1)n (a + n)−s . 2 2 2

(5)

n=0

Hurwitz’s Formula for ζ (s, a) The series expression ζ (s, a) was originally meaningful for σ > 1 (s = σ + it). Hurwitz obtained another series representation for ζ (s, a) valid in the half-plane σ < 0: o 1 0(s) n − 1 πis e 2 L(a, s) + e 2 π is L(−a, s) s (2π) (0 < a 5 1, σ = 1; 0 < a < 1, σ > 0), ζ (1 − s, a) =

(6)

where the function L(x, s) is defined by

L(x, s) :=

∞ 2π inx X e ns

(x ∈ R; σ = 1),

(7)

n=1

which is often referred to as the periodic (or Lerch) Zeta function. We note that the Dirichlet series in (7) is a periodic function of x with period 1 and that L(1, s) = ζ (s), the Riemann Zeta function (see Section 2.3). The series in (7) converges absolutely for σ > 1. Yet, if x ∈ / Z, the series can also be seen to converge conditionally for σ > 0. So, the formula (6) is also valid for σ > 0, if a 6= 1. We observe that the function L(x, s) in (7) is a linear combination of the Hurwitz Zeta functions, when x is a rational number. Indeed, setting x = p/q (1 5 p 5 q; p, q ∈ N) in (7), the terms in (7) can be rearranged according to the residue classes mod q, by letting n = kq + r (1 5 r 5 q; k ∈ N0 ),

The Zeta and Related Functions

157

which gives us, for σ > 1, X X q X ∞ ∞ exp 2π irp q 2πinp p 1 exp ,s = = L q ns q (kq + r)s n=1 r=1 k=0 ∞ q 1 X 2πirp X 1 s = s exp r q q r=1 k=0 k + q q r 2πirp 1 X ζ s, . exp = s q q q

r=1

Therefore, if we take a = p/q in the Hurwitz formula (6), we obtain:

p ζ 1 − s, q

q 2 0(s) X πs 2πrp r = cos − ζ s, (2πq)s 2 q q r=1

(8)

(1 5 p 5 q; p, q ∈ N), which holds true, by the principle of analytic continuation, for all admissible values of s ∈ C.

Hermite’s Formula for ζ (s, a) We, first, recall Plana’s summation formula: n X k=m

1 f (k) = [ f (m) + f (n)] + 2

Zn

f (τ ) dτ − 2

m

Z∞ 0

q(m, t) − q(n, t) dt, e2πt − 1

(9)

where f (z) is a bounded analytic function in m 5 a); |q(x, y)| 5 (a + x) + y y sinh 2 h i− 1 σ 2 sinh y|s| (|y| < a). |q(x, y)| 5 (a + x)2 + y2 x+a

(11)

Making use of (11), it is easily seen that the integral: Z∞

−1 q(x, y) e2π y − 1 dy (σ > 0)

0

converges when x = 0 and tends to 0 as x → ∞. Also, the improper integral: Z∞ (a + x)−s dx

(σ = 1)

0

converges. Therefore, if σ > 1, it is valid to make n → ∞ (m = 0) in (9) with the function f (z) in (10). Thus, we readily obtain Hermite’s formula for ζ (s, a): 1 a1−s ζ (s, a) = a−s + +2 2 s−1

Z∞ − 1 s n y o dy 2 a2 + y2 sin s arctan . a e2π y − 1

(12)

0

We note that the integral involved in (12) converges for all admissible values of s ∈ C. Moreover, the integral is an entire function of s. A special case of the formula (12) when a = 1 is attributed to Jensen. Setting s = 0 in (12), we have ζ (0, a) =

1 − a, 2

(13)

which is also obtained from 2.1(17) in view of 1.7(8). If we set z = s in 1.3(30) and differentiate the resulting equation with respect to s, we find that 1 ψ(s) = log s − − 2 2s

Z∞ 0

t 2 + s2

t dt e2π t − 1

( 0).

(14)

Taking the limit in (12) as s → 1, by virtue of the uniform convergence of the integral in (12), we get Z∞ 1 a1−s − 1 1 y dy , lim ζ (s, a) − = lim + +2 2 2 s→1 s→1 s − 1 s−1 2a a + y e2π y − 1

0


159

which, in view of (14), yields 0 0 (a) 1 =− = −ψ(a). lim ζ (s, a) − s→1 s−1 0(a)

(15)

Differentiating (12) with respect to s and setting s = 0 in the resulting equation, we have

Z∞ arctan ay 1 d ζ (s, a) log a − a + 2 = a− dy, ds 2 e2π y − 1 s=0

(16)

0

which, by virtue of 1.2(30), yields d 1 ζ (s, a) = log 0(a) − log (2π), ds 2 s=0

(17)

which is equivalent to the identity 2.1(29). In addition to (17), it is easy to find from the definition (1) of ζ (s, a) that ∂ ζ (s, a) = −s ζ (s + 1, a). ∂a

(18)

The respective special cases of (15) and (17) when a = 1, by means of 1.2(4) and 1.1(13), become

1 1 lim ζ (s, a) − = lim ζ (1 + , a) − =γ s→1 →0 s−1

(19)

1 ζ 0 (0) = − log (2π), 2

(20)

and

where ζ (s) is the Riemann Zeta function (see Section 2.3).

Further Integral Representations for ζ (s, a) In addition to (12), some known integral representations of ζ (s, a) are recalled here: 0(s) ζ (s, a) =

Z∞ 0

Z1 = 0

t s−1 e−at dt = 1 − e−t ta−1 1−t

Z∞ 0

t s−1 e−(a−1)t dt et − 1

1 s−1 log dt t

(21) 1; 0 ,

160


which is the same as (2); 1 a1−s 1 ζ (s, a) = a−s − + 2 1 − s 0(s)

Z∞ 0

1 1 1 − + t e −1 t 2

e−at t s−1 dt

(22)

( −1; 0); h i t Z∞ h 1 i s−2 cos (s − 1) arctan (1−s) 2a−1 π2 2 ζ (s, a) = t2 + (2a − 1)2 dt 2 1 s−1 cosh πt (23) 2 0 1 ; 2 Z∞ 1 t−s ζ (s, a) = cos π s sin(2πa) dt 2 cosh(2πt) − cosh(2πa) 0

Z∞ −s t cosh(2πa) − e−2π t 1 πs dt + sin 2 cosh(2πt) − cosh(2πa)

(24)

0

( 0), 2 2 da da ds (k + a)2 s=0 k=0


161

which implies that f (a) is logarithmically convex on (0, ∞). Thus, by appealing to the Bohr-Mollerup theorem (Theorem 1.1), we obtain, for some constant C, f (a) = C 0(a), which, for a = 1, yields 1

C = exp[ζ 0 (0)] = (2π)− 2 . This completes our second proof of the derivative formula (17). Many authors gave seemingly different proofs of Stirling’s formula 1.1(52) (see e.g., Blyth and Pathak [135], Choi [261], Diaconis and Freeman [380] and Patin [889]). Here, by taking the limit in (16) as a → ∞, we have lim

a→∞

1 ζ (0, a) + a + log a − a log a = 0, 2 0

(27)

which, upon taking the exponential and using (17), immediately yields Stirling’s formula 1.1(33). Combining formulas (17) and 1.1(42), we obtain a formula for the Beta function B(α, β) : 1 B(α, β) = (2π) 2 exp ζ 0 (0, α) + ζ 0 (0, β) − ζ 0 (0, α + β) , where ζ 0 (s, a) = identities:

(28)

∂ ∂s ζ (s, a). Applying the formula (16) and the following trigonometric

a+b 1 − ab a−b arctan a − arctan b = arctan 1 + ab arctan a + arctan b = arctan

(ab < 1), (29) (ab > −1)

to (28), we can readily deduce an integral representation of B(α, β) (cf. Choi and Nam [276]): α α− 2 β β− 2 1

1

B(α, β) =

α+β− 12

(α + β)

1

(2π) 2 eI(α,β)

(α > 0, β > 0),

where, for convenience, I(α, β) := 2ρ

Z∞

arctan

0

(t3 + t)ρ 3 dt 2π tρ αβ(α + β) e −1

ρ 2 := α 2 + αβ + β 2 .

(30)

162


Differentiating both sides of (22) with respect to s and letting s = 0 in the resulting equation, we obtain

1 ζ (0, a) = a − 2 0

Z∞

log a − a +

0

1 1 1 − + t e −1 t 2

e−at dt t

(31)

( 0). In a similar manner, (31) leads us to another integral representation of B(α, β) (cf. Choi and Nam [276]): α α− 2 β β− 2 1

B(α, β) =

1

α+β− 12

(α + β)

1

(2π) 2 eJ(α,β)

(α > 0; β > 0),

(32)

where, for convenience,

J(α, β) :=

Z∞ 0

1 1 −αt 1 −βt −(α+β)t dt − + e + e − e . et − 1 t 2 t

Another Form for 02 (a) From 2.1(24) and 2.1(30), by virtue of (17), we obtain another form for the double Gamma function 02 (a) : 1 02 (a) = A {0(a)}1−a exp − + ζ 0 (−1, a) 12

(a > 0),

(33)

∂ ζ (s, a). where ζ 0 (s, a) = ∂s In addition to the integral representation 1.4(78), we can express log 02 (a) as improper integrals in many ways. For example, we give two integral representations for log 02 (a):

1 a2 1 2 1 log 02 (a) = − + log A − + a − a log a + (1 − a) log 0(a) 12 4 2 2 ∞ Z 1 2 2 1 t +2 (a + t ) 2 sin arctan log(a2 + t2 ) (34) 2 a 0 t t dt 2 2 12 + (a + t ) cos arctan arctan ( 0); 2π a a e t −1


163

2 a2 a a 1 log 02 (a) = log A − + − + log a + (1 − a) log 0(a) 4 2 2 12 Z∞ 1 1 t e−at 1 − + − dt ( 0). − et − 1 t 2 12 t2

(35)

0

Indeed, by differentiating Hermite’s formula (12) for ζ (s, a), with respect to s, letting s → −1 in the resulting equation and applying (17) and the identity for ζ 0 (−1, a), we readily obtain (34). Conversely, setting n = 2 in (25), we have Z∞

sa−s−1 a−s a1−s 1 ζ (s, a) = + + + 12 2 s − 1 0(s)

0

1 1 1 t − + − t s−1 e−at dt et − 1 t 2 12 (36) ( −3; 0).

Employing the same technique as in getting (34), by making use of (36) and considering the following identities: d 1 = −1 ds 0(s) s=−1

and

1 = 0, 0(s) s=−1

we obtain (35). Glaisher [484, p. 47] expressed the Glaisher-Kinkelin constant A given in 1.3(2) as an integral: 1

 7 1 1 2 A = 2 36 π − 6 exp  + 3 3

Z2

  log 0(t + 1) dt.

(37)

0

By setting a = 1 in (34) and (35), we can also obtain integral representations of log A: 1 log A = − 2 3

Z∞

1 1 (1 + t2 ) 2 sin (arctan t) log(1 + t2 ) 2

0 1

+ (1 + t2 ) 2 cos (arctan t) arctan t

o

(38)

dt e2π t − 1

and 1 log A = + 4

Z∞ 0

1 1 1 t − + − et − 1 t 2 12

e−t dt. t2

(39)

164


The formula (35) can be used to obtain an asymptotic formula for log 02 (a) by first observing that, for some M > 0 and for all a > 0, Z∞ 1 1 1 t e−at M dt − + − < a . et − 1 t 2 12 t2 0

Thus, by employing 1.1(34), we have 2 1 a 5 3a2 −a− − −a+ log a log 02 (a) = log A + 4 12 2 12 1 + (1 − a) log(2π) + O a−1 (a → ∞; a > 0), 2

(40)

which may be compared with 1.3(7).

2.3 The Riemann Zeta Function The Riemann Zeta function ζ (s) is defined by P ∞ ∞ P 1 1 1  = ( 1)  s −s  n (2n−1)s 1−2 n=1 n=1 ζ (s) := ∞ P  (−1)n−1   11−s ( 0; s 6= 1). ns 1−2

(1)

n=1

It is easy to see from the definitions (1) and 2.2(1) that −1 1 ζ s, ζ (s) = ζ (s, 1) = 2s − 1 = 1 + ζ (s, 2) 2

(2)

and m−1 j 1 X ζ s, ζ (s) = s m −1 m

(m ∈ N \ {1}).

(3)

j=1

In view of (2), we can deduce many properties of ζ (s) from those of ζ (s, a) given in P −s in (1) represents an analytic function Section 2.2. In fact, the series ζ (s) = ∞ n=1 n of s in the half-plane 1. Setting a = 1 in 2.2(2), we have an integral representation of ζ (s) in the form: 0(s) ζ (s) =

Z∞ 0

Z1 = 0

xs−1 e−x dx = 1 − e−x 1 1−x

log

1 x

Z∞ 0

xs−1 dx ex − 1 (4)

s−1 dx

( 1).


165

Furthermore, just as ζ (s, a), ζ (s) can be continued meromorphically to the whole complex s-plane (except for a simple pole at s = 1 with its residue 1) by means of the contour integral representation: ζ (s) = −

0(1 − s) 2πi

(−z)s−1 e−z dz, 1 − e−z

Z C

(5)

where the contour C is the Hankel loop of Theorem 2.5. Now 2.2(19) and (5), together, imply that the Laurent series of ζ (s) in a neighborhood of its pole s = 1 has the form: ∞

ζ (s) =

X 1 +γ + an (s − 1)n , s−1

(6)

n=1

where γ is the Euler-Mascheroni constant given in 1.1(3) and an is also expressed as (see Ivic´ [586, pp. 4–6]): ) ( m X (log k)n (log m)n+1 − an = lim m→∞ k n+1

(n ∈ N).

(7)

k=1

The Riemann Zeta function ζ (s) in (1) plays a central roˆ le in the applications of complex analysis to number theory. The number-theoretic properties of ζ (s) are exhibited by the following result, known as Euler’s formula, which gives a relationship between the set of primes and the set of positive integers: ζ (s) =

Y

1 − p−s

−1

( 1),

(8)

p

where the product is taken over all primes. From 2.2(4), we have [cf. Equation (2) for the special cases when n = 0 and n = 1]

ζ (s) = ζ (s, n + 1) +

n X

k−s

(n ∈ N0 ).

(9)

k=1

The connection between ζ (s) and the Bernoulli numbers is given as follows:

ζ (−n) =

 1  − 2

(n = 0)

  Bn+1 − n+1

(n ∈ N),

(10)

which is deduced by setting x = 1 in 2.1(17) and using 1.6(5).

166


Riemann’s Functional Equation for ζ (s) The special case of 2.2(8), when p = 1 = q, yields Riemann’s functional equation for ζ (s): ζ (1 − s) = 2(2π)

−s

1 π s ζ (s) 0(s) cos 2

(11)

or, equivalently, ζ (s) = 2(2π)s−1 0(1 − s) sin

1 πs ζ (1 − s). 2

Taking s = 2n + 1 (n ∈ N) in (11), the factor cos ζ (−2n) = 0

(12)

1 2πs

vanishes, and we find that

(n ∈ N),

(13)

which are often referred to as the trivial zeros of ζ (s). The equation (13) can also be proven by combining (10) and 1.7(7). By using the Legendre duplication formulas 1.1(29) and 1.1(12), it is not difficult to see that the functional equations (11) or (12) can be written in a simpler form: 8(s) = 8(1 − s),

(14)

where the function 8(s) is defined by 1

8(s) := π − 2 s 0

1 s ζ (s). 2

(15)

The function 8(s) has simple poles at s = 0 and s = 1. According to Riemann, to remove these poles, we multiply 8(s) by 12 s(1 − s) and define ξ(s) :=

1 s(1 − s) 8(s), 2

(16)

which is an entire function of s and satisfies the functional equation: ξ(s) = ξ(1 − s).

(17)

Setting s = 2n (n ∈ N) in (11) and applying (10), we have the well-known identity: ζ (2n) = (−1)n+1

(2π)2n B2n 2 (2n)!

(n ∈ N0 ),

(18)


167

which, in view of 1.7(7), enables us to list the following special values: π4 π6 π2 , ζ (4) = , ζ (6) = , 6 90 945 π8 π 10 ζ (8) = , ζ (10) = , ... . 9450 93555

ζ (2) =

(19)

We may recall here a known recursion formula for ζ (2n) (see also Section 4.1): n−1

ζ (2n) =

2 X ζ (2k)ζ (2n − 2k) 2n + 1

(n ∈ N \ {1}),

(20)

k=1

which can also be used to evaluate ζ (2n) (n ∈ N \ {1}). We get no information about ζ (2n + 1) (n ∈ N) from Riemann’s functional equation, since both members of (11) vanish upon setting s = 2n + 1 (n ∈ N). In fact, until now, no simple formula analogous to (18) is known for ζ (2n + 1) or even for any special case, such as ζ (3). It is not even known whether ζ (2n + 1) is rational or irrational, except that the irrationality of ζ (3) was proven recently by Ape´ ry [56]. Instead, a known integral formula for ζ (2n + 1) is recalled here: (−1)n+1 (2π)2n+1 ζ (2n + 1) = 2(2n + 1)!

Z1

B2n+1 (t) cot(π t) dt

(n ∈ N).

(21)

0

It is readily seen that ζ (s) 6= 0 ( 0).

(96)

Repeated applications of this process lead us to Spence’s formula: Tin (x) + (−1)

n−1

1 (log x)n−1 1 = π Tin x 2 (n − 1)! h

1 2 (n−1)

i

X (log x)n − 2k−1 Ti2k+1 (1) +2 (n − 2k − 1)!

(97) (x > 0).

k=1

Next, we find from (73) that Li2n+1 (i) − Li2n+1 (−i) = 2i Ti2n+1 (1). Setting x = exp

1 2 πi

(98)

and writing 2n + 1 for n in (89), we obtain

Li2n+1 (−i) − Li2n+1 (i) " # n X B2k 22k−1 − 1 π 2n+1 n 2n+1 = (−1) i − + 2π . (2n + 1)! 22n+1 (2k)! (2n − 2k + 1)! 22n−2k+1 k=1

(99) Consider 1

1

1

e 2 z − e− 2 z 2e 2 z = 2 ez + 1 2

1 2z z − , ez − 1 2 e2z − 1

(100)

which can be expanded by 1.6(2) and 1.6(40) in powers of z. Upon equating the coefficients of z2n , we find that " # n X B2k 22k−1 − 1 E2n 1 = −2 . (101) (2n)! 22n+1 (2n + 1)! 22n+1 (2k)! (2n − 2k + 1)! 22n−2k+1 k=1

Now, a combination of (98), (99) and (101) leads us to the well-known Euler series: Ti2n+1 (1) =

∞ X k=1

(−1)k+1

1 (−1)n π 2n+1 = E2n , (2k − 1)2n+1 (2n)! 22n+2

(102)

190


which, in view of 1.6(59), gives Ti3 (1) =

π3 , 32

Ti5 (1) =

5π 5 , 1536

Ti7 (1) =

61 π 7 , .... 184320

(103)

Starting with the known formula for Li2 (x, θ): Li2 (r, θ) + Li2

1 1 , θ = 2 Gl2 (θ) − (log r)2 r 2

(104)

and following a procedure analogous to that in proving (89), one obtains Lin (r, θ) + (−1) Lin n

1 1 , θ = − (log r)n r n! h

+2

i

(105)

1 2n

X (log r)n−2k k=1

(n − 2k)!

Gl2k (θ)

(n ∈ N \{1}),

which, upon setting r = eiπ , noting that Lin (−r, θ) = Lin (r, π − θ) and writing 2n for n, gives n

Gl2n (π − θ) =

(−1)n−1 π 2n X (−1)n−k π 2n−2k + Gl2k (θ). 2(2n)! (2n − 2k)!

(106)

k=1

The following relationship between Gl2n (θ) and Gl2n (π − θ) can readily be obtained from the series definition (85): Gl2n (π − θ) + Gl2n (θ) =

1 22n−1

Gl2n (2θ).

(107)

The generating function 1.7(1) for the Bernoulli polynomials can be rewritten in the form: ∞

X 2t cos(2nπx) − 4nπ sin(2nπx) text = 1+t . t e −1 t2 + 4n2 π 2 n=1

Making use of the expansion: 2k ∞ 1 1 X t k = 2 2 (−1) 2nπ t2 + 4n2 π 2 4n π k=0

(|t| < 2nπ; n ∈ N)

(108)


191

and rearranging the double series in (108) in powers of t, the coefficients are easily expressible in terms of Gln (2πx), defined by (85). Thus, employing 1.7(1) on the lefthand side of (108) and equating the coefficients of the same powers on both sides, we find that h i 1+ 21 n n−1

Gln (2πx) = (−1)

2

πn

Bn (x) n!

(0 5 x 5 1; n ∈ N \ {1}),

(109)

which gives us the following special cases: π2 1 , Gl2 (θ) = (π − θ)2 − 4 12 1 Gl3 (θ) = θ(π − θ)(2π − θ), 12 1 2 1 4 1 2 2 Gl4 (θ) = π − θ π − πθ + θ , 90 12 4 1 Gl5 (θ) = θ(π − θ)(2π − θ) 4π 2 + 6πθ − 3θ 2 . 720

(110)

The Log-Sine Integrals A recurrence relationship for Lsn (π), defined by (82), is given by (−1)m Lsm+2 (π) = π 1 − 2−m ζ (m + 1) m! m−1 X (−1)k+1 + 1 − 2k−m ζ (m − k + 1) Lsk+1 (π) k!

(111) (m ∈ N).

k=2

To derive the recurrence relation (111), we let Zπ I :=

1 exp x log 2 sin θ dθ 2

0

n ∞ Zπ n X x 1 = log 2 sin θ dθ n! 2 n=0 0

=−

∞ n X x Lsn+1 (π). n! n=0

(112)

192


Alternatively, in view of 1.1(44), we have x Zπ 1 dθ I= exp log 2 sin θ 2 0

Zπ =

2x sinx

1 θ dθ 2

0

1 1 0 + x √ 2 2 . = 2x π 1 0 1 + 2x By the Legendre duplication formula 1.1(29) for 0, we, thus, obtain I = π y,

(113)

where, for convenience, 0(1 + x) y := h i2 . 1 0 1 + 2x Making use of the following notations: Dn f (x) =

dn f (x) = f (n) (x) dxn

and Dn0 f (x) =

dn f (x) = f (n) (0), n dx x=0

differentiating (112) n times and setting x = 0, we find that Lsn+1 (π) = −Dn0 I = −π Dn0 y, which, by virtue of 1.3(53), yields Dn0 log y = (−1)n (n − 1)! 1 − 21−n ζ (n)

(114)

(n ∈ N \ {1}),

(115)

so that we have log y =: f (x) =

∞ n X x (−1)n 1 − 21−n ζ (n), n n=2

since log y|x=0 = 0 = D10 log y.

(116)


193

From (114), we get Lsn+1 (π) = −π Dn0 exp(f (x)). Let ym := Dm e f . Then, ym+1 = Dm+1 e f = Dm De f = Dm f 0 e f . Hence, using Leibniz’s rule for differentiation, we have

ym+1 =

m X m

k

k=0

f (m−k+1) yk ,

(117)

which, upon considering ym |x=0 = −

1 Lsm+1 (π) π

in (114), yields (m+1) Lsm+2 (π) = −π f0 +

m X m (m−k+1) f Lsk+1 (π). k 0

(118)

k=1

Now, setting 0

Ls2 (π) = 0,

f0 = 0

(m)

and f0

= (−1)m (m − 1)! 1 − 21−m ζ (m)

(m ∈ N \ {1})

in (118), we immediately arrive at the desired recurrence relation (111). Some simple consequences of (111) are presented below:

Ls2 (π) = −

Zπ

1 log 2 sin θ dθ = 0, 2

0

2 Zπ 1 1 Ls3 (π) = − dθ = − π 3 , log 2 sin θ 2 12 0

3 Zπ 1 3 Ls4 (π) = − log 2 sin θ dθ = π ζ (3), 2 2 0

4 Zπ 1 19 5 Ls5 (π) = − log 2 sin θ dθ = − π , 2 240 0

194


5 Zπ 45 5 1 Ls6 (π) = − dθ = π ζ (5) + π 3 ζ (3), log 2 sin θ 2 2 4 0

6 Zπ 1 45 275 7 Ls7 (π) = − log 2 sin θ dθ = − π {ζ (3)}2 − π , 2 2 1344 0

7 Zπ 2835 315 3 133 5 1 dθ= π ζ (7)+ π ζ (5)+ π ζ (3), Ls8 (π) = − log 2 sin θ 2 4 8 20 0

8 Zπ 1 log 2 sin θ Ls9 (π) = − dθ 2 0

24177 9 105 3 =− π − 1890 π ζ (3) ζ (5) − π {ζ (3)}2 . 26880 2

2.5 Hurwitz–Lerch Zeta Functions The Hurwitz–Lerch Zeta function 8(z, s, a) is defined by 8(z, s, a) :=

∞ X

zn (n + a)s

n=0 a ∈ C \ Z− ; 0 s∈C

(1)

when |z| < 1; 1 when |z| = 1 ,

which satisfies the obvious functional relation: 8(z, s, a) = zn 8(z, s, n + a) +

n−1 X k=0

zk (k + a)s

n ∈ N; a ∈ C \ Z− 0 .

(2)

By writing the Eulerian integral 1.1(1) in the form:

0(z) = s

z

Z∞

e−st tz−1 dt

( 0; 0),

(3)

0

we can deduce the following integral representation from (1): 1 8(z, s, a) = 0(s)

Z∞ 0

t s−1 e−at 1 dt = 1 − ze−t 0(s)

Z∞ 0

t s−1 e−(a−1)t dt et − z

( 0; |z| 5 1, z 6= 1, 0; z = 1, 1),

(4)


195

by noting that 1 zn = (n + a)s 0(s)

Z∞

e−at t s−1 ze−t

n

dt

( 0; 0).

(5)

0

If use is made of the infinite-series version of Plana’s summation formula 2.2(9) (cf. Lindelo¨ f [769, p. 61]; see also Erde´ lyi et al. [421, p. 22]) and the definition (1), another definite integral representation of 8(z, s, a) is obtained in the form: 1 8(z, s, a) = s + 2a

Z∞ 0

Z∞ −2 0

zt dt (t + a)s (6)

− 1 s t dt 2 sin t log z − s tan−1 t2 + a2 2π a e t −1

( 0),

which, for z = 1, immediately reduces to Hermite’s formula 2.2(12) for ζ (s, a). By setting z = eiθ in (4) and using (3), we get Lipschitz’s formula: 2 0(s)

∞ X n=1

einθ = (n + a)s

Z∞

e−at t s−1

0

eiθ − e−t dt cosh t − cos θ

(7)

(0 < θ < 2π; −1; 0). Several contour integral and series representations of 8(z, s, a) include (see Erde´ lyi et al. [421, pp. 28–31]): Z 0(1 − s) (−t)s−1 e−at 8(z, s, a) = − dt 2π i 1 − z e−t (8) C ( 0; |arg(−t)| 5 π), where the contour C is the Hankel loop of Theorem 2.5, which, obviously, does not enclose any pole of the integrand, that is, t = log z ± 2nπi (n ∈ N0 ); 8(z, s, a) =

∞ 0(1 − s) X (− log z + 2nπ i)s−1 e2nπ ia za n=−∞

(9)

(0 < a 5 1; 0

when ξ ∈ R \ Z; R(s) > 1

when ξ ∈ Z ,

which was first studied by Rudolf Lipschitz (1832–1903) and Matya´ sˇ Lerch (1860– 1922) in connection with Dirichlet’s famous theorem on primes in arithmetic progressions. Replacing s by 1 − s in 2.2(6), we obtain another equivalent form of Hurwitz’s formula 2.2(6): ζ (s, a) = 2(2π)s−1 0(1 − s)

∞ X n=1

πs ns−1 sin 2nπa + 2

(12)

( 0, ζ (3) − p > 1 q qθ +

and

θ = 13.41782 · · · ,

which proves the irrationality of ζ (3). (Cf. van der Poorten [1180, p. 195]; see also Apéry [56]) 27. Prove the following inequalities: For n, p ∈ N, ! n ! n p−1 X n X X 1 1 X 1 8 1 0 ≤ p+ − < 2 n k2p k2p−2j k2j k=1

j=1

k=1

k=1

and ! X n k+1 2 X n 1 (−1) < 6. −2 n 2 2k − 1 (2k − 1) k=1 k=1 (Hovstad [571, p. 93]) 28. Show that the first inequality of Problem 27 implies that ∞ ∞ X X 1 1 = R p k2p k2 k=1

!p (p ∈ N),

k=1

where Rp is a rational number, satisfying the recurrence formula:

R1 = 1

and

p−1 X 1 p+ Rp = Rj Rp−j 2

(p ∈ N \ {1}).

j=1

(Hovstad [571, p. 93])


233

29. Prove that ζ (s) has the factorization ζ (s) =

Y s s ebs 1− eρ , 1 ρ 2 (s − 1) 0 1 + 2 s ρ

where b = −1 −

γ + log(2π ); 2

γ is the Euler-Mascheroni constant, and the product is taken over all the so-called nontrivial zeros of ζ (s). (Cf. Titchmarsh [1151, pp. 30–31]; see also Melzak [821, p. 111]) 30. Prove that X ∞ 1 (2n)! 1 ζ ,a + 1 = ζ n + (−a)n (|a| < 1) 2 2 22n (n!)2 n=0

and ζ

n−1 1 X 1 k 1 , na = n− 2 ,a + ζ 2 2 n

(n ∈ N).

k=0

(Powell [911, p. 117]) 31. Let (s − 1) ζ (s, a) = 1 +

∞ X

γn (a) (s − 1)n+1

(0 < a ≤ 1; n ∈ N0 ).

n=0

Show that ! m X {log(k + a)}n {log(m + a)}n+1 (−1)n lim − γn (a) = , n! m→∞ k+a n+1 k=0

which, for a = 1, gives the coefficients in the Laurent expansion of ζ (s) about s = 1 (cf. Equation 2.3(7)). (Berndt [118, p. 152]) 32. Prove that lim sup t→∞

|ζ (1 + it)| ≥ eγ , log (log t)

where γ is the Euler-Mascheroni constant defined by 1.1(3). (Titchmarsh [1150, p. 79]) 33. Prove that ∞ X zn Lim ez = ζ (m − n) n! n=0 n6=m−1

m−1 1 z 1 + 1 + + ··· + − log(−z) 2 m−1 (m − 1)!

(m ∈ N; |z| < 2π). (Cohen et al. [336, p. 26])

234


34. Prove that ∞ n−1 4 1 5X X 1 − ζ (5) = 2 k 2 5 n2 n=1

!

k=1

(−1)n . n3 2n n

(van der Poorten [1181, p. 274]) 35. Let K be a normal extension field of degree n over the rational number field Q. Denote by OK the integer ring of K. Let I(K) be the set of all nonzero ideals of OK , Na the absolute norm of an ideal a ∈ I(K) and Tr α the trace of α ∈ K over Q. We assume that [K : Q] = n > 1, let OK,0 = {α ∈ OK | Tr α = 0}; TK = min{Tr α | α ∈ OK , Tr α > 0}. For a ∈ I(K) and a0 = OK,0 ∩ a, define T(a) = Min

Trα | α ∈ a, Trα > 0 ; T(K)

N0 (a) = #{OK,0 /a0 }, where OK,0 /a0 is the quotient of Z-module OK,0 by submodule a0 . For any x ∈ R+ , let jK (x) = #{a ∈ I(K) | Na ≤ x}; jK,0 (x) = #{a ∈ I(K) | N0 (a) ≤ x} Na = # a ∈ I(K) | ≤x , T(a) and q ≡ 1 (mod n). Suppose that K is the subfield of the qth cyclotomic field Cq , such that K/Q is a tamely ramified cyclic extension of degree n. Show that ζ (2k + 1) =

2 qn − qn−1 − q jk (x) ζ (2k) lim . x→+∞ jK,0 (x) qn − 1 (Lan [728, p. 273])

36. Prove that ∞ ∞ (2) X 469 3 (Hn )3 11 X Hn − = ζ (8) − 16 ζ (3) ζ (5) + ζ (2) {ζ (3)}2 5 4 32 2 n n6 n=1

n=1

and ∞ ∞ (2) X 561 47 (Hn )3 13 X Hn − = ζ (10) − {ζ (5)}2 7 8 4 20 4 n n n=1

n=1

49 15 − ζ (7) ζ (3) + 3 ζ (2) ζ (3) ζ (5) + {ζ (3)}2 ζ (4). 2 4 (Flajolet and Salvy [454, p. 27])


235

37. Prove that, for an odd weight m = p + q, ∞ X 1 (−1)p m − 1 (−1)p m − 1 H (p) (n) − = ζ (m) − p q nq 2 2 2 n=1

+

[p/2] X m − 2k − 1 1 − (−1)p ζ (p) ζ (q) + (−1)p ζ (2k) ζ (m − 2k) q−1 2 k=1

+ (−1)p

[q/2] X k=1

m − 2k − 1 ζ (2k) ζ (m − 2k), p−1

where ζ (1) should be interpreted as 0 wherever it occurs. (Cf. Borwein et al. [146, p. 278]; see also Flajolet and Salvy [454, p. 22]) 38. Prove that ∞ 2n−1 X X (−1)n+k 27 = πG− ζ (3) 2 16 n k n=1 k=1

and ∞ 2n−1 X X (−1)n−1 29 ζ (3), = πG− 16 n2 k n=1 k=1

where G denotes the Catalan constant defined by 1.3(16). (Sitaramachandrarao [1034, p. 13]) 39. Prove that, for m ∈ N and 0, ∂ ζ 0 (−m, a) = ζ (z, a) ∂z z=−m 1 1 1 = am+1 log a − am+1 − am log a m+1 2 (m + 1)2 ∞ 1 m−1 X m m−1 a log a + a + α2k a−(2k−m+1) , + 12 12 k=1

where

α2k :=

   2k  j X  B m m (−1) 2k+2     log a +    2k + 2 2k + 1 j 2k − j + 1

(2k 5 m − 1),

j=0

m   B2k+2 X m (−1)j      2k + 2 j 2k − j + 1

(2k = m).

j=0

(Elizalde [410, p. 349]) 40. Define constants 8k , for k > 1 an odd integer, by 8k = −

k−2 d d−1 π 2 X (−1) 2 ζ (k − d + 1). π d! d=1 d odd

236


Show that, for real r ≥ 1 and s > 1, ∞ k−1 X X 1 k ζ (r, s) = − ζ (r + s) + 8k η(r − j) η(s − k + j), 2 j j=0 j even

k=3 k odd

where the Eta function is defined by η(s) := 1 − 21−s ζ (s). (Crandall and Buhler [344, p. 279]) 41. Let f (s) be a function defined by the Dirichlet series as follows: f (s) :=

∞ X

n−s

n=2

X

k−1

( 1).

k 0, show that Z(s) = ζ (2s) a−s +

22s−1 as−1

√

π 1

0(s) (−d)s− 2

1 ζ (2s − 1) 0 s − + Q(s), 2

where 1

2π s ·2s− 2

∞ X

nπb Q(s) = √ n σ1−2s (n) cos s 1 − a a 0(s)·(−d) 2 4 n=1 ! √ Z∞ 3 πn −d u + u−1 du. · us− 2 exp − 2a s− 12

0

(Selberg and Chowla [1017, p. 87])


237

43. Let r, s ∈ N0 with r > s. Show that Z1 Z1 0 0

x r ys dxdy 1 − xy

is a rational number whose denominator is a divisor of dr2 , where dr denotes the lowest common multiple of 1, 2, . . . , r. (Beukers [126, p. 268]) 44. Prove that ζ (3) =

∞ 1X 56 n2 − 32 n + 5 (−1)n−1 4 (2n − 1)2 n=1

1 3n 2n 3 n n n

and ζ (3) =

∞ X n=0

(−1)n 5265 n4 + 13878 n3 + 13761 n2 + 6120 n + +1040 . 3n (4n + 1)(4n + 3)(n + 1)(3n + 1)2 (3n + 2)2 72 4n n n (Amdeberhan [34, p. 2])

45. Prove that n ∞ X (−1)n X 1 1 1 4 = − 24 Li + 21 ζ (3) log 2 + (log 2) 4 6 2 n2 k2 n=1

k=1

+

17 4 π2 (log 2)2 + π , 6 480

where Li4 (z) denotes the Tetralogarithm (see Section 2.4). (Daudé et al. [367, p. 421]) 46. Prove that   h i 2 (log(2n + 1)) π (−1)n 2 ζ 00 0, 14 − ζ 00 0, 34 = 2n + 1 4 n=1    2 0 41 π −4[γ + log(2π )] log   + [γ + log(2π)]2 + 12  0 43

∞ X

and Z∞ 0

 i (log t)2 π  h 00 1 dt = 2 ζ 0, 4 − ζ 00 0, 34 cosh t 2     0 14 1 2 2 −4 log(2π ) log   + [log(2π)] + π . 4  0 34 (Shail [1021])

238


47. For an integer a = −1, prove the following asymptotic expansion: Ga (x) :=

∞ X

ζ (n − a)

n=a+1

(−x)n n!

(−x)a+1 {log x − ψ(a + 2) + γ } (a + 1)! ! a n X X (a−k)! 1 1 n + (−x) − +O (a+1)! (n−k)! n! (a + 1 − n) x

=−

n=0

(x → ∞).

n=0

(Buschman and Srivastava [195, p. 296]) 48. For any multi-index k = (k1 , k2 , . . . , kr )) (ki ∈ N), the weight wt(k) and depth dep(k) of k are defined by |k| =k1 + k2 + · · · + kr and r, respectively. The height of the index k is also defined by ht k = # j | kj = 2 . Denote, by I(k, r), the set of multi-indices k of weight k and depth r and, by I0 (k, r), the subset of I(k, r) with admissible indices, that is indices with the additional requirement that k1 = 2. For (k1 , . . . , kr ) ∈ I0 (k, r), the multiple zeta value (MZV) and the non-strict multiple zeta value (MZSV) can often be defined, respectively, as follows: ζ (k1 , k2 , . . . , kr ) :=

X

1

k1 kr n1 >···>nr >0 n1 · · · nr

and ζ ∗ (k1 , k2 , . . . , kr ) :=

X

1

nk1 · · · nkr r n1 =···=nr =1 1

.

Prove the following formulas: (a) Sum Formula. For r < k (r, k ∈ N), there hold X

ζ (k) = ζ (k) and

k∈I0 (k, r)

X

ζ ∗ (k) =

k∈I0 (k, r)

k−1 ζ (k). r−1

(b) Cyclic Sum Formula. For (k1 , . . . , kr ) ∈ I0 (k, r), r kX i −2 X

ζ ∗ (ki − j, ki+1 , . . . , kr , k1 , . . . , ki−1 , j + 1) = k ζ (k + 1),

i=1 j=0

where the empty sum means zero. (Ohno and Okuda [873, p. 3030]) 49. Kamano [624] investigated the following multiple zeta function: ζn (s1 , . . . , sn ; a) =

X 05m1 0; (m1 , . . . , mn ) ∈ Z ; (s1 , . . . , sn ) ∈ C , n

n

where Z denotes the set of integers. The special case n = 1 of ζn (s1 , . . . , sn ; a) in (a) reduces to the Hurwitz (or generalized) zeta function ζ (s, a). Also ζn (s1 , . . . , sn ; 1)


239

becomes the Euler-Zagier multiple zeta function denoted by ζn (s1 , . . . , sn ). Matsumoto [806] proved the analytic continuation of a more general class of multiple zeta functions, including (a) as a special case. Kamano [624] presented three kinds of limiting values of ζn (s1 , . . . , sn ; a) in (a) at nonpositive integers. Show that ζn0 (0 ; a) =

n−1 (−1)n−1 Y 1 0(a) k+a− log √ (n − 1)! 2 2π k=1 ∂ ∂s ζn (s ; a), ζn (s ; a) := ζn (s, . . . , s ; a)

where ζn0 (s ; a) = stood to be nil.

(n ∈ N), in (a) and an empty sum is under(Kamano [624, Theorem 3])

50. Show that z γ (z) =

∞ X

(−1)k

k=2

Li(z) k

(|z| 5 1),

where γ (z) denotes the generalized-Euler-constant function, defined by γ (z) =

∞ X

zn−1

n=1

Z1 Z1 = 0 0

1 n+1 − log n n

(|z| 5 1)

1−x dxdy (1 − xyz)(− log xy)

(C \ [1, ∞)).

(Sondow and Hadjicostas [1053, Theorem 1]) 51. Let {αn }n∈N0 be a positive sequence, such that the following infinite series: ∞ X

e−αn t

n=0

converges for any t ∈ R+ . Then, for the generalized Hurwitz–Lerch Zeta function (ρ ,...,ρ ,σ ,...,σ )

8λ11,...,λpp;µ11,...,µqq (z, s, a), defined by 2.6(44), show that (ρ ,...,ρ ,σ ,...,σ )

8λ11,...,λpp;µ11,...,µqq (z, s, a) =

∞ Z∞

1 X 0(s)

t s−1 e−(a−α0 +αn )t 1 − e−(αn+1 −αn )t

n=0 0

 · p 9q∗ 

(λ1 , ρ1 ), . . . , (λp , ρp ); (µ1 , σ1 ), . . . , (µq , σq );

 ze−t  dt

min{ 0 .

(Srivastava et al. [1107]) 53. Show that each of the following integral representations holds true:

(ρ ,...,ρ ,σ ,...,σ ) 8λ11,...,λpp;µ11,...,µqq (z, s, a) =

1 0(s)

  Z∞ (λ1 , ρ1 ), . . . , (λp , ρp ); t s−1 e−at p 9q∗  ze−t  dt (µ , σ ), . . . , (µ , σ ); q q 1 1 0 min{ 0),

(55)

in conjunction with (51), we obtain (52). Furthermore, if the various alreadydeveloped integral representations of S(r) are applied to the relationship (42), several ˜ can be obtained. For example, by applying (47) to (42), integral representations of S(r) we obtain the last identity (53). Each of the following identities holds true: T(r) :=

∞ X n=1

8n n2 + r2

3

i i 0 1 h (2) 0 (2) ψ (1 + ir) − ψ (1 − ir) + ψ (1 + ir) + ψ (1 − ir) (56) 2 r3 2 r2 i 1 = 3 [ζ (2, 1 + ir) − ζ (2, 1 − ir)] − 2 [ζ (3, 1 + ir) + ζ (3, 1 − ir)] 2r r √ 0 < |r| < 1; i = −1

=

Series Involving Zeta Functions

389

and Z∞ 2 3 + r2 1 1 1 1 − + T(r) = 3 + 3 et − 1 t 2 r 1 + r2 0

· t e−t [sin(rt) − rt cos(rt)] dt

0 .

(57)

Proof. Upon differentiating (27) and (40) with respect to r, if we combine the resulting identities and make use of 1.3(53), we find a new series T(r), which is expressed in terms of the Polygamma functions and the Hurwitz Zeta function as in (56). By applying (51) to (56), we obtain (57).

Problems 1. Evaluate the following special values of the G-function given in Section 1.4:

G

√ 1 √ n o− 2 3 ζ 2, 3 1 4 1 π 3 3  · 3 72 · A− 3 · 0 1 = exp  + − ; 3 9 54 36π

G

√ 1 √ n o 1 3 ζ 2, 3 1 13 1 4 1 π 3  · 2− 3 · 3 72 · π − 3 · A− 3 · 0 1 3 ; = exp  − + 3 9 54 36π



1 3



2 3

√ 1 n o− 5 3 ζ 2, 3 19 61 5 5 5 π 3 3  · 2 72 · 3− 144 · π 12 · A− 6 · 0 1 − ; G 16 = exp  + 3 72 36 24π  √ 1 √ n o 1 3 ζ 2, 3 17 11 1 5 π 3 5  · 2− 72 · 3 144 π − 4 · A− 6 · 0 1 3 . G 56 = exp  − + 3 72 36 24π 

√

(Gosper [498]; Choi and Srivastava [289, 294]) 2. Applying the results of Problem 1, derive the following closed-form evaluations of series involving the Riemann Zeta function: √ √ 1 ζ (k) γ π 3 3 (−1) = 1 + + − ζ 2, 6 18 12π 3 (k + 1) · 3k k=2 1 1 1 + log 2− 2 · 3 24 · π − 2 · A−4 · 0 13 ; √ √ ∞ X ζ (k) − 1 11 γ π 3 3 1 (−1)k = + + − ζ 2, 6 6 18 12π 3 (k + 1) · 3k k=2 13 73 1 + log 2− 2 · 3 24 · π − 2 · A−4 · 0 13 ; ∞ X

k

390

Zeta and q-Zeta Functions and Associated Series and Integrals ∞ X k=2

√ √ γ 312π ζ (k) π 3 1 = − + − 2, 6 18 ζ 3 (k + 1) · 3k n o−1 13 1 1 ; + log 2 2 · 3− 24 · π 2 · A4 · 0 13

√ √ ∞ X ζ (k) − 1 7 γ π 3 3 1 = − + − ζ 2, 6 6 18 12π 3 (k + 1) · 3k k=2 n o−1 85 1 7 + log 2 2 · 3− 24 · π 2 · A4 · 0 13 ; √ √ 1 π 3 1 1 ζ (2k) 3 = + − ζ 2, − log 3; 2 18 12π 3 4 (2k + 1) · 32k k=1 √ √ ∞ 3 X 3 1 ζ (2k) − 1 11 π 3 −2 − 14 + − ζ 2, + log 2 · 3 ; = 6 18 12π 3 (2k + 1) · 32k ∞ X

k=1 ∞ X k=1

n o−6 ζ (2k + 1) 3 24 3 − 74 · π · A · 0 31 = −3 − γ + log 2 · 3 ; (k + 1) · 32k

∞ n o−6 X ζ (2k + 1) − 1 30 − 79 4 · π 3 · A24 · 0 1 ; = −2 − γ + log 2 · 3 3 (k + 1) · 32k k=1 √ √ ∞ X ζ (k) 2 k 3 γ π 3 1 (−1)k = 1+ − + ζ 2, k+1 3 3 36 24π 3 k=2 n o−1 1 1 23 + log 2 2 · 3− 48 · π 2 · A−2 · 0 13 ; ∞ X

(−1)k

k=2

ζ (k) − 1 k+1

√ √ k 2 5 γ π 3 3 1 = + − + ζ 2, 3 3 3 36 24π 3 n o−1 1 49 3 1 + log 2 2 · 3 48 · 5− 2 · π 2 · A−2 · 0 13 ;

√ √ ∞ X 3 ζ (k) 2 k γ π 3 1 =− − + ζ 2, k+1 3 3 36 24π 3 k=2 1 1 1 + log 2− 2 · 3− 48 · π − 2 · A2 · 0 13 ; √ √ ∞ X ζ (k) − 1 2 k 4 γ π 3 3 1 = − − + ζ 2, k+1 3 3 3 36 24π 3 k=2 1 73 1 + log 2− 2 · 3− 48 · π − 2 · A2 · 0 13 ; √ √ ∞ X ζ (2k) 2 2k 1 π 3 3 1 1 = − + ζ 2, − log 3; 2k + 1 3 2 36 24π 3 4 k=1 √ √ ∞ 1 X ζ (2k) − 1 2 2k 3 π 3 3 3 1 = − + ζ 2, + log 3− 4 · 5− 4 ; 2k + 1 3 2 36 24π 3 k=1


391

∞ n o3 X 3 11 3 3 ζ (2k + 1) 2 2k = − − γ + log 2− 2 · 3 16 · π − 2 · A6 · 0 13 ; k+1 3 2 k=1

∞ n o3 X 61 9 3 3 ζ (2k + 1) − 1 2 2k 1 = − − γ + log 2− 2 · 3− 16 · 5 4 · π − 2 · A6 · 0 13 ; k+1 3 2 k=1

√ √ ζ (k) 1 γ π 3 3 (−1) = 1+ + − ζ 2, 12 6 4π 3 (k + 1) · 6k k=2 n o2 11 11 + log 2− 12 · 3 24 · π −1 · A−5 · 0 13 ;

∞ X

∞ X

k

(−1)k

k=2

∞ X k=2

√ √ γ π 3 3 1 ζ (k) − 1 23 + + − ζ 2, = 12 12 6 4π 3 (k + 1) · 6k n o2 61 155 + log 2 12 · 3 24 · 7−6 · π −1 · A−5 · 0 13 ; √ √ γ ζ (k) π 3 3 1 = − + − ζ 2, 12 6 4π 3 (k + 1) · 6k n o−2 11 11 + log 2 12 · 3− 24 · π · A5 · 0 31 ;

√ √ ∞ X ζ (k) − 1 13 3 γ π 3 1 = − + − ζ 2, 12 12 6 4π 3 (k + 1) · 6k k=2 n o−2 − 155 6 5 − 61 12 24 ·3 · 5 · π · A · 0 13 ; + log 2 ∞ X k=1

√ √ ζ (2k) 1 π 3 3 1 = + − ζ 2, ; 2 6 4π 3 (2k + 1) · 62k

√ √ ∞ X ζ (2k) − 1 3 π 3 3 1 = + − ζ 2, + log 53 · 7−3 ; 2k 2 6 4π 3 (2k + 1) · 6 k=1

∞ n o−24 X ζ (2k + 1) 11 − 11 2 · π 12 · A60 · 0 1 = −6 − γ + log 2 · 3 ; 3 (k + 1) · 62k k=1

∞ X k=1

∞ X k=2

ζ (2k + 1) − 1 = −5 − γ (k + 1) · 62k n o−24 155 + log 2−61 · 3− 2 · 536 · 736 · π 12 · A60 · 0 13 ;

ζ (k) (−1) k+1 k

√ √ k 5 5 π 3 3 1 = 1+ γ − + ζ 2, 6 12 30 20π 3 n o−2 49 61 + log 2 60 · 3− 120 · π · A−1 · 0 13 ;

392

Zeta and q-Zeta Functions and Associated Series and Integrals ∞ X

(−1)

√ √ k 5 19 3 5 π 3 1 = + γ− + ζ 2, k+1 6 12 12 30 20π 3 n o−2 121 83 6 + log 2 60 · 3 120 · 11− 5 · π · A−1 · 0 31 ;

k ζ (k) − 1

k=2

√ √ ∞ X ζ (k) 5 k 3 5 π 3 1 =− γ − + ζ 2, k+1 6 12 30 20π 3 k=2 n o2 61 −1 − 49 60 120 ; · 3 · π · A · 0 31 + log 2 √ √ ∞ X ζ (k) − 1 5 k 17 5 π 3 1 3 − γ− + ζ 2, = k+1 6 12 12 30 20π 3 k=2 n o2 83 121 ; + log 2− 60 · 3− 120 · π −1 · A · 0 13 √ √ ∞ X ζ (2k) 5 2k 1 π 3 3 1 = − + ζ 2, ; 2k + 1 6 2 30 20π 3 k=1

∞ X k=1

ζ (2k) − 1 2k + 1

√ √ 2k 5 3 π 3 3 1 3 = − + ζ 2, − log 11; 6 2 30 20π 3 5

∞ n o 24 X 61 12 12 49 ζ (2k + 1) 5 2k 6 5 = − − γ + log 2− 25 · 3 50 · π − 5 · A 5 · 0 13 ; k+1 6 5 k=1

∞ X k=1

∞ X k=2

ζ (2k + 1) − 1 k+1

2k 5 1 = − −γ 6 5 n o 24 83 12 121 36 12 5 ; + log 2− 25 · 3− 50 · 11 25 · π − 5 · A 5 · 0 13

√ √ k 3 4 19 2 π 3 1 = + γ+ − ζ 2, 3 12 3 72 48π 3 1 1 3 1 + log 2− 2 · 3 96 · 7− 4 · π − 2 · A−1 · 0 13 ; √ √ ∞ X ζ (k) − 1 4 k 17 2 3 π 3 1 = − γ+ − ζ 2, k+1 3 12 3 72 48π 3 k=2 n o −1 1 1 − 49 1 2 96 2 + log 2 · 3 ·π ·A· 0 3 ;

(−1)k

ζ (k) − 1 k+1

√ √ ∞ 1 X 3 ζ (2k) − 1 4 2k 3 π 3 3 1 = + − ζ 2, + log 3− 4 · 7− 8 ; 2k + 1 3 2 72 48π 3 k=1

∞ n o− 3 X 3 25 9 3 3 ζ (2k + 1) − 1 4 2k 1 2 = − − γ + log 2 4 · 3− 64 · 7 16 · π 4 · A 2 · 0 13 ; k+1 3 8 k=1

Series Involving Zeta Functions ∞ X

(−1)

k=2

393

√ √ k 5 47 5 3 π 3 1 = + γ− + ζ 2, k+1 3 30 6 90 60π 3 n o−1 7 59 1 4 + log 2− 10 · 3− 120 · π 2 · A− 5 · 0 13 ;

k ζ (k) − 1

√ √ ∞ X ζ (k) − 1 5 k 43 5 3 π 3 1 = − γ− + ζ 2, k+1 3 30 6 90 60π 3 k=2 1 4 1 1 + log 2− 2 · 3− 120 · π − 2 · A 5 · 0 13 ; √ √ ∞ 3 X 1 ζ (2k) − 1 5 2k 3 π 3 3 1 = − + ζ 2, + log 2− 5 · 3− 4 ; 2k + 1 3 2 90 60π 3 k=1

∞ n o 6 X 29 3 24 3 ζ (2k + 1) − 1 5 2k 2 5 = − − γ + log 2 25 · 3 100 · π − 5 · A 25 · 0 13 ; k+1 3 25 k=1

∞ X

(−1)

k=2

√ √ k 7 131 7 π 3 1 3 = + γ+ − ζ 2, k+1 6 84 12 42 28π 3 n o2 83 − 71 − 67 −1 − 57 84 168 + log 2 · 3 · 13 · π · A · 0 13 ;

k ζ (k) − 1

√ √ ∞ X 7 π 3 1 ζ (k) − 1 7 k 121 3 = − γ+ − ζ 2, k+1 6 84 12 42 28π 3 k=2 n o −2 83 5 71 ; + log 2 84 · 3− 168 · π · A 7 · 0 31 √ √ ∞ X ζ (2k) − 1 7 2k 3 π 3 1 3 3 − ζ 2, = + − log 13; 2k + 1 6 2 42 28π 3 7 k=1

∞ X k=1 ∞ X

ζ (2k + 1) − 1 k+1

(−1)k

k=2

ζ (k) − 1 k+1

2k n o− 24 71 83 12 60 7 5 7 = − − γ + log 2 49 · 3− 98 · π 7 · A 49 · 0 13 ; 6 49

11 6

k =

√ √ 203 11 π 3 3 1 + γ− + ζ 2, 132 12 66 44π 3 n o−2 133 6 5 6 109 ; + log 2 132 · 3− 264 · 5 11 · 17− 11 · π · A− 11 · 0 31

√ √ ∞ X ζ (k) − 1 11 k 193 11 π 3 3 1 = − γ− + ζ 2, k+1 6 132 12 66 44π 3 k=2 n o2 133 5 − 109 −1 132 264 11 + log 2 · 3 · π · A · 0 31 ; √ √ ∞ 3 X 3 ζ (2k) − 1 11 2k 3 π 3 3 1 = − + ζ 2, + log 5 11 · 17− 11 ; 2k + 1 6 2 66 44π 3 k=1

394


∞ X ζ (2k+1)−1 11 2k k=1

k+1

6

5 −γ 121 n o 24 133 36 36 12 60 109 11 . + log 2− 121 · 3 242 · 5− 121 · 17 121 · π − 11 · A 121 · 0 13

=−

(Choi and Srivastava [289, 294]) 3. Prove that ∞ λ X ζ (m, a) m+λ X λ 0 z = ζ (−k, a − z) zλ−k − ζ 0 (−λ, a) m+λ k

m=2

−

k=0

λ−1 X `=0

ζ (−`, a) λ−` zλ+1 z − [ψ(λ + 1) − ψ(a) + γ ] λ−` λ+1

(λ ∈ N0 ; |z| < |a|). (Kanemitsu et al. [628])

4. Prove the following identity: 1

∞ X k=2

γ 1 5 1 4 (−1)k ζ (k) = + + log 2 + log π − (k + 1)(k + 2) 6 2 18 3 3

Z2

log 0(t) dt,

0

where γ denotes the Euler-Mascheroni constant defined by 1.1(2). (Janos and Srivastava [606]) 5. Prove the following companion of the summation formulas deduced in Problems 46 and 47 (Chapter 1): ζ (k − 1) =

∞ X n=k

(−1)n+k s(n, k). (n − 1)(n − 2) · (n − 1)!

(cf. Jordan [614, p. 339]; see also Equation 3.5(16) and Hansen [531, p. 348]) 6. For 8(z, s, a) defined by 2.5(1), derive the following generalization of the expansion formula 3.2(7): ∞ X (λ)n 8(z, λ + n, a) tn = 8(z, λ, a − t) n!

(|t| < |a|; λ 6= 1).

n=0

(cf. Equation 2.5(33)) 7. For every nonnegative integer `, prove the following formula: ∞ X

8n (z, k, a)

k=2

−

` X tk+` ` 0 = 8 (z, −k, a − k)t`−k k+` k n k=0

`−1 X k=0

8n (z, −k, a)

t`−k

t`+1 − H` Ln,1 (z, a) + Ln,2 (z, a) − 80n (z, −`, a) `−k `+1 (|t| < a ; |z| < 1 ; ` ∈ N0 ),


395

where Ln,1 (z, a) and Ln,2 (z, a) are given, respectively, as Ln,1 (z, a) := lim {(s + `)8n (z, s + ` + 1, a)} s→−`

and Ln,2 (z, a) := lim {8n (z, s + ` + 1, a) + (s + l)80n (z, s + ` + 1, a)}. s→−`

(See Choi et al. [273, Theorem 2]) 8. Show that ∞ X

(−1)n−1

2 Hn−1 + Hn2 On n

n=1

=

π4 π2 7 + log2 2 − log 2 ζ (3). 96 8 4 (See Zheng [1264])

9. Show that, for λ, µ ∈ N0 , n X n + µn n + λn

Hλn+k n−k 2n + λn + µn = Hλn+n + Hλn+µn+n − Hλn+µn+2n . n k

k=0

(See Chu and de Donno [316, Theorem 1]) 10. Show that ∞ X k=1

22k−1 − 1 ζ (2k) (2p)2k (m + k) p 2m 1 − γ (p) + (−1)m (2m)! 1 − 2−2m−1 ζ (2m + 1) 4m π p−1 m (2m)! X (−1)k (2π )−2k X (2` + 1)π 2` + 1 − cos ζ 2k + 1, 2p (2m − 2k)! p 2p

=−

k=1

`=0

p−1 m X (−1)k (2π )−2k X (2` + 1)π 2` + 1 π sin ζ 2k, − (2m)! p (2m − 2k + 1)! p 2p k=1

`=0

(m ∈ N; p ∈ N \ {1}) and ∞ X k=1

22k−1 − 1 1 1 ζ (2k) = − − γ (p) 4(2m + 1) 2 (2p)2k (2m + 2k + 1)

+

p−1 m (2m + 1)! X (−1)k (2π )−2k X (2` + 1)π 2` + 1 sin ζ 2k + 2, 8pπ (2m − 2k)! p 2p k=0

`=0

p−1 m (2m + 1)! X (−1)k (2π )−2k X (2` + 1)π 2` + 1 − cos ζ 2k + 1, 4p (2m − 2k + 1)! p 2p k=1

`=0

(m ∈ N0 ; p ∈ N \ {1}),

396


where γ (p) is given by γ (p) =

p−1 ∞ X X

(2` + 1)π 1 cos . 2pj + 2` + 1 p

j=0 `=0

(Cho et al. [257, Equations (40) and (41)]) 11. Show that n X

1 − 5 j Hj + 5 j Hn−j

j=0

n 2 X n 5 n n+j = (−1)n , j j j j=0

where Hn is the harmonic number. (Paule and Schneider [890, Eq. (5)]) 12. For n ∈ N, define F(n) =

n X n+j 2 n 2 j=1

j

j

1 + 2 j Hn+j + 2 j Hn−j − 4 j Hj .

Show that F(n) = 0. (Ahlgren and Ono [14, Theorem 7]) 13. A regularized version of Riemann Zeta function is ζ (s) −

∞ X s 1 (−1)n bn . = n s−1 n=0

Show that bn = n (1 − γ − Hn−1 ) −

∞ 1 X n + (−1)k ζ (k). 2 k k=2

(Flajolet and Vepstas [455, Eq. (2)]) 14. Show that Zt

log(sin θ ) dθ = t(log t − 1) −

∞ X (−1)k+1 B2k (2t)2k+1 4k(2k + 1)!

(|t| < π).

k=1

0

(Monegato and Strozzi [839, Eq. (17)]) 15. Show that  3 n k X X n   = n 8n + 8n − 3n 2n 2n . j 2 4 n k=0

j=0

(Zhang and Wang [1256, Eq. (1)]) 16. Let {n }n∈N0 be a suitably bounded sequence of complex numbers. Also let the parameters λ and µ be constrained by 0

(λ ∈ / Z− 0 ).


397

Then show that ∞ ∞ ∞ X (ν)n X k zk X k zk = [ψ(λ + k) − ψ(λ − ν + k)] n(λ)n (λ + n)k k! (λ)k k! n=1

k=0

k=0

and ∞ ∞ ∞ X zk (ν)n X (ν + n)k k zk X (ν)k k = [ψ(λ + k) − ψ(λ − ν)] , n(λ)n (λ + n)k k! (λ)k k! n=1

k=0

k=0

provided that the series involved converge absolutely. (Nishimoto and Srivastava [863, p. 104]; see also 3.5(12) and R. Srivastava [1115] for further results of these classes) 17. Derive the following hypergeometric identity: 1 + a, 2 + 12 a, 1 + b, 1 + c, 2 + 2a − b − c + N, 1 − N, 1, 1;

 8 F7 

2 + a, 1 +

=

1 2 a, 2 + a − b, 2 + a − c, 1 + b + c − a − N, 2 + a + N, 2;

 1

(1 + a)(1 + a − b)(1 + a − c)(b + c − a − N)(1 + a + N) (2 + a)bc(1 + 2a − b − c + N)N · ψ(1 + a) − ψ(1 + a + N) + ψ(1 + a − b + N) − ψ(1 + a − b) + ψ(1 + a − c + N) − ψ(1 + a − c) + ψ(1 + a − b − c) − ψ(1 + a − b − c + N)

(N ∈ N),

provided that no zeros appear in the denominator of either member. (Srivastava [1079, p. 80, Eq. (1.3)]) 18. Prove the following identity: n X k=1

k

n m k k n+k m+k k k

1 3 1 1 2 Hk−1 + Hk2 = e + e0 e1 + e2 , 12 0 4 6

where m, n ∈ N; n 5 m and (`+1)

(`+1) e` := Hn(`+1) + Hm − Hn+m

(` ∈ N0 ). (Choi [266, Eq. (3.20)])

4 Evaluations and Series Representations

Based essentially on some of the most recent works on the subject by Srivastava [1084, 1086], Srivastava and Tsumura [1111] and others, this chapter aims at investigating rather systematically several interesting evaluations and representations of the Riemann Zeta function ζ (s) when s ∈ N \ {1}. We begin by presenting some of the methods of evaluation of ζ (2n) (n ∈ N). We then proceed to develop various families of rapidly convergent series representations for ζ (2n + 1) (n ∈ N). Finally, in one of the many computationally useful special cases considered here, we observe that ζ (3) can be represented by means of a series that converges much more rapidly than that in Euler’s celebrated formula 3.1(8), as well as the series 4.2(2) used by Ape´ ry [56] in his proof of the irrationality of ζ (3). Symbolic and numeric computations using Mathematica (Version 4.0) for Linux show, among other things, that only 50 terms of this series are capable of producing an accuracy of seven decimal places.

4.1 Evaluation of ζ (2n) The solution of the so-called Basler problem (cf., e.g., Spiess [1061, p. 66]): ζ (2) =

∞ X 1 π2 = 6 k2

(1)

k=1

was first found in 1736 by Leonhard Euler (1707–1783), although Jakob Bernoulli (1654–1705) and Johann Bernoulli (1667–1748) did their utmost to sum the series in (1). In fact, the former of these Bernoulli brothers did not live to see the solution of the problem, and the solution became known to the latter soon after Euler found it (see, for details, Knopp [676, p. 238]). The Basler problem (1) has been solved in the mathematical literature in many different ways. In addition to numerous papers containing elementary proofs of (1), many of which are referenced by Stark [1125], there are a fairly large number of books on complex analysis and advanced calculus in which (1) is proven by using Cauchy’s residue calculus, Weierstrass’s product theorem, Parseval’s theorem, Fourier series expansions and so on. Some of these books were referred to by Choe [258], who seems to have rediscovered one of Euler’s remarkably elementary proofs of (1) Zeta and q-Zeta Functions and Associated Series and Integrals. DOI: 10.1016/B978-0-12-385218-2.00004-9 c 2012 Elsevier Inc. All rights reserved.

400


(cf., e.g., Ayoub [81, p. 1079]), which is based on the integration of the following Taylor series expansion of arcsin x near x = 0: ∞ X 1 · 3 · 5 · · · (2k − 1) x2k+1 arcsin x = x + 2 · 4 · 6 · · · (2k) 2k + 1

(−1 5 x 5 1)

(2)

k=1

with respect to t (where x = sin t), by means of Wallis’ integral formula in the form: Zπ/2 sin2k+1 t dt = 0

2 · 4 · 6 · · · (2k) 1 · 3 · 5 · · · (2k + 1)

(k ∈ N).

(3)

Motivated largely by the aforementioned (Euler’s) proof of (1) detailed by Choe [258], Choi and Rathie [279] gave an essentially analogous derivation of (1), by appealing to the Gauss summation theorem 1.4(7). Subsequently, Choi, Rathie and Srivastava [281] showed how the derivation of (1) by Choi and Rathie [279] can be accomplished without using the Gauss summation theorem 1.4(7). They also presented several other evaluations of ζ (2) and related sums, including (for example) a fairly straightforward evaluation based on the theory of hypergeometric series.

Elementary and Hypergeometric Evaluations of ζ (2) Starting from the elementary integral: 1 (arcsin x)2 = 2

Zx 0

arcsin t dt √ 1 − t2

(4)

and replacing arcsin t by its Taylor series expansion given by (2), it is not difficult to obtain the representation: ∞ −2k Z x X 1 2k 2 t2k+1 2 (arcsin x) = dt √ 2 k 2k + 1 1 − t2 k=0

(−1 5 x 5 1)

(5)

0

or, equivalently, 1 (arcsin x)2 = 2

∞ X k=0

1 2 k

k!

1 2k + 1

Zx 0

t2k+1 dt √ 1 − t2

(−1 5 x 5 1).

(6)

Choi and Rathie [279, p. 396] express the second member of (6) as a hypergeometric 2 −1/2 to evaluate the integral, 2 F1 series, by using the binomial expansion of 1 − t set x = 1 and then apply the Gauss summation theorem 1.5(7).

Evaluations and Series Representations

401

The use of the Gauss summation theorem 1.5(7) in the above derivation of (1) by Choi and Rathie [279] √ can easily be avoided. Indeed, in the special case of (6) when x = 1, if we set t = u and apply the known integral formula 1.1(39) and 1.1(42), we readily find from (6) that ∞ X k=0

π2 1 = , 8 (2k + 1)2

(7)

which, in view of 2.3(1), is equivalent to (1). By virtue of the relationships 2.5(26) and 1 1 8 1, s, = ζ s, = 2s − 1 ζ (s), 2 2

(8)

2.5(17) immediately yields the hypergeometric representations: ζ (n, a) = a−n n+1 Fn (1, a, . . . , a; a + 1, . . . , a + 1; 1)

(n ∈ N \ {1})

(9)

(n ∈ N \ {1}).

(10)

and ζ (n) =

2n n 2 −1

1 1 3 3 F 1, , . . . , ; , . . . , ; 1 n+1 n 2 2 2 2

In particular, by setting n = 2 in (10), we have 4 1 1 ζ (2) = 3 F2 1, , ; 23 , 32 ; 1 , 3 2 2

(11)

which can be evaluated by applying Dixon’s summation theorem 3.5(4) with a=1

and b = c =

1 2

or by applying Whipple’s summation theorem 3.5(6) with 1 a=b= , 2

c=1

3 and d = e = . 2

We, thus, find from (11) and either one of the above applications that 1 π2 4 n 3 o3 0 2 0 = . ζ (2) = 3 2 6

(12)

Next, by appealing directly to the definition 2.3(1) or (9) with a = 1, we obtain the hypergeometric representation: ζ (n) = n+1 Fn (1, . . . , 1; 2, . . . , 2; 1)

(n ∈ N \ {1}) .

(13)

402


The representations (10) and (13), together, yield the hypergeometric transformation: 1 3 1 3 , . . . , ; 1 F , . . . , ; 1, n+1 n 2 2 2 2 (14) −n = (1 − 2 ) n+1 Fn (1, . . . , 1; 2, . . . , 2; 1) (n ∈ N \ {1}) , which, for n = 2, assumes the form: 1 1 3 3 3 3 F2 1, , ; 2 , 2 ; 1 = 3 F2 (1, 1, 1; 2, 2; 1). 2 2 4

(15)

We remark in passing that, by making use of contiguous analogues of Dixon’s summation theorem 3.5(4) and Whipple’s summation theorem 3.5(6), which were given recently by Lavoie et al. [731, p. 268, Eq. (2); 732, p. 294, Eq. (4)] we can easily express the sums of many series [analogous to those occurring in (1) and (7)] in terms of ζ (2). The details are fairly straightforward. Since ζ (n) =

Z1

Z1 ···

0

0

dt1 . . . dtn 1 − t1 . . . tn

(n ∈ N \ {1}),

(16)

it is not difficult to provide yet another solution of the Basler problem (1), by using the following change of variables: v+u t1 = √ 2

v−u and t2 = √ 2

in the double integral resulting from (16) when n = 2 (cf., e.g., Ojha and Singh [874]).

The General Case of ζ (2n) Following Srivastava [1086], let us begin, by recalling the following summation formula for the Bernoulli polynomials (cf. Hansen [531, p. 342, Entry (50.11.10)]): [n/2] X k=0

(−n)2k 1 B2k (x)Bn−2k (y) = (x + y − 1)nBn−1 (x + y) (2k)! 2 + (−1)n (x − y)nBn−1 (x − y) − (n − 1) Bn (x + y) + (−1)n Bn (x − y) ,

(17)

which, upon setting x = 1 and y = 0 (and replacing n by 2n), yields n X 2n k=0

2k

B2k B2n−2k = −nB2n−1 − (2n − 1)B2n

(n ∈ N0 ) .

(18)


403

Finally, by transposing the terms for k = 0 and k = n in (18) to the right-hand side and making use of 1.7(7), we obtain n−1 X 2n

2k

k=1

B2k B2n−2k = −(2n + 1)B2n

(n ∈ N \ {1}),

(19)

which is, in view of the relationship 2.3(18), equivalent to 2.3(20), which (as noted there) can also be used to evaluate ζ (2n) (n ∈ N \ {1}) with (1). To give a direct proof of the summation formula 2.3(20), let us put (cf. Prudnikov [917]) (x) :=

∞ X

exp −π 2 k2 x

(x > 0),

(20)

k=1

so that Zx 0

∞ 3 X exp −π 2 k2 x , (x − t)(t)dt = x(x) + 2 π 2 k2

(21)

k=1

which, by virtue of (1), readily yields Zx

(x − t)(t)dt =

lim

x→0

1 4

(22)

0

and Zx

(x − t)(t)dt = O xe−x

(x → ∞).

(23)

0

Making use of (21) and (23), it is easily seen that Z∞Z∞ Is := (x + y)s−2 (x)(y)dx dy 0 0

Z∞ Zx s−2 = x dx (x − t)(t)dt 0

0

0(s − 1) = s + 12 ζ (2s) π 2s

( 1).

(24)

404


Conversely, since Z∞ 0(s) ts−1 (t)dt = 2s ζ (2s) π

0; s 6= 21 ,

(25)

0

which is an immediate consequence of the definition (20) and 2.3(1), an appeal to the Binomial theorem for (x + y)n−2 in the form: (x + y)n−2 =

n−1 X n − 2 k−1 n−k−1 x y k−1

(n ∈ N \ {1})

(26)

k=1

would lead us also to n−1

(n − 2)! X ζ (2k)ζ (2n − 2k) In = π 2n

(n ∈ N \ {1}).

(27)

k=1

Now, the summation formula 2.3(20) follows upon equating the two values of In given by (24) (with s = n) and (27). Remark 1 The recursion formula (19) is a well-known (rather classical) result for Bernoulli numbers. It appears (for example) in Nielsen’s book [861] and was also derived independently by Underwood [1177]. The equivalent result 2.3(20) was proven, by elementary methods by Williams [1229, p. 20, Theorem I] (see also Apostol [63, p. 427]). Remark 2 In terms of the L-function defined by [cf. Eq. 2.5(1)] L(s) =

∞ X (−1)k −s 1 = 2 8 −1, s, 2 (2k + 1)s

( 0),

(28)

k=0

Williams [1229, p. 22, Theorem II] gave an interesting companion of the result 2.3(20) in the form: n X

L(2k − 1)L(2n − 2k + 1) = n − 12 1 − 2−2n ζ (2n),

(29)

k=1

which appears erroneously in Hansen [531, p. 357, Entry (54.7.1)]. Since L(1) is the π well-known Gregory series for (with L(2) being the familiar Catalan constant G), 4 by setting n = 1 in (29), we immediately obtain ζ (2) =

8 π2 {L(1)}2 = . 3 6

(30)


405

Remark 3 In view of the constraint 0 s 6= 12 associated with the Mellin transform in (25), an earlier attempt by Kalla and Villalobos [623] to extend the summation formula 2.3(20), by expressing Is as a non-terminating (infinite) sum analogous to (27), cannot be justified. Thus, the main result of Kalla and Villalobos [623, p. 17, Eq. (17)] holds true only in the finite form given already by the well-known result 2.3(20).

4.2 Rapidly Convergent Series for ζ (2n + 1) On the subject of series representations for ζ (3), in addition to Euler’s result 3.1(8), Chen and Srivastava [252] gave many series expressions for ζ (3), which are more rapidly convergent than that in 3.1(8), including ζ (3) = −

∞ ζ (2k) 8π 2 X . 5 (2k + 1)(2k + 2)(2k + 3) 22k

(1)

k=0

And, as pointed out by (for example) Chen and Srivastava [252, pp. 180–181], another interesting series representation: ∞

ζ (3) =

5 X (−1)k−1 , 2k 2 k=1 k3 k

(2)

which played a key roˆ le in Ape´ ry’s proof of the irrationality of ζ (3) (see Ape´ ry [56]), was proven independently by (among others) Hjortnaes [559], Gosper [496] and Ape´ ry [56]. Furthermore, in their recent work on the Ray-Singer torsion and topological field theories, Nash and O’Connor [855, 856] obtained a number of remarkable integral expressions for ζ (3), including (for example) the following result [856, p. 1489 et seq.]: 8 2π 2 log 2 − ζ (3) = 7 7

Zπ/2 z2 cot z dz.

(3)

0

Since (Erde´ lyi et al. [421, p. 51, Eq. 1.20(3)]) z cot z = −2

∞ X k=0

ζ (2k)

z 2k π

(|z| < π ),

(4)

the result (3) is obviously equivalent to the series representation (cf. Da¸browski [362, p. 202]; see also Chen and Srivastava [252, p. 191, Eq. (3.19)]): ! ∞ X 2π 2 ζ (2k) ζ (3) = log 2 + . (5) 7 (k + 1)22k k=0

406


Moreover, by integrating by parts, it is easily seen that Zπ/2 Zπ/2 2 z cot z dz = −2 z log sin z dz, 0

(6)

0

so that the result (3) is equivalent also to the integral representation: 16 2π 2 log 2 + ζ (3) = 7 7

Zπ/2 z log sin z dz,

(7)

0

which was proven in the aforementioned 1772 paper by Euler (cf., e.g., Ayoub [81, p. 1084]; see also Srivastava [1086]). If follows from 3.2(7) that (cf., e.g., Hansen [531, p. 359]; see also Srivastava [1084] where other references are also cited) ∞ X (s)2k 1 ζ (s + 2k, a) t2k = [ζ (s, a − t) + ζ (s, a + t)] (2k)! 2

(|t| < |a|) .

(8)

k=0

In view of 2.3(3), the special case of the identity (8) when a = 1 and t = rewritten in the form:

1 m

can be

∞ X (s)2k ζ (s + 2k) (2k)! m2k k=0

 s (2 − 1) ζ (s) − 2s−1       m−2 = 1 X j  s  s   ζ s,   2 (m − 1)ζ (s) − m − m

(m = 2)

(9)

(m ∈ N \ {1, 2}),

j=2

where (as usual) an empty sum is interpreted as nil. In addition to the case m = 2, the formula (9) simplifies also in the cases when m = 3, 4 and 6, and we, thus, obtain the identities: ∞ X (s)2k ζ (s + 2k) 1 s = (3 − 1) ζ (s) − 3s , (2k)! 2 32k

(10)

∞ X (s)2k ζ (s + 2k) 1 s = (4 − 2s ) ζ (s) − 4s 2k (2k)! 2 4

(11)

k=0

k=0


407

and ∞ X (s)2k ζ (s + 2k) 1 s = (6 − 3s − 2s + 1) ζ (s) − 6s , 2k (2k)! 2 6

(12)

k=0

respectively. Identities of this kind seem to have first appeared in the work of Ramaswami [966], who actually proved the cases m = 2, 3 and 6 of the general result in (9). Each of these three identities of Ramaswami [966] can also be found in the work of Hansen [531, p. 357], who referred to Apostol [61] as his source for the identities (10) and (12) only. As a matter of fact, Apostol [61] reproduced the identities (10) and (12) from Ramaswami’s work [966] and then proved an interesting arithmetical generalization of these identities (see also Klusch [673, p. 520]). In its slightly variant form: ∞ X (s + 1)2k ζ (s + 2k) = (2s − 2) ζ (s), (2k)! 22k

(13)

k=1

the case m = 2 of the general result (9) was applied by Zhang and Williams [1251] (and, subsequently, by Cvijovic´ and Klinowski [351]) with a view to finding two seemingly different series representations for ζ (2n + 1) (n ∈ N). Srivastava [1084] obtained much more rapidly converging series representations for ζ (2n + 1) (n ∈ N), chiefly by appealing appropriately to each of the aforementioned cases (m = 2, 3, 4 and 6) of the general result (9). Following his work (Srivastava [1084]), we begin, here, with the case m = 2 of the general result (9). Upon separating the first n + 1 terms of the series occurring on the left-hand side, if we transpose the terms for k = 0 and k = n to the right-hand side, we obtain n−1 ∞ X X (s)2k ζ (s + 2k) (s)2k ζ (s + 2k) + 2k (2k)! (2k)! 2 22k k=1

k=n+1

(14)

(s)2n ζ (s + 2n) = (2s − 2)ζ (s) − 2s−1 − , (2n)! 22n which readily yields the identity: n−1 ∞ X X (s)2k 2(n−k) (s)2n+2k ζ (s + 2n + 2k) 2 ζ (s + 2k) + (2k)! (2n + 2k)! 22k k=1

k=1

(s)2n = 22n (2s − 2) ζ (s) − 2s+2n−1 − ζ (s + 2n) (2n)!

(n ∈ N).

(15)

408


Now, we apply the functional equation 2.3(12) in the first term on the right-hand side of (15) and divide both sides by s + 2n. We, thus, find that X ∞ n−1 X (s)2n (s + 2n +1)2k−1 ζ (s + 2n + 2k) (s)2k 2(n−k) ζ (s + 2k) 2 + (2k)! s + 2n (2n + 2k)! 22k k=1 k=1    sin 12 πs  (16) = 2s+2n (2s − 2) π s−1 0(1 − s) ζ (1 − s)  s + 2n  s+2n−1 2 + (s)2n (2n)! ζ (s + 2n) (s 6= −2n; n ∈ N). − s + 2n It is easy to see from 1.6(10) that d {(s)2n } = −(2n)! H2n ds s=−2n

(n ∈ N),

(17)

where Hn denotes the familiar harmonic numbers, defined by 3.2(36). We observe also that the following limit formula: lim

s→−2n

ζ (s + 2k) (−1)n−k = (2n − 2k)! ζ (2n − 2k + 1) s + 2n 2(2π)2(n−k) (k = 1, . . . , n − 1; n ∈ N \ {1})

(18)

is needed in the first sum on the left-hand side of (16) only when this sum is nonzero (that is, only when n ∈ N \ {1}). Furthermore, by l’Hoˆ pital’s rule, we have    s+2n−1 + (s)2n ζ (s + 2n)    2   (2n)! lim  s→−2n  s + 2n     (19) ζ (s + 2n) (s)2n 0 d s+2n−1 + ζ (s + 2n) = 2 log 2 + {(s)2n } · ds (2n)! (2n)! s=−2n 1 = (H2n − log π) (n ∈ N). 2 Finally, by letting s → −2n in (16) and making use of the limit relationships (18) and (19), we obtain Srivastava’s first series representation for ζ (2n + 1): ζ (2n + 1) = (−1)

n−1

" n−1 (2π)2n H2n − log π X (−1)k ζ (2k + 1) + (2n)! (2n − 2k)! 22n+1 − 1 π 2k k=1 (20) # ∞ X (2k − 1)! ζ (2k) +2 (n ∈ N). (2n + 2k)! 22k k=1


409

In precisely the same manner, we can apply the identities (10), (11) and (12) to prove the following additional series representations for ζ (2n + 1) (Srivastava [1084, p. 389]):  2 n−1 2n H − log X 2n 3π (−1)k ζ (2k + 1) 2(2π)  + ζ (2n + 1) = (−1)n−1 2n+1  (2n)! (2n − 2k)! 2 2k 3 −1 k=1 3π  ∞ X (2k − 1)! ζ (2k)  +2  (n ∈ N); (2n + 2k)! 32k k=1

(21) 

ζ (2n + 1) = (−1)

n−1

+

n−1 X k=1

H − log 12 π 2(2π)2n  2n  (2n)! 24n+1 + 22n − 1

 ∞ X ζ (2k + 1) (2k − 1)! ζ (2k)   +2 (2n − 2k)! 1 2k (2n + 2k)! 42k k=1 π 2 (−1)k

(n ∈ N); (22)



ζ (2n + 1) = (−1)

n−1

+

n−1 X k=1

H − log 13 π 2(2π)2n  2n  (2n)! 32n (22n + 1) + 22n − 1 ∞ X



(2k − 1)! ζ (2k)  (−1)k ζ (2k + 1)  2k + 2 (2n − 2k)! 1 (2n + 2k)! 62k k=1 π 3

(n ∈ N). (23)

Remarks and Observations The series representation (20) is markedly different from each of the series representations for ζ (2n + 1), which were given earlier by Zhang and Williams [1251, p. 1590, Eq. (3.13)] and (more recently) by Cvijovic´ and Klinowski [351, p. 1265, Theorem A]. Since ζ (2k) → 1 as k → ∞, the general term in the series representation (20) has the order estimate: O 2−2k · k−2n−1 (k → ∞; n ∈ N), whereas the general term in each of these earlier series representations has the order estimate: O 2−2k · k−2n (k → ∞; n ∈ N).

410


By suitably combining (20) and (22), it is fairly straightforward to obtain the series representation:

ζ (2n + 1) = (−1)

n−1

+

" log 2 2(2π)2n (22n − 1)(22n+1 − 1) (2n)!

n−1 X (−1)k (22k − 1) ζ (2k + 1) (2n − 2k)! π 2k k=1 ∞ X

−2

k=1

(24)

(2k − 1)! (22k − 1) ζ (2k) (2n + 2k)! 24k

# (n ∈ N).

Making use of the relationship given in Problem 11 of Chapter 2, the series representation (24) can immediately be put in the form:

ζ (2n + 1) = (−1)

n−1

+

" 2(2π)2n log 2 2n 2n+1 (2 − 1)(2 − 1) (2n)!

n−1 X (−1)k (22k − 1) ζ (2k + 1) (2n − 2k)! π 2k

(25)

k=1

# ∞ 1 X (−1)k−1 π 2k + E2k−1 (0) 2 (2n + 2k)! 2

(n ∈ N),

k=1

which is a slightly modified (and corrected) version of a result proven in a significantly different way by Tsumura [1169, p. 383, Theorem B]. Another interesting combination of the series representations (20) and (22) leads to the following variant of Tsumura’s result (24) or (25):

ζ (2n + 1) = (−1)n−1

+

π 2n



22n+1 − 1



H2n − log

1 4π

(2n)!

n−1 X (−1)k (22k+1 − 1) ζ (2k + 1) (2n − 2k)! π 2k k=1 ∞ X

−4

k=1

(2k − 1)! (22k−1 − 1) ζ (2k) (2n + 2k)! 24k

(26) # (n ∈ N),

which is essentially the same as the determinate expression for ζ (2n + 1), derived recently by Ewell [438, p. 1010, Corollary 3] by employing an entirely different technique from Srivastava’s [1084].


411

Other similar combinations of the series representations (20) to (23) would yield the following companions of Ewell’s result (26):

ζ (2n + 1) = (−1)n−1

+

2(2π)2n



(22n+1 − 1)(32n + 1)



H2n − log

1 6π

(2n)!

n−1 X (−1)k (22k+1 − 1) ζ (2k + 1) 2k (2n − 2k)! 2 k=1 3π

(27)

 ∞ X (2k − 1)!(22k−1 − 1) ζ (2k)  −4 (2n + 2k)! 62k

(n ∈ N);

k=1

ζ (2n + 1) = (−1)n−1

+

2(2π)2n



2H2n − log

(22n + 1)(32n+1 − 1)



(2n)!

π2 27

n−1 X (−1)k (32k+1 − 1) ζ (2k + 1) (2n − 2k)! π 2k

(28)

k=1

 ∞ 2k−1 X (2k − 1)! (3 − 1) ζ (2k)  −6 (2n + 2k)! 62k

(n ∈ N);

k=1

ζ (2n + 1) = (−1)n−1

+

2(2π)2n



H2n − log

32n+2 − 22n+3 + 1



(2n)!

8π 27

n−1 X (−1)k (32k+1 − 22k+1 ) ζ (2k + 1) (2n − 2k)! (2π)2k

(29)

k=1

 ∞ X (2k − 1)! (32k−1 − 22k−1 ) ζ (2k)  − 12 (2n + 2k)! 62k

(n ∈ N);

k=1

 2n H2n − log 27π 128 2(2π)  ζ (2n + 1) = (−1)n−1 4n+3 (2n)! 2 + 22n+2 − 32n+2 − 1 n−1 X (−1)k (42k+1 − 32k+1 ) ζ (2k + 1) + (2n − 2k)! (2π)2k

(30)

k=1

 ∞ 2k−1 2k−1 X (2k − 1)! (4 −3 ) ζ (2k)  − 24 (2n + 2k)! 122k k=1

(n ∈ N),

412


and

ζ (2n + 1) = (−1)n−1

+

2(2π)2n



H2n − log

32n+1 (22n + 1) − 24n+2 + 22n − 1



(2n)!

n−1 X (−1)k (32k+1 − 22k+1 ) ζ (2k + 1) (2n − 2k)! π 2k

4π 27

(31)

k=1

 ∞ X (2k − 1)! (32k−1 − 22k−1 ) ζ (2k)  − 12 (2n + 2k)! 122k

(n ∈ N).

k=1

Next, we recall another identity similar to (8) (see Srivastava [1084, p. 386, Eq. (1.7)]): ∞ X 1 (s)2k+1 ζ (s + 2k + 1, a) t2k+1 = [ζ (s, a − t) − ζ (s, a + t)] (2k + 1)! 2

(|t| < |a|) .

k=0

(32) By setting t = 1/m and differentiating both sides with respect to s, we find from (32) that ∞ X k=0

  2k X 1  (s)2k+1  0 ζ (s + 2k + 1, a) + ζ (s + 2k + 1, a) s+j (2k + 1)! m2k j=0

m d 1 1 = ζ s, a − − ζ s, a + 2 ds m m

(33)

(m ∈ N \ {1}).

In particular, when m = 2, (33) immediately yields ∞ X k=0

  2k X (s)2k+1 1 ζ 0 (s + 2k + 1, a) + ζ (s + 2k + 1, a)  s+j (2k + 1)! 22k j=0

(34)

1 1 −s = − a− log a − . 2 2 By letting s → −2n − 1 (n ∈ N) in the further special of this last identity (34) when a = 1, Wilton [1233, p. 92] obtained the following series representation for ζ (2n + 1)


413

(see also Hansen [531, p. 357, Entry (54.6.9)]): " n−1 X (−1)k ζ (2k+1) n−1 2n H2n+1 − log π ζ (2n + 1) = (−1) π + (2n + 1)! (2n−2k+1)! π 2k k=1 # ∞ X (2k − 1)! ζ (2k) +2 (n ∈ N), (2n + 2k + 1)! 22k

(35)

k=1

which may be compared with the series representation (20). As a matter of fact, since (2k)! (2k − 1)! (2k − 1)! = − 2n (2n + 2k)! (2n + 2k − 1)! (2n + 2k)!

(n, k ∈ N),

it is not difficult to deduce from (20) and (35) (with n replaced by n − 1) that " n−1 X (−1)k−1 k ζ (2k + 1) (2π)2n n ζ (2n + 1) = (−1) (2n − 2k)! n(22n+1 − 1) π 2k k=1 # ∞ X (2k)! ζ (2k) + (n ∈ N), (2n + 2k)! 22k

(36)

(37)

k=0

which is precisely the aforementioned main result of Cvijovic´ and Klinowski [351, p. 1265, Theorem A]. As a matter of fact, in view of the derivative formula 2.3(22), the series representation (37) is essentially the same as a result given earlier by Zhang and Williams [1251, p. 1590, Eq. (3.13)] (see also Zhang and Williams [1251, p. 1591, Eq. (3.16)]), where an obviously more complicated (asymptotic) version of (37) was proven, by applying the same identity (13) above. Observing also that (2k)! (2k − 1)! (2k − 1)! = − (2n + 1) (2n + 2k + 1)! (2n + 2k)! (2n + 2k + 1)!

(n, k ∈ N),

(38)

we obtain yet another series representation for ζ (2n + 1), by applying (20) and (35): " n−1 X (−1)k−1 k ζ (2k + 1) 2(2π)2n n ζ (2n + 1) = (−1) 2n (2n − 2k + 1)! r(2n − 1) 2 + 1 π 2k k=1 # ∞ X (2k)! ζ (2k) + (n ∈ N), (2n + 2k + 1)! 22k k=0

(39) which provides a significantly simpler (and much more rapidly convergent) version of the other main result of Cvijovic´ and Klinowski [351, p. 1265, Theorem B]: ζ (2n + 1) = (−1)n

∞ 2(2π)2n X ζ (2k) n,k 2k (2n)! 2 k=0

(n ∈ N),

(40)

414


where the coefficients n,k are given explicitly by n,k :=

2n X 2n

j

j=0

B2n−j (j + 2k + 1)(j + 1) 2j

(n ∈ N; k ∈ N0 ),

(41)

in terms of the Bernoulli numbers (see Section 1.6). The definition (41) can be rewritten at once in the form: n X 2n

n,k =

2j

j=0

B2n−2j 1 − (2j + 2k + 1)(2j + 1) 22j (2n + 2k) 22n

(n ∈ N; k ∈ N0 ), (42)

or, equivalently, n,k = (−1)

n−1

n 2(2n)! X (−π 2 ) j ζ (2n − 2j) (2j + 1)!(2j + 2k + 1) (2π)2n j=0

1 − (2n + 2k) 22n

(43)

(n ∈ N; k ∈ N0 ),

by virtue of the relationship 2.3(18). Combining the partial fractions occurring in (42) or (43), it is easily seen that n  Q (2k + 2` + 1)−1 n X 2n 2n + 2k 2n−2j `=0  n,k = 2 B2n−2j  2n 2j 2j + 1 (2n + 2k) 2 j=0

(44)

 ·

n Y

(2k + 2` + 1) −

`=0 (`6=j)

n Y `=0

 (2k + 2` + 1) 

(n ∈ N; k ∈ N0 ).

In view of the identity: n 2n−2j X B2n−2j 2n 2 j=0

2j

2j + 1

=1=

n X 2n j=0

22j B2j , 2j 2n − 2j + 1

(45)

which is due essentially to Euler (cf., e.g., Riordan [978, p. 123, Problem 12]), the expression inside brackets in (44) is a polynomial in k of degree n (not n + 1), and, therefore, n,k = O(k−2 )

(k → ∞; n ∈ N).

(46)


415

It follows from (46) that the general term in (40) has the order estimate: O(2−2k · k−2 )

(k → ∞),

(47)

whereas the general term in our series representation (39) has precisely the same order estimate: O(2−2k · k−2n−1 )

(k → ∞; n ∈ N)

(48)

as that in (20). Thus, even in the special case when n = 1, the series representing ζ (3) converges faster in (39) than in (40). Various known series representations for ζ (2n + 1) (n ∈ N) of other types include those given (for example) by Ramanujan [965] (see also Berndt [121]), Glaisher [489] (see also Hansen [531, p. 359]), Koshliakov [695], Leshchiner [747], Grosswald [514, 515], Terras [1147], Cohen [334], Butzer et al. [197, 198], Da¸browski [362] and others (see, e.g., Berndt [122, pp. 275 and 276]). We conclude this section by remarking that a particular case of the series representation (23) when n = 1 was proven, by an entirely different method by Zhang and Williams [1251, p. 707, Theorem 9]. Furthermore, the following particular case of (37) when n = 1 (see 3.1(8)): ζ (3) = −

∞ ζ (2k) 4π 2 X , 7 (2k + 1)(2k + 2) 22k

(49)

k=0

which is contained in a 1772 paper entitled Exercitationes Analyticae by Euler (see, e.g., Ayoub [81, pp. 1084–1085]), was rediscovered by Ramaswami [966] and (more recently) by Ewell [436]. In fact, Euler’s formula (49) was reproduced by Srivastava [1072, p. 7, Eq. (2.23)] from the work of Ramaswami [966]. In the current mathematical literature, however, Euler’s formula (49) is being attributed to Ewell [436].

4.3 Further Series Representations There are many further known series representations for ζ (2n + 1) (n ∈ N), which converge much more rapidly than those given by the defining series in 2.3(1). For example, we have the series representation (see also Equation 4.2(35)): " ζ (2n + 1) = (−1)

n−1

π

2n

n−1

H2n+1 − log π X (−1)k ζ (2k + 1) + (2n + 1)! (2n − 2k + 1)! π 2k k=1 # ∞ X (2k − 1)! ζ (2k) +2 (n ∈ N), (2n + 2k + 1)! 22k k=1

(1)

416


which was given over seven decades ago by Wilton [1233, p. 92] (see also Hansen [531, p. 357, Entry (54.6.9)]), and the following result given recently by Srivastava [1083, p. 10, Eq. (42)] (see also Srivastava [1082, p. 5, Eq. (3.4)]): π 2n



H2n+1 − log

1 2π

2(4n − 1) B2n+2 log 2 2 (2n + 1)! (2n + 2)! 1 22n+1 − 1 0 24n+3 0 − ζ (−2n − 1) − ζ −2n − 1, (2n + 1)! (2n + 1)! 4  n−1 ∞ k X X ζ (2k+1) (−1) (2k−1)! ζ (2k) +  (n ∈ N), +2 (2n−2k+1)! 1 2k (2n+2k+1)! 42k k=1 k=1 π 2

ζ (2n +1) = (−1)n−1



+

(2) where (and in what follows) a prime denotes the derivative of ζ (s) or ζ (s, a) with respect to s, an empty sum is to be interpreted as nil and Hn denotes the familiar harmonic numbers defined by 3.2(36). Of the two seemingly analogous representations (1) and (2), the infinite series in (2) would obviously converge more rapidly with their general terms having the order estimates: O k−2n−2 · m−2k (3) (k → ∞; n ∈ N; m = 2 and 4). Srivastava and Tsumura [1111] derived three (presumably new) members of the class of the series representations (1) and (2). The general terms of the infinite series occurring in these three members [given in (29), (30) and (31) below] have the order estimates: O k−2n−2 · m−2k (4) (k → ∞; n ∈ N; m = 3, 4, 6), which exhibit the fact that each of the three series representations derived here for ζ (2n + 1) converges more rapidly than Wilton’s result (1) and two of them (cf. Equations (30) and (31) below) at least as rapidly as Srivastava’s result (2). Srivastava and Tsumura [1111] begin, by defining the sequence {γn (x)}∞ n=0 by means of the generating function (cf. Eq. 1.6(2)): ∞

F(x, t) :=

t − log x X tn = γ (x) n et − x n!

1 5 x 5 1 + c; c > 0 ,

(5)

n=0

so that, clearly, γn (1) = Bn

(n ∈ N0 ) .

(6)


417

Since the zeros of et − x are given by t = 2nπi + log x

(n ∈ Z) ,

(7)

the radius of convergence of the series in (5) is at least 2π. Hence, by the CauchyHadamard theorem for absolute convergence (cf., e.g., Whittaker and Watson [1225, p. 30]), we have Lemma 4.1 (Srivastava and Tsumura [2000a]) Let the sequence {γn (x)}∞ n=0 be defined by (5). Then there exists some nonnegative real number κ, such that lim inf n→∞

|γn (x)| n!

−1/n

= 2π + κ

(κ = 0).

(8)

Following Srivastava and Tsumura [1111], we now consider the following Dirichlet series (cf. Eq. 2.3(1)):

ω(s, x) :=

∞ ∞ −n−1 X log x X x−n−1 x + ns s−1 ns−1

1 5 x 5 1 + c; c > 0 ,

(9)

n=1

n=1

so that, obviously, ω(s, 1) = ζ (s).

(10)

In case 1 < x 5 1 + c (c > 0), we can see that the function ω(s, x) is meromorphic, that is, holomorphic on the whole complex s-plane, except for a simple pole at s = 1 with residue log x . x−1 Lemma 4.2, below, provides an interesting generalization of the familiar relationship 2.3(10). Lemma 4.2 (Srivastava and Tsumura [1111]) Let γn (x) and ω(s, x) be defined by (5) and (9), respectively. Then ω(1 − n, x) = −

γn (x) n

(n ∈ N \ {1}) .

Making use of the generating function (5), we also have

(11)

418


Lemma 4.3 (Srivastava and Tsumura [1111]) For n ∈ N and |θ| < 2π (θ ∈ R),

(2n + 1)

∞ −k−1 ∞ −k−1 X X x x [sin(kθ) log x − θ cos(kθ)] sin(kθ) + 2n+2 k k2n+1 k=1

k=1

n−1

X (−1)k θ 2k+1 (2k + 1)! (−1)n θ 2n+1 log x = + (2n + 1)! x(x − 1) (2n − 2k) ω(2n − 2k + 1, x)

(12)

k=0

∞ X (−1)k θ 2k+1 γ2k−2n (x) (2k + 1)!

+

(1 < x 5 1 + c, c > 0).

k=n+1

By applying Lemma 4.3, Srivastava and Tsumura [1111] first proved that

(2n + 1)

∞ X sin(kθ) k=1

k2n+2

= 2(−1)n θ 2n+1

−θ

k=1 n−1 X k=1

−

∞ X k=0

∞ X cos(kθ)

k2n+1

− 2nθ ζ (2n + 1)

(−1)k · k ζ (2k + 1) (2n − 2k + 1)! θ 2k

(2k)! ζ (2k) (2n + 2k + 1)! (2π/θ)2k

(13)

! (|θ| < 2π ; θ ∈ R; n ∈ N).

Since each series in (13) is uniformly convergent with respect to θ on the open interval (−2π, 2π), by executing termwise differentiation in (13) with respect to θ , they showed that (Srivastava and Tsumura [1111, p. 328, Theorem 2])

2n

∞ ∞ X X cos(kθ) sin(kθ) + θ − 2n ζ (2n + 1) 2n+1 k k2n k=1

k=1

= 2(−1) θ

n 2n

n−1 ∞ X (−1)k · k ζ (2k + 1) X (2k)! ζ (2k) − 2k (2n − 2k)! (2n + 2k)! (2π/θ)2k θ k=1

!

(14)

k=0

(|θ | < 2π ; θ ∈ R; n ∈ N). Formula (14) was proven independently by Katsurada [638, p. 81, Theorem 1], by using the Melin transform techniques.


419

In their special cases when θ = π, Equations (13) and (14) readily yield a result given by Srivastava [1084, p. 393, Eq. (3.20)]: ζ (2n + 1) = (−1)n

X n−1 (−1)k−1 · k ζ (2k + 1) 2(2π)2n 2n (2n − 2k + 1)! (2n − 1) 2 + 1 π 2k k=1

(15) +

∞ X k=0

ζ (2k) (2k)! (2n + 2k + 1)! 22k

(n ∈ N)

and one of the two main results of Cvijovic´ and Klinowski [351, p. 1265, Theorem A]: X n−1 (2π)2n (−1)k−1 · k ζ (2k + 1) ζ (2n + 1) = (−1) (2n − 2k)! n(22n+1 − 1) π 2k n

k=1

(16) +

∞ X k=0

ζ (2k) (2k)! (2n + 2k)! 22k

(n ∈ N).

Remark 1 The series representation (15) was given by Srivastava [1082, p. 4, Eq. (2.5)] (see also Srivastava [1084, p. 393, Eq. (3.20)]. Indeed, as already observed by Srivastava [op. cit.], (15) provides a significantly simpler (and much more rapidly convergent) version of one of the two main results of Cvijovic´ and Klinowski [351, p. 1265, Theorem B]. Remark 2 The series representation (16) is the other main result of Cvijovic´ and Klinowski [351, p. 1265, Theorem A], whose companion was referred to in Remark 1 above. Remark 3 Since 1 (2k)! (2k − 1)! (2k − 1)! = − (2n + 2k + 1)! 2n + 1 (2n + 2k)! (2n + 2k + 1)!

(n, k ∈ N),

(17)

it is not difficult to obtain Wilton’s result (1), by combining the series representation (15) with another result 4.2(20) of Srivastava [1082, p. 1, Eq. (1.3)] (see also Srivastava [1084, p. 389, Eq. (2.9)]). With a view to applying (13) and (14) in their other special cases when 2 θ = π, 3

1 π 2

and

1 π, 3

(18)

420


we, now, recall several trigonometric sums given by Lemma 4 (Srivastava and Tsumura [1111]). For 1, ∞ cos 2 nπ X 3 31−s − 1 = ζ (s), (19) ns 2 n=1 −s ∞ sin 2 nπ X √ 3 3 −1 1 −s = 3 ζ (s) + 3 ζ s, , (20) ns 2 3 n=1 ∞ cos 1 nπ X 2 = 2−s (21−s − 1) ζ (s), (21) ns n=1 ∞ sin 1 nπ X 2 1 −s 1−2s = (2 − 1) ζ (s) + 2 ζ s, , (22) ns 4 n=1 ∞ cos 1 nπ X 3 1 1−s = 6 − 31−s − 21−s + 1 ζ (s), (23) s n 2 n=1

and ∞ sin X n=1

1 3 nπ ns

=

√ 3−s − 1 1 1 3 ζ (s) + 6−s ζ s, + ζ s, . 2 6 3

(24)

Srivastava and Tsumura [1111] first proved the following results: 2n 2n (3 − 1)π (−1)n π 1 n 2(2π) B2n + √ ζ 2n, ζ (2n + 1) = (−1) √ 3 n(32n+1 − 1) 4 3 (2n)! 3 (2π)2n  n−1 ∞ k−1 X X (−1) · k ζ (2k + 1) (2k)! ζ (2k)  +  (n ∈ N) + (2n − 2k)! 2 2k (2n + 2k)! 32k k=1 k=0 3π (25) " 2n 2(2π) 22n−3 (22n − 1)π B2n ζ (2n + 1) = (−1)n (2n)! n(24n+1 + 22n − 1)

+

+

X n−1 (−1)n π 1 (−1)k−1 · k ζ (2k + 1) ζ 2n, + 2n 4 (2n − 2k)! 1 2k 2(2π) k=1 2π ∞ X k=0

(2k)! ζ (2k) (2n + 2k)! 42k

# (n ∈ N),

(26)


421

and " 2(2π)2n 22n−3 (32n − 1)π B2n ζ (2n + 1) = (−1) √ 2n 2n 2n n(6 + 3 + 2 − 1) 3 (2n)! (−1)n π 1 1 + √ + ζ 2n, ζ 2n, 3 6 2 3 (2π)2n n

 n−1 ∞ k−1 X X (−1) · k ζ (2k + 1) (2k)! ζ (2k)  + +  (2n − 2k)! 1 π 2k (2n + 2k)! 62k k=1

3

(27)

(n ∈ N),

k=0

in terms of the Bernoulli numbers Bn , defined by 1.7(2). By comparing the series representation (26) with a known result, due to Srivastava [1083, p. 9, Eq. (41)] (see also Srivastava [1082, p. 5, Eq. (3.3)]), Srivastava and Tsumura [1111] derived an interesting identity involving the Zeta functions ζ (s) and ζ (s, a) (and their derivatives with respect to s), which is given below: n o B (−1)n π 1 2n = (22n−2 − 1) log 2 − 22n−3 (22n − 1)π ζ 2n, 2n 4 (2n)! 2(2π) (28) 1 22n−1 − 1 0 42n−1 0 − ζ (1 − 2n) − ζ 1 − 2n, (n ∈ N). 2(2n − 1)! (2n − 1)! 4 By applying (25), (26) and (27) and Srivastava’s series representations 4.2(21), 4.2(22) and 4.2(23), Srivastava and Tsumura [1111] proved their main results given by  2n H2n+1 − log 2 π 3 (32n+2 − 1)π 2π  + √ ζ (2n + 1) = (−1)n−1 B2n+2 3 (2n + 1)! 2 3 (2n + 2)! X n−1 1 (−1)k ζ (2k+1) ζ 2n+2, + +√ 2k 2n+1 3 (2n−2k+1)! 3 (2π) 2 k=1 3π # ∞ X (2k − 1)! ζ (2k) +2 (n ∈ N), (2n + 2k + 1)! 32k (−1)n−1

(29)

k=1

 π 2n H2n + 1 − log 12 π 22n (22n + 2 − 1)π  ζ (2n+1) = (−1)n − 1 + B2n + 2 2 (2n + 1)! (2n + 2)! nX −1 ( − 1)n − 1 1 ( − 1)k ζ (2k + 1) ζ 2n + 2, + 2n + 1 4 (2n − 2k + 1)! 1 2k 2(2π) k=1 2π  ∞ X (2k − 1)! ζ (2k)  +2 (n ∈ N), (2n + 2k + 1)! 42k

+

k=1

(30)

422


and 

H2n+1 − log

1 3π

22n (32n+2 − 1)π + √ B2n+2 3 (2n + 1)! 3 (2n + 2)! 1 1 (−1)n−1 ζ 2n + 2, + ζ 2n + 2, + √ 3 6 2 3 (2π)2n+1 (31) n−1 X (−1)k ζ (2k + 1) + (2n − 2k + 1)! 1 2k k=1 3π  ∞ X (2k − 1)!(2n + 2k + 1)! ζ (2k)  +2 (n ∈ N). 62k

ζ (2n + 1) = (−1)n−1

π 2n



k=1

Each of the main series representations (29), (30) and (31) belongs to the class containing Wilton’s formula (1) and Srivastava’s formula (2). If, for convenience, we denote the summands of the infinite series in the representations (1), (2), (29), (30) (j) and (31), by S k (j = 1, 2, 3, 4, 5), respectively, and apply Stirling’s formula 1.1(34) and the fact that ζ (2k) → 1 as k → ∞, we easily obtain the following order estimates (cf. Equations (3) and (4)): (j) S k = O k−2n−2 · m−2k (k → ∞; n ∈ N) (32) (m = 2 when j = 1; m = 3 when j = 3; m = 4 when j = 2 and j = 4; m = 6 when j = 5). Clearly, therefore, of these five series representations for ζ (2n + 1), the result (31) involves the most rapidly convergent series. Conversely, the rate of convergence of the series involved in each of the Srivastava-Tsumura results (29), (30) and (31) is obviously much better than that of the series involved in Wilton’s result (1). And, in particular, the rate of convergence of the series involved in the Srivastava-Tsumura result (30) is as good as that of the series involved in Srivastava’s result (2).

4.4 Computational Results Srivastava [1086] investigated rather systematically several interesting evaluations and representations of ζ (s) when s ∈ N \ {1}. In one of many computationally useful special cases considered by him, it is observed that ζ (3) can be represented by means of a series which converges much more rapidly than that in Euler’s celebrated formula 3.1(8), as well as the series 4.2(2) used recently by Ape´ ry [56] in his proof of the irrationality of ζ (3). Symbolic and numeric computations using Mathematica (Version 4.0) for Linux show, among other things, that only 50 terms of the series (38) below are capable of producing an accuracy of seven decimal places. Since √ 2 (1) i cot iz = coth z = 2z + 1 i := −1 , e −1


423

by replacing z in the known expansion 4.2(4) by 21 iπz, it is easily seen that (cf., e.g., Koblitz [681, p. 25]; see also Erde´ lyi et al. [421, p. 51, Eq. 1.20(1)]) ∞

πz πz X (−1)k+1 ζ (2k) 2k + = z eπ z − 1 2 22k−1

(|z| < 2) .

(2)

k=0

By setting z = it in (2), multiplying both sides by tm−1 (m ∈ N) and then integrating the resulting equation from t = 0 to t = τ (0 < τ < 2), if we apply the (readily derivable) integral formula:   Zτ m m m!eλτ j X (−1) (−λτ )  tm eλt dt = − e−λτ  (λ 6= 0), (3) j! λm+1 j=0

0

we obtain ) m (X ∞ i cos(kπτ ) + i sin(kπτ ) m! − ζ (m + 1) πτ km+1 k=1 ! ∞ ∞ X X πτ cos(kπτ ) sin(kπτ ) +i + + k 2(m + 1) k k=1

k=1

+

m−1 X j=1

m! (m − j)!

i πτ

∞ X ζ (2k) τ 2k = k + 21 m 2

j X ∞ k=1

cos(kπτ ) + i sin(kπτ ) kj+1

(4)

(m ∈ N; 0 < τ < 2),

k=0

which, in the special case when τ = 1, would immediately simplify to the form: i m iπ m+1 − 2 −1 m! ζ (m + 1) 2(m + 1) 2π m−1 X m! i j j 2 −1 ζ (j + 1) = log 2 + (m − j)! 2π j=1

+

∞ X

ζ (2k) 1 2k k=0 k + 2 m 2

(5)

(m ∈ N) ,

where we have also applied the last part of the definition 2.3(1). Furthermore, in (4), (5) and elsewhere in this section, an empty sum is interpreted (as usual) to be nil. Now, for a given sequence {n }∞ n=1 , it is easily verified that m−1 X j=1

j =

[(m−1)/2] X j=1

2j +

[m/2] X j=1

2j−1 ,

(6)

424


where, just as in 4.1(17), [κ] denotes the greatest integer in κ. Thus, we find from (5) that i m iπ − 2m+1 − 1 m! ζ (m + 1) 2(m + 1) 2π [(m−1)/2] 2j X j m (2j)! 2 − 1 = log 2 + (−1) ζ (2j + 1) 2j (2π)2j j=1

+i

[m/2] X

(−1)

j−1

j=1

+

(2j − 1)! 22j−1 − 1 m ζ (2j) 2j − 1 (2π)2j−1

∞ X

ζ (2k) 1 2k k=0 k + 2 m 2

(7)

(m ∈ N).

Setting m = 2n (n ∈ N) in (7) and then equating the real and imaginary parts in the resulting equation, we obtain (Srivastava [1086]) ζ (2n + 1) = (−1)n−1

(2π)2n (2n)! 22n+1 − 1

 ∞ n−1 2j −1 X X (2j)! 2 ζ (2k) 2n  ζ (2j + 1) + · log 2 + (−1) j 2j (2π)2j (k + n)22k 

j=1

(n ∈ N)

k=0

(8) and ζ (2n) = (−1)n−1

(2π)2n−1 (2n)!(22n−1 − 1)



 n−1 2j−1 −1 X (2j−1)! 2 2n π · + (−1) j ζ (2j) 2(2n + 1) 2j−1 (2π)2j−1

(n ∈ N).

j=1

(9) Similarly, if we set m = 2n + 1 (n ∈ N0 ) in (7), we shall obtain the following results, which can provide series representations for ζ (2n + 1) and ζ (2n) when n ∈ N : n 2j X j−1 2n + 1 (2j)! 2 − 1 (−1) ζ (2j + 1) 2j (2π)2j j=1 (10) ∞ X ζ (2k) = log 2 + (n ∈ N0 ) 1 2k k=0 k + n + 2 2


425

and (2π)2n−1 (2n − 1)! 22n − 1   n−1 2j−1 − 1 X (2j − 1)! 2 2n − 1 π ζ (2j) (−1) j · + 2j − 1 4n (2π)2j−1

ζ (2n) = (−1)n−1

(n ∈ N).

j=1

(11) Indeed, in its particular cases when n = 0 and n ∈ N, the summation formula (10) immediately yields the known sum 3.4(493) and (Srivastava [1086]) (2π)2n (2n + 1)! 22n − 1  n−1 2j X j 2n + 1 (2j)! 2 − 1  · log 2 + ζ (2j + 1) (−1) 2j (2π)2j j=1  ∞ X ζ (2k)  (n ∈ N), + 1 2k k + n + 2 k=0 2

ζ (2n + 1) = (−1)n−1

(12)

respectively. Each of the recursion formulas (9) and (11) can be used to evaluate ζ (2n) (n ∈ N). Formula (9) was proven, in a markedly different way, by (for example) Stark [1121, p. 199, Eq. (5)]. Formulas (9) and (11), together, yield the series identity (Srivastava [1086]): n−1 2j−1 − 1 X π (2n − 1)22n−1 − 2n j−1 2n − 1 (2j − 1)! 2 ζ (2j) = (−1) 2j − 1 (2π)2j−1 4n(2n + 1) 22n−1 − 1 j=1 n−1 X 22n − 1 2n (2j − 1)! 22j−1 − 1 j−1 + (−1) ζ (2j) (n ∈ N) 2j − 1 (2π)2j−1 2n 22n−1 − 1 j=1

(13) or, equivalently, n−1 o (2j − 2)! 22j−1 − 1 X 2n − 1 n (−1) j−1 2(n − j) 22n−1 − 1 − 22n−1 ζ (2j) 2j − 2 (2π)2j−1 j=1 π (2n − 1)22n−1 − 2n = (n ∈ N), (14) 4n(2n + 1) which can also be used as a recurrence relation for evaluating ζ (2n) (n ∈ N).

426


An integral representation for ζ (2n + 1), which is equivalent to the series representation (8), was given earlier by Da¸browski [362, p. 203, Eq. (16); p. 206], who also mentioned the existence of (but did not fully state) the series representation (12). The series representation (8) is derived also in a forthcoming paper by Borwein et al. [153, Eq. (57)]. For n = 1, (8) immediately yields the series representation 4.2(5) which, in conjunction with the known sum 3.4(493), would lead us readily to Euler’s formula 3.1(8). Conversely, by setting n = 1 in the series representation (12), we obtain 4π 2 ζ (3) = 9

! ∞ X 1 ζ (2k) log 2 + , 2 (2k + 3)22k

(15)

k=0

which was derived independently by (for example) Glasser [491, p. 446, Eq. (12)], Zhang and Williams [1251, p. 1585, Eq. (2.13)] and Da¸browski [362, p. 206] (see also Chen and Srivastava [252, p. 183, Eq. (2.15)]). By suitably combining the series occurring in 4.2(5), 3.4(493) and (15), it is not difficult to derive several other series representations for ζ (3), which are analogous to Euler’s formula 3.1(8). More generally, since λk2 + µk + ν (2k + 2n − 1)(2k + 2n)(2k + 2n + 1)

(16)

A B C = + + , 2k + 2n − 1 2k + 2n 2k + 2n + 1

where, for convenience, 1 1 λn2 − (λ + µ)n + (λ + 2µ + 4ν) , 2 4 B = Bn (λ, µ, ν) := − λn2 − µn + ν ,

A = An (λ, µ, ν) :=

(17) (18)

and 1 1 2 C = Cn (λ, µ, ν) := λn + (λ − µ)n + (λ − 2µ + 4ν) , 2 4

(19)

by appealing to (8), (10) with n replaced by n − 1 and (12), we can derive the following unification of a large number of known (or new) series representations, including (for


427

example) Euler’s result 3.1(8) (Srivastava [1086]): (−1)n−1 (2π)2n (2n)! B + (2n + 1) 22n − 1 C  n−1 X 1 2n − 1 ·  λ log 2 + 2j(2j − 1)A + [λ(4n − 1) − 2µ] nj (−1) j 4 2j − 2

ζ (2n + 1) =

22n+1 − 1

j=1

(2j − 2)! 22j − 1 1 + λn n + ζ (2j + 1) 2 (2π)2j  ∞ X λk2 + µk + ν ζ (2k)  + (2k + 2n − 1)(k + n)(2k + 2n + 1)22k

(20)

(n ∈ N; λ, µ, ν ∈ C),

k=0

where A, B and C are given by (17), (18) and (19), respectively. For λ = 0, the series representation (20) simplifies to the form: ζ (2n + 1) =

n

(−1)n−1 (2π)2n h io 22n+1 − 1 (µn − ν) − 22n − 1 n + 12 µ(n + 21 ) − ν

(2n)!  n−1 X 1 j 2n − 1  − 2µnj · j(2j − 1) ν − µ n − (−1) 2 2j − 2

j=1

 ∞ X (2j−2)! 22j −1 (µk+ν)ζ (2k)  · ζ (2j+1) + (2π)2j (2k+2n−1)(k+n)(2k+2n+1)22k k=0

(n ∈ N; µ, ν ∈ C).

(21)

Furthermore, by setting λ=µ=0

and ν = 1

in (20) or (alternatively) by setting µ=0

and ν = 1

in (21), we immediately obtain the series representation (Srivastava [1086]): ζ (2n + 1)  n−1 X (−1)n−1 (2π)2n 2n − 1 (2j)! 22j − 1 j  = (−1) ζ (2j + 1) 2j − 2 (2π)2j (2n)! 22n (2n − 3) − 2n + 1 j=1 (22)  ∞ X ζ (2k)  (n ∈ N), +2 (2k + 2n − 1)(k + n)(2k + 2n + 1)22k k=0

428


which, in the special case when n = 1, was given by Chen and Srivastava [252, p. 189, Eq. (2.45)]. Of the three representations (20), (21) and (22) for ζ (2n + 1) (n ∈ N), the infinite series in (22) converges most rapidly. Many other families of rapidly convergent series representations for ζ (2n + 1) (n ∈ N) can be found in the recent works of Srivastava [1083, 1084] (and, indeed, also in the numerous references already cited in each of these earlier works), including the results presented in previous sections (especially Section 4.2). For various suitable special values of the parameters λ, µ, and ν, we can easily deduce from (20) and (21) several known (or new) series representations for ζ (2n + 1) (n ∈ N). For example, if we set µ=2

and ν = 2n + 1

in our series representation (21), we shall obtain (Srivastava [1086])  n−1 2n X 2n − 1 (2π)  (−1) j ζ (2n + 1) = (−1)n−1 2j − 1 (2n)! 22n+1 − 1 j=1 # ∞ X (2j)! 22j − 1 ζ (2k) · ζ (2j + 1) − (2π)2j (2k + 2n − 1)(k + n)22k

(23) (n ∈ N),

k=0

which, in the special case when n = 1, immediately yields Euler’s formula 3.1(8). The following additional series representations for ζ (2n + 1) (n ∈ N), which are analogous to (23), can also be deduced similarly from (21) (Srivastava [1086]):

ζ (2n + 1) = (−1)n−1



(2π)2n

2n

(2n)! (2n − 1)22n − 2n

n−1 X 2n − 1 (−1) j 2j − 2 j=1 #

∞ X (2j)! 22j − 1 ζ (2k) · ζ (2j + 1) − (2π)2j (k + n)(2k + 2n + 1)22k

(24)

(n ∈ N)

k=0

and  n−1 X 4nj − 2j + 1 2n − 1  (−1) j ζ (2n + 1) = (−1)n−1 2j − 1 2j − 2 (2n + 1)! 22n − 1 j=1 # ∞ X (2j)! 22j − 1 ζ (2k) · ζ (2j + 1) − 4 (n ∈ N). (2π)2j (2k + 2n − 1)(2k + 2n + 1)22k (2π)2n

k=0

(25)


429

The special case of each of the last two series representations (24) and (25) when n = 1 was given by Zhang and Williams [1251, p. 1586]. For n = 2, Srivastava’s formulas (22) to (25) would readily yield the following series representations for ζ (5): ∞

ζ (5) = ζ (5) =

8π 4 X π2 ζ (2k) ζ (3) − , 13 13 (2k + 3)(2k + 4)(2k + 5)22k k=0 ∞ 4X

3π 2 4π ζ (3) + 31 93

k=0 ∞

ζ (2k) , (2k + 3)(2k + 4)22k

π4 X ζ (2k) π2 ζ (3) + ζ (5) = , 11 33 (2k + 4)(2k + 5)22k

(26)

(27)

(28)

k=0

and ∞

ζ (5) =

7π 2 8π 4 X ζ (2k) ζ (3) + . 75 225 (2k + 3)(2k + 5)22k

(29)

k=0

Next, with a view to further improving the rate of convergence in the most rapidly convergent series representations (22) and (26) considered in this work, we observe that 1 (2k + 2n − 1)(2k + 2n)(2k + 2n + 1)(2k + 2n + 2) 1 1 1 1 1 = − − . 6 2k + 2n − 1 2k + 2n + 2 2 (2k + 2n)(2k + 2n + 1)

(30)

Thus, by applying the series representations (10) with n replaced by n − 1, (8) with n replaced by n + 1 and (24), we obtain (Srivastava [1086]) 2π 2 22n+2 + n(2n − 3) 22n − 1 − 1 ζ (2n + 1) ζ (2n + 3) = (n + 1)(2n + 1) 22n+3 − 1  n−1 2n+2 X (2π) 2n − 1 n−1 j  + (−1) (−1) 2j (2n + 2)! 22n+3 − 1 j=1 (2j)! 22j − 1 2n + 2 2n − 1 − + 6n ζ (2j + 1) 2j 2j − 2 (2π)2j + 12

∞ X k=0

 ζ (2k)  (2k + 2n − 1)(2k + 2n)(2k + 2n + 1)(2k + 2n + 2)22k

where the series converges faster than that in (22).

(31)

(n ∈ N),

430


In its special case when n = 1, (31) readily yields the following improved version of the series representation (26) above (cf. Zhang and Williams [1251, p. 1590, Eq. (3.14)]: ∞

ζ (5) =

8π 4 X ζ (2k) 4π 2 ζ (3) + 31 31 (2k + 1)(2k + 2)(2k + 3)(2k + 4)22k

(32)

k=0

in which ζ (3) can be replaced by its known value −4π 2 ζ 0 (−2) given by 2.3(22) for n = 1. Yet another rapidly convergent series representation for ζ (2n + 3) (n ∈ N), analogous to (31), can be derived by means of the identity: 1 (2k + 2n)(2k + 2n + 1)(2k + 2n + 2)(2k + 2n + 3) 1 1 1 1 1 = − − , 6 2k + 2n 2k + 2n + 3 2 (2k + 2n + 1)(2k + 2n + 2)

(33)

together with the series representations (8), (12) with n replaced by n + 1 and (23) with n replaced by n + 1. We, thus, obtain the series representation (Srivastava [1086]): o n 2π 2 13 (2n + 1) 2n2 − 4n + 3 22n − 1 − 22n+1 + 1 ζ (2n + 1) ζ (2n + 3) = (n + 1)(2n + 1) (2n − 3)22n+2 − 2n  n−1 2n+2 X (2π) 2n  (−1) j + (−1)n−1 2j (2n + 2)! (2n − 3)22n+2 − 2n j=1 (2j)! 22j − 1 2n + 3 2n + 1 ζ (2j + 1) − +3 2j 2j − 1 (2π)2j # ∞ X ζ (2k) + 12 (n ∈ N), (2k + 2n)(2k + 2n + 1)(2k + 2n + 2)(2k + 2n + 3)22k k=0

(34) which, in the special case when n = 1, yields ∞

ζ (5) =

4π 4 X ζ (2k) 2π 2 ζ (3) − , 27 9 (2k + 2)(2k + 3)(2k + 4)(2k + 5)22k

(35)

k=0

where the series obviously converges faster than that in (26). Lastly, by applying the identity: 1 1 1 = 2k(2k + 2n − 1)(2k + 2n)(2k + 2n + 1) 2n(2n − 1)(2n + 1) 2k 1 1 1 1 1 1 − + − 2(2n − 1) 2k + 2n − 1 2n 2k + 2n 2(2n + 1) 2k + 2n + 1

(36)


431

in conjunction with the series representations (10) with n replaced by n − 1, (8), (12) and the known result 3.4(26), we arrive at the following series representation for ζ (2n + 1) (n ∈ N) (Srivastava [1086]): " 12n2 − 1 (2π)2n n−1 ζ (2n + 1) = (−1) 2 · (2n − 1)! n − 1 − (n − 2)22n 2n2 4n2 − 1 2 n−1 2j X log π j 2n − 2 (2j − 1)! 2 − 1 − (−1) ζ (2j + 1) − 2j − 3 (2π)2j n 4n2 − 1 j=2 # ∞ X ζ (2k) (n ∈ N), + k(k + n)(2k + 2n − 1)(2k + 2n + 1)22k k=1

(37) where we have also applied the fact that ζ (0) = − 12 . For n = 1, (37) reduces immediately to Wilton’s formula (cf. Hansen [531, p. 357, Entry (54.5.9)]; see also Chen and Srivastava [252, p. 181, Eq. (2.1)]): ! ∞ X π 2 11 1 ζ (2k) ζ (3) = − log π + . (38) 2 18 3 k(k + 1)(2k + 1)(2k + 3)22k k=1

Furthermore, in its special case when n = 2, (37) would yield the following companion of the series representations (32) and (35): ! ∞ X 2π 4 47 ζ (2k) ζ (5) = log π − − 30 , (39) 45 60 k(k + 2)(2k + 3)(2k + 5)22k k=1

which does not contain a term involving ζ (3) on the right-hand side. By eliminating ζ (2n + 3) between the results (31) and (34), we can obtain a series representation for ζ (2n + 1) (n ∈ N), which would converge as rapidly as the series in (37). For n ∈ N, we, thus, find that (Srivastava [1086])  n n−1 o 2n (2π) X ζ (2n + 1) = (−1)n−1 (−1) j (2n − 3)22n+2 − 2n (2n)!1n j=1

2n − 1 2n + 2 2n − 1 − + 6n − 22n+3 − 1 2j 2j 2j − 2 (2j)! 22j − 1 2n 2n + 3 2n + 1 ζ (2j + 1) · − +3 2j 2j 2j − 1 (2π)2j + 12

∞ X k=0

(40)

# (ξn k + ηn ) ζ (2k) , (2k + 2n − 1)(2k + 2n)(2k + 2n + 1)(2k + 2n + 2)(2k + 2n + 3)22k

432


where, for convenience, 1 2n+3 2 2n 2n+1 1n := 2 −1 (2n + 1) 2n − 4n + 3 2 − 1 − 2 +1 3 n on o − (2n − 3)22n+2 − 2n 22n+2 + n(2n − 3) 22n − 1 − 1 , n o ξn := 2 (2n − 5)22n+2 − 2n + 1 ,

(41)

(42)

and ηn := 4n2 − 4n − 7 22n+2 − (2n + 1)2 .

(43)

In its special case when n = 1, (40) yields the following series representation (Srivastava [1086]): ∞

ζ (3) = −

6π 2 X (98k + 121)ζ (2k) , 23 (2k + 1)(2k + 2)(2k + 3)(2k + 4)(2k + 5)22k

(44)

k=0

where the series obviously converges much more rapidly than that in each of the celebrated results 3.1(8) and 4.2(2). We conclude this section by summarizing below the results of some symbolic and numeric computations with the series in (44) using Mathematica (Version 4.0) for Linux: In[1] := (98k + 121)Zeta[2k] / ((2k + 1)(2k + 2)(2k + 3)(2k + 4)(2k + 5)2(2k)) (121 + 98k)Zeta[2k] Out[1] = 2k 2 (1 + 2k)(2 + 2k)(3 + 2k)(4 + 2k)(5 + 2k) In[2] := Sum [%, {k, 1, Infinity}] //Simplify 121 23 Zeta[3] Out[2] = − 240 6 Pi2 In[3] := N[%] Out[3] = 0.0372903 In[4] := Sum [N[%1]//Evaluate, {k, 1, 50}] Out[4] = 0.0372903 In[5] := N Sum [%1//Evaluate, {k, 1, Infinity}] Out [5] = 0.0372903 Since ζ (0) = − 12 , Out [2] evidently validates the series representation (44) symbolically. Furthermore, our numeric computations in Out [3], Out [4] and Out [5], together, exhibit the fact that only 50 terms (k = 1 to k = 50) of the series in (44) can produce an accuracy of seven decimal places (see, for details, Srivastava [1086]).


433

Problems 1. Prove that ∞ ∞ ∞ X X X 1 Hn Hn = 2 = 2 , n2 n3 (n + 1)2 n=1

n=1

n=1

where Hn denotes the harmonic numbers, defined by 3.2(36). (Cf. Equations 2.3(54) and 3.5(18); see also Klamkin [669, p. 195]) 2. Prove that ζ (4) =

∞ 36 X 1 . 4 2k 17 k k=1 k

(Comtet [337, p. 89]; Cohen [334, p. 280]) 3. Prove that π2 ζ (3) = 6

(

∞

3 1 π X ζ (2k) − log + 4 2 3 k(2k + 1)(2k + 2) 62k

) .

k=1

(Zhang and Williams [1251, Section 2]; Chen and Srivastava [252, p. 184]) 4. Prove that π2 ζ (3) = 14

(

∞

X (22k−1 − 1)ζ (2k) 3 π − log − 4 2 4 k(2k + 1)(2k + 2) 24k

) .

k=1

(Ewell [438, p. 1011]; Chen and Srivastava [252, p. 187]) 5. Prove the following series representations for ζ (3): 2π 2 ζ (3) = 35 ζ (3) =

2π 2 65

ζ (3) =

4π 2 25

ζ (3) =

4π 2 31

) ∞ X 3 π (22k−1 − 1)ζ (2k) − log − 4 ; 2 6 k(2k + 1)(2k + 2) 62k k=1 ( ! ) ∞ X π2 (32k−1 − 1)ζ (2k) 3 − log −6 ; 27 k(2k + 1)(2k + 2) 62k k=1 ( ) ∞ X 3 1 8π (32k−1 − 22k−1 )ζ (2k) − log −6 ; 4 2 27 k(2k + 1)(2k + 2) 62k k=1 ( ) ∞ X 3 1 27π (42k−1 − 32k−1 )ζ (2k) − log − 12 ; 4 2 128 k(2k + 1)(2k + 2) 122k (

k=1

and 4π 2 ζ (3) = 37

(

) ∞ X 3 1 4π (32k−1 − 22k−1 )ζ (2k) − log −6 . 4 2 27 k(2k + 1)(2k + 2) 122k k=1

(Chen and Srivastava [252, p. 187])

434


6. For n ∈ N \ {1}, show that ζ (2n + 1) =

(n−1 (−1)n (2π )2n X 22k 2n+1 2 − 1 n k=1 (2k − 1)! (2n − 2k)! (2k + 1)2 +

n−1 X k=1

·

∞

X (−1) j (2j − 1)! 22k (2k − 1)! (2n − 2k)! 22j−1 π 2j j=1

min(2j−2,2k) X i=0

2k i

∞ X (2k)! ζ (2k) + . 2j − i − 1 (2k + 4)! 22k

(−1)i ζ (2j)

k=0

(Zhang and Williams [1251, p. 1591]) 7. Let

Cp,q :=

j−1 ∞ X X 1 jp k q

(p ∈ N \ {1}; q ∈ N)

j=2 k=1

and

Cp,q,r :=

j−1 X ∞ X k−1 X j=3 k=2 i=1

1 jp kq ir

(p ∈ N \ {1}; q, r ∈ N).

Show that  p−2 q+n−1 X X (q)n (r)m Cp−n,r+m,q+n−m Cp,q,r = (−1)q  n!m! n=0 m=0 ) r−1 X (q)n+m Cp−n,q+n+m,r−m + n!m! m=0

+

q−2 X n=0

·

( r−2 X

(p)n (p + q − 2)! (−1)n ζ (p + n) Cq−n,r + (−1)q−1 n! (p − 1)! (q − 1)! (−1)m ζ (r − m) Cp+q−1,m+1 + ζ (p + q + m)

m=0

+ (−1)r−1 Cp+q−1,r,1 + Cp+q+r−1,1 + Cp+q−1,r+1 + ζ (p + q + r) , where empty sums are interpreted (as usual) to be nil. (Cf. Berndt [122, p. 253]; Markett [801, p. 122]) 8. Prove that ζ (2n + 1) = (−1)n

∞ 4(2π )2n X R2n+1,k ζ (2k) (2n + 1)! k=0

(n ∈ N),


435

where the coefficients R2n+1,k are rational and given by R2n+1,k :=

2n X 2n

m

m=0

(2n + 1) B2n−m 2k+m+1 2 (2k + m + 1) (m + 1)

(k ∈ N0 )

and by the family of generating functions ∞ X

R2n+1,k

n=0

t2n+1 (2n + 1)!

(2k − 1)! = 2k−1 t t (e − 1)

1−e

1 2t

2k−1 X m=0

(−1)m m!

m ! t 2

(k ∈ N; |t| < 2π).

(Cvijović and Klinowski [351, p. 1265]; see also 4.2(39) et seq.) 9. Prove that ζ (n) =

n−2 X ∞ X

1

j=1 p, q=1

pj (p + q)n−j

(n ∈ N \ {1, 2}).

(cf. Briggs [177]; see also Markett [801, p. 115]) 10. Let a, b, c be real numbers greater than 1 and P(a, b, c) =

∞ X

r−a

r=1

r X

k−b

k=1

k X

l−c .

l=1

Then show that P(a, b, c) + P(a, c, b) + P(b, c, a) + P(b, a, c) + P(c, a, b) + P(c, b, a) = ζ (a) ζ (b) ζ (c) + ζ (a) ζ (b + c) + ζ (b) ζ (c + a) + ζ (c) ζ (a + b) + 2 ζ (a + b + c), where ζ is the Riemann Zeta function, defined by 2.3(1). (Sitaramachandrarao and Subbarao [1037, p. 471]) 11. For each pair (m, n) ∈ N0 × N, one defines A2m (n) as follows: (i) A2m (1) := B2m , and (ii) for n ∈ N \ {1}, A2m (n) :=

2m 2j1 , ..., 2jn

X

{2j1 + 1} {2 (j1 + j2 ) + 1} · · · {2 (j1 + · · · + jn−1 ) + 1}

· B2j1 · · · B2jn ,

where the sum is taken over all (j1 , . . . , jn ) ∈ N0 n , such that j1 + · · · + jn = m and 2m 2j1 , ..., 2jn is a multinomial coefficient; Bj are Bernoulli numbers, defined by 1.6(2). Then show that ∞

ζ (n) =

2n−2 2 X π 2k π (−1)k A2k (n − 2) n 2 −1 (2k + 2)!

(n ∈ N \ {1, 2}).

k=0

(Ewell [437, p. 61])

436


12. Show that ζ (3) =

∞ X 1 2 3 π −4 e−2πn σ−3 (n) 2 π 2 n2 + π n + 45 2 n=1

and ∞

ζ (3) =

X 7 3 π −2 e−2π n σ−3 (n), 180 n=1

where σν (n) is given in Problem 20 of Chapter 2. (Cf. Terras [1147, p. 181] and Grosswald [514]; see also Problem 21 of Chapter 2) 13. Consider the cosecant integral: Is (ω) :=

π/ω Z ts csc2 t dt

( 1;

ω > 1) .

0

Prove the following identities: Is (ω) = −

Is (ω) = −

π s ω π s ω

cot

cot

π ω π ω

π/ω Z +s ts−1 cot t dt

( 1;

ω > 1) ;

0

− 2s

∞ π s−1 X

ω

k=0

ζ (2k) (s + 2k − 1)ω2k ( 1;

Zπ/2 ∞ π s−1 X ζ (2k) ts csc2 t dt = −2s 2 (s + 2k − 1)22k

ω > 1) ;

( 1) ;

k=0

0

Zπ/4 ∞ π s−1 X π s ζ (2k) − 2s ts csc2 t dt = − 4 4 (s + 2k − 1)42k

( 1) ;

k=0

0

Zπ/3 ∞ π s−1 X ζ (2k) 1 π s ts csc2 t dt = − √ − 2s 3 (s + 2k − 1)32k 3 3 k=0

( 1) ;

0

Zπ/6 ∞ π s−1 X √ π s−1 ζ (2k) ts csc2 t dt = − 3 − 2s 6 6 (s + 2k − 1)62k

( 1) ;

k=0

0

Is (ω) = −

∞ π s h π i X γ (s, −2kπ i/ω) cot + i −2is ω ω (−2ki)s

( 1; ω > 1) ,

k=1

where γ (z, α) denotes the incomplete Gamma function, defined by 1.1(76);   m j X α  (m ∈ N0 ) ; γ (m + 1, α) = m! 1 − e−α j! j=0


I2n (ω) = −

π 2n ω

cot

437

π ω

− (2n)!

n−1 X

π 2n−1 ω

"

∞

X cos(2kπ/ω) 1 (2n − 1)! k k=1

∞

ω 2j X cos(2kπ/ω) (−1) j + (2n − 2j − 1)! 2π k2j+1 j=1 k=1  n ∞ X (−1) j ω 2j−1 X sin(2kπ/ω)  + (2n − 2j)! 2π k2j j=1

(n ∈ N; ω > 1) ;

k=1

π 2n+1

π

(2n + 1)! ζ (2n + 1) 22n ∞ n π 2n X cos (2kπ/ω)) X (−1) j ω 2j  1 − (2n + 1)! + ω (2n)! k (2n − 2j)! 2π j=1 k=1  ∞ n ∞ ω 2j−1 X X cos(2kπ/ω) X (−1) j sin(2kπ/ω)  · + (2n − 2j + 1)! 2π k2j+1 k2j

I2n+1 (ω) = −

ω

cot

ω

j=1

k=1

∞ X k=0

ζ (2k) (2k + 2n − 1)ω2k

+ (−1)n

k=1

(n ∈ N; ω > 1) ; h i π 1 = − log 2 sin 2 ω  n−1 ω 2j (2n − 1)! X (−1) j 2π + C`2j+1 2 (2n − 2j − 1)! 2π ω j=1  n X 2π  (−1) j ω 2j−1 + C`2j (2n − 2j)! 2π ω j=1

and ∞ X

2n h π i ζ (2k) 1 n−1 (2n)! ω = (−1) ζ (2n + 1) − log 2 sin 2 ω 22n+1 π (2k + 2n)ω2k k=0 n ω 2j (2n)! X (−1) j + 2 (2n − 2j + 1)! 2π j=1 2π 2π 2π · (2n − 2j + 1)C`2j+1 + C`2j , ω ω ω

in terms of the generalized Clausen functions C`2n and C`2n+1 , defined by 2.4(80) and 2.4(81) and n ∈ N and ω > 1. (Srivastava, Glasser, and Adamchik [1097]) 14. Continuing Problem 13, show that  ∞ X (2π )2n ζ (2k) n−1  log 2 + 2 ζ (2n + 1) = (−1) (2k + 2n + 1)22k (2n + 1)! 22n − 1 k=0  ! n−1 j 2j X (−1) 2 −1 + (2n + 1)! ζ (2j + 1) (2n − 2j + 1)! (2π)2j j=1

438


and  ∞ 2n X (2π ) ζ 2k) log 2 + ζ (2n + 1) = (−1)n−1 2n+1 (k + n)22k (2n)! 2 −1 k=0  ! n−1 X (−1) j 22j − 1 + (2n)! ζ (2j + 1) , (2n − 2j)! (2π )2j j=1

for n ∈ N, respectively; ∞ X sin(kx + y) k=1

ks

=

1 x (2π )s csc(π s) cos y − π s ζ 1 − s, 20(s) 2 2π x 1 , − cos y + π s ζ 1 − s, 1 − 2 2π

for 1 and 0 < x < 2π; ∞ X sin(2kπ/ω) k=1

ks

=

(2π )s 1 1 1 csc πs ζ 1 − s, − ζ 1 − s, 1 − , 40(s) 2 ω ω

for 1 and ω > 1;  ∞ 2n X ζ (2k) (2π ) log 3 + 4 ζ (2n + 1) = (−1)n−1 2n (2k + 2n + 1)32k (2n + 1)! 3 − 1 k=0 ! n−1 X (−1) j 32j − 1 + (2n + 1)! ζ (2j + 1) (2n − 2j + 1)! (2π )2j j=1  n+1 2ζ 2j, 31 − 32j − 1 ζ (2j) (2n + 1)! X (−1) j  − √ (2n − 2j + 2)! (2π)2j−1 3

(n ∈ N)

j=1

and " ∞ X (2π )2n ζ (2k) log 3 + 2 ζ (2n + 1) = (−1) (k + n)32k (2n)! 32n+1 − 1 k=0 ! n−1 X (−1) j 32j − 1 + (2n)! ζ (2j + 1) (2n − 2j)! (2π )2j j=1  n 2ζ 2j, 13 − 32j − 1 ζ (2j) (2n)! X (−1) j  − √ (2n − 2j + 1)! (2π)2j−1 3 n−1

j=1

" ∞ X π2 ζ (2k) log 3 + 4 ζ (3) = 12 (2k + 3)32k k=0 1 2j 2 √ X (−1) j−1 2ζ 2j, 3 − 3 − 1 ζ (2j) +2 3 (4 − 2j)! (2π )2j−1 j=1

(n ∈ N) ;


439

and " # ∞ X ζ (2k) 2 π2 1 log 3 + 2 + √ ζ 2, − 4ζ (2) ; ζ (3) = 13 3 (k + 1)32k π 3 k=0 " ∞ X (2π )2n ζ (2k) n−1 log 2 + 4 ζ (2n + 1) = (−1) (2k + 2n + 1)42k (2n + 1)! 22n − 1 k=0 ! n−1 j 2j X (−1) 2 −1 + (2n + 1)! ζ (2j + 1) (2n − 2j + 1)! (2π)2j j=1  nX +1 ζ 2j, 41 −22j−1 22j − 1 ζ (2j) (−1) j  − (2n + 1)! (2n − 2j + 2)! (2π)2j−1

(n ∈ N)

j=1

and " ∞ X ζ (2k) (2π )2n log 2 + 2 ζ (2n + 1) = (−1) (k + n)42k (2n)! 24n+1 + 22n − 1 k=0 ! n−1 X (−1) j 22j − 1 + (2n)! ζ (2j + 1) (2n − 2j)! (2π )2j j=1  1 2j−1 22j − 1 ζ (2j) n j ζ 2j, X 4 −2 (−1)  − (2n)! (2n − 2j + 1)! (2π)2j−1 j=1  ∞ X 2π 2  ζ 2k) ζ (3) = log 2 + 4 9 (2k + 3)42k k=0  1 2 X (−1) j−1 ζ 2j, 4 − 22j−1 22j − 1 ζ (2j)  +6 (4 − 2j)! (2π )2j−1 n−1

(n ∈ N);

j=1

and " # ∞ X ζ (2k) 1 1 2π 2 log 2 + 2 + ζ 2, − 6ζ (2) ; ζ (3) = 35 4 (k + 1)42k π k=0 " ∞ X (2π )2n 4 ζ (2k) n−1 − ζ (2n + 1) = (−1) 2n 2n (2n + 1)! (2k + 2n + 1)62k 2 −1 3 −1 k=0 ! n−1 X 22j − 1 32j − 1 (−1) j + ζ (2j + 1) (2n − 2j + 1)! (2π )2j j=1  n+1 ζ 2j, 13 + ζ 2j, 16 − 22j−1 32j −1 ζ (2j) 1 X (−1) j  +√ (2n−2j + 2)! (2π)2j−1 3 j=1

(n ∈ N)

440


and " ∞ 2 X ζ (2k) (2π )2n ζ (2n + 1) = (−1) 22n + 32n + 62n − 1 (2n)! (k + n)62k k=0 ! n−1 X 22j − 1 32j − 1 (−1) j − ζ (2j + 1) (2n − 2j)! (2π )2j j=1  n ζ 2j, 13 +ζ 2j, 16 − 22j−1 32j −1 ζ (2j) 1 X (−1) j  −√ (2π)2j−1 3 j=1 (2n−2j+1)! " ∞ X ζ (2k) π2 ζ (3) = − 2 18 (2k + 3)62k k=0  1 1 2j−1 32j − 1 ζ (2j) 2 j−1 ζ 2j, + 2j, √ X 3 6 −2 (−1)  + 3 (4 − 2j)! (2π )2j−1 n−1

(n ∈ N);

j=1

and "∞ # π 2 X ζ (2k) 1 1 1 ζ (3) = + + ζ 2, − 16ζ (2) . √ ζ 2, 12 3 6 (k + 1)62k 2π 3 k=0 (Srivastava, Glasser and Adamchik [1097]) 15. Continuing Problems 13 and 14, show that ∞ X k=0

2j Z∞ n ζ (2k) 1 X (−1) j−1 2n + 1 2 t2j+1 sech2 t dt, = 2k 2 2j + 1 2j π (2k + 2n + 1)2 j=0

0

for n ∈ N0 ; ∞ X

ζ (2k) (2k + 2n + 1)22k k=0 n 2n + 1 (2j)! 22j − 1 1X 1 (−1) j−1 = ζ (2j + 1) − log 2, 2 2j 2 (2π )2j j=1

for n ∈ N0 ; " ∞ X (2π )2n ζ (2k) log 2 + 2 ζ (2n + 1) = (−1) 2n (2k + 2n + 1)22k (2n + 1)! 2 − 1 k=0  n−1 2j − 1 X (2j)! 2 2n + 1 + (−1) j−1 ζ (2j + 1) , 2j (2π )2j n−1

j=1

for n ∈ N. (Srivastava, Glasser and Adamchik [1097, p. 842 et seq.])


441

16. Continuing Problems 13, 14 and 15, define Sp :=

∞ X k=0

π ζ (2k) =− 2k 2ω (2k + p)ω

Z1

tp cot

0

πt dt ω

(p ∈ N; |ω| > 1) .

Show that Sp =

Z πi 1 iω p (log z)p + − dz 2(p + 1)ω 2 2π 1−z

(p ∈ N; ω| > 1) ;

1

Z1 πi 1 iω p dt {log (1 − (1 − )t)}p , Sp = − − 2(p + 1)ω 2 2π t 0

where, for convenience, 2π i := exp (|ω| > 1) ; ω or, equivalently, ∞ X k=0

πi p! ζ (2k) = − 2(p + 1)ω 2 (2k + p)ω2k

iω 2π

p

S1,p 1 − e2π i/ω

in terms of Nielsen’s generalized Polylogarithmic function Sn,p (z), defined by (cf., e.g., Kölbig [686, p. 1233, Eq. (1.3)]) (−1)n+p−1 Sn,p (z) := (n − 1)!p! S1,p (z) = ζ (p + 1) +

Z1

0 p X k=0

(log t)n−1 {log (1 − zt)}p

dt t

(n, p ∈ N; z ∈ C) ;

(−1)k−1 {log(1 − z)}k Lip−k+1 (1 − z); k!

∞ X

ζ (2k) πi 1 2π i/ω = − log 1 − e 2(p + 1)ω 2 (2k + p)ω2k k=0 p 1X p p! iω p iω k ζ (p + 1) + Lik e2π i/ω , − k! 2 2π 2 k 2π k=1

for p ∈ N and |ω| > 1; Z1

tp cot(νt)dt = −

0

+

i 1 + log 1 − e2νi p+1 ν p! ν

i 2ν

p

ζ (p + 1) −

p 1X p i k k! Lik+1 e2νi , ν k 2ν k=1

for p ∈ N and ν ∈ C \ {0}. (Srivastava, Glasser and Adamchik [1097, p. 843 et seq.])

442


17. Prove the following series representations for the values of L (s, χ ): (a) If χ (−1) = 1 and χ 6= 1, then ∞ ∞ X 2lπx 2lπx χ (l) πx X χ (l) cos sin − p p p l2n+1 l2n l=1 l=1 " n−1 2π x 2n X (−1)k−1 · k L (2k + 1, χ )

nL (2n + 1, χ ) − n = (−1)n

p

k=1

(2n − 2k) !

∞

+

τ (χ ) X (2k) ! L (2k, χ ) x2k p (2n + 2k) !

(2π x/p)2k # (n ∈ N) ;

k=1

(b) If χ (−1) = −1, then

L (2n, χ ) −

∞ X χ (l) l=1

l2n

cos

2lπx p

2n−1 "X n−1

L (2k, χ ) (−1)k−1 (2n − 2k) ! (2πx/p)2k−1 k=1 # ∞ 2τ (χ ) i X (2k) ! 2k+1 + L (2k + 1, χ ) x (n ∈ N) , p (2n + 2k) !

= (−1)

n

2π x p

k=0

where L (s, χ ) is the Dirichlet L-function associated with a nontrivial Dirichlet character χ of modulus p and τ (χ ) denotes the Gauss sum defined by

τ (χ ) :=

p X k=1

2kπi χ (k) exp p

i :=

√

−1 .

(Katsurada [638, p. 82, Theorem 3]) 18. Prove the following series representation: ∞ 1 X sin(2π lx) 2π x l2n+2 l=1 " 1 = (−1)n−1 (2π x)2n (2n + 1)!

ζ (2n + 1) +

+

n−1 X k=1

2n+1 X m=1

1 − log(2π x) m ∞

!

X (2k − 1)! ζ (2k) ζ (2k + 1) (−1) + 2 x2k (2n + 2k + 1)! (2n − 2k + 1)! (2π x)2k

#

k

k=1

(n ∈ N; x ∈ R; |x| ≤ 1). (Katsurada [638, p. 81, Theorem 2])


443

19. Prove the following series representations for the values of L (s, χ ): (a) If χ (−1) = 1 and χ 6= 1, then ∞ 2lπx p X χ (l) sin 2π x p l2n+2 l=1 2n "X n−1 2π x L (2k + 1, χ ) (−1)k = (−1)n+1 p (2n − 2k + 1) ! (2π x/p)2k k=0 # ∞ X 2τ (χ ) (2k − 1) ! + L (2k, χ ) x2k (n ∈ N) ; p (2n + 2k + 1) !

L (2n + 1, χ ) −

k=1

(b) If χ (−1) = −1, then ∞ p X χ (l) 2lπ x sin 2π x p l2n+1 l=1 2n−1 "X n−1 2π x L (2k, χ ) (−1)k = (−1)n+1 p (2n + 2k + 1) ! (2π x/p)2k−1 k=1 # ∞ X 2τ (χ ) i (2k) ! 2k+1 − L (2k + 1, χ ) x (n ∈ N) , p (2n + 2k + 1) !

L (2n, χ ) −

k=0

where L (s, χ ) and τ (χ ) are as those given in Problem 17. (Srivastava and Tsumura [1112]) 20. Prove the following series representations for ζ (3): 120 2 π 1573 ∞ X

ζ (3) = − ·

k=0

ζ (3) = −

8576k2 + 24286k + 17283 ζ (2k) ; (2k + 1) (2k + 2) (2k + 3) (2k + 4) (2k + 5) (2k + 6) (2k + 7) 22k

6 2 1 π (80H4 − 31H5 − 49 log π) 17 120

+2

∞ X k=1

32k + 49 ζ (2k) (n ∈ N) , (2k) (2k + 1) (2k + 2) (2k + 3) (2k + 4) (2k + 5) 22k

where Hk denotes the harmonic numbers, defined by 3.2(36). (Srivastava and Tsumura [1112, pp. 20 and 22])

5 Determinants of the Laplacians During the last two decades, the problem of evaluation of the determinants of the Laplacians on Riemann manifolds has received considerable attention by many authors, including (among others) D’Hoker and Phong [377, 378], Sarnak [1004] and Voros [1201], who computed the determinants of the Laplacians on compact Riemann surfaces of constant curvature in terms of special values of the Selberg Zeta function. Although the first interest in the determinants of the Laplacians arose mainly for Riemann surfaces, it is also interesting and potentially useful to compute these determinants for classical Riemannian manifolds of higher dimensions, such as spheres. In this chapter, we are particularly concerned with the evaluation of the functional determinant for the n-dimensional sphere Sn with the standard metric. In computations of the determinants of the Laplacians on manifolds of constant curvature, an important roˆ le is played by the closed-form evaluations of the series involving the Zeta function given in Chapter 3 (cf., e.g., Choi and Srivastava [291, 292], Choi et al. [269]), as well as the theory of the multiple Gamma functions presented in Section 1.3 (cf., e.g., Voros [1201], Vardi [1190], Choi [260] and Quine and Choi [954]).

5.1 The n-Dimensional Problem Let {λn } be a sequence, such that 0 = λ0 < λ1 5 λ2 5 · · · 5 λn 5 · · · ;

λn ↑ ∞

(n → ∞);

(1)

hence, we consider only such nonnegative increasing sequences. Then, we can show that Z(s) :=

∞ X 1 , λsn

(2)

n=1

which is known to converge absolutely in a half-plane σ for some σ ∈ R. Definition 5.1 (cf. Osgood et al. [881]). The determinant of the Laplacian 1 on the compact manifold M is defined to be Y det0 1 := λk , (3) λk 6=0 Zeta and q-Zeta Functions and Associated Series and Integrals. DOI: 10.1016/B978-0-12-385218-2.00005-0 c 2012 Elsevier Inc. All rights reserved.

446


where {λk } is the sequence of eigenvalues of the Laplacian 1 on M. The sequence {λk } is known to satisfy the condition as in (1), but the product in (3) is always divergent; so, for the expression (3) to make sense, some sort of regularization procedure must 0 be used. It is easily seen that, formally, e−Z (0) is the product of nonzero eigenvalues of 1. This product does not converge, but Z(s) can be continued analytically to a neighborhood of s = 0. Therefore, we can give a meaningful definition: det0 1 := e−Z (0) , 0

(4)

which is called the Functional Determinant of the Laplacian 1 on M. Definition 5.2 The order µ of the sequence {λk } is defined by ) ( ∞ X 1 0 λαk

(5)

k=1

The analogous and shifted analogous Weierstrass canonical products E(λ) and E(λ, a) of the sequence {λk } are defined, respectively, by !) ( ∞ Y λ λ[µ] λ λ2 1− + ··· + (6) E(λ) := exp + [µ] λk λk 2λ2k [µ]λk k=1 and E(λ, a) :=

∞ Y 1− k=1

λ λ λ[µ] , exp + ··· + λk + a λk + a [µ] (λk + a)[µ]

(7)

where [µ] denotes the greatest integer part in the order µ of the sequence {λk }. There exists the following relationship between E(λ) and E(λ, a) (see Voros [1201]):   [µ] m X λ E(λ − a) E(λ, a) = exp  Rm−1 (−a)  , (8) m! E(−a) m=1

where, for convenience, R[µ] (λ − a) :=

d[µ]+1 {− log E(λ, a)} . dλ[µ]+1

(9)

The shifted series Z(s, a) of Z(s) in (2) by a is given by Z(s, a) :=

∞ X k=1

1 . (λk + a)s

(10)

Determinants of the Laplacians

447

Formally, indeed, we have Z 0 (0, −λ) = −

∞ X

log(λk − λ),

k=1

which, if we define D(λ) := exp −Z 0 (0, −λ) ,

(11)

immediately implies that D(λ) =

∞ Y

(λk − λ).

k=1

In fact, Voros [1201] gave the relationship between D(λ) and E(λ) as follows:   [µ] m X λ D(λ) = exp[−Z 0 (0)] exp − FPZ(m)  m m=1

 · exp −

[µ] X

C−m

m=2

m−1 X k=1

1 k

!

(12)

 λm  E(λ), m!

where an empty sum is interpreted to be nil and the finite part prescription is applied (as usual) as follows (cf. Voros [1201, p. 446]): ( f (s), if s is not a pole, FPf (s) := (13) Residue , if s is a simple pole, lim f (s + ) − →0

and Z(−m) = (−1)m m! C−m .

(14)

Now consider the sequence of eigenvalues on the standard Laplacian 1n on Sn . It is known from the work of Vardi [1190] (see also Terras [1148]) that the standard Laplacian 1n (n ∈ N) has eigenvalues µk := k(k + n − 1)

(15)

with multiplicity k+n k+n−2 (2k + n − 1) (k + n − 2)! βkn := − = n n k! (n − 1)! n−2 2k + n − 1 Y = (k + j) (n − 1)! j=1

(16) (k ∈ N0 ) .

448


2 From now on, we consider the shifted sequence {λk } of {µk } in (15) by n−1 as a 2 fundamental sequence. Then, the sequence {λk } is written in the following simple and tractable form:

n−1 λk = µk + 2

2

n−1 2 = k+ 2

(17)

with the same multiplicity as in (16). We will exclude the zero mode, that is, start the sequence at k = 1 for later use. Furthermore, with a view to emphasizing n on Sn , we choose the notations Zn (s), Zn (s, a), En (λ), En (λ, a) and Dn (λ), whereas instead of Z(s), Z(s, a), E(λ), E(λ, a) and D(λ), respectively. We readily observe from (11) that Dn

n−1 2

2 !

= det0 1n ,

(18)

where det0 1n denote the determinants of the Laplacians on Sn (n ∈ N). Choi [260] computed the determinants of the Laplacians on the n-dimensional unit sphere Sn (n = 1, 2, 3) by factorizing the analogous Weierstrass canonical product of a shifted sequence {λk } in (17) of eigenvalues of the Laplacians on Sn into multiple Gamma functions, whereas Choi and Srivastava [291, 292] and Choi et al. [269] made use of some closed-form evaluations of the series involving Zeta function given in Chapter 3 for the computation of the determinants of the Laplacians on Sn (n = 2, 3, 4, 5, 6, 7). Quine and Choi [954] made use of zeta regularized products to compute det0 1n and the determinant of the conformal Laplacian, det (1Sn + n(n − 2)/4). In following three sections, we compute the determinants of the Laplacians on Sn (n = 1, 2, 3, 4, 5, 6, 7) and det0 (1Sn + n(n − 2)/4), by using the aforementioned methods.

5.2 Computations Using the Simple and Multiple Gamma Functions Factorizations Into Simple and Multiple Gamma Functions We begin by expressing En (λ) (n = 1, 2, 3) as the simple and multiple Gamma functions. Our results are summarized in the following proposition (see Choi [260]; see also Voros [1201])


449

Proposition 5.1 The analogous Weierstrass canonical products En (λ) (n = 1, 2, 3) of the shifted sequence {λk } in 5.1(17) can be expressed in terms of the simple and multiple Gamma functions as follows: 1 √ √ 2 , 0(1 − λ)0(1 + λ) √ √ 4 02 ( 12 ) ec1 λ 0 21 − λ 0 12 + λ E2 (λ) = √ √ √ √ 2 , π 1 − 2 λ 1 + 2 λ 02 12 − λ 02 12 + λ E1 (λ) =

(1)

(2)

and √ √ ec2 λ 02 (1 − λ)02 (1 + λ) E3 (λ) = , √ √ 1 − λ {03 (1 − λ)03 (1 + λ)}2

(3)

where, for convenience, c1 := 2(γ − 1 + 2 log 2)

and

c2 := log(2π) − 32 ,

and γ is the Euler-Mascheroni constant, defined by 1.1(3). Proof. Here we will prove only the representation (2). First of all, it follows from 5.1(15) and 5.1(16) with n = 2 that the eigenvalues of 12 on S2 have the sequence µk = k(k + 1) with multiplicity

2k + 1

(k ∈ N0 )

and the corresponding shifted sequence {λk } of {µk } in (4) by is 1 2 λk = k + 2

with multiplicity

(4) 1 4

2k + 1 (k ∈ N0 ).

[or 5.1(17) with n = 2]

(5)

It is easily seen from the definition 5.1(5) that both of the sequences (4) and (5) are of order µ = 1. From 5.1(6) we find that the analogous Weierstrass canonical product of the shifted sequence {λk } in (5) is  ∞ Y  E2 (λ) = 1 − k=1

 λ



2k+1

λ    2  exp  2  k + 12 k + 12

.

(6)

450


To express E2 (λ) in terms of simple and multiple Gamma functions, we consider the associated product E2+, defined by

E2+ (z) :=

∞ Y

   1−

k=1

z k+

 exp 

1 2

2k+1



!

z2

z k + 12

 + 2  1 2 k+ 2

.

(7)

Letting λ = −w2 in (6) and using the definition (7), we obtain E2 (λ) = E2+ (iw) E2+ (−iw).

(8)

It is seen from 1.4(3) that the G-function itself is related to the Barnes product EB (λ) given by EB (λ) : =

k ∞ Y λ2 λ λ exp + 2 1− k k 2k

(9)

k=1

λ = (2π) 2

e

1 2

(1+γ )λ2 −λ

G(1 − λ).

Thus, E2+ (λ) can be further decomposed to involve the Barnes G-function, or rather the shifted Barnes product:

EB z, −

1 2

=

∞ Y k=1

   1−

z k−

1 2

k



!

 exp 

z k − 21

z2

 + 2  . 1 2 k− 2

(10)

It readily follows from (7) and (10) that

E2+ (z) =

o2 n EB z, − 12 (1 − 2z)2 exp 4z + 4z2    −1 !   ∞ Y  2 z z   z  · 1 − exp + .    1 1 2   k+ 2 k+ 2 2 k+ 1 k=1  2

(11)

Obviously, the sequence {λk } with λk = k is of order µ = 1. It follows from 1.3(2) and 5.1(6) that the analogous Weierstrass canonical product E(λ) of this sequence is ∞ Y λ eγ λ λ E(λ) = 1− ek = , k 0(1 − λ) k=1

(12)


451

which, in view of 5.1(9), yields R0 (λ) = −

d {log E(λ)} = −ψ(1 − λ) − γ , dλ

(13)

where ψ is the Psi (or Digamma) function defined by 1.3(1). Now, the shift equation (8) applied to (12) with a = − 12 and µ = 1 yields the classical result: ! ! 1 ∞ 0 Y z z (γ +2 log 2)z 2 , (14) 1− exp = e k + 12 k + 12 0 12 − z k=0 which, by virtue of 1.1(14), immediately becomes √ (γ +2 log 2)z ∞ Y z 1 πe z 1 . 1− + exp + = k 2 k 2 (1 − 2z) e2z 0 21 − z k=1 If we apply 1.1(14), 1.3(53), 2.2(4), 2.3(2) and (15) to (11), we easily find that n o2 0 12 − z EB z, − 12 h i. E2+ (z) = √ 2 π (1 − 2z) 22z exp (2 + γ )z + 2 + π4 z2

(15)

(16)

Similarly, the shift equation 5.1(8) applied to (9) with a = − 21 and µ = 2 yields 1 π2 2 EB z, − 12 = (2π) 2 z exp (log 2)z + log 2 + z 8 (17) h n oi G 12 − z . · exp 12 (1 + γ )z2 + γ z G 21 A combination of (16) and (17) gives  2 h i (2π)z 0 12 − z  G 12 − z  E2+ (z) = √ exp −2z + (γ − 1 + 2 log 2) z2 .  π (1 − 2z)  G 1 2 (18) Finally, it follows from (8) and (18) that n o2 0 12 − iw 0 21 + iw G 12 − iw G 12 + iw E2 (λ) = n o2 π G 12 (1 − 2iw)(1 + 2iw) h i · exp −2(γ − 1 + 2 log 2) w2 .

(19)

452


√ Since λ = −w2 , we have w = ±i λ.√We see that √ the second member of (19) remains the same with w replaced by either i λ or −i λ. In view of this observation and the relationship G(z) =

1 , 02 (z)

(19) is seen to be equivalent to (2). This completes the proof of the proposition (2). Propositions (1) and (3) can be proven similarly.

Evaluations of det0 1n (n = 1, 2, 3) By making use of Proposition 5.1, we can compute the determinants of the Laplacians on Sn (n = 1, 2, 3) explicitly. The results of these computations (as well as our computation of det0 14 ) are summarized by Proposition 5.2 The following evaluations hold true: det0 11 = 4π 2 , i h 1 det0 12 = exp 12 − 4 ζ 0 (−1) = e 6 A4 = 3.195311496 . . . , ζ (3) 0 det 13 = π exp = 3.338851215 . . . , 2π 2

(20) (21) (22)

and 1 35, 639, 301 13 0 2 0 det 14 = exp − ζ (−1) − ζ (−3) 3 3 3 217 · 5 · 7 13 2 183, 758, 875 1 , = · A 3 · C 3 · exp 3 217 · 33 · 7 0

(23)

where A is the Glaisher-Kinkelin’s constant, defined by 1.4(2), and C is the mathematical constant, defined by 1.4(70). Proof. det0 11 : It follows from 5.1(15) and 5.1(16) with n = 1 that the sequence of eigenvalues of 11 on S1 is µk = k2

with multiplicity

2 (k ∈ N0 ),

(24)

which corresponds to the shifted sequence {λk } in 5.1(17). It is obvious from the definition 5.1(5) that the sequence {µk } in (24) (and also its shifted sequence {λk }) is of order µ = 12 . Thus, its corresponding Zeta series Z1 (s) is given by Z1 (s) =

∞ X 2 = 2 ζ (2s), k2s k=1

(25)


453

where ζ (s) is the Riemann Zeta function, defined by 2.3(1). It is seen from 5.1(4) and 2.2(20) that det0 11 = exp[−4 ζ 0 (0)] = 4π 2 , which proves the evaluation (20).

Alternatively, even though it is not necessary to do so in the case of det0 11 , we can also prove (20), by making use of 5.1(12), 5.1(18) and (1). Indeed, since the shifted sequence considered above is of order µ = 12 , it is found from 5.1(12), 5.1(18) and 1.1(13) that det0 11 = D1 (0) = exp −Z10 (0) lim E1 (λ) = exp −Z10 (0) , λ→0

where Z1 (s) is the same as in (25) and the evaluation (20) follows immediately. det0 12 : In this case, we saw that the shifted sequence {λk } in (5) is of order µ = 1. Therefore, we find from 5.1(12) and 5.1(18) that det0 12 = D2

1 4

h i = exp −Z2 0 (0) − 14 FPZ2 (1) lim E2 (λ). λ→ 14

(26)

From (5) and 5.1(2), we have Z2 (s) =

∞ X k=1

2k + 1 2s s 2s = 2 − 2 ζ (2s − 1) − 4 , 1 k+ 2

(27)

where ζ (s) is the Riemann Zeta function, defined by 2.3(1). Since Z2 (s) has a simple pole only at s = 1 with its residue 1 (see Section 2.3), it is seen from 5.1(13), 2.2(19) with a = 1 and (27) that 1 FPZ2 (1) = lim Z2 (1 + ) − →0 1 = −4 + lim 22+2 − 2 ζ (2 + 1) − →0 1 21+2 − 2 2+2 = −4 + lim 2 − 2 ζ (2 + 1) − + →0 2 1+2 2 −2 = −4 + 2γ + lim , →0 which, upon employing l’Hoˆ spital’s rule, immediately yields FPZ2 (1) = 2γ + 4 log 2 − 4.

(28)

454


It is easy to see from 2.3(10), 1.7(7) and 2.1(31) that Z2 0 (0) = −

13 1 13 log 2 − 2 ζ 0 (−1) = − − log 2 + 2 log A. 6 6 6

It follows from (2) and 1.1(12) (with z replaced by n o4 02 12

1 2

(29)

+ z) that

1 lim E2 (λ) = exp (γ − 1 + 2 log 2) 1 4 2 λ→ 4 √ √ G 12 − λ G 12 − λ √ , · lim √ 1 λ→ 41 2 − λ cosh π i λ

which, upon considering the following limit relationships: √ − λ 1 √ = lim π λ→ 14 cosh πi λ G

1 2

G and

√ − λ = 1, √ 1 2− λ

lim

λ→ 14

1 2

(30)

yields

lim E2 (λ) =

λ→ 14

n o4 02 12 4π

exp

1 (γ − 1 + 2 log 2) . 2

(31)

Finally, if we substitute from (28), (29) and (31) into (26) and make use of 2.1(31), we are led at once to the evaluation (21). det0 13 : From 5.1(17) with n = 3, we find that the shifted sequence {λk } (by 1) of eigenvalues of the Laplacian 13 on S3 is λk = (k + 1)2

with multiplicity (k + 1)2

(k ∈ N0 ),

(32)

which is easily seen to be of order µ = 23 . Therefore, the analogous Weierstrass canonical product and its accompanying Zeta series for this sequence are ∞ Y E3 (λ) = 1− k=1

λ (k + 1)2

λ

exp

(k + 1)2

(k+1)2 (33)

and Z3 (s) =

∞ X (k + 1)2 k=1

(k + 1)2s

= ζ (2s − 2) − 1.

(34)


455

It follows from 5.1(12) and 5.1(18) that det0 13 = D3 (1) = exp −Z3 0 (0) − FPZ3 (1) lim E3 (λ). λ→1

(35)

It is also seen that Z3 (s) has a simple pole only at s = 32 with its residue 21 . Thus, we have from 5.1(13), 2.2(13) (with a = 1) and 2.3(22) (with n = 1) that FPZ3 (1) = Z3 (1) = ζ (0) − 1 = − 23

(36)

and Z3 0 (0) = 2 ζ 0 (−2) = −

ζ (3) . 2π2

(37)

If we substitute from (36) and (37) into (35), we obtain 3 ζ (3) + det 13 = exp 2 2π2

0

lim E3 (λ).

(38)

λ→1

By taking the limit as λ → 1 on both sides of (3) and applying 1.4(6) and Theorem 1.6(a), we obtain lim E3 (λ) = 2π exp − 32 A,

(39)

λ→1

where, for convenience,

A = lim

 √ n √ o−2     02 1 − λ 03 1 − λ 

λ→1  

1−λ

.

(40)

 

It is readily seen from 2.1(27), 2.1(28), 2.1(24) and 2.2(17) that A = exp ζ 0 (0) + 2 ζ 0 (−1) + ζ 0 (−2) · B · C , where, for convenience, B and C are defined by   λ    1  1 (2π) 2    B := lim √  λ→1  1 − λ 0 1 − λ  and n h √ √ √ io C := lim exp −ζ 0 −2, 1 − λ − 2 λ ζ 0 −1, 1 − λ . λ→1

(41)

456


First, to determine the limit value B , consider h  i 1 λ    (2π) 2  1 B = lim √ · √ oλ  , √ n λ→1   1+ λ  1− λ 0 1− λ √ λ and using 1.1(9) and 1.1(13), immediately becomes ( 1 2 ) π 1 1 π 2 1 z −2z 2 lim = lim . B= z→0 z {0(z)}1−2z+z2 z→0 2 2 0(z)

which, upon setting z = 1 −

We, thus, find from 1.1(2) that π 12 π 12 2 B= lim zz −2z = . z→0 2 2

(42)

Next, to evaluate the limit value C , by setting n = 1 in 2.2(4), we have ζ (s, a) = ζ (s, 1 + a) + a−s ,

(43)

which, upon differentiating with respect to s, yields ζ 0 (s, a) = ζ 0 (s, 1 + a) − a−s log a.

(44)

Therefore, we see from (44) that √ √ √ √ ζ 0 −1, 1 − λ = ζ 0 −1, 2 − λ − 1 − λ log 1 − λ and √ √ √ 2 √ ζ 0 −2, 1 − λ = ζ 0 −2, 2 − λ − 1 − λ log 1 − λ , and so √ lim ζ 0 −1, 1 − λ = ζ 0 (−1) and

λ→1

√ lim ζ 0 −2, 1 − λ = ζ 0 (−2),

λ→1

which yield C = exp −2 ζ 0 (−1) − ζ 0 (−2) .

(45)

Finally, by substituting from (42) and (45) into (41) and using 2.2(20), we find from (39) that lim E3 (λ) = π exp − 23 , (46) λ→1

which, when substituted into (38), proves the evaluation (22).


457

A proof of the evaluation (23) will be presented in Section 5.3 in which each of the results (21), (22) and (23) is obtained by means of the summation formulas of Chapter 3 for series of Zeta functions (see also Problem 3 at the end of this chapter).

5.3 Computations Using Series of Zeta Functions In this section, we compute det0 1n (n = 2, 3, 4, 5, 6, 7) in a markedly different way from that detailed in Section 5.2 for n = 1, 2, 3. Our evaluations, here, are based largely on the summation formulas of Chapter 3 for series of Zeta functions. det0 12 : In view of 5.2(26), 5.2(28) and 5.2(29), it suffices to evaluate E2 ( 14 ). It follows from 5.2(6) that " Y 2k+1 # ∞ 1 1 1 1− E2 = exp , (1) 4 2k + 1 (2k + 1)2 k=1

which, upon taking logarithms and applying the Maclaurin series expansion of log(1 + x), yields ! ∞ ∞ X 1 X 1 1 =− −1 log E2 4 n (2k − 1)2n−1 n=2 n=2 (2) ∞ ∞ X ζ (2n − 1) − 1 X ζ (2n − 1) + . =− n n · 22n−1 n=2

n=2

If we apply 3.4(520) and 3.4(529) on the right-hand side of (2), we obtain 1 7 γ log E2 = −1 + − log 2 + 6 log A, 4 2 6

(3)

which, upon combining 5.2(26) with 5.2(8), 5.2(29) and 2.1(31), proves the evaluation 5.2(21). det0 13 : By virtue of 5.2(38), it suffices to compute E3 (1). Indeed, letting λ = 1 in 5.2(33) and taking logarithms of the resulting equation with the aid of the Maclaurin series expansion of log(1 + x), we get ! ∞ X ∞ ∞ X X 1 ζ (2n) − 1 log E3 (1) = − = − , (4) 2n n+1 (n + 1)k k=2

n=1

n=1

which, in view of 3.4(571), immediately yields 3 log E3 (1) = log π − . 2 Now, if we make use of (5) in 5.2(38), we are led to the evaluation 5.2(22).

(5)

458


det0 14 : Letting n = 4 in the shifted sequence 5.1(17) of eigenvalues of 14 on S4 , we obtain the following sequence: 2 λk = k + 23

with multiplicity

1 6 (k + 1)(k + 2)(2k + 3)

(k ∈ N0 ),

(6)

which obviously has the order µ = 2. Now, it follows from 5.1(12) and 5.1(18) that 9 det0 14 = D4 4 (7) 9 81 81 9 . = exp −Z40 (0) − FPZ4 (1) − FPZ4 (2) − C−2 E4 4 32 32 4 We can express Z4 (s) for the sequence (6) in terms of the Riemann Zeta function as follows: ∞

1 X (k + 1)(k + 2)(2k + 3) Z4 (s) = 2s 6 k=1 k + 32 ∞

∞

22s X (k + 1)(k + 2) 22s X k(k + 1) = 6 6 (2k + 3)2s−1 (2k + 1)2s−1 k=1 k=2 ! ∞ ∞ X 22s X 1 1 = − 24 (2k + 1)2s−3 (2k + 1)2s−1 =

=

22s 24

k=2

k=2

∞ X

1

∞ X

1

k=1

(2k − 1)2s−3

k=1

(2k − 1)2s−1

−

−

(8)

1 32s−3

+

1 32s−1

! ,

which, in view of 2.3(1), becomes Z4 (s) =

1 2s−3 1 2s−3 1 2 − 1 ζ (2s − 3) − 2 − ζ (2s − 1) 3 3 4 1 2 2s−3 1 2 2s + . − 3 3 8 3

(9)

We observe from Section 2.3 that Z4 (s) has simple poles at s = 1 and s = 2 with 1 and 16 , respectively. their residues − 24 Using 2.3(10) and 5.1(14), we obtain 1 9, 801, 047 C−2 = Z4 (−2) = − 12 3 2 2 ·3 ·5·7

(10)

2869 1 7 Z40 (0) = log 2− 1440 · 32 + ζ 0 (−1) − ζ 0 (−3). 12 12

(11)

and


459

Now, we evaluate FPZ4 (1) and FPZ4 (2). Since Z4 (s) has simple poles at s = 1 and s = 2, we have to use the second case of the definition of FPf (s) in 5.1(13) to compute the finite parts of Z4 (s) for s = 1 and s = 2. Using the expression in (9) for Z4 (s) and 2.3(6) (or 2.2[19] with a = 1), we easily see that 1 FPZ4 (1) = lim Z4 (1 + ) + →0 24 1 1 22 − 1 31 1 2−1 = − − lim 2 − ζ (2 + 1) − + 72 3 →0 4 2 4 γ 1 31 − log 2. =− − 72 12 6

(12)

Similarly, we have

1 FPZ4 (2) = lim Z4 (2 + ) − →0 6 h 16 7 1 = − − ζ (3) + lim 22+1 − 1 81 12 3 →0 1 22+1 − 2 + · ζ (2 + 1) − 2 2 2 7 16 γ = − + + log 2 − ζ (3). 81 3 3 12

(13)

Since the sequence in (6) is of order µ = 2, its analogous Weierstrass canonical product E4 (λ) is

∞ Y



 1 (k+1)(k+2)(2k+3)

λ   E4 (λ) = 1 − 2  3 k=1 k+ 2 

6





(14)

λ2

λ 1   · exp  (k + 1)(k + 2)(2k + 3)  2 + 4  . 6 k + 32 2 k + 32

Upon setting λ = 49 in (14) and taking the logarithms on both sides of the resulting equation, if we make use of 2.3(1) and the Maclaurin series expansion of log(1 + x),

460


we obtain (∞ ) ∞ X 1X 9 32n =− (k + 1)(k + 2) log E4 4 6 n(2k + 3)2n−1 n=3 k=1 " # ∞ ∞ ∞ X 1 X 32n X 1 1 1 1 =− − − + 24 n (2k − 1)2n − 3 (2k − 1)2n−1 32n−3 32n−1 n=3 k=1 k=1 " 2n ∞ 1 X 1 2n 3 =− 3 ζ (2n − 3) − 8 ζ (2n − 3) − 32n ζ (2n − 1) 24 n 2 n=3 # 2n 3 +2 ζ (2n − 1) − 24 . (15) 2 Now, let α1 , α2 , α3 , and α4 denote the sums of the Zeta series occurring in 3.4(726), 3.4(729), 3.4(732) and 3.4(736), respectively. We then find from (15) that 1 9 = − (2α1 − α2 + α3 − 8α4 ) log E4 4 24 4 21 189 =− + γ − ζ (3) 9 32 128 979 17 5 + log A + log C + log 2− 1440 · 3 . 4 4

(16)

Finally, in view of 2.1(31) and 2.3(27), we are easily led from (7) and (9) through (16) to the evaluation 5.2(23). det0 15 : To evaluate det0 15 , we begin by setting a = 3, n = 1, n = 2 and t = 2 in 3.2(64), and then use 2.2(4), 2.2(20) and 2.3(22). We, thus, obtain ∞ X ζ (2n, 3) 2n+2 2 = 10 + log 3 · π −4 n+1

(17)

∞ X ζ (2n, 3) 2n+4 13ζ (3) 288 −16 2 = 20 − + log 2 · 3 · π . n+2 π2

(18)

n=1

and

n=1

By setting n = 5 in 5.1(17), we find that the shifted sequence of eigenvalues of 15 of S5 is given as follows: (k + 2)2

with multiplicity

1 (k + 1)(k + 2)2 (k + 3) 12

(k ∈ N).

(19)


461

It is seen that the sequence in (19) has the order µ = 25 . We also have Z5 (s) =

∞ 1 X (k + 1) (k + 2)2 (k + 3) 12 (k + 2)2s k=1

1 1 = [ζ (2s − 4) − ζ (2s − 2)] + 12 3

(20)

1 1 . − 22s 22s−2

It is observed that Z5 (s) has simple poles at s = FPZ5 (1) = Z5 (1) = −

5 24

and

3 2

and s = 52 . We, therefore, have

FPZ5 (2) = Z5 (2) = −

1 5 ζ (2) − . 12 48

We also have C−2 =

1 Z5 (−2) = −8 2

and

Z50 (0) =

ζ (5) ζ (3) + + 2 log 2. 8π 4 24π 2

We, thus, find that ∞ Y E5 (λ) = 1− k=1

· exp

λ (k + 2)2

1

12

(k+1)(k+2)2 (k+3)

λ2 1 λ , (k + 1)(k + 2)2 (k + 3) + 12 (k + 2)2 2(k + 2)4

which, upon setting λ = 4 and taking logarithms on each side of the resulting equation and using (17) and (18), yields ∞ 1 X 22n 1 1 log E5 (4) = − ζ (2n − 4) − ζ (2n − 2) + 2n−2 − 2n−4 12 n 2 2 n=3 # "∞ ∞ X 22n+2 1 X 22n+4 =− ζ (2n, 3) − ζ (2n, 3) 12 n+2 n+1 n=1

n=2

π 2 13 ζ (3) −24 =− + + log 2 · π . 9 12 π 2 If we set n = 5 in 5.1(18) and use 5.1(12), we finally have det0 15 = D5 (4) = exp[−Z5 0 (0) − 4 FPZ5 (1) − 8 FPZ5 (2) − 8 C−2 ] E5 (4) π 197 ζ (3) ζ (5) = 26 exp − − . 3 2 24 π 2 8 π 4

(21)

det0 16 : Next, by setting n = 6 in 5.1(17), we obtain the shifted sequence of eigenvalues of 16 on S6 as follows: 5 2 1 k+ with multiplicity (2k + 5)(k + 4)(k + 3)(k + 2)(k + 1). (22) 2 120

462


We see that ∞ 1 X (2k + 5)(k + 4)(k + 3)(k + 2)(k + 1) 120 (k + 25 )2s k=1 1 h 2s = 2 − 32 ζ (2s − 5) − 10 22s − 8 ζ (2s − 3) 1920 i + 9 22s − 2 ζ (2s − 1) − ( 25 )2s .

Z6 (s) =

(23)

1 3 , − 48 It is observed that Z6 (s) has simple poles at s = 1, 2 and 3 with its residues 640 1 and 120 , respectively. It is also seen that the sequence in (22) has the order µ = 3. Now, we can find that

31 0 7 3 0 log 2 + 2 log 5 − ζ (−5) + ζ 0 (−3) − ζ (−1), 960 96 320 3 FPZ6 (1) = lim Z6 (1 + ) − →0 640 Z60 (0) = −

484051

28 · 33 · 5 · 7

3 9323 3 γ+ log 2, + 160 28 · 32 · 52 320 4483 1 21 FPZ6 (2) = − 5 2 4 + ζ (3) − (γ + 2 log 2), 320 24 2 ·3 ·5 6 2 1 7 93 γ FPZ6 (3) = − 6 + + log 2 − ζ (3) + ζ (5), 60 30 24 320 5 1 2217581021 , C−2 = Z6 (−2) = − 15 2 2 2 · 3 · 5 · 7 · 11 =−

and 56 1 62451523 + 7 . C−3 = − Z6 (−3) = 18 4 2 6 2 · 3 · 5 · 7 · 11 · 13 2 · 3 We also see that ∞ 5 9 1 X λn 7 7 7 log E6 (λ) = − ζ 2n − 5, 2 − 2 ζ 2n − 3, 2 + ζ 2n − 1, 2 , 60 n 16 n=4

(24) which, for λ =

25 4 ,

yields "∞ ∞ 25 1 X ζ (2n + 1, 72 ) 5 2n+6 5 X ζ (2n + 1, 72 ) 5 2n+4 log E6 =− − 4 60 n+3 2 2 n+2 2 n=1 n=2 # ∞ 9 X ζ (2n + 1, 72 ) 5 2n+2 + . (25) 16 n+1 2 n=3


463

By setting n = 0, 1, 2, a = 72 and t = 25 in 3.2(67) and using some identities recorded in previous and present chapters, we have ∞ X ζ (2n + 1, 72 ) 5 2n+2 155 25 11 = − γ− log 2 n+1 2 12 4 12 n=1

−

5 5 log 3 − log 5 + 3 ζ 0 (−1); 2 2

(26)

∞ X ζ (2n + 1, 72 ) 5 2n+4 3325 625 3561553 = 6 − 5 γ+ 5 log 2 n+2 2 2 2 .5.3831 2 n=1

−

53 125 15 log 3 − log 5 + 6 ζ 0 (−1) + ζ 0 (−3); (27) 8 8 8

and ∞ X ζ (2n + 1, 27 ) 5 2n+6 623005 15625 253849 485 = 8 2 − 6 log 3 γ+ 6 2 log 2 − n+3 2 32 2 ·3 2 ·3 2 ·3 ·7 n=1

238627 3133 0 3 log 5 + ζ (−1) + ζ 0 (−2) 32 8 2 15 0 63 0 507 0 ζ (−3) + ζ (−4) + ζ (−5). + 4 8 16

−

(28)

Applying (26), (27) and (28) to (25), we obtain 25 4639 1385 246717677 log E6 = − 10 3 + 8 2 γ − 8 2 log 2 4 2 ·3 2 ·3 2 · 3 · 5 · 7 · 3831 6053 0 1 0 651 118711 log 5 − 6 ζ (−1) − ζ (−2) − 6 ζ 0 (−3) + 6 40 2 ·3·5 2 ·3·5 2 ·5 21 413875 96875 1 − 5 ζ 0 (−4) − 6 ζ 0 (−5) − 11 2 ζ (3) + 12 ζ (5). 2 2 ·5 2 ·3 2 (29) If we set n = 6 in 5.1(18) and use 5.1(12), we get 25 25 0 det 16 = D6 = exp −Z6 0 (0) − FPZ6 (1) 4 4 625 15625 625 25 25 − FP Z6 (2) − FP Z6 (3) − C−2 + C−3 · E6 , 32 192 32 8 4 which, upon using the above computations, yields 116791 38441354615245651 1511 0 0 − 3990625 det 16 = 2 735552 · 5 960 · exp − − ζ (−1) 5441253801984 240 1 0 2023 0 1 0 11 0 − ζ (−2) − ζ (−3) − ζ (−4) − ζ (−5) . 40 960 32 480

(30)

464


det0 17 : By setting n = 7 in 5.1(17), we see that the shifted sequence of eigenvalues of 17 on S7 is given as follows: λk = µk + 9 = (k + 3)2 with multiplicity 1 (k + 1)(k + 2)(k + 3)2 (k + 4)(k + 5). 360

(31)

It is found that Z7 (s) =

∞ 1 X (k + 1)(k + 2)(k + 3)2 (k + 4)(k + 5) 360 (k + 3)2s

(32)

k=1

1 = [ζ (2s − 6) − 5 ζ (2s − 4) + 4 ζ (2s − 2)] − 3−2s , 360 1 which shows us that Z7 (s) has simple poles at s = 23 , 52 and 72 with their residues 180 , 1 1 − 144 and 720 , respectively. It is also observed that the sequence in (31) has the order µ = 27 . If we set n = 7 in 5.1(18) and use 5.1(12), we get 81 det0 17 = D7 (9) = exp −Z7 0 (0) − 9FPZ7 (1) − FP Z7 (2) 2 (33) 9 81 C−2 + C−3 · E7 (9). −243 FP Z7 (3) − 2 2

We can also easily verify each of the following evaluations: 243 7 , FP Z7 (1) = − , 2 60 π2 π4 161 FP Z7 (3) = − 4 6 − 4 3 + 2 4 2 , 2 ·3 ·5 2 ·3 2 ·3 ·5

C−2 = −

81 , 2

C−3 =

FP Z7 (2) = −

and Z7 0 (0) = −

ζ (7) ζ (5) ζ (3) − − + 2 log 3. 32 π 6 48 π 4 180 π 2

It is observed that ( ∞ Y E7 (λ) = 1− k=1

1 (k+1)(k+2)(k+3)2 (k+4)(k+5) 360 λ 2 (k + 3) 1 · exp (k + 1)(k + 2)(k + 3)2 (k + 4)(k + 5) 360 ) λ λ2 λ3 · + + , (k + 3)2 2(k + 3)4 3 (k + 3)6

7 24 · 34

+

π2 , 540


465

which, upon setting λ = 9 and taking logarithms on each side of the resulting equation and considering the Taylor-Maclaurin expansion of log(1 − x), yields log E7 (9) = −

∞ ∞ 1 X ζ (2n, 4) 2n+4 1 X ζ (2n, 4) 2n+6 ·3 + ·3 360 n+3 72 n+2 n=1

n=1

∞ 1 X ζ (2n, 4) 2n+2 39 2 3 4 259 − ·3 − π + π + . 90 n+1 80 100 240

(34)

n=1

By setting a = 4, n = 1, 2, 3 and t = 3 in 3.2(64) and using some identities already recorded in this presentation, we have ∞ X ζ (2n, 4) 2n+2 63 ·3 = − 5 log 2 + 9 log 3 + 4 log 5 − 9 log π, n+1 2 n=1 ∞ X n=1

(35)

ζ (2n, 4) 2n+4 1071 ·3 = − 29 log 2 + 243 log 3 + 16 log 5 n+2 4 − 81 log π −

27 ζ (3) , π2

(36)

and ∞ X ζ (2n, 4) 2n+6 32211 ·3 = − 125 log 2 + 3645 log 3 − 503936 log 5 n+3 10 n=1

1215 ζ (3) 405 ζ (5) − 729 log π − + . 2π2 2π4

(37)

If we apply (35), (36) and (37) to (34), we get 137 789 − log 3 + 1400 log 5 5 2 ·5 20 21 ζ (3) 9 ζ (5) 39 2 3 4 + log π + π + π . − − 2 4 80 100 16 π 16 π

log E7 (9) = −

(38)

Finally, from (34) and (38), and other previously recorded results, we obtain 1230367 949 ζ (3) 13 ζ (5) ζ (7) 0 − 177 1400 20 det 17 = 3 ·5 · π · exp − + − + . 60 720 π 2 24 π 4 32 π 6 (39)

5.4 Computations using Zeta Regularized Products Quine and Choi [954] made use of the zeta regularized products to compute det0 1n and the determinant of the conformal Laplacian, det (1Sn + n(n − 2)/4), which is

466


introduced here. In general, the conformal Laplacian is defined to be 1+

(n − 2) K , 4(n − 1)

where 1 is the Laplacian and K is the scalar curvature. For the sphere Sn , K = n(n − 1). Computation of the above determinants is equivalent to computing derivatives at s = 0 of the zeta function ∞ X k=1

βkn [k(k + n − 1)]s

for the Laplacian and ∞ X k=0

βkn [(k + n/2)(k + n/2 − 1)]s

for the conformal Laplacian and βkn being given in 5.1(16). For simplicity, we restrict our discussion of the conformal Laplacian to the case when n is even. We consider the more general zeta function as in Section 5.1, Zn (s, a) =

∞ X k=1

βkn [(k + a)(k + n − 1 − a)]s

for integers a, 0 5 a 5 n − 1, with a = 0 corresponding to the Laplacian and a = n/2 to the conformal Laplacian. If a(n − 1 − a) 6= 0, then det (1Sn + a(n − 1 − a)) a (n − 1 − a) exp −Zn0 (0, a) , and, if a(n − 1 − a) = 0, then det0 1n = exp −Zn0 (0, a) . We show below (Theorem 5.4) that for integers a, det(1Sn + a(n − 1 − a)), if a(n − 1 − a) 6= 0 and (n − 1) det0 1n are both of the form exp τn (a) +

n X

! τkn (a) ζ 0 (−k + 1)

,

(1)

k=1

where ζ is the Riemann Zeta function defined by 2.3(1) and the numbers τn (a) and τkn (a) are rational numbers for which we give explicit expressions in terms of coefficients of the Taylor expansion of βkn about k = −a and k = −(n − 1)/2. Using (1) and the functional equation for the Riemann Zeta function 2.3(11) or 2.3(12), it is easy


467

to compute numerical values for the constants involved. Our computations show that τn (a) = 0 for n odd and τkn (a) = 0, if n and k have opposite parity. The technique that we introduce here for dealing with this computation, which is different from other approaches, is a lemma on zeta regularized products (Lemma 5.3), which may be useful in simplifying and understanding other computations of this type. This gives us, in addition, a way to factor our functional determinant det(1Sn + a(n − 1 − a)) into multiple Gamma functions, as shown in Section 5.2, generalizing an equation of Voros [1201] and giving an alternate approach to computing it for integral a, 0 5 a 5 n − 1 (Section 5.2). We remark that choosing a to be an integer makes things considerably simpler because βkn = 0 for k = −1, . . . , −(n − 1) and the computation reduces to computation of derivatives of the Riemann Zeta function. The same techniques could be used for noninteger values of a but would involve a more complicated expression involving derivatives of the Hurwitz zeta function.

A Lemma on Zeta Regularized Products and a Main Theorem Let {λk } be a sequence of nonzero complex numbers. If Z(s) in 5.1(2) converges for s0 and has meromorphic continuation to a function meromorphic in σ , for some σ < 0, with at most simple poles, then we say that the sequence is zeta regularizable. We define λ−s k = exp(−s log λk ), and the definition of Z depends on the choice of arg λk . For a zeta regularized sequence, we define the zeta regularized product (see 5.1(4)) Y

z λk

= exp −Z 0 (0) .

We also define the product (see 5.1(11)) Y

z

(λk − λ) = exp −Z 0 (0, −λ) ,

where, for λ 6= λk , we adopt the convention that arg (λk − λ) ∼ = arg λk for |λk | large. Q For λk , the sequence of nonzero eigenvaluesQof the Laplacian on a manifold, z λk is called the determinant of the Laplacian and z (λk + λ) the functional determinant, det0 (1 + λ), as in Section 5.1. Information on the formal properties of zeta regularized products can be found in [955] and [1201]. The use of zeta regularized products can be traced back as far as Barnes [97]. Our approach is to try to product Q factor a zeta regularized Q Q into simpler ones. One sees that the equality of z (λ2k − λ2 ) and z (λk − λ) z (λk + λ) is untrue without the introduction of an exponential factor. This factor can be computed from the relationship between the zeta regularized product and the Weierstrass product found in the above-cited references. We give a confirmation independent of this in the proof below. Lemma 5.3 Let λj be a zeta regularizable sequence, and let h be an integer, such that P∞ −h−1 < ∞. For 2, they are conjugate with 0 to a Möbius transformation of the form z → N(τ ) z, 1 < N(τ ) < ∞, where N(τ ) is called the norm of τ . (τ1 , τ2 ∈ 0 are conjugate within 0, if there exists a τ3 ∈ 0, such that τ1 = τ3 τ2 τ3−1 .) The class of all elements in 0 that are conjugate to a given τ is called the conjugacy class of τ in 0 and denoted by {τ }. The number N(τ ) is, of course, the same within a conjugacy class and measure the “magnification.” N(τ ) has, however, another striking geometric interpretation, since there exists a unique relationship between the conjugacy classes in 0 and the homotopy classes of closed paths on the surface M. In each class, one defines a length l(τ ) by the length of the shortest closed path measured by means of the Poincaré distance. One, then, obtains N(τ ) = exp[l(τ )], l(τ ) > 0. Thus, the conjugacy classes in 0 can be uniquely parameterized by their length spectrum {l(τ )}. Given any τ ∈ 0, there is a unique τ0 , such that τ = τ0n , n ∈ N; τ0 is called primitive element of 0, since it cannot be expressed as a power of any other element of 0. The corresponding closed orbit with length l(τ0 ) is called a prime geodesic on M. Obviously, l(τ ) = l(τ0n ) = n l(τ0 ), since in this case the prime geodesic is traversed n times. For the length spectrum of M, one has Huber’s law ν(x) ∼ ex /x (x → ∞), where ν(x), is the number of inconjugate primitive τ ’s with l(τ ) ≤ x. (a) Prove the Selberg trace formula: ∞ X n=0

A h(τn ) = 2π +

Z∞ t tanh(π t) h(t) dt −∞

∞ XX {τ }p n=1

l(τ ) g(n l(τ )), sinh 21 nl(τ )

where all series and the integral converges absolutely under the following conditions on the function h(t): (i) h(−t) = h(t), (ii) h(t) is analytic in a strip |= t| ≤ 12 + ( > 0), (iii) |h(t)| ≤ a(1 + |t|2 )−1− (a > 0); the function g(u) is the Fourier transform of h(t): Z∞

1 g(u) = 2π

exp(−iut) h(t) dt; −∞

the sum on the left-hand side runs over the eigenvalues of 1 parameterized in the form λn = 14 + τn2 , that is, over the pairs (τn , −τn ), τn ∈ C (τ = 0 has to be counted twice, if 41 happens to be an eigenvalue); the sum on the right-hand side is taken over all primitive conjugacy class in 0, denoted by {τ }p . By choosing 1

h(t) = t2 +

s−

1 2

2 −

1 t2 +

z−

1 2

2

( −1, c > 0. Introduce the function of the complex variable z: φ(z, s, a, b, c) =

∞ X n=1

Show that: For z(s, a, b, c) =

Pd (z) . (cz + a)s (cz + b)s

d+1 2 ,

1 φ(1, s, a, b, c) + 2

Z∞

φ(z, s, a, b, c) dz + I(s),

1

where I(s) is an entire function of s. (See Spreafico [1063, Proposition 1])

6 q-Extensions of Some Special Functions and Polynomials

The theory of hypergeometric functions in one, two and more variables leads to a natural unification of much of the material of concern to the mathematical analysts from the seventeenth century to the present day. Functions of this type may be generalized along the lines of basic (or q-) number, resulting in the formation of q-extensions (or q-analogues). This was first systematically effected by E. Heine (1821–1881) in the middle of the nineteenth century, and the work was subsequently greatly extended by F. H. Jackson (1870–1960), W. N. Bailey (1893–1961), L. J. Slater, G. E. Andrews and many others up to the present day. In fact, in recent years, various families of basic (or q-) series and basic (or q-) polynomials have been investigated rather widely and extensively due mainly to their having been found to be potentially useful in such wide variety of fields as (for example) theory of partitions, number theory, combinatorial analysis, finite vector spaces, Lie theory, particle physics, nonlinear electric circuit theory, mechanical engineering, theory of heat conduction, quantum mechanics, cosmology and statistics (see, for details, [1099, pp. 350–351]). The books and monographs by (among others) W. N. Bailey [87], L. J. Slater [1040], H. Exton [439], H. M. Srivastava and P. W. Karlsson [1099, Chapter 9] and G. Gasper and M. Rahman [468] discussed extensively basic (or q-) hypergeometric functions in one, two and more variables. Here, in the present chapter, we choose to investigate some remarkable qdevelopments around (especially) the Zeta and related functions, which are reported in recent years. We also present the background material, involving (for example) Jackson’s q-integral, the q-Gamma function and the q-Beta function, multiple q-Gamma functions, q-Bernoulli numbers and q-Bernoulli polynomials, q-Euler numbers and q-Euler polynomials, q-Zeta functions, multiple q-Zeta functions and so on.

6.1 q-Shifted Factorials and q-Binomial Coefficients The q-shifted factorial (a; q)n is defined by  1 Q (a; q)n := n−1  (1 − a qk )

(n = 0) (n ∈ N),

k=0

Zeta and q-Zeta Functions and Associated Series and Integrals. DOI: 10.1016/B978-0-12-385218-2.00006-2 c 2012 Elsevier Inc. All rights reserved.

(1)

480


where a, q ∈ C and it is assumed that a 6= q−m (m ∈ N0 ). It is noted that some other notations that have been used in the literature for the product (a; q)n in (1) are (a)q,n , [a]n , and even (a)n , when the Pochhammer symbol 1.1(5) is not used and the base q is understood. The q-shifted factorial for negative subscript is defined by 1 (1 − a q−1 ) (1 − a q−2 ) · · · (1 − a q−n )

(a; q)−n :=

(n ∈ N0 ),

(2)

which yields n (−q/a)n q(2) (a; q)−n = = (a q−n ; q)n (q/a; q)n

1

(n ∈ N0 ).

(3)

We also write ∞ Y

(a; q)∞ :=

(1 − a qk )

(a, q ∈ C; |q| < 1).

(4)

k=0

It is noted that, when a 6= 0 and |q| = 1, the infinite product in (4) diverges. So, whenever (a; q)∞ is involved in a given formula, the constraint |q| < 1 will be tacitly assumed. It follows from (1), (2) and (4) that (a; q)n =

(a; q)∞ (a qn ; q)∞

(n ∈ Z),

(5)

which can be extended to n = α ∈ C as follows: (a; q)α =

(a; q)∞ (a qα ; q)∞

(α ∈ C; |q| < 1),

(6)

where the principal value of qα is taken. A list of easily-verified useful identities is given below: n q1−n ; q (−a)n q(2) (n ∈ Z); a n −1 1−n n n a q ; q = (a; q)n −a−1 q−(2) (n ∈ Z); a n q a n n (a q−n ; q)n = ;q − q−(2) (n ∈ Z); a q n

(a; q)n =

(a; q)n−k =

(a; q)n −1 (a q1−n ; q)k

(−q a−1 )k q(2)−n k k

(n, k ∈ Z);

(7) (8) (9) (10)

q-Extensions of Some Special Functions and Polynomials

(q−n ; q)k =

k (q; q)n (−1)k q(2)−n k (q; q)n−k

481

(n, k ∈ Z);

(a; q)k (q a−1 ; q)n −n k q (n, k ∈ Z); (a−1 q1−k ; q)n a n−k (k)−(n) (q/a; q)n −n − q 2 2 (n, k ∈ Z); (a q ; q)n−k = (q/a; q)k q (a q−n ; q)k =

(a qk ; q)n−k =

(a; q)n (a; q)k

(n, k ∈ Z);

(11) (12) (13) (14)

(a; q)n+k = (a; q)n (a qn ; q)k

(n, k ∈ Z);

(15)

(a; q)k (a qk ; q)n (a; q)n

(n, k ∈ Z);

(16)

(a qn ; q)k =

(a q2k ; q)n−k =

(a; q)n (a qn ; q)k (a; q)2k

(a; q)2n = (a; q2 )n (a q; q2 )n

(n, k ∈ Z);

(n ∈ Z);

(18)

(a2 ; q2 )n = (a; q)n (−a; q)n (n ∈ Z); a n −3 (n) (q/a; q)2n −2n − 2 q 2 (a q ; q)n = (q/a; q)n q (a q−kn ; q)n = (a qkn ; q)n =

(19) (n ∈ Z);

n (q/a; q)kn 2 (−a)n q(2)−k n (q/a; q)(k−1)n

(a; q)(k+1)n (a; q)kn

(a qj k ; q)n−k =

(17)

(n, k ∈ Z);

(n, k ∈ Z);

(a; q)n (a qn ; q)( j−1) k (a; q)j k

(20) (21) (22)

(n, k, j ∈ Z).

(23)

We, now, introduce some more q-notations, which would appear in this chapter quite frequently. The notation [z]q is defined by [z]q :=

qz − 1 1 − qz = 1−q q−1

(z ∈ C; q ∈ C \ {1}; qz 6= 1).

(24)

A special case of (24) when z ∈ N is [n]q =

qn − 1 = 1 + q + · · · + qn−1 q−1

(n ∈ N),

which is called the q-analogue (or q-extension) of n ∈ N, since lim [n]q = lim (1 + q + · · · + qn−1 ) = n.

q→1

q→1

(25)

482


The q-analogue of n! is then defined by [n]q ! :=

1 [n]q [n − 1]q · · · [2]q [1]q

if n = 0, if n ∈ N,

from which the q-binomial coefficient (or the Gaussian polynomial analogous to is defined by [n]q ! n := [n − k]q ! [k]q ! k q

n, k ∈ N0 ; 0 5 k 5 n .

(26) n k )

(27)

It is easily seen from (1) and (27) that (q; q)n = (1 − q)n [n]q !

(n ∈ N0 ).

(28)

The q-binomial coefficient in (27) can be generalized in a similar way as in 1.1(20): [α]q;k α := [k]q ! k q

(α ∈ C; k ∈ N0 ),

(29)

where [α]q;k is defined by [α]q;k := [α]q [α − 1]q · · · [α − k + 1]q

(α ∈ C; k ∈ N0 ).

(30)

The following notations are also frequently used: (a1 , a2 , . . . , am ; q)n := (a1 ; q)n (a2 ; q)n · · · (am ; q)n

(31)

(a1 , a2 , . . . , am ; q)∞ := (a1 ; q)∞ (a2 ; q)∞ · · · (am ; q)∞ .

(32)

and

Now, the generalized binomial coefficient in (29) can be further generalized as follows: qβ+1 , qα−β+1 ; q ∞ α := β q q, qα+1 ; q ∞

(α, β ∈ C; |q| < 1).

(33)

It is noted that, whenever there is no confusion, the notations [z]q , [n]q ! and βα q are simply written as [z], [n]! and βα , respectively. We record some known identities


of [z]q and

α

β q

[−n]q = − [n] 1 = q

as follows: 1 [n]q qn

(n ∈ Z);

1 [n]q qn−1

(n ∈ Z);

q

1 qn(n−1)/2

(34) (35)

(m, n ∈ Z);

(36)

[n]q ! (n ∈ Z);

(37)

[m n]q = [m]q [n]qm [n] 1 ! =

483

n n (q; q)n n, k ∈ N0 ; 0 5 k 5 n = = k q n − k q (q; q)k (q; q)n−k [m]q m − 1 m = (m ∈ Z; n ∈ N); n q [n]q n − 1 q 1 m m = n(m−n) (m ∈ Z; n ∈ N0 ); n 1 n q q q m k m m−j = (m ∈ Z; k, j ∈ N0 ; 0 5 j 5 k); k q j q j q k−j q q−α ; q k k k α −qα q−(2) (α ∈ C; k ∈ N0 ) ; = (q; q)k k q qα+1 ; q k k+α = (α ∈ C; k ∈ N0 ) ; k q (q; q)k k −α α+k−1 = (−q−α )k q−(2) (α ∈ C; k ∈ N0 ); k q k q α+1 α α k = q + k q k q k−1 q α α = + qα+1−k (α ∈ C; k ∈ N0 ) ; k q k−1 q n X k n (z; q)n = (−z)k q(2) z ∈ C; n, k ∈ N0 ; 0 5 k 5 n . k q

(38) (39) (40)

(41)

(42)

(43) (44)

(45)

(46)

k=0

6.2 q-Derivative, q-Antiderivative and Jackson q-Integral We begin by noting that F. J. Jackson was the first to develop q-calculus in a systematic way.

484


q-Derivative The q-derivative of a function f (t) is defined by Dq {f (t)} :=

dq f (qt) − f (t) {f (t)} = . dq t (q − 1)t

(1)

It is noted that lim Dq {f (t)} =

q→1

d {f (t)}, dt

if f (t) is differentiable. We record some easily derivable q-derivative formulas: Dq {a f (t) + b g(t)} = a Dq {f (t)} + b Dq {g(t)}; Dq {f (t) g(t)} = f (qt) Dq {g(t)} + g(t) Dq {f (t)},

(2) (3)

where the functions f (t) and g(t) are obviously interchangeable; Dq

g(qt) Dq {f (t)} − f (t) Dq {g(t)} f (t) = , g(t) g(t) g(q t)

(4)

so that, clearly, g(q t) Dq f (t) − f (q t) Dq g(t) d f (t) f (t) = lim = . lim Dq q→1 q→1 g(t) g(t) g(q t) dt g(t)

(5)

It is known (see [618, pp. 3–4]) that there does not exist a general chain rule for q-derivatives. There is an exception given by Dq {f (u(x))} = Dqβ f (u(x)) · Dq {u(x)},

(6)

where u = u(x) = α xβ , α and β being constants.

q-Antiderivative and Jackson q-Integral The function F(t) is a q-antiderivative of f (t), if Dq {F(t)} = f (t). It is denoted by Z

f (t) dq t.

(7)

In ordinary calculus, an antiderivative is unique up to an additive constant. However, in the case of a q-antiderivative, it is known (see [618, p. 65, Proposition 18.1]) that, for 0 < q < 1, up to an additive constant, any function f (t) has at most one q-antiderivative that is continuous at t = 0.


485

The Jackson integral of f (t) is, thus, defined, formally, by Z

∞ X

f (t) dq t := (1 − q)t

qj f qj t ,

(8)

j=0

which can be easily generalized as follows: Z

f (t) dq g(t) =

∞ X

f qj t

g q j t − g qj+1 t .

(9)

j=0

The following theorem gives a sufficient condition under which the formal series in (8) actually converges to a q-antiderivative. Theorem 6.1 (618, p. 68, Theorem 19.1) Suppose 0 < q < 1. If |f (t) tα | is bounded on the interval (0, A] for some 0 5 α < 1, then the Jackson integral, defined by (7), converges to a function F(t) on (0, A], which is a q-antiderivative of f (t). Moreover, F(t) is continuous at t = 0 with F(0) = 0. Suppose that 0 < a < b. The definite q-integral is defined as follows: Zb

f (t) dq t := (1 − q)b

∞ X

q j f (q j b)

(10)

j=0

0

and Zb a

f (t) dq t =

Zb

f (t) dq t −

0

Za

f (t) dq t.

(11)

0

A more general version of (10) is given by Zb

f (t) dq g(t) =

∞ X

f q j b g q j b − g qj+1 b .

(12)

j=0

0

The improper q-integral of f (t) on [0, ∞) is defined by Z∞ 0

f (t) dq t : =

∞ Zq X

j

f (t) dq t

j=−∞ j+1 q

= (1 − q)

∞ X j=−∞

(13) f (q j ) q j

(0 < q < 1)

486


and Z∞

j+1

f (t) dq t =

q ∞ Z X j=−∞

0

=

q−1 q

f (t) dq t

qj ∞ X

(14) f (q j ) q j

(q > 1).

j=−∞

As noted earlier (see [618, p. 71, Proposition 19.1]), the improper q-integral, defined above, converges, if tα f (t) is bounded in a neighborhood of t = 0 with some α < 1 and for sufficiently large t with some α > 1. The formula for q-integration by parts, the fundamental theorem of q-calculus and q-Taylor formula are given as follows (see [618, Chapter 20]): Zb

f (t) dq g(t) = f (b) g(b) − f (a) g(a) −

a

Zb

g(q t) dq f (t)

(0 5 a < b 5 ∞).

a

(15) Theorem 6.2 (Fundamental Theorem of q-Calculus) If F(t) is an antiderivative of f (t) and if F(t) is continuous at t = 0, then Zb

f (t) dq t = F(b) − F(a)

(0 5 a < b 5 ∞).

(16)

a

The q-analogue of (t − a)n is defined by the polynomial (t − a)nq :=

1 (n = 0) (t − a) (t − q a) · · · (t − qn−1 a) (n ∈ N).

(17)

It is easy to see that Dq {(t − a)nq } = [n]q (t − a)n−1 q

(n ∈ N).

(18)

Theorem 6.3 (q-Taylor Formula with the Cauchy Remainder Term) Suppose that j Dq {f (t)} is continuous at t = 0 for any j, n ∈ N0 ( j 5 n + 1). Then a q-analogue of Taylor’s formula with the Cauchy remainder is given as follows:

f (b) =

n X j=0

Dqj f

j

(b − a)q 1 (a) + [j]q ! [n]q !

Zb a

n Dn+1 q {f (t)}(b − q t)q dq t.

(19)


487

6.3 q-Binomial Theorem We begin by recalling the well-known Ramanujan’s 1 91 -sum: q b ∞ ; q (q; q) ; q (az; q) X ∞ ∞ az a (a; q)k k ∞ ∞ z = 1 91 (a; b; q, z) := q b (b; q)k (z; q) ; q (b; q) ; q ∞ az ∞ a k=−∞ ∞

(1)

∞

(|q| < 1; |a| > |q|; |b| < 1; |b/a| < |z| < 1). A simple proof of (1) is given in [584] (see also [1080]). A special case of (1) when b = q yields the q-binomial theorem: 1 80 (a; −; q, z) :=

∞ X (a; q)k k (az; q)∞ z = (q; q)k (z; q)∞

(|q| < 1; |z| < 1),

(2)

k=0

which were proven by several mathematicians, such as Cauchy [225] and Heine [549]. Two special cases of (2) when a = 0 and when z is replaced by z a−1 and a → ∞ yield Euler’s formulas: ∞ X k=0

zk 1 = (q; q)k (z; q)∞

(|q| < 1; |z| < 1)

(3)

and k ∞ X (−1)k q(2) k z = (z; q)∞ (q; q)k

(|q| < 1; |z| < 1),

(4)

k=0

respectively. It is observed that lim q↓1

(qa z; q)∞ = lim 1 80 qa ; −; q, z = 1 F0 (a; −; z) = (1 − z)−a q↓1 (z; q)∞ (|z| < 1; a ∈ C),

(5)

which, by the principle of analytic continuation, holds true for z ∈ C cut along the positive real axis from 1 to ∞, with (1 − z)−a positive when z is real and less than 1. The special case of (2) when a = q−n (n ∈ N0 ) gives 1 80 (q

−n

; −; q, z) = (z q

−n

; q)n = (−z) q

n −n(n+1)/2

q ;q z n

(n ∈ N0 ),

(6)

which, by the principle of analytic continuation, holds true for z ∈ C and is seen to be equivalent to 6.1(9).

488


A q-analogue of the classical exponential function ez is defined by ezq :=

∞ X zk , [k]q !

(7)

k=0

and another q-analogue of the classical exponential function ez is defined by Eqz :=

∞ X

qk(k−1)/2

k=0

zk = (1 + (1 − q)z)∞ q . [k]q !

(8)

It is easily seen, by applying (3) and (4), that ezq Eq−z = 1.

(9)

From 6.1(37), we also see that ez1/q = Eqz .

(10)

These two q-exponential functions under q-differentiation are given by and Dq {Eqz } = Eqqz .

Dq {ezq } = ezq

(11)

In general, we have z+w ezq ew q 6= eq .

However, the following additive property does hold true: z+w ezq ew q = eq

(wz = qzw).

(12)

Corresponding to the above-defined q-exponential functions ezq and Eqz , the q-trigonometric functions are defined as follows: sinq x := cosq x :=

−ix eix q − eq

2i ix eq + e−ix q 2

and and

Sinq x :=

Eqix − Eq−ix

; 2i Eqix + Eq−ix Cosq x := . 2

(13) (14)

It is easy to see from (10) that Sinq x = sin1/q x

and

Cosq x = cos1/q x.

(15)

We can use (9) to obtain sinq x · Sinq x + cosq x · Cosq x = 1,

(16)


489

which is the q-analogue of the elementary identity: sin2 x + cos2 x = 1. q-analogues of other trigonometric identities can also be obtained (see [468, p. 23]). We apply the chain rule 6.2(6) and use (11) to find the following derivative formulas for the q-trigonometric functions: Dq {sinq x} = cosq x

and Dq {Sinq x} = Cosq qx

(17)

and Dq {cosq x} = − sinq x

and Dq {Cosq x} = −Sinq x.

(18)

Ramanujan’s 1 91 -sum (1) includes the Jacobi triple product identity [602] as a limiting case: ∞ X

2

(−1)k qk zk = lim 1 91 (−1/c; 0; q2 , −qzc) c→0

k=−∞

(19)

q = zq; q2 ; q2 q2 ; q2 , ∞ z ∞ ∞ which can be used to express the theta functions as follows (see, e.g., Whittaker and Watson [1225, Chapter 21], Gasper and Rahman [468, Section 1.6] and Bellman [113]): θ1 (x) = 2 θ2 (x) = 2

∞ X k=0 ∞ X

(−1)k q(k+1/2) sin(2k + 1) x,

(20)

q(k+1/2) cos(2k + 1) x,

(21)

2

2

k=0

θ3 (x) = 1 + 2

∞ X

2

qk cos 2kx

(22)

k=1

and θ4 (x) = 1 + 2

∞ X

2

(−1)k qk cos 2kx.

(23)

k=1

In terms of infinite products, upon replacing z in (19) by q e2ix , −q e2ix , −e2ix and e2ix , respectively, we have θ1 (x) = 2 q1/4 sin x

∞ Y 1 − q2k 1 − 2 q2k cos 2x + q4k , k=1

(24)

490


θ2 (x) = 2 q1/4 cos x

∞ Y 1 − q2k 1 + 2 q2k cos 2x + q4k ,

(25)

k=1

θ3 (x) =

∞ Y 1 − q2k 1 + 2 q2k−1 cos 2x + q4k−2

(26)

k=1

and θ4 (x) =

∞ Y 1 − q2k 1 − 2 q2k−1 cos 2x + q4k−2 .

(27)

k=1

We conclude this section by giving a widely-investigated generalization r 8s of the function 1 80 (a; −; q, z) in (2), which is defined by   a1 , . . . , ar ; q, z = r 8s (a1 , . . . , ar ; b1 , . . . , bs ; q, z) r 8s  b1 , . . . , bs ; :=

∞ X

(−1)(1−r+s)k q(1−r+s)(2) k

k=0

·

(a1 ; q)k · · · (ar ; q)k zk , (b1 ; q)k · · · (bs ; q)k (q; q)k

(28)

provided that the generalized basic (or q-) hypergeometric series in (28) converges.

6.4 q-Gamma Function and q-Beta Function q-Gamma Function The classic Gamma function 0(z) (see Section 1.1) was found by Euler, while he was trying to extend the factorial n! = 0(n + 1) (n ∈ N0 ) to real numbers. The q-factorial function [n]q ! (n ∈ N0 ), defined by 6.1(26), can be rewritten as follows: (1 − q)

−n

∞ Y (q; q)∞ (1 − qk+1 ) = (1 − q)−n := 0q (n + 1) (1 − qk+1+n ) (qn+1 ; q)∞

(0 < q < 1).

k=0

(1) Replacing n by x − 1 in (1), Jackson [592] defined the q-Gamma function 0q (x) by 0q (x) :=

(q; q)∞ (1 − q)1−x (qx ; q)∞

(0 < q < 1).

(2)

Note that it is not completely obvious that Jackson’s q-Gamma function is the most natural extension of the above-defined 0q (n + 1) (n ∈ N). Askey [73] has found analogues of many of the known facts about the Gamma function, and these strongly


491

indicate that (2) is the natural extension of 0q (n + 1) (n ∈ N), which reduces to 0(x) as q → 1. Also note that, throughout this section, 0 < q < 1 is assumed. Setting x = 1 in (2) gives 0q (1) = 1.

(3)

The q-Gamma function in (2) satisfies the fundamental functional relation 0(x + 1) = x 0(x) (see 1.1(9)): 0q (x + 1) =

1 − qx 0q (x) = [x]q 0q (x). 1−q

(4)

Indeed, by observing that (q x+1 ; q)∞ =

(qx ; q)∞ , 1 − qx

(5)

we find that (q; q)∞ (1 − q)−x q x+1 ; q ∞ 1 − qx 1 − qx (q; q)∞ 0q (x). (1 − q)1−x = = x 1 − q (q ; q)∞ 1−q

0q (x + 1) =

The q-Gamma function in (2) also satisfies the q-analogue of the Bohr-Mollerup theorem (see Theorem 1.1): Theorem 6.4 (73, Theorem 3.1) Let f (x) be a function that satisfies each of the following properties: (a) f (1) = 1; (b) for some q ∈ (0, 1), f (x + 1) =

1 − qx f (x); 1−q

(6)

(c) log f (x) is convex for x > 0.

Then, f (x) = 0q (x) =

(q; q)∞ (1 − q)1−x (qx ; q)∞

(x > 0).

Proof. It is proven from (3) and (4) that 0q (x) satisfies (a) and (b). Observe that, for 0 < q < 1, ∞ d X d log 0q (x) = − log(1 − q) − log 1 − qn+x dx dx n=0 ∞ X

= − log(1 − q) + log q

n=0

qn+x 1 − qn+x

492


and ∞

X qn+x d2 2 log 0 (x) = (log q) q . n+x 2 dx2 n=0 1 − q Thus, log 0q (x) is convex for x > 0.

Let 0 < x < 1. The convexity of log f (x) implies log f (x + n) 5 log f (n) + x [log f (n + 1) − log f (n)] or f (n + x) 5 f (n)

1 − qn 1−q

x

.

The functional equation (6) then gives (1 − q)n 1 − qn x (1 − q)n f (x + n) 5 x f (n) f (x) = x 1−q (q ; q)n (q ; q)n (1 − q)n (q; q)n−1 (1 − qn )x = x (q ; q)n (1 − q)n−1 (1 − q)x so f (x) 5

(q; q)n−1 1−x n x (1 − q) 1 − q (qx ; q)n

(0 < q < 1, 0 < x < 1).

(7)

To find a lower bound for f (x), apply the convexity of log f (x) at n + x, n + 1 and n + x + 1 to obtain log f (n + 1) 5 log f (n + x) + (1 − x) [log f (n + x + 1) − log f (n + x)] or f (n + 1) 5 f (n + x)

1 − qn+x 1−q

1−x

.

As above, the functional equation (6) gives (1 − q)n (1 − q)n f (x + n) = x f (x) = x (q ; q)n (q ; q)n =

(1 − q)1−x (q; q)n , 1−x 1 − qn+x (qx ; q)n

1−q 1 − qn+x

1−x

f (n + 1)


493

so f (x) =

(q; q)n (1 − q)1−x 1−x (qx ; q)n 1 − qn+x

(0 < q < 1, 0 < x < 1).

(8)

Setting n → ∞ in (7) and (8) gives f (x) =

(q; q)∞ (1 − q)1−x = 0q (x). (qx ; q)∞

The functional equation (6) then gives f (x) = 0q (x) for x > 0. Remark 1 The q-gamma function has simple poles at x = 0, −1, −2, . . . and the residues Res 0q (x) =

x=−n

(1 − q)n+1 q−n ; q n log q−1

(n ∈ N0 ).

(9)

Indeed, lim (x + n) 0q (x) = lim

x→−n

x→−n

(x + n) (q; q)∞ · · · (1 − q)1−x (1 − qx ) 1 − q x+1

=

(1 − q)n+1 x+n lim 1 − q−n · · · 1 − q−1 x→−n 1 − q x+n

=

(1 − q)n+1 . q−n ; q n log q−1

The q-gamma function has no zeros, so its reciprocal is an entire function with zeros at x = −n (n ∈ N0 ). ∞ Y 1 1 − qn+x x−1 = (1 − q) , 0q (x) 1 − qn+1

(10)

n=0

which also has zeros at x = −n + (2πik/ log q) (k ∈ Z; n ∈ N0 ). Applying 6.3(3) and 6.3(4) to the definition of the q-Gamma function (2) yields 0q (x) = (q; q)∞ (1 − q)1−x

∞ X k=0

qkx (q; q)k

(11)

and k ∞ 1 (1 − q)x−1 X (−1)k q(2) qkx = . 0q (x) (q; q)∞ (q; q)k

k=0

(12)

494


Now, the generalized binomial coefficient βα , defined by 6.1(33), can, in terms of q q-Gamma function, be rewritten like 1.1(41) as follows: 0q (α + 1) α = . (13) β q 0q (β + 1) 0q (α − β + 1) The q-factorial [n]q !, defined by 6.1(26), is a monotone increasing function of q for q > 0, so it is natural to ask if something happens for the q-Gamma function. Since 0q (1) = 0q (2) = 1, something different may happen for 1 < x < 2 than for other x > 0. Theorem 6.5 (73, Theorem 4.1) The following inequalities hold: 0r (x) 5 0q (x) 5 0(x)

(0 < x 5 1 or x = 2, 0 < r < q < 1),

(14)

0(x) 5 0q (x) 5 0r (x)

(1 5 x 5 2, 0 < r < q < 1).

(15)

Also lim 0q (x) = 0(x).

q→1−

(16)

Proof. Only the proof of (16) is provided. By (14) and (15), 0q (x) is a monotone function of q, which is bounded, and so has a limit. By Theorem 6.4, the limit satisfies the assumptions of the Bohr-Mollerup theorem (see Theorem 1.1) and so is 0(x). A q-analogue of Legendre duplication formula for the Gamma function 1.1(29) is given as follows: 1 1 0q (2x) 0q2 = 0q2 (x) 0q2 x + (1 + q)2x−1 . (17) 2 2 Indeed, we have 1−x+1/2−x 0q2 (x) 0q2 x + 21 q2 ; q2 ∞ q; q2 ∞ 1 − q2 = 1/2 q2x ; q2 ∞ q2x+1 ; q2 ∞ 1 − q2 0q2 21 1−2x (q; q)∞ = 2x 1 − q2 = 0q (2x) (1 + q)1−2x . q ;q ∞ Also, a q-analogue of Gauss multiplication formula for the Gamma function 1.1(51) is similarly proven: 1 2 n−1 0q (nx) 0qn 0qn · · · 0qn n n n (18) 1 n−1 = 0qn (x) 0qn x + · · · 0qn x + [n]nx−1 (n ∈ N). q n n


495

q-Beta Function A special case of 6.2(10) when b = 1 yields Z1

f (t) dq t = (1 − q)

∞ X

qn f qn .

(19)

n=0

0

Now, try to get a q-analogue of the Beta function, defined by 1.1(59), and keep the integrand f (t) = tα−1 (1 − t)β−1 of 1.1(59) in mind. Since qn already appears in (19), it is natural to use tα−1 in itself. Even though there is no hope of evaluating a sum that contains a factor that (1 − qn )β−1 , in view of 6.3(2) and 6.3(5), there seems to be an alternative of a q-analogue of (1 − t)β−1 as follows: ∞ X q1−β ; q n n t q1−β ; q ∞ t = . (20) (q; q)n (t; q)∞ n=0

(20) is almost the right q-analogue of (1 − t)β−1 . The one thing wrong is that the β−1 function being integrated in 1.1(39) is not (1 − t)β−1 but (1 − t)+ , since the range of integration in 1.1(39) stops at t = 1. The first point in the sequence t = qn , which lies to the right of t = 1, is t = q−1 . So, by shifting the variable t by t qβ , Askey [73, Eq. (5.7)] chose a natural candidate for the q-Beta function: Bq (α, β) : =

Z1

tα−1

0

= (1 − q)

(t q; q)∞ dq t (t qβ ; q)∞ ∞ X

qnα

n=0

( 0; β ∈ C \ Z− 0) (21)

qn+1 ; q ∞ . qn+y ; q ∞

We use 6.1(5) and 6.3(2) to prove a relationship between the q-Gamma function and the q-Beta function (see 1.1(42)), which shows a natural choice in (21). Theorem 6.6 We have Bq (α, β) =

0q (α) 0q (β) . 0q (α + β)

(22)

Proof. ∞ (q; q)∞ X qβ ; q n nα Bq (α, β) = (1 − q) β q q ; q ∞ n=0 (q; q)n (1 − q) (q; q)∞ qα+β ; q ∞ = qβ ; q ∞ (qα ; q)∞ =

(q;q)∞ (qα ;q)∞

(1 − q)1−α (q;q)∞ (qα+β ;q)∞

(q;q)∞

(qβ ;q)∞

(1 − q)1−β

(1 − q)1−α−β

=

0q (α) 0q (β) . 0q (α + β)

496


Although it is not possible to change variables in a sum (and so in a q-integral), there are times when a change of variables in an ordinary integral will lead to another integral that can be approximated by a q-integral. For example, setting u = ct in 1.1(41) yields B(x, y) = cx

Z∞ 0

tx−1 dt (1 + ct)x+y

( 0; 0).

(23)

Askey [73] observed that Ramanujan’s sum 6.3(1) gives a q-integral extension of (23). Indeed, we use 6.1(5) to rewrite 6.3(1) as follows: q b ∞ X (bqn ; q)∞ n (ax; q)∞ ax ; q ∞ (q; q)∞ a ; q ∞ x = , b (aqn ; q)∞ (x; q)∞ ax ;q (a; q)∞ aq ; q ∞ n=−∞ ∞

α+β and using the q-binomial theowhich, upon setting x = qα , a = −c, b = −cnq α+β+n rem 6.3(2) to replace −c q ; q ∞ / (−c q ; q)∞ , gives

"∞ # X (qα+β ; q)k n k (−c q ) qαn (q; q) k n=−∞ ∞ X

=

k=0 (−cqα ; q)∞ (−c−1 q1−α ; q)∞ (q; q)∞ (qα+β ; q)∞ . (−c; q)∞ (−c−1 q; q)∞ (qα ; q)∞ (qβ ; q)∞

(24)

If we use 6.2(13) and 6.3(2), (24) can be rewritten as Z∞ X ∞ 0

k=0

(qα+β ; q)k (−c x)k xα−1 dq x (q; q)k

(−cqα ; q)∞ (−c−1 q1−α ; q)∞ 0q (α) 0q (β) (−c; q)∞ (−c−1 q; q)∞ 0q (α + β) Z∞ (−cqα+β x; q)∞ α−1 = x dq x. (−cx; q)∞ =

0

In view of 6.3(5), it is seen that (−cqα ; q)∞ −c−1 q1−α ; q lim q→1 (−c; q)∞ −c−1 q; q ∞

∞

= (1 + c)

and −cqα+β x; q

lim

q→1

(−cx; q)∞

∞

= (1 + cx)−α−β .

−α

1 α 1+ = c−α c

(25)


497

6.5 A q-Extension of the Multiple Gamma Functions Motivated by Theorem 6.4 and Theorem 1.5, Nishizawa [865, 866] and Ueno and Nishizawa [1176] presented a q-analogue of the multiple Gamma functions (see Section 1.4) and its properties. Here, we introduce their works without proof. We assume 0 < q < 1 throughout this section. We begin by giving q-analogues of Gauss’s and Euler’s product forms of the Gamma function 1.1(4) and 1.1(7) (see [865, Equations (3.2) and (3.3)]): 0q (z + 1) = lim

N→∞

[1][2] · · · [N] [N + 1]z [z + 1][z + 2] · · · [z + N]

(1)

and ( ) ∞ Y [n + 1] z [z + n] −1 0q (z + 1) = , [n] [n]

(2)

n=1

where [z] = [z]q throughout this section. Nishizawa [865, Definition 4.1] defines the q-analogue of Vigne´ ras’s Gr -function (see Theorems 1.4 and 1.5) as follows: Definition 6.7 For z ∈ C with 0, G0 (z + 1; q) := [z + 1], 

z

Gr (z + 1; q) := (1 − q)− r

∞   Y

(−1) 1 − qz+n

 n=1 

1 − qn

r(n+r−2 r−1 )

(1 − qn )gr (z,n)

  

(r ∈ N),

 

where g1 (z, n) := 0, gr (z, n) :=

r−1 X k=1

(−1)k−1

z r−k

n+k−2 (r ∈ N \ {1}). k−1

We note that G1 (z; q) = 0q (z). The infinite products of these functions are absolutely convergent. Nishizawa [865, Theorem 4.2] (see also [866, Theorem 3.1]) proves a q-analogue of Theorem 1.5: Theorem 6.8 A unique hierarchy of functions exists, which satisies (i) Gr (z + 1; q) = Gr−1 (z; q) Gr (z; q), (ii) Gr (1; q) = 1, dr+1 (iii) dz r+1 log Gr+1 (z + 1; q) = 0 (z = 0), (iv) G0 (z; q) = [z],

498


where

z−u −u gr (z, u) = − . r−1 r−1 Gr (z + 1; q) is expressed as the following infinite product representation:   −k ∞  gr (z,k)  z+k (r−1) Y 1 − q 1 − qk (r ∈ N). (3) Gr (z + 1; q) := (1 − q)−( )   1 − qk z r

k=1

The case of Theorem 6.8 when r = 1 corresponds to Askey’s theorem (see Theorem 6.4). So, the sequence {Gr (z; q)} includes a q-Gamma function. We call an element of the sequence a multiple q-Gamma function. The expression (3) can be regarded as a q-analogue of the Weierstrass product form for the function Gr (z) given in Theorem 1.4. Nishizawa [865, Proposition 4.4] (see also [866, Proposition 3.2]) derives the following counterpart of (1) and (2) for the function Gr (z; q): Theorem 6.9 If 0, then Gr (z + 1; q) = lim

N→∞

Gr−1 (1; q) · · · Gr−1 (N; q) Gr−1 (z + 1; q) · · · Gr−1 (z + N; q) ) r Y z ( ) · Gr−m (N + 1; q) m m=1

and

Gr (z + 1; q) =

∞ Y n=1

(

z ) r Gr−1 (n; q) Y Gr−m (n + 1; q) (m) . Gr−1 (z + n; q) Gr−m (n; q) m=1

Koornwinder [694, Theorem B.2] proved rigorously the following result: lim 0q (z + 1) = 0(z + 1) q↑1

(z ∈ C \ {−1, −2, . . .}).

(4)

By making use of the Euler-Maclaurin summation formula (see 2.7(21)), Ueno and Nishizawa [1176, Proposition 4.1] (see also Nishizawa [866, Proposition 3.3]) obtain an expansion formula of log Gn (z + 1; q) as follows:


499

Theorem 6.10 Suppose that −1 and m > n. They have ( ) X n 1 − qz+1 d r−1 z z+1 Br log log Gn (z + 1; q) = − + n r! dz n−1 1−q r=1

+

( n X r=1

+

n−1 X j=0

d − dz

r−1

) Zz+1

z n−1

1

ξ r qξ log q dξ r! 1 − qξ

(5)

m X Br Gn,j (z) Cj (q) + Fn,r−1 (z; q) − Rn,m (z; q), r! r=1

where −t 1 − qz+t log , n−1 1 − qz+1 t=1 Z∞ n+1 X (−1)n e Bn+1 (t) Br fj+1,r−1 (1; q) + fj+1,n+1 (t; q) dt, Cj (q) := − r! (n + 1)!

Fn,r−1 (z; q) :=

dr−1 dtr−1

r=1

dr−1

1

1 − qt j fj+1,r−1 (t; q) := r−1 t log , 1−q dt Z∞ −t 1 − qz+t (−1)m−1 dm e log dt. Rn,m (z; q) := Bm (t) m n−1 m! dt 1 − qz+1

1

The formula (5) is a generalization of Moak’s [838, Theorem 2]. As remarked by Daalhuis [361], the formula (5) is not an asymptotic expansion. Yet, Ueno and Nishizawa [1176] (see also Nishizawa [866]) observes that each term of (5) converges uniformly as q ↑ 1. Thus, they obtain the following theorem (see Nishizawa [866, Theorem 3.4]): Theorem 6.11 As q → 1 − 0, Gn (z + 1; q) converges to Gn (z + 1) uniformly on any compact set in the domain C \ {−1, −2, . . .}.

6.6 q-Bernoulli Numbers and q-Bernoulli Polynomials Carlitz ([215] and [217]) introduced q-extensions of the classic Bernoulli numbers and polynomials. Since then, many authors have studied this and related subjects (see, e.g., [227, 228, 255, 363, 493, 626, 649, 652, 653, 659, 660, 682, 971, 992, 1006, 1100, 1168, 1170] and [1174]). In the frequently cited [215], Carlitz introduced the relation (qη + 1)k = ηk

(k ∈ N \ {1}),

η0 = 1,

η1 = 0

(1)

500


as an inductive definition for a certain sequence {ηk }∞ k=0 of functions ηk = ηk;q , depending rationally on the parameter q. This definition is interpreted according to the umbral calculus convention that, after expanding the left side into monomials, one replaces each power ηj by the corresponding sequence element ηj . The same paper exhibits polynomials ηk (x) = ηk;q (x) in qx , determined by a difference equation analogous to that governing the Bernoulli polynomials (see 1.7(4)), namely ηk;q (x + 1) − ηk;q (x) = k qx [x]k−1 q ,

ηk (0) = ηk ,

(2)

where [x]q is defined by 6.1(24). In what follows, we require, for convenience, that 0 < q < 1. Because ηk and ηk (x) do not remain finite at q = 1, Carlitz [215] used ηk and ηk (x) in (1) and (2) to define a set of numbers βk = βk;q , by βk = ηk + (q − 1) ηk+1 ,

(3)

and a set of polynomials βk (x) = βk;q (x), by the recurrence relation qx βk (x) = ηk (x) + (q − 1) ηk+1 (x),

βk := βk (0)

(k ∈ N0 ).

(4)

In the limit q → 1, βk reduces to the Bernoulli number Bk . These numbers βk;q and polynomials βk;q (x) are q-analogues of the ordinary Bernoulli numbers Bk and polynomials Bk (x) (see Section 1.6). Recently, a variety of works on q-Bernoulli numbers and polynomials and q-Euler numbers and polynomials have appeared, for example, Cenkci and Can [227], Ryoo [992] and Kim [654]. Kim, in particular, has published 50 papers on this and related subjects; a few are cited here. Although many authors, including those whose works are mentioned above, have treated more general generating functions of q-eta and q-epsilon polynomials than those discussed here, we, first, revisit the seminal work of Carlitz [215], substituting formal generating series for the difference equations employed by him (see (1.2)). In fact, Carlitz himself noted [215, p. 988, lines 4–6] that it is easy to define numbers and polynomials, as well as generating functions, of higher order. We do propose certain minor corrections to [215]; corrected statements appear here, for example, as Equations (74) and (80). Here, we will recover the identities for ηk and ηk (x) given by Carlitz [215]. In place of the difference relations (1) and (2), defining ηk and ηk (x), we apply the (nowstandard) formal generating series Gq (x, t) := −t

∞ X

qk+x e[k+x]q t ≡

k=0

∞ X

ηk;q (x)

k=1

tk k!

(5)

and its specialization Gq (t) := Gq (0, t) = −t

∞ X k=0

qk e[k]q t ≡

∞ X k=1

ηk;q

tk . k!

(6)


501

It is evident from (5) and (6) that ηk;q = ηk;q (0)

(k ∈ N).

(7)

To retain as close as possible a correspondence with Carlitz [215], we observe that the absence of any term involving η0;q in the generating series permits us to define η0;q = 1 as in (1). It follows from 6.1(24) that [x + y]q = [x]q + qx [y]q .

(8)

Using (8), we find that ηk;q (x) =

k X k j=0

j

xj [x]k−j q q ηj;q

(k ∈ N0 ).

(9)

Indeed, by (5), (6) and (8), we have ∞ X

∞

ηk;q (x)

k=0

X tk x = −t qk+x e([x]q +[k]q q ) t k! k=0

[x]q t

=e

x

−q t

∞ X

! k [k]q qx t

q e

k=0

   ∞ ∞ j j X X t t ηj;q q xj  = e[x]q t Gq qx t =  [x]qj  j! j! j=0 j=0   k ∞ X X 1 xj   tk , [x]k−j = q ηj;q q j!(k − j)!

k=0

j=0

which, upon equating the coefficients of tk , yields (9). From (5), we get q-difference equation for Gx;q (t): et Gx;q (qt) = t qx e[x]q t + Gx;q (t),

(10)

which easily yields k+1 X k+1 j=0

j

q j ηj;q (x) = (k + 1) qx [x]kq + ηk+1;q (x)

(k ∈ N0 ).

(11)

The special cases of (10) and (11) when x = 0, respectively, give the q-difference equation for Gq (t): et Gq (qt) = Gq (t) + t

(12)

502


and

ηk;q =

 1    q−1 k

if k = 1, (13)

k j j ηj;q q

P   

j=0

if k ∈ N \ {1}.

We find from (9) and (13) that ηk;q (1) = ηk;q (0) = ηk;q

(k ∈ N \ {1}).

(14)

It is seen from (5) that Gx+1;q (t) − Gx;q (t) = t qx e[x]q t ,

(15)

which gives ηk+1;q (x + 1) − ηk+1;q (x) = (k + 1) qx [x]kq

(k ∈ N0 ).

(16)

Considering t

qj+x

e[j+x]q t = e 1−q e q−1

t

(17)

in (5), we find that ∞ X

ηk;q (x)

∞ ∞ ∞ X tk q(`+1)x t` X (`+1)j X (−1)` t` = −t q k! (q − 1)` `! (q − 1)` `! `=0

k=0

`=0

j=0

q(`+1)x t`

∞ X (−1)` t` = −t 1 − q`+1 (q − 1)` `! `=0 (q − 1)` `! `=0   ∞ X k  ( j+1)x X q tk (−1)k−j = −t ,  j!(k − j)! 1 − qj+1  (q − 1)k ∞ X

k=0

j=0

which, upon equating the coefficients of tk+1 and assuming 0/[0]q = 1, after a little simplification, yields an explicit expression of ηk;q (x): (q − 1)k ηk;q (x) =

k X

(−1)k−j

j=0

k j jx q j [j]q

(k ∈ N),

(18)

which can be equivalently written as: k X k j=0

j

(q − 1)j ηj;q (x) =

k kx q [k]q

(k ∈ N),

(19)


503

by using the inverse relation technique in combinatorial analysis. Indeed, using a wellknown series manipulation k X ` X

k X k X

A`,j

(20)

k ` k k−j = , ` j j `−j

(21)

A`,j =

j=0 `=j

`=0 j=0

and a binomial identity

we have k X k `=0

k k X k j jx X `−j k − j q (−1) (q − 1) η`;q (x) = `−j ` j [j]q `

`=j

j=0

=

k X j=0

k kx k j jx q (1 − 1)k−j = q . [k]q j [j]q

The special cases of (18) and (19) when x = 0, respectively, gives (q − 1) ηk;q = k

k X

(−1)

k−j

j=0

k j j [j]q

(k ∈ N),

(22)

or, equivalently, k X k j=0

j

(q − 1)j ηj;q =

k [k]q

(k ∈ N).

(23)

We give multiplication formula for ηk;q (x): [m]k−1 q

m−1 X

ηk;qm

η

j = ηk;q (mx) x+ m

j=0

x+j m

= ηk;q (x)

(k ∈ N0 ; m ∈ N),

(24)

or, equivalently, [m]k−1 q

m−1 X j=0

k;qm

(k ∈ N0 ; m ∈ N).

(25)

504


Indeed, we have ∞ X

[m]k−1 q

m−1 X

ηk;qm

j=0

k=0

x+j m

m−1 ∞ [m]q t tk 1 XX x+j = ηk;qm k! [m]q m k!

k

j=0 k=0

m−1 ∞ XX

= −t

qms+j+x e[ms+j+x]q t

j=0 s=0 ∞ X `+x [`+x]q t

= −t

q

e

`=0

=

∞ X

ηk;q (x)

k=0

tk . k!

It follows from (5) and (8) that Gx+y;q (t) = e[y]q t Gx+y;q qy t ,

(26)

which yields ηk;q (x + y) =

k X k j=0

j

If we replace q and x by

ηj;q (x) qyj [y]k−j q 1 q

(k ∈ N0 ).

(27)

and 1 − x, respectively, in (18) and use 6.1(35), we get

ηk;q−1 (1 − x) = (−1)k qk−1 ηk;q (x)

(k ∈ N),

(28)

the special case x = 1 of which, in view of (12), yields ηk;q−1 (0) = (−1)k qk−1 ηk;q (1) = (−1)k qk−1 ηk;q

(k ∈ N \ {1}).

(29)

q-Stirling Numbers of the Second Kind From the initial condition 10 f (x) = f (x) and the recurrence relation 1n+1 f (x) = 1n f (x + 1) − qn 1n f (x)

(n ∈ N0 ),

(30)

one may define (see [593], [869, Chapter 1] and [1209]) a sequence {1n }∞ n=0 of q-difference operators in which the index identifies the position of the pertinent operator within the sequence but is not directly interpreted as an operator exponent. It is easy to prove that 1n f (x) =

n X

1

(−1)j q 2 j( j−1)

j=0

where

n j q

is defined by 6.1(29).

n f (x + n − j), j q

(31)


505

If f (x) is a polynomial in qx of degree 5 n, it is obvious that we may put f (x) =

n X

αj [x]q;j ,

(32)

j=0

where [x]q;j is defined by 6.1(30) and αj is independent of x. To determine the coefficients αj , by making use of the easily proved formula 1n [x]q;j = [j]q;n [x]q;j−n qn(x−j+n) ,

(33)

apply 1n to both members of (32) and put x = 0 to finally get αj =

1j f (0) . [j]q !

(34)

q-Stirling numbers of the second kind Sq (n, j) are defined by taking f (x) = [x]nq in (32) and writing, for convenience, as [x]nq =

n X

1

q 2 j( j−1) Sq (n, j) [x]q;j ,

(35)

j=0

where Sq (n, j) is independent of x. It follows from (35) that 1 q 2 j( j−1) Sq (n, j) [x]q;j [x − j]q q j + [j]q X 1 = q 2 j( j−1) Sq (n, j − 1) + [j]q Sq (n, j) [x]q;j ,

[x]n+1 = q

X

which yields a recursion formula Sq (n + 1, j) = Sq (n, j − 1) + [j]q Sq (n, j).

(36)

It follows immediately from (36) that Sq (n, j) is a polynomial in q with integral coefficients. We use (32) and (31) to express Sq (n, j) explicitly: Sq (n, j) =

1 j 1 q− 2 j( j−1) X j (−1)` q 2 `(`−1) [j − `]nq [j]q ! ` q

(n, j ∈ N0 ),

(37)

`=0

the right side of which vanishes for n < j and is equivalently written as j 1 X j−` 21 `(`+1−2j) j Sq (n, j) = (−1) q [`]n [j]q ! ` q q `=0

(n, j ∈ N0 ).

(38)

506


Note that, upon taking the limit q → 1 on (38), Sq (n, j) reduces to the familiar Stirling numbers of the second kind (see 1.6(21)). A formula of a different sort may also be introduced: n X n n = (q − 1)`−j Sq (`, j), j q `

(39)

`=j

or, equivalently, (q − 1)n−j Sq (n, j) =

n X

(−1)n−`

`=j

n ` . ` j q

(40)

Note that (38) and (40) are, in fact, equivalent. We define a slightly-generalized form of Sq (n, j) by means of [x + y]nq =

n X

q 2 (y+j)(y+j−1) Sq (n, j)(y) [x]q;j , 1

(41)

j=0

so that Sq (n, j)(0) = Sq (n, j). As in getting (36) and (37), we obtain a recursion formula Sq (n + 1, j)(y) = Sq (n, j − 1)(y) + [y + j]q Sq (n, j)(y),

(42)

which shows that Sq (n, j)(y) is a polynomial in qy ; an explicit formula q

1 2 (y+j)(y+j−1)

j 1 X ` 21 `(`−1) j (−1) q Sq (n, j)(y) = [y + j − `]nq . [j]q ! ` q

(43)

`=0

It is easily seen that 1

q 2 y(y−1+2j) Sq (n, j)(y) =

n X n

`

`=j

`y [y]n−` q q Sq (`, j).

(44)

It follows from either (43) or (44) that q

1 2 y(y−1)

n X n

`

`=j

`−j

(q − 1)

Sq (`, j)(y) = q

(n−j)y

n . j q

(45)

The Polynomial βk (x) = βk;q (x) ∞ Let aj j=0 be an arbitrary sequence of numbers and define fn (x) =

n X n j=0

j

jx aj [x]n−j q q .

(46)


507

Then, fn (0) = an and fn (x + y) =

n X n

j

j=0

jx fj (y) [x]n−j q q ,

(47)

the special case y = 0 of which reduces to (46). The polynomial ηn (x) = ηn;q (x) and the number ηn = ηn;q (0) do not remain finite when q = 1. Carlitz [215], therefore, introduces a polynomial βn (x) = βn;q (x) and number βn = βn;q that will approach a finite limit for q = 1. We, first, define βn = ηn + (q − 1) ηn+1

(n ∈ N0 ).

(48)

Repeated application of (48) leads to k X k j=0

j

(q − 1) ηn−k+j = j

k−1 X k−1 j

j=0

(q − 1)j βn−k+j ,

which, for n = k, becomes n X n j=0

j

(q − 1) ηj = j

n−1 X n−1 j

j=0

(q − 1)j βj ,

which, upon comparing with (23) and (22), leads at once to n X n j=0

j

(q − 1) j βj =

n+1 [n + 1]q

(49)

and (q − 1)n βn =

n X

n j+1 . j [j + 1]q

(−1)n−j

j=0

(50)

Carlitz [215] defines (q − 1) βn (x) = n

n X j=0

(−1)

n−j

n j + 1 jx q , j [j + 1]q

(51)

which implies (4). It follows from (50) and (51) that βn (x) =

n X n j=0

j

n x βj qjx [x]n−j q := q β + [x]q ,

(52)

508


which, in view of (47), generalizes the following result: βn (x + y) =

n X n j=0

j

n x βj (y) qjx [x]n−j q := q β(y) + [x]q .

(53)

Analogous to (16), we find that q x+1 βn (x + 1) − q x βn (x) = n qx [x]n−1 + (q − 1) (n + 1) qx [x]nq , q so that q βn (x + 1) − βn (x) = n qx [x]n−1 + (q − 1) [x]nq , q which, for x = 0, becomes ( q(qβ + 1)n − βn =

1 0

(n = 1), (n ∈ N \ {1}).

(54)

(55)

Analogous to (25) and (28), we have [k]n−1 q

k−1 X j=0

j = βn;q (kx) q j βn;qk x + k

(k ∈ N)

(56)

and βn;q−1 (1 − x) = (−q)n βn;q (x),

(57)

which, for x = 1, implies βn;q−1 (0) = (−q)n βn;q (1) = (−1)n qn−1 βn

(n ∈ N \ {1}).

(58)

It follows from (35) that k−1 X `=0

q` [`]nq =

n X

1

q 2 j( j+1) Sq (n, j)

j=0

[k]q;j+1 . [j + 1]q

(59)

Conversely, in view of (16), the left member of (59) is n+1 1 1 X n+1 [k]qj q(n+1−j)k ηn+1−j (ηn+1 (k) − ηn+1 ) = n+1 n+1 j j=1 η n+1 (n+1)k + q −1 n+1 n η X n n+1 (n−j)k ηn−j = [k]j+1 + q(n+1)k − 1 . q q j j+1 n+1 j=0


509

Comparison with (59) yields an identity in k. Considering [k]q;j+1 = [k]q [k − 1]q;j

and

lim

k→0

q(n+1)k − 1 = (n + 1)(q − 1), [k]q

to divide both members of this identity by [k]q and then let k → 0, we obtain ηn + (q − 1)ηn+1 =

n X

1

q 2 j( j+1) Sq (n, j)

j=0

[−1]q;j , [j + 1]q

which, in view of (48), yields βn =

n X

(−1)j Sq (n, j)

j=0

[j]q ! . [j + 1]q

(60)

It is obvious from (60) and (37) that βn remains finite for q = 1. Indeed, lim βn =

q→1

n X j=0

j j 1 X (−1)j−` ( j − `)n = Bn , j+1 `

(61)

`=0

where Bn is the n-th Bernoulli number in No¨ rlund’s notation (see 1.5(30)). Substitution from (37) in (60) leads to the explicit formula βn =

n X j=0

j X 1 ` 12 `(`+1−2j) j (−1) q [`]n , [j + 1]q ` q q

(62)

`=0

which can, upon using (52), be more generally expressed as βn (x) =

n X j=0

j X 1 ` 21 `(`+1−2j) j (−1) q [x + `]nq . [j + 1]q ` q

(63)

`=0

Comparison with (63) and (43) yields a generalized formula of (60): βn (x) =

n X

1

(−1)j q 2 x(x−1+2j) Sq (n, j)(x)

j=0

[j]q ! . [j + 1]q

(64)

6.7 q-Euler Numbers and q-Euler Polynomials We begin by defining a set polynomials k;q (x) = k (x) and a set of numbers k;q (0) = k;q = k by means of the following generating functions given by formal series: Hx;q (t) ≡ [2]q

∞ X j=0

(−1) q e

j j [j+x]q t

:=

∞ X k=0

k;q (x)

tk k!

(1)

510


and Hq (t) ≡ [2]q

∞ X

(−1)j q j e[j]q t :=

j=0

∞ X

k;q

k=0

tk . k!

(2)

It is easy to see from (1) that q Hx+1;q (t) + Hx;q (t) = [2]q e[x]q t ,

(3)

which immediately yields q k (x + 1) + k (x) = [2]q [x]kq

(k ∈ N0 ).

(4)

It is observed that k (x) is uniquely determined by the functional equation (4). From (4) follows a summation formula (−1)k−1 qk n (x + k) + n (x) = [2]q

k−1 X

(−1)j q j [x + j]nq .

(5)

j=0

Considering 6.6(17) in (1), analogous to 6.6(18), we obtain an explicit formula for k;q (x): (q − 1)k k;q (x) = (q + 1)

k X

(−1)k−j

j=0

k q xj , j 1 + qj+1

(6)

or, equivalently, k X k j=0

j

(q − 1)j j;q (x) =

(q + 1) qkx . 1 + qk+1

(7)

The special cases of (6) and (7) when x = 0, respectively, give (q − 1)k k;q =

k X

(−1)k−j

j=0

k q+1 j 1 + qj+1

(8)

and k X k j=0

j

(q − 1)j j;q =

q+1 . 1 + qk+1

(9)

Replacing x and q, respectively, by 1 − x and q−1 in (6) leads to k;q−1 (1 − x) = (−1)k qk k;q (x).

(10)


511

From 6.6(8) we find that Hx+y;q (t) = e[x]q t Hy;q qx t ,

(11)

which gives k;q (x + y) =

k X k

j

j=0

k jx x [x]k−j q q j;q (y) := q (y) + [x]q .

(12)

The special case of (12) when y = 0 becomes k;q (x) =

k X k j=0

j

k jx x [x]k−j q q j;q := q + [x]q .

(13)

Rearranging the series in (1) as even and odd terms, we see that t qx Hx;q (t) = G x+1 ;q2 [(q + 1)t] − G x ;q2 [(q + 1)t],

(14)

2

2

which yields (k + 1) q k;q (x) = (q + 1) x

k+1

x x+1 ηk+1;q2 − ηk+1;q2 2 2

(k ∈ N0 ), (15)

serves to express qx k;q (x) in terms of η-polynomials. Corresponding to 6.6(25), we get the multiplication formulas [m]kq

m−1 X j=0

qm + 1 j = (−q)j k;qm x + k;q (m x) m q+1

(m is odd)

(16)

and [2]q [m]kq

m−1 X

(−1)

j+1

η

k+1;qm

j=0

j x+ = (k + 1)qmx k;q (m x) m

(m is even).

(17)

Note that, for m = 2, (17) reduces to (15). To get explicit formulas of a simpler kind, Carlitz [215] starts by noting the function (cf. [861, Chapter VII]) Yn;q (x) :=

n X k=0

x qn−k , j+1 +1 k q j=k q

(−1)n−k Qn

which is straightforwardly proven to satisfy x q Yn;q (x + 1) + Yn;q (x) = . n q

(18)

(19)

512


In view of (4) and 6.6(35), we have n X

q n;q (x + 1) + n;q (x) = [2]q

1 q 2 k(k−1) Sq (n, k) [k]q ! q Yk;q (x + 1) + Yk;q (x) ,

k=0

which yields n;q (x) = [2]q

n X

1

q 2 k(k−1) Sq (n, k) [k]q ! Yk;q (x).

(20)

k=0

The special case of (20) when x = 0 implies n;q =

n X

1 Sq (n, k) [k]q ! (−1)k q 2 k(k+1) Qk , j+1 + 1 j=1 q k=0

(21)

which, using 6.6(37), becomes n;q =

n X

k X

(−1)k qk Qk

k=0

qj+1 + 1

j=1

(−1)` q 2 `(`−1) 1

`=0

k [k − `]nq . ` q

(22)

Conversely, if we use 6.6(41), we obtain q n;q (x + y + 1) + n;q (x + y) = [2]q

n X

1 q 2 (y+k)(y+k−1) Sq (n, k)(y) [k]q ! q Yk;q (x + 1) + Yk;q (x) ,

k=0

which gives n;q (x + y) = [2]q

n X

q 2 (y+k)(y+k−1) Sq (n, k)(y) [k]q ! Yk;q (x). 1

(23)

k=0

The special case of (23) when x = 0 becomes n;q (y) =

n X

1 Sq (n, k)(y) [k]q ! (−1)k q 2 (y+k)(y+k−1)+k Qk , j+1 + 1 j=1 q k=0

(24)

which, upon using 6.6(43), gives n;q (y) =

n X

Qk k=0

k X

(−1)k qk j=1

qj+1 + 1

`=0

`

(−1) q

1 2 `(`−1)

k [y + k − `]nq . ` q

(25)


513

Remark 2 It is, in view of (22), observed that the product n Y qj+1 + 1 n;q j=1

is a polynomial in q, which, for q = 1, reduces to the number Cn (see [869, p. 27]). Likewise, (25) indicates that the product −n

2

n Y 1 2( j+1) (q + 1) q + 1 n;q2 2 n

j=1

is a polynomial in q, which, for q = 1, reduces to the Euler number En .

6.8 The q-Apostol-Bernoulli Polynomials Bk(n) (x; λ) of Order n We begin this section by setting a = qn (n ∈ N) in the q-binomial theorem 6.3(2) to obtain 1 = n−1 (z; q)n Q

1

=

1 − z qk

∞ X [n]q;k k=0

[k]q !

zk .

(1)

k=0 (n)

A q-extension of the Apostol-Bernoulli polynomials Bk (x; λ; q) of order n ∈ N (see Section 1.7) is defined, here, by means of the following generating function: (n)

Gx;λ;q (t) ≡ (−t)n

∞ X [n]q;k k=0

:=

∞ X

[k]q !

λk qk+x e[k+x]q t

(n) Bk (x; λ; q)

k=0

(2)

tk , k! (n)

and a q-extension of the Apostol-Bernoulli numbers Bk (λ; q) of order n ∈ N is defined, here, by means of the following generating function: (n)

Gλ;q (t) ≡ (−t)n

∞ X [n]q;k k=0

:=

∞ X k=0

[k]q !

(n) Bk (λ; q)

λk qk e[k]q t

tk . k!

(3)

514


Remark 3 Here, for convenience, only the case of order α = n ∈ N is considered. It is (n) (n) easy to see that, when q → 1, the generating functions of Bk (x; λ; q) and Bk (λ; q) (n) (n) in (2) and (3) would tend, respectively, to those of Bk (x; λ) and Bk (λ) given by 1.8(13) and 1.8(14). It is also observed that (n)

(n)

Bk (0; λ; q) = Bk (λ; q)

(k ∈ N0 )

(4)

and (1)

Bk (x; λ; q) = Bk (x; λ; q)

(1)

and Bk (λ; q) = Bk (λ; q),

(5)

which were, very recently, introduced and investigated by Cenki and Can [227, Eqs. (4) and (6)]. We note that (1)

Bk (1; q) = Bk (q)

are the familiar q-Bernoulli numbers that were considered by many authors (see, e.g., [215], [228], [649], [659], [660], [682] and [1100]). (n) A q-extension of the Apostol-Euler polynomials Ek (x; λ; q) of order n ∈ N is defined, here, by means of the following generating function: (n)

Hx;λ;q (t) ≡ 2n

∞ X [n]q;k k=0

:=

∞ X

[k]q !

(−λ)k qk+x e[k+x]q t

(n) Ek (x; λ; q)

k=0

(6)

tk , k! (n)

and a q-extension of the Apostol-Euler numbers Ek (λ; q) of order n is defined here by means of the following generating function: (n)

Hλ;q (t) ≡ 2n

∞ X [n]q;k k=0

:=

∞ X

[k]q !

(−λ)k qk e[k]q t

(n) Ek (λ; q)

k=0

(7)

tk . k! (n)

Remark 4 It is easy to see that, when q → 1, the generating functions of Ek (x; λ; q) (n) (n) (n) and Ek (λ; q) in (6) and (7) would tend, respectively, to those of Ek (x; λ) and Ek (λ) given by 1.8(15) and 1.8(16). It is also observed that (n)

(n)

Ek (0; λ; q) = Ek (λ; q)

(k ∈ N0 ).

(8)


515

We, first, examine (2) and (3) to get the following relationship between these two generating functions: (n) (n) Gx;λ;q (t) = Gλ;q qx t e[x]q t q(1−n)x ,

(9)

which, upon considering (2) and (3) again, yields (n)

Bk (x; λ; q) =

k X k

j

j=0

(n)

( j+1−n)x Bj (λ; q) [x]k−j q q

(k ∈ N0 ).

(10)

Next, by considering the following identity: t

k

q − 1−q t

e[k]q t = e 1−q e

,

we obtain (n)

t

Gλ;q (t) = (−t)n e 1−q t

= (−t)n e 1−q

∞ ∞ X [n]q;k λk qk X (−1)j qj k j t [k]q ! j! (1 − q)j j=0 ∞ (−1)j tj X [n]q;k [k]q ! j! (1 − q)j j=0 k=0

k=0 ∞ X

k λ qj+1 .

In view of (1), the last sum is equal to 1/ λ qj+1 ; q n , and we, thus, find that ∞ X k=0

(n)

Bk (λ; q)

k ∞ X X 1 (−1)j tk tk = (−t)n , k! (1 − q)k λ qj+1 ; q n j! (k − j)! j=0 k=0

which, upon equating the coefficients of tn+k on both sides, yields the following (n) explicit expression for Bk (λ; q): (n)

Bn+k (λ; q) = (−1)n

k (k + 1)n X k (−1)j k j λ qj+1 ; q n (1 − q) j=0

(k ∈ N0 ; n ∈ N).

(11)

In light of (5), a special case of (11) when n = 1 reduces to the following result: k X 1 k (−1)j j Bk (λ; q) = j 1 − λ qj (1 − q)k−1

(k ∈ N0 ),

(12)

j=0

which provides a corrected version of the corresponding formula given in [227, p. 215].

516

Zeta and q-Zeta Functions and Associated Series and Integrals (n)

We, now, give the following q-difference equation for Gλ;q (t): (n)

(n)

(n−1)

λ et q1−n Gλ;q (qt) = Gλ;q (t) + t Gqλ;q (t)

(n ∈ N),

(13)

where (0)

Gqλ;q (t) := 1. Indeed, we have (n)

λ et q1−n Gλ;q (qt) = (−t)n

∞ X k=0

[n]q;k k k [k] t [k]q λ q e q, [n + k − 1]q [k]q !

which, upon considering the following relationship: [n − 1]q [k]q = 1− qk , [n + k − 1]q [n + k − 1]q leads us to (13). The special case of (13) when n = 1 reduces immediately to a known result [227, p. 216, Eq. (5)]. From (3) and (13) it is easy to derive the following q-difference equation for (n) Bk (λ; q): λ q1−n

k+1 X k+1 j

j=0

(n)

(n−1)

(n)

q j Bj (λ; q) = Bk+1 (λ; q) + (k + 1) Bk

(qλ; q)

(14)

(k ∈ N0 ; n ∈ N). (n)

Remark 5 Since the q-difference equation for Gλ;q (t) in (13) holds true, if and only (n)

if the q-difference equation for Bk (λ; q) in (14) holds true and also since the coefficients of a power series are uniquely determined, we conclude that the generating (n) (n) function Gλ;q (t) of the q-extension of the Apostol-Bernoulli numbers Bk (λ; q) of (n)

order n can be determined as a solution of the q-difference equation for Gλ;q (t) in (13). (n)

Remark 6 The following q-distribution relation holds true for Bk (x; λ; q): [m]k−1 q

m−1 X j=0

λ

j

(n) Bk

x+j m m (n) ; λ ; q = Bk (x; λ; q) m

(k ∈ N0 ; m, n ∈ N).

(15)


517

The special case of (15) when n = 1 was proven by Cenki and Can [227, p. 216, Lemma 5]. Here we give an equivalence statement that (15) holds true for n ∈ N \ {1}. Indeed, we have (n;m) Ix;λ;q (t) ≡

∞ X

[m]k−1 q

=

1 [m]q 1 [m]q

λ

j

(n) Bk

j=0

k=0

=

m−1 X

m−1 X

λj

∞ X

j=0 m−1 X

(n)

Bk

k=0 (n)

λj G x+j m

j=0

;λm ;qm

x+j m m ;λ ;q m

x+j m m ;λ ;q m

tk k! [m]q t k!

k

[m]q t .

By applying (9) and the identity 6.5(8) to the last equation, we get (n;m)

Ix;λ;q (t) =

m−1 q(1−n)x e[x]q t X j (1−n)j (n) x λq Gλm ;qm q x+j [m]q t e[j]q q t . [m]q

(16)

j=0

Moreover, in view of (2) and (9), we have ∞ X

(n)

Bk (x; λ; q)

k=0

tk (n) = Gλ;q qx t e[x]q t q(1−n)x . k!

(17)

Now, it is easily seen from (16) and (17) that (15) holds true, if and only if the following result holds true: (n) Gλ;q (t) =

m−1 1 X j (1−n)j (n) λq Gλm ;qm qj [m]q t e[j]q t [m]q

(m, n ∈ N).

(18)

j=0

(n)

Let Pλ;q (t) be the right-hand side of (18). In view of Remark 5, it suffices to show that (n)

Pλ;q (t) satisfies the following q-difference equation: (n)

(n)

(n−1)

λ et q1−n Pλ;q (qt) = Pλ;q (t) + t Pqλ;q (t)

(n ∈ N \ {1}).

Indeed, if we begin with (n)

m 1 X j (1−n)j (n) λq Gλm ;qm qj [m]q t e[j]q t [m]q j=1 1 (n) (n) = Pλ;q (t) + λm q(1−n)m Gλm ;qm qm [m]q t e[m]q t [m]q (n) − Gλm ;qm [m]q t

λ et q1−n Pλ;q (qt) =

(19)

518


and apply (13) to the last equation, we obtain (n) (n) (n−1) λ et q1−n Pλ;q (qt) = Pλ;q (t) + t G(qλ)m ;qm [m]q t .

(20)

Now, it is seen from (19) and (20) that an equivalence condition for the validity of the (n) q-distribution relation for Bk (x; λ; q) in (15) can be written as in Remark 7 below. Remark 7 The q-difference equation (15) holds true, if and only if (n)

(n)

Pqλ;q (t) = G(qλ)m ;qm [m]q t

(m, n ∈ N),

that is, if and only if the following result holds true for n, m ∈ N: X (n) m−1 (n) [m]q − 1 G(qλ)m ;qm [m]q t = (qλ)j q(1−n)j G(qλ)m ;qm qj [m]q t e[j]q t ,

(21)

j=1

where an empty sum is understood (as usual) to be nil.

6.9 The q-Apostol-Euler Polynomials Ek(n) (x; λ) of Order n By applying the methodology and techniques used above in getting some identities for the generating functions of the q-extensions of the Apostol-Bernoulli polynomials and numbers, we can derive the following corresponding identities involving the generating functions of the q-extensions of the Apostol-Euler polynomials and numbers: (n) (n) Hx;λ;q (t) = Hλ;q qx t e[x]q t qx . k X k (n) (n) ( j+1)x Ek (x; λ; q) = E (λ; q) [x]k−j q q j j

(1) (k ∈ N0 ; n ∈ N).

(2)

j=0

(n) Ek (λ; q) =

k X k (−1)j 1 j −λ qj+1 ; q n (1 − q)k j=0

(k ∈ N0 ; n ∈ N).

(3)

(n)

The q-difference equation for Hλ;q (t) is given by (n)

(n−1)

(n)

Hλ;q (t) − 2 Hqλ;q (t) = −λ et q Hλ;q (qt) where (0)

Hqλ;q (t) := 1.

(n ∈ N),

(4)


519

(n)

The q-difference equation for Ek (λ; q) is given by (n−1) (n) Ek (λ; q) − 2 Ek (qλ; q) = −λ

k X k j=0

j

(n)

Ej (λ; q) qj+1

(k ∈ N0 ; n ∈ N). (5)

(n)

(n)

The following relationship holds true between Bk (x; λ; q) and Ek (x; λ; q): (n)

(n)

(−2)n Bn+k (x; −λ; q) = (k + 1)n Ek (x; λ; q)

(k ∈ N0 ; n ∈ N),

(6)

which follows immediately, by using the following generating-function relationship: (n) Gx;−λ;q (t) =

t n (n) − Hx;λ;q (t) 2

(n ∈ N).

(7)

Remark 8 We conjecture that the following q-distribution relation holds true for (n) Ek (x; λ; q): [m]k−1 q

m−1 X

(−λ)

j

j=0

(n) Ek

x+j m m (n) ; λ ; q = Ek (x; λ; q) m

(k ∈ N0 ; m, n ∈ N). (8)

When n = 1, (8) is easily verified. The other cases of our conjectured relationship (8) when n ∈ N \ {1} remain to be proven.

6.10 A Generalized q-Zeta Function Many authors have tried to give q-analogues of the Riemann Zeta function ζ (s), defined by 2.3(1), and its related functions (see, e.g., [227, 255, 363, 493, 626, 652, 1006, 1100, 1168–1170] and [1174]). Here, we give a q-analogue of the generalized zeta function ζ (s, a), defined by 2.2(1), among other things, by just following the method of Kaneko et al. [626], who mainly used Euler-Maclaurin summation formula (see 2.7(21) or 2.7(22)) to present and investigate a q-analogue of the Riemann zeta function ζ (s), defined by 2.3(1), and gave a good and reasonable explanation that their q-analogue may be a best choice. They [626] also commented that q-analogue of ζ (s, a) can be achieved by modifying their method.

An Auxiliary Function Defining Generalized q-Zeta Function We begin this section by presenting a function gq (a; s, t), defined by gq (a; s, t) :=

∞ X q(n+a)t [n + a]sq n=0

(0 < q < 1; 0 < a 5 1).

(1)

520


Remark 1 It is easy to see that, by the ratio test, the series in (1) converges absolutely for 0 and any s ∈ C. The series in (1) also converges to the generalized Zeta function ζ (s, a), defined by 2.2(1), as q ↑ 1, if 1 and 0. So, we can define gq (a; s, t) in (1) as a q-analogue of the generalized Zeta function ζ (s, a). However, Kaneko et al. [626] took several examples to contend that t = s − 1 in gq (1; s, t) seems the best possible choice to define a q-analogue of the Riemann Zeta function ζ (s), defined by 2.3(1). Theorem 6.12 For 0 < q < 1, as a function of (s, t) ∈ C2 , gq (a; s, t) in (1) is continued meromorphically, by means of the series expansion: ∞ X s + r − 1 qa(t+r) s (2) gq (a; s, t) = (1 − q) 1 − qt+r r r=0

with simple poles at t = −r +

2`π log q

(r ∈ N0 ; ` ∈ Z).

(3)

Proof. It follows immediately from the binomial theorem that gq (a; s, t) = (1 − q)s

∞ X

q(n+a)t 1 − qn+a

−s

n=0

= (1 − q)s

∞ X

q(n+a)t

n=0

= (1 − q)s = (1 − q)s

∞ X s+r−1 r

r=0

∞ ∞ X s+r−1 X r=0 ∞ X r=0

r

q(n+a)r

q(n+a)(t+r)

n=0

s + r − 1 qa(t+r) . r 1 − qt+r

In view of Remark 1, we also choose to define a q-analogue of the generalized Zeta function ζ (s, a) in 2.2(1) by ζq (s, a) := gq (a; s, s − 1) :=

∞ X q(n+a)(s−1) [n + a]sq

(0 < q < 1; 0 < a 5 1).

n=0

We list, here, some properties that easily follow from Theorem 6.12 as Theorem 6.13 (a) The function ζq (s, a) has simple poles at points in 1 + 2πi Z/ log q and in the set 2π i m `+ ` ∈ Z , m ∈ Z \ {0} , 50 log q

(4)


521

where Z50 := Z \ N. In particular, s = 1 is a simple pole of ζq (s, a) with its residue (q − 1)/ log q. (b) For m ∈ N0 , the limiting value lim ζq (s, a) =: ζq (−m, a)

s→−m

exists and is given explicitly by ( ζq (−m, a) = (1 − q)

−m

m X r=0

) (1−a)(m+1−r) (−1)m+1 m q (−1) + . (m + 1) log q r qm+1−r − 1

(5)

(m ∈ N0 ),

(6)

r

Now, we have Theorem 6.14 lim ζq (−m, a) = − q↑1

Bm+1 (a) m+1

where Bm+1 (a) denotes the Bernoulli polynomials (see Section 1.7). Proof. In view of (5), it is equivalent to show that

lim (1 − q)

−m

( m X

q↑1

r=0

) (1−a)(m+1−r) m q (−1)m+1 Bm+1 (a) (−1) + =− , m+1−r r (m + 1) log q m+1 q −1 r

which, upon multiplying (−1)m+1 (m + 1) and setting m + 1 − r = r0 and then dropping the prime on r in the resulting equation, yields ( lim (1 − q) q↑1

−m

(m + 1)

m+1 X r=1

(−1)

r

) (1−a)r 1 m q + = (−1)m Bm+1 (a). r − 1 qr − 1 log q

(7) Using 1.6(1) q(1−a)r e(1−a)r log q 1 r log q e(1−a)r log q 1 = r log q = r q −1 r log q e −1 er log q − 1 ∞ k 1X (r log q) 1 = Bk (1 − a) , r k! log q k=0

522


we have (m + 1)

m+1 X

(−1)

r

r=1

(1−a)r m q r − 1 qr − 1

m+1 X

X ∞ m 1 (r log q)k 1 Bk (1 − a) = (m + 1) (−1) r−1 r k! log q r=1 k=0 ! ∞ m+1 X X m+1 k (log q)k−1 = r Bk (1 − a) . (−1)r k! r r

r=1

k=0

The inner sum of the last expression can be evaluated as m+1 X

(−1)r

r=1

o m+1 k d kn r = x (1 − x)m+1 − 1 x=1 r dx  (k = 0), −1 0 (0 < k < m + 1), =  m+1 (−1) (m + 1)! (k = m + 1),

and we find (m + 1)

m+1 X

(−1)

r

r=1

=−

(1−a)r m q r − 1 qr − 1

1 + (−1)m+1 Bm+1 (1 − a) (log q)m + O (log q)m+1 log q

(q → 1).

From this observation, the expansion log q = q − 1 + O (q − 1)2 (q → 1) and the known relation 1.6(10), we prove the desired result (7): ( lim (1 − q)−m (m + 1)

q→1

m+1 X

(−1)r

r=1

= (−1)

m+1

Bm+1 (1 − a) lim

q→1

) (1−a)r m q 1 + r − 1 qr − 1 log q

(log q)m = −Bm+1 (1 − a) = (−1)m Bm+1 (a). (1 − q)m

Definition 6.8 In view of Theorem 6.14, it is natural to define the q-Bernoulli polynomials Bn;q (a) by Bn;q (a) := −n ζq (1 − n, a)

(n ∈ N).

(8)


523

If we use (8), then we get some interesting properties and relations, which are summarized as in the following theorem: Theorem 6.15 (a) Bn;q (a) can be expressed in an explicit form: (q − 1)n Bn;q (a) =

n X

(−1)r

r=0

n r (1−a)r q r [r]q

(n ∈ N),

(9)

where the term with r = 0 is understood to be lim

r→0

1 r = . qr − 1 log q

(10)

So B0;q (a) may be defined as B0;q (a) :=

q−1 . log q

(11)

(9) can be written equivalently as follows: n X r=0

(−1)r

n n q(1−a)n . (q − 1)r Br;q (a) = [n]q r

(12)

(b) We have a relationship between the Carlitz’s q-Bernoulli polynomials βn (a) in 6.6(51) and Bn;q (a) here: qa βn (a) = (−1)n Bn;q (1 − a) + (1 − q) Bn+1;q (1 − a)

(n ∈ N).

(13)

(c) We have a relationship between βk (a) and ζq (−k, a): For n ∈ N, qa βn (a) = (−1)n (q − 1) (n + 1) ζq (−n, 1 − a) − n ζq (1 − n, 1 − a) .

(14)

(d) We have a relationship between n;q (2a) in Section 6.7 and ζq2 (−n, a): 1 q2a n;q (2a) = (−1)n (q + 1)n+1 ζq2 −n, − a − ζq2 (−n, 1 − a) 2

(n ∈ N0 ). (15)

Proof. In view of (5), Bn;q (a) can be expressed as ( n X

) n r 1 (1−a)r Bn;q (a) = (q − 1) (−1) q + r qr − 1 log q r=1 ( n ) X n r q(1−a)r = (q − 1)−n+1 (−1)r (n ∈ N), r qr − 1 −n+1

r

r=0

which is seen equal to (9). As in getting 6.6(19), (9) can be written as (12).

524


Using the familiar binomial identity

n+1 n n = + , r r r−1

we find from 6.6(51) that (q − 1) βn (a) = (−1) q n

n −a

( n X

(−1)r

r=1 n+1 X

−

r=1

n r ra q r [r]q

) n + 1 r ra q , (−1) [r]q r r

which, in view of (9), yields the relationship (13). (14) follows easily from (8) and (13). Note that, if we also assume the convention of (10) in the term with j = 0 of Equation 6.6(18), it is easy to see from 6.6(18) and (9) the following relation: ηn;q (a) = (−1)n Bn;q (1 − a)

(n ∈ N0 ).

(16)

Now, an application of (8) and (16) to 6.7(15) yields (15). Remark 2 Substitution of (16) for (13) is seen to come back to the defining relation 6.6(4). Taking the limit on each side of (14) as q ↑ 1, together with (6), reduces immediately to a familiar identity 1.7(10). Replacing q and a by q2 and 14 , respectively, in (15) and letting q ↑ 1 in the resulting equation, together with (6), in view of Remark 1, we obtain 1 22n+1 3 1 En = 2 En = Bn+1 − Bn+1 2 n+1 4 4 n

(n ∈ N0 ),

(17)

which is a special case of the well-known relationship between Euler polynomials En (x) and Bernoulli polynomials Bn (x) (see [1094, p. 65, Eq. (60)]; see also [869, p. 25, Eq. (25)]).

Application of Euler-Maclaurin Summation Formula Here, we mainly show Theorem 6.16 lim ζq (s, a) = ζ (s, a). q↑1

(18)


525

We present, for convenience, a special case of the Euler-Maclaurin summation formula 2.7(21) when a = 0, b = N ∈ N, K = M + 1 ∈ N: N X

f (n) =

n=0

ZN

f (x) dx +

M X Bk+1 (k) 1 f (N) − f (k) (0) ( f (0) + f (N)) + 2 (k + 1)! k=1

0

(−1)M+1 − (M + 1)!

ZN

(19) e BM+1 (x) f (M+1) (x) dx,

0

where f ∈ C∞ [0, ∞), N ∈ N and M ∈ N0 . The Hurwitz (or generalized) Zeta function ζ (s, a), defined by 2.2(1), is restricted as ζ (s, a) :=

∞ X

( 1; 0 < a 5 1).

(k + a)−s

(20)

k=0

Applying f (x) = (x + a)−s ( 1) to (19) and taking the limit of each side of the resulting equation as N → ∞ gives M

ζ (s, a) =

X Bk+1 a−s+1 1 + s+ (s)k a−s−k s−1 2a (k + 1)! k=1

(s)M+1 − (M + 1)!

Z∞

(21)

e BM+1 (x) (x + a)−s−M−1 dx.

0

Remark 3 It is observed that the integral involved in (21) converges for −M. So, ζ (s, a) can be continued analytically to −M. Since M can be arbitrarily large, ζ (s, a) can be continued analytically to the whole s-plane, except for an obvious simple pole at s = 1. The special case of (19) when M = 1 is given as N X n=0

f (n) =

ZN

f (x) dx +

1 1 f 0 (N) − f 0 (0) ( f (0) + f (N)) + 2 12

0

−

1 2

ZN

(22) e B2 (x) f (2) (x) dx.

0

Consider a function fq (x; s, a) :=

q(x+a)(s−1) s 1 − q x+a

( 1; 0 < a 5 1)

(23)

526


ready for application to (22) to compute the following: ∂ s − 1 + q x+a fq (x; s, a) = (log q) q(x+a)(s−1) s+1 ; ∂x 1 − q x+a x+a + 1 − q x+a 2 ∂2 2 (x+a)(s−1) s(s + 1) − 3s 1 − q ; fq (x; s, a) = (log q) q s+2 ∂x2 1 − q x+a and, in general, −s−2 ∂k pq (x; s, a), fq (x; s, a) = (log q)k q(x+a)(s−1) 1 − q x+a k ∂x where pq (x; s, a) is a polynomial in s and q x+a . Applying (23) to (22), taking the limit of the resulting equation as N → ∞ and considering Z∞ 0

Z∞

q(x+a)(s−1) s dx = 1 − q x+a

q−x−a

s dx q−x−a − 1 0 " 1−s #∞ q−x−a − 1 qa(s−1) (1 − qa )1−s = =− , (s − 1) log q (s − 1) log q

(24)

0

we readily find that, for 1, ∞ (n+a)(s − 1) X q qa(s − 1) (1 − q)s (1 − qa )1 − s s = − ζq (s, a) = (1 − q) (s − 1) log q 1 − qn+a n=0 s

a (1 − q)s qa(s − 1) 1 a(s − 1) s s−1 + q − (log q)q (1 − q) − (1 − q)s 2 (1 − qa )s 12 (1 − qa )s+1 2 Z∞ s(s + 1) − 3s 1 − q x+a + 1 − q x+a (log q)2 e · B2 (x)q(x+a)(s − 1) dx. s+2 2 1 − q x+a

+

0

(25) Remark 4 Unlike in the classic case represented by (21), the integral in (25) cannot be made to converge by simply choosing M sufficiently large, instead of M = 1, because (M+1) the presence of the factor q(x+a)(s−1) in fq (x; s, a) implies that necessary 1. Therefore, in this case, we use the known Fourier expansion of the periodic Bernoulli polynomials (see [1225, Chapter IX, Miscellaneous Exercise 12]): e Bk (x) = −k!

X n∈Z\{0}

e2π inx . (2πin)k

(26)


527

The equality in (26) is valid for all real numbers x when k = 2, in which case the sum is absolutely and uniformly convergent. Substituting (26) with k = 2 into (25) and interchanging the summation and integration, we find that qa(s−1) (1 − q)s (1 − qa )1−s (1 − q)s qa(s−1) + (s − 1) log q 2 (1 − qa )s s − 1 + qa 1 − (log q) qa(s−1) (1 − q)s 12 (1 − qa )s+1 X 1 + (1 − q)s (log q)2 In (x; s, a, q), (2πin)2

ζq (s, a) = −

(27)

n∈Z\{0}

where In (x; s, a, q) :=

Z∞

2π inx (x+a)(s−1)

e

q

0

2 s(s + 1) − 3s 1 − q x+a + 1 − q x+a dx. s+2 1 − q x+a

By letting q x+a = u in In (x; s, a, q), we obtain e−2π ina s(s + 1) Bqa (s + δn − 1, −s − 1) log q − 3s Bqa (s + δn − 1, −s) + Bqa (s + δn − 1, −s + 1) ,

In (x; s, a, q) = −

(28)

where 2πi δ := log q

and Bt (α, β) =

Zt

uα−1 (1 − u)β−1 du ( 0)

(29)

0

is the incomplete Beta function given in 1.1(61). It follows from (27) and (28) that qa(s−1) (1 − q)s (1 − qa )1−s (1 − q)s qa(s−1) + (s − 1) log q 2 (1 − qa )s 1 s − 1 + qa − (log q)qa(s−1) (1 − q)s 12 (1 − qa )s+1 (30) X e−2π ina s − (1 − q) log q s(s + 1)Bqa (s + δn − 1, −s − 1) (2πin)2 n∈Z\{0} −3sBqa (s + δn − 1, −s) + Bqa (s + δn − 1, −s + 1) .

ζq (s, a) = −

528


Note that each of the incomplete Beta functions in (30) converges absolutely for 1 and is uniformly bounded with respect to n. Indeed, we see that a

Bqa (s − 1 + δn, −s + ν) 5

Zq

|u| 1): " n X 1 1 log Cp := lim k− log k − n→∞ p p k=1 2 1 1 1 1 1 n2 n 1 n + − n+ 2 − + log n − + − , − 2 2 p 2p 12 p 4 2p 2p

(44)

where (obviously) Cp = A−p .

(45)

In terms of the mathematical constants Ap and Cp , the following log-Gamma integrals were evaluated earlier by Choi et al. [310, p. 114, Eqs. (39) and (40)]: Z1/p 1 1 3 log 0(1 + t) dt = log A − log Ap + log(2π) − − 2 2p 2p 4p

(p > 0)

(46)

(p > 1).

(47)

0

and Z1/p 1 1 3 log 0(1 − t) dt = log Cp − log A + log(2π) − + 2 2p 2p 4p 0

With a view to deriving a relationship between 1 0 Ap and ζ −1, (p > 0) p

564


analogous to that in 2.1(31), we set

1 f (x) := x + p

−s

(p > 0)

in the Euler-Maclaurin summation formula 1.4(68) with a = 0. We, then, obtain n X k=1

1 k+ p

−s ∼ C(s, p) +

1−s n + 1p − ps−1

+

1 2

n+

1 p

−s

1−s m X B2r 1 −s−2r+1 − (s)2r−1 n + + R(s, p; n; m) (2r)! p

(48)

r=1

( −2m − 1; m ∈ N0 ), where C(s, p) is a constant depending on s and p, (s)n denotes the Pochhammer symbol defined by 1.1(5) and the remainder part R(s, p; n; m) satisfies the following limit relationship: lim R(s, p; n; m) = 0

n→∞

( −2m − 1; m ∈ N0 ) .

In fact, C(s, p) can be expressed explicitly as follows, by applying the same f (x) to 1.4(68) and reducing the domain to 1:   1−s 1 s−1 n −s n + − p p 1 X  C(s, p) = lim  k+ −  n→∞ p 1−s k=1 (49) s−1 1 p = ζ s, − ps − ( 1). p s−1 Now, it follows from (48) and (49) that  1−s −s n n + 1p X 1 1 1 1 −s  ζ s, = lim  k+ − − n+ n→∞ p p 1−s 2 p k=1

 −s−2r+1 m X 1 B2r  s + (s)2r−1 n + +p (2r)! p r=1

(m ∈ N, −2m − 1, and s 6= 1; m = 0 and 1). We note that the right-hand side of (50) is analytic for −2m − 1

(m ∈ N0 )

and s 6= 1.

(50)

Miscellaneous Results

565

Thus, we can differentiate (50) with respect to s under the limit sign and set s = −1 in the resulting equation. We, therefore, obtain (see [296, Eq. (3.24)]) 1−s " n 1 −s n + X p 1 1 1 k+ = lim − log k + − ζ 0 s, n→∞ p p p (s − 1)2 k=1 1−s n + 1p 1 1 1 −s 1 + log n + + n+ log n + 1−s p 2 p p   2r−1 m X B2r 1 −s−2r+1  X 1 1  (s)2r−1 n + − log n + +  (2r)! p s+j−1 p  r=1 s

+ p log p

j=1

(m ∈ N, −2m − 1, and s 6= 1; m = 0 and 1), (51)

where we have also used the following elementary identity: ! n X d 1 (s)n = (s)n (n ∈ N0 ) . ds s+k−1

(52)

k=1

By setting s = −1 and s = −2 in (51), we find that " n X 1 1 1 0 = − lim k+ log k + ζ −1, n→∞ p p p k=1 2 1 n 1 1 1 1 1 (53) − + + n+ 2 + + log n + 2 p 2 2p 12 p 2p # n 1 1 1 n2 + − 2+ + log p (p > 0) + 4 2p 12 p 4p and " n 3 X 1 1 1 2 1 n 1 2 ζ −2, = − lim log k + − + + n k+ n→∞ p p p 3 p 2 k=1 1 1 1 1 1 1 1 n3 + 2+ + n+ 3 + 2 + log n + + p 6 6p p 9 p 3p 2p 2 n 1 1 1 1 log p + + − n − 3+ + 2 (p > 0). (54) 2 3p 12 12p 3p 9p p 0

Comparing (42) and (53), we finally obtain the desired relationship: 1 1 1 1 0 log Ap = −ζ −1, + − 2 + log p (p > 0), p 12 4p p which, in view of (43), immediately yields 2.1(31) when p = 1.

(55)

566


Next, in view of the well-known reflection formula (see 1.1(12)): 0(1 + z) 0(1 − z) =

πz , sin(π z)

we can evaluate the following integral (cf. [310, p. 114, Eq. (43)]): Zπ/p 1 log sin t dt = π log Ap − log Cp − log(2p) p

(p > 1),

(56)

0

which, by virtue of another known result [289, p. 95, Eq. (2.2)]:     π 1 Zπ/p sin G 1 + p p π  + π log   log sin t dt = log  1 p 2π G 1 − p 0

(57)

yields the following relationship between the mathematical constants Ap and Cp : 1/p G 1 + 1 p π p Cp sin Ap = 1 π p G 1− p

(p > 1).

(58)

To get a more general class of mathematical constants (see [296]) than those given by (42), we begin by differentiating the function 1 1 q log x + f (x) := x + p p

(p > 0; q ∈ N)

(59)

l times (l ∈ N). We, thus, obtain    q−l  Y l  1  (q − j + 1) log x + 1 + Pl (q) f (l) (x) = x +   p p

(l ∈ N),

j=1

(60) where Pl (q) is a polynomial of degree l − 1 in q satisfying the following recurrence relation:

Pl (q) :=

 l−1 Y    (q − l + 1) Pl−1 (q) + (q − j + 1)

(l ∈ N \ {1}),

j=1

  

1

(l = 1).

(61)


567

In fact, by mathematical induction on l ∈ N, we can give an explicit expression for Pl (q) as follows:   l l X Y Pl (q) =  (q − j + 1)  j=1

j=1

 1  q−j+1

(l ∈ N).

(62)

By substituting from (59) and (60) into the Euler-Maclaurin summation formula 1.4(68) with a = 0, we get a class of mathematical constants Cp,q , defined by "

n X

1 q 1 1 1 q+1 log Cp,q := lim k+ log k + + n+ n→∞ p p p (q + 1)2 k=1 q+1 q 1 1 1 1 1 1 − n+ log n + − n+ log n + q+1 p p 2 p p [(q+1)/2] X B2r 1 q−2r+1 n+ − (2r)! p r=1     2r−1  Y 1 ·  (q − j + 1) log n + + P2r−1 (q)  (p > 0; q ∈ N),   p

(63)

j=1

where [x] denotes (as usual) the greatest integer less than or equal to x. Setting q = 1 in (63) and comparing the resulting equation with (42), it is easy to get a relationship between Ap and Cp,1 as follows:

1 1 Cp,1 = exp − 2 12 4p

Ap

(p > 0).

(64)

If we set s = −q (q ∈ N) in (51), we obtain " n X 1 1 q 1 1 1 q+1 −ζ −q, = lim log k + + n + k+ n→∞ p p p p (q + 1)2 k=1 1 q+1 1 1 1 q 1 1 n+ log n + − n+ log n + − q+1 p p 2 p p   m 2r−1 X B2r 1 q−2r+1  Y 1 − n+ (q − j + 1) log n +  (2r)! p p r=1 j=1    2r−1 2r−1  Y X 1   − p−q log p + (q − j + 1)  q−j+1  0

j=1

j=1

(m, q ∈ N; q < 2m + 1).

(65)

568


Thus, by comparing (63) and (65) and applying (62), we get the following relationship 0 between Cp,q and ζ −q, p1 : 1 log Cp,q = −ζ −q, + p−q log p (p > 0; q ∈ N). p 0

(66)

We conclude this section by remarking that, in a mild sense, the constants Cp,q are generalizations of the Bendersky-Adamchik constants Dk in (24), because, in view of (25) and (66), there is a relationship between Dk and C1,k : log C1,k = log Dk −

Bk+1 Hk k+1

(k ∈ N).

(67)

7.2 Log-Sine Integrals Involving Series Associated with the Zeta Function and Polylogarithms Motivated essentially by their potential for applications in a wide range of mathematic and physical problems, the Log-Sine integrals have been evaluated, in the existing literature on the subject, in many different ways. Very recently, Choi et al. [270] showed how nicely some general formulas analogous to the generalized Log-Sine integral (m) Lsn π3 can be obtained using the theory of Polylogarithms. Relevant connections of the results presented here with those obtained in earlier works are also indicated precisely. For the Log-Sine integrals Lsn (θ) of order n, defined by (see 2.4(82)) Zθ x n−1 Lsn (θ) := − log 2 sin dx 2

(n ∈ N \ {1}),

(1)

0

the recurrence relation 2.4(111) holds true when θ = π (see, e.g., [752, p. 218, Eq. (7.112)]). By using an idea analogous to that of Shen [1024], Beumer [128] presented a recursion formula for (−1)n−1 D(n) := 2 · (n − 1)!

Zπ h x in−1 dx log sin 2

(n ∈ N)

0

in the following form: 2n−1 X k=1

(−1)k−1 D(k) D(2n − k) = (−1)n+1

22n − 1 2n π B2n (2n)!

(n ∈ N),

(2)


569

where Bn are the Bernoulli numbers (see Section 1.7), and D(1) =

π 2

and D(2) =

π log 2. 2

In fact, by mainly analyzing the generalized binomial theorem and the familiar Weierstrass canonical product form of the Gamma function 0(z) (see 1.1(2)), Shen [1024, p. 1396, Eq. (19)] evaluated the Log-Sine integral Lsk+1 (2π) as follows: 1 2π

Z2π 0

∞ x k k! X n log 2 sin dx = (−1)k k σk 2 2 k

(k, n ∈ N),

(3)

n= 2

where σkn are given in terms of the Stirling numbers s(n, k) of the first kind (see, for details, Section 1.5; see also [969]), by σkn =

k−1 X s(n, k − m) s(n, m) . n! n!

m=1

More recently, Batir [102] presented integral representations, involving Log-Sine terms, for some series associated with −1 −2 2k 2k −n k and k−n , k k and for some closely-related series, by using a number of elementary properties of Polylogarithms. Lewin [752, pp. 102–103; p. 164] presented the following integral formulas: π

Z2

x dx = −G log 2 sin 2

(4)

35 1 x dx = ζ (3) − π G, x log 2 sin 2 32 2

(5)

0

and π

Z2 0

where G denotes the Catalan constant, defined by 1.4(16). Several other authors have concentrated on the problem of evaluation of the Log(m) Sine integral Lsn (θ) and the generalized Log-Sine integral Lsn (θ) of order n and index m, defined by Ls(m) n (θ) := −

Zθ 0

x n−m−1 xm log 2 sin dx 2

(6)

570


with the argument θ given by θ = π3 . (Throughout this section, we choose the principal branch of the logarithm function log z in case z is a complex variable.) In particular, van der Poorten [1181] proved that π

Z3

x 7 3 log2 2 sin π dx = 2 108

(7)

x 17 4 x log2 2 sin dx = π . 2 6480

(8)

0

and π

Z3 0

Zucker [1267] established the following two integral formulas: π

Z3 x 3 2 x 253 5 4 2 log 2 sin − x log 2 sin dx = π 2 2 2 3240

(9)

0

and π

Z3 x x3 x 313 x log4 2 sin − log2 2 sin dx = π 6. 2 2 2 408240

(10)

0

(m) Zhang and Williams [1253] extensively investigated Lsn π3 and Lsn π3 along with other integrals to present two general formulas (see [1253, p. 272, Eqs. (1.6) and (1.7)]), which include the integral formulas (7) to (10) as special cases. We choose to recall here one more explicit special case of the Zhang-Williams integral formulas as follows: π

Z3 x 15 2 x log6 2 sin − x log4 2 sin 2 4 2 0 15 x 77821 + x4 log2 2 sin dx = 6 6 π 7 . 16 2 2 ·3 ·7

(11)

The following well-known formula is recorded (see [531, p. 334, Entry (50.5.16)]):

x · cot x = 1 +

∞ X n=1

(−1)n

22n B2n 2n x (2n)!

(|x| < π).

(12)


571

Analogous Log-Sine Integrals Choi et al. [270] showed how nicely some general formulas analogous to the gener (m) alized Log-Sine integral Lsn π3 can be obtained, by using the theory of Polylogarithms. Indeed, by carrying out repeated integration by parts in 2.4(71) in conjunction with 2.4(3), we obtain

Lin (z) − Lin (w) =

Zz

Lin−1 (t)

dt t

w n−2 X (−1)n−1 (−1)k−1 = (log t)k Lin−k (t) + (log t)n−1 log(1 − t) k! (n − 1)! k=1

+

(−1)n−1 (n − 1)!

Zz

(log t)n−1

w

dt 1−t

(n ∈ N \ {1}),

! z

t=w

(13)

where (and elsewhere in this section) an empty sum is understood to be nil; in particular, we have (−1)n−1 ζ (n) = Lin (1) = (n − 1)!

Z1

(−1)n−1

Z1

=

(n − 1)!

(log t)n−1

0

dt 1−t (14)

[log(1 − t)]n−1

dt t

(n ∈ N \ {1}),

0

which, by substituting t = u−1 , yields

(−1)

n−1

Ze−iθ

(log t)

1

n−1

dt = 1−t

Zeiθ 1

(log t)n−1

dt 1 + (iθ)n 1−t n

(15)

(0 5 θ 5 π). Furthermore, it is easily observed, by setting t = 1 − eix , that 1−e Z iθ 0

(log t)

n−1

dt = −i 1−t

Zθ 0

1 x n−1 i (x − π) + log 2 sin dx, 2 2

(16)

572


or, equivalently, that 1−e Z iθ

(log t)

n−1

1

dt = −i 1−t

Zθ

1 x n−1 i (x − π) + log 2 sin dx 2 2

(17)

0

+ (−1)n (n − 1)! ζ (n)

(n ∈ N \ {1})

in view of (14). In its special case when n = 2m + 1 (m ∈ N) and w = 1, (13) yields Zz

(log t)2m

1

dt = (2m)! Li2m+1 (z) − (2m)! ζ (2m + 1) 1−t + (2m)!

2m−1 X k=1 2m

− (log z)

(18)

(−1)k (log z)k Li2m+1−k (z) k!

log (1 − z)

(m ∈ N).

π

Putting z = ei 3 in (18) and using the following elementary identity: π

π

1 − ei 3 = e−i 3 ,

(19)

we get i π3

Ze

(log t)2m

1

π 2m+1 dt = i (−1)m − (2m)! ζ (2m + 1) 1−t 3 + (2m)!

2m−1 X k=0

(−1)k k!

π k π i Li2m+1−k ei 3 3

(20) (m ∈ N).

We, now, separate the even and odd parts of the sum occurring in (20) and make use of 2.4(79) and 2.4(85). We, thus, obtain 2m−1 X k=0

π 2m−1 π X (−1)[k/2] π k (−1)k π k i Li2m+1−k ei 3 = Cl2m+1−k k! 3 k! 3 3

+i

k=0

2m−1 X k=0

h

(−1) k!

k+1 2

i

π k 3

Gl2m+1−k

π 3

.

(21)

Upon substituting from (21) into (20), and equating the real and imaginary parts on each side of the resulting equation, we obtain


573

i π3



Ze



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