(June 16, 2007)
Asymptotics of integrals Paul Garrett
[email protected] http://www.math.umn.edu/˜garrett/
• Heuristic for Stirling’s asymptotic • Watson’s lemma • Watson’s lemma illustrated on B(s, a) • Simple form of Laplace’s method • Laplace’s method illustrated on Bessel functions
Here some standard methods in asymptotic expansions [1℄ of integrals are illustrated. These methods are known as Watson’s lemma and Laplace’s method. Watson’s lemma dates from at latest [Watson 1918a], and Laplace’s method at latest from [Laplace 1774]. The exposition here is revisionist, in the sense that we prove a version of Laplace’s method by reducing it to Watson’s lemma. For example, we obtain a robust argument for Stirling’s formula for Γ(s) Γ(s) ∼
√ 1 2π e−s ss− 2
(as |s| −→ ∞, with Re (s) ≥ δ > 0)
and obtain a useful result about ratios of gamma functions, Γ(s + a) ∼ sa Γ(s)
(as |s| −→ ∞, for fixed a, for Re(s) ≥ δ > 0)
The latter is awkward to obtain as a corollary from Stirling’s formula. Laplace’s method is further illustrated by an application to asymptotics of functions closely related to Bessel functions, namely, √ y
Z
∞ 1
e−(u+ u )y uiν
0
√ du ∼ π · e−2y u
(as y −→ +∞)
One point is avoidance of standard but immemorable arguments special to the gamma function. Of course, these special arguments do bear more forcefully upon gamma itself: see [Whittaker-Watson 1927] or [Lebedev 1963]. However, to the extent possible, we want to understand the asymptotics of gamma and other important special functions on general principles.
[1℄ The simplest notion of asymptotic F (s) for f (s) as s goes to +∞ on
R
C
, or in a sector in , is a simpler function F (s) such that lims f (s)/F (s) = 1, written f ∼ F . One might require an error estimate, for example, f ∼ F ⇐⇒ f (s) = F (s) · (1 + O
“ 1” ) |s|
A more precise form is to say that f (s) ∼ f0 (s) ·
“c
0 sα
+
c1 sα+1
+
c2 sα+2
+ ...
”
“
1
(with any auxiliary function f0 ) is an asymptotic expansion for f when f = f0 (s) ·
“c
0 sα
+
c1 sα+1
+ ... +
1
cn sα+n
+O
|s|α+n+1
””
Paul Garrett: Asymptotics of integrals (June 16, 2007)
1. Heuristic for Stirling’s asymptotic First we give a heuristic and mnemonic for the main term of Stirling’s formula, namely 1
Γ(s) ∼ e−s · ss− 2 ·
√ 2π
Using Euler’s integral, s · Γ(s) = Γ(s + 1) =
Z
∞
e
−u
u
s+1
0
du = u
Z
∞
e
−u
s
u du =
Z
∞
e−u+s log u du
0
0
The trick is to replace the exponent −u + s log u by the quadratic polynomial in u best approximating it near its maximum, and evaluate the resulting integral. This replacement can be justified via Watson’s lemma and Laplace’s method, below, but the heuristic is simpler than the justification. More precisely, the exponent takes its maximum where its derivative vanishes, at the unique solution uo = s of s −1 + = 0 u The second derivative in u of the exponent is −s/u2 , which takes value −1/s at uo = s. Thus, near uo = s, the quadratic Taylor-Maclaurin polynomial in t approximating the exponent is 1 · (u − s)2 2! s
−s + s log s − Thus, we imagine that s · Γ(s) ∼
Z
∞
0
2
1
e−s+s log s− 2s ·(u−s) du = e−s · ss ·
Z
∞ 1
2
e− 2s ·(u−s) du
−∞
Note that the latter integral is taken over the whole real line. [2℄ To simplify the remaining integral, replace u by su and cancel a factor of s from both sides, Γ(s) ∼ e Then replace u by u + 1, and then u by u · Z
∞
−∞
e
−s(u−1)2 /2
du =
Z
−s
s
·s ·
Z
∞
e−s(u−1)
2
/2
du
−∞
p 2π/s, obtaining ∞
−∞
e
−su2 /2
√ Z ∞ √ 2π 2π −πu2 du = √ e du = √ s −∞ s
We obtain
1
Γ(s) ∼ e−s · ss− 2 ·
√ 2π
Several aspects of this heuristic are dubious, so it is striking that it can be made rigorous, as below.
[2℄ Evaluation of the integral over the whole line, and simple estimates on the integral over (−∞, 0], show that the
integral over (−∞, 0] is of a lower order of magnitude than the whole. Thus, the leading term of the asymptotics of the integral over the whole line is the same than the integral from 0 to +∞.
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Paul Garrett: Asymptotics of integrals (June 16, 2007)
2. Watson’s lemma The often-rediscovered Watson’s lemma [3℄ gives an asymptotic expansion for certain Laplace transforms, valid in half-planes in . For example, let h be a smooth function on (0, +∞) all whose derivatives are of polynomial growth, and expressible for small x > 0 as
C
h(x) = xα · g(x)
for some α ∈
C, where g(x) is differentiable [4℄ h(x) = xα ·
∞ X
R near 0. Thus, h(x) has an expression
on
cn xn
(for 0 < x sufficiently small)
n=0
Then there is an asymptotic expansion of the Laplace transform of h, Z
∞
e−xs h(x) 0
Γ(α + 1) c1 Γ(α + 2) c2 Γ(α) c0 dx + + + ... ∼ x sα sα+1 sα+2
(for Re (s) > 0)
A simple corollary of the error estimates given below is that, letting Re(α) + 1 − ε be the greatest integer less than or equal Re(α) + 1, Z
∞
e−xs h(x) 0
dx = x
Z
∞
e−xs xα g(x)
0
dx Γ(α) g(0) 1 +O = α Re (α)+1−ε x s |s|
Since Re(α) + 1 − ε > Re(α) the error term is of strictly smaller order of magnitude in s. The idea of the proof is straightforward: the expansion is obtained from Z
0
∞
e
−xs
dx = h(x) x
Z
∞
e
−xs
α
x
n
c0 + . . . + c n x
0
dx x
+
Z
0
∞
dx e−xs xα g(x) − c0 + . . . + cn xn x
The first integral gives the asymptotic expansion, and for Re(s) > 0 the second integral can be integrated by parts essentially Re (α) + n times and trivially bounded to give a O(1/sα+n−ε ) error term for some small ε ≥ 0. Note that for the integration by parts the denominator x in the measure must be moved into the integrand proper, accounting for a slight reduction of the order of vanishing of the integrand at 0. To understand the error, let ε ≥ 0 be the smallest such that N = Re(α) + n − ε ∈
Z
The subtraction of the initial polynomial and re-allocation of the 1/x from the measure makes xα−1 (g(x) − (c0 + . . . + cn xn ) vanish to order N at 0. This, with the exponential e−sx and the presumed [3℄ This lemma appeared in the treatise [Watson 1922] on page 236, citing [Watson 1918a], page 133. Curiously,
the aggregate bibliography of [Watson 1922] omitted [Watson 1918a], and the footnote mentioning it gave no title. Happily, [Watson 1918a] is mentioned by title in [Blaustein-Handelsman 1975]. In the bibliography at the end, we mention [Watson 1917], [Watson 1918a], [Watson 1918b], for perspective. [4℄ We do not need to assume that g is real-analytic near 0, only that it and its derivatives have finite Taylor
expansions approximating it well as x → 0+ .
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Paul Garrett: Asymptotics of integrals (June 16, 2007) polynomial growth of h and its derivatives, allows integration by parts N times without boundary terms, giving Z ∞ Γ(α) c0 Γ(α + 1) c1 Γ(α + n) cn e−xs h(x) dx = + + ...+ α α+1 s s sα+n 0 Z ∞ ∂ N 1 xα · g(x) − (c0 + . . . + cn xn ) dx e−sx + N s ∂x 0
The last error-like term is O(s−[Re (α)+n−ε] ). That is, computing in this fashion, the error term swallows up the last term in the asymptotic expansion. Visibly, this argument applies to more general sorts of expansions near 0.
3. Watson’s lemma illustrated on B(s, a) Here is an important example of an asymptotic result non-trivial to derive from Stirling’s formula for Γ(s), but easy to obtain from Watson’s lemma. Euler’s beta integral is [5℄ Z 1 Γ(s) Γ(a) xs−1 (1 − x)a−1 dx = B(s, a) = Γ(s + a) 0 Fix a with Re(a) > 0, and consider this integral as a function of s. Letting x = e−u gives an integrand fitting Watson’s lemma, Z ∞ Z ∞ u2 −su −u a−1 + . . .)a−1 du e−su (u − e (1 − e ) du = B(s, a) = 2! 0 0 Z ∞ u du Γ(a) = e−su ua · (1 − + . . .)a−1 ∼ 2! u sa 0 taking just the first term in an asymptotic expansion, using Watson’s lemma. Thus, Γ(a) Γ(s) Γ(a) ∼ Γ(s + a) sa giving Γ(s) 1 ∼ a Γ(s + a) s
(for a fixed)
4. Simple form of Laplace’s method A simple version of Laplace’s method [6℄ obtains asymptotics in s for certain integrals of the form Z ∞ e−s·f (u) du 0
[5℄ We recall how to obtain the expression for beta in terms of gamma. With x = u/(u + 1) in the beta integral,
B(s, a) =
Z
∞
us−1 (u + 1)−(s−1)−(a−1)−2 du = 0
1 = Γ(s + a) using
R∞ 0
e
−vy
b
Z
0
∞Z ∞
Z
∞
us−1 (u + 1)−s−a du 0
us e−v(u+1) v s+a
0
b
dv du v u
v dv/v = Γ(b)/y . Replacing u by u/v gives B(s, a) = Γ(s)Γ(a)/Γ(s + a).
[6℄ Perhaps the first appearance of this is in [Laplace 1774].
4
Paul Garrett: Asymptotics of integrals (June 16, 2007) with f real-valued. The idea is that the minimum values of f (u) should dominate, and the leading term of the asymptotics should be √ Z ∞ 2π 1 −s·f (u) −sf (uo ) e du ∼ e ·p ·√ (for |s| −→ ∞, with Re(s) ≥ δ > 0) ′′ s f (uo ) 0 To reduce this to Watson’s lemma, break the integral at points where the derivative f ′ changes sign, and change variables to convert each fragment to a Watson-lemma integral. For Watson’s lemma to be legitimately applied, we will find that f must be smooth with all derivatives of at most polynomial growth and at most polynomial decay, as u −→ +∞. For simplicity assume that there is exactly one point uo at which f ′ (uo ) = 0, and that f ′′ (uo ) > 0. Further, assume that f (u) goes to +∞ at 0+ and at +∞. Since f ′ (u)p> 0 for u > uo and f ′ (u) < 0 for 0 < u < uo , on each of these two intervals there is a smooth square root f (u) − f (uo ) and there are smooth functions F, G such that p (for uo < u < +∞) F ( f (u) − f (uo )) = u Then
Z
∞
p G( f (u) − f (uo )) = Z
uo
Z
∞
e−s f (u) du = e−sf (uo )
u
(for 0 < u < uo )
e−s (f (u)−f (uo )) du + e−sf (uo )
0
0
= e−sf (uo )
Z
∞
e−s (f (u)−f (uo )) du
uo
2
e−s x F ′ (x) dx + 0
Z
∞
0
e−s x G′ (x) dx 2
p √ by letting x = f (u) − f (uo ) in the two intervals. In both integrals, replacing x by x gives Watson’s-lemma integrals Z ∞ Z ∞ Z ∞ √ dx √ dx e−s f (u) du = e−sf (uo ) e−sx 12 x1/2 F ′ ( x) + e−sx 12 x1/2 G′ ( x) x x 0 0 0 At this point the needed conditions on F , hence, on f , become clear: since F must be smooth with all derivatives of at most polynomial growth, direct chain-rule computations show that it suffices that no derivative of f increases or decreases faster than polynomially as u −→ +∞. The assumptions f ′ (uo ) = 0 and f ′′ (uo ) > 0 assure that F has a Taylor series expansion near 0, giving a suitable expansion 1 1/2 ′ F (x) 2x
=
1/2 1 ′ 2 F (0)x
+
1 (2) (0) 3/2 2F
1!
x
+
1 (3) (0) 5/2 2F
2!
x
+
1 (4) (0) 7/2 2F
3!
x
+ ...
(small x > 0)
From this, the main term of the Watson’s lemma asymptotics for the integral involving F would be Z ∞ √ dx Γ( 12 ) F ′ (0) 1 ∼ ·√ e−sx 12 x1/2 F ′ ( x) s x 2 0 dx To determine F ′ (0), or any higher coefficients, from F (x) = u, we have F ′ (x) · du = 1. Since r r p f ′′ (uo ) f ′′ (uo ) f (u) − f (uo ) = (u − uo )2 · x = + ... = · (u − uo ) + . . . 2! 2
the derivative is
dx = du
r
Thus, F ′ (x) =
1 dx du
f ′′ (uo ) · 1 + O(u − uo ) 2
=
s
2 · 1 + O(u − u ) o f ′′ (uo ) 5
Paul Garrett: Asymptotics of integrals (June 16, 2007) which allows evaluation at x = 0, namely s
′
F (0) =
2 f ′′ (uo )
The same argument applied to G gives G′ (0) = F ′ (0). Thus, Z
∞
0
q √ 2 Γ( 12 ) · 2 · f ′′ (u 2π 1 o) −sf (uo ) −s f (u) −sf (uo ) √ = e ·p e du ∼ e · ·√ ′′ 2 s s f (uo )
Last, we verify that this outcome is what would be obtained by replacing f (u) by its quadratic approximation f ′′ (0) · (u − uo )2 2!
f (uo ) +
in the exponent in the original integral, integrated over the whole line. The latter would be Z
∞
es·
f (uo )+ 12 f ′′ (uo )(u−uo )2
−∞
= e
sf (uo )
du = esf (uo )
∞
Z
e
s· 12 f ′′ (uo )u2
du = e
sf (uo )
−∞
Z
∞ 1
es· 2 f
′′
(uo )(u−uo )2
du =
−∞
√ 2π 1 1 sf (uo ) ·p ·√ = e ·p ·√ ′′ 1 ′′ s s f (uo ) 2 f (uo ) √ π
This does indeed agree. Last, verify that the integral of the exponentiated quadratic approximation over (−∞, 0] is of a lower order of magnitude. Indeed, for u ≤ 0 and uo > 0 we have (u − uo )2 ≥ u2 + u2o , and f ′′ (uo ) < 0 by assumption, so esf (uo )
Z
0
es·
2 1 ′′ 2 f (uo )(u−uo )
−∞
≤ e
sf (uo )
·e
s· 12 f ′′ (uo )·u2o
Z
1
du ≤ esf (uo ) · es· 2 f
′′
(uo )·u2o
Z
0
e
du = e
sf (uo )
−∞
′′
(uo )u2
du
−∞
∞ s· 21 f ′′ (uo )u2
1
es· 2 f
·e
s· 21 f ′′ (uo )·u2o
√ 2π 1 ·p ·√ ′′ s f (uo )
Thus, the integral over (−∞, 0] has an additional exponential decay by comparison to the integral over the whole line, so the leading-term of the asymptotics of the integral from 0 to +∞ is the same as those of the integral from −∞ to +∞. The case of Γ(s) can be converted to this situation as follows. For real s > 0, in the integral s · Γ(s) = Γ(s + 1) =
Z
∞
e
−u
s
u du =
0
Z
∞
e−u+s log u du
0
can replace u by su, to put the integral into the desired form s · Γ(s) =
Z
0
∞
e−su+s log u+s log s s du = s · es log s
Z
∞
e−s(u+log u) du
0
R∞ For complex s with Re (s) > 0, both s · Γ(s) and the integral s · es log s 0 e−s(u+log u) du are holomorphic in s, and they agree for real s. Thus, by the identity principle, they are equal for Re(s) > 0.
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Paul Garrett: Asymptotics of integrals (June 16, 2007)
5. Laplace’s method illustrated on Bessel functions Consider the standard integral [7℄ f (y) =
√ y
∞
Z
1
e−(u+ u )y uiν
0
du u
The exponent −(u + u1 )y is of the desired form, with the earlier s replaced by y, but the uiν in the integrand does not fit into the simpler Laplace’ method. Thus, consider integrals Z ∞ e−sf (u) g(u) du 0
where f is real-valued, but g may be complex-valued. The idea still is that the minimum values of f (u) should dominate, and the leading term of the asymptotics should be (assuming a unique minimum at uo ) √ Z ∞ 2π · g(uo ) 1 −s·f (u) −sf (uo ) e g(u) du ∼ e · p ·√ s f ′′ (uo ) 0
As in the simpler case, reduce this to Watson’s lemma by breaking the integral where f ′ changes sign, and change variables to convert each fragment to a Watson-lemma integral. Thus, the first part of the story is much as the simple case of Laplace’s method. As in the simple case of Laplace’s method, the course of the argument reveals conditions on f and g.
For simplicity assume that there is exactly one point uo at which f ′ (uo ) = 0, that f ′′ (uo )p > 0, and that f (u) goes to +∞ at 0+ and at +∞. Thus, on these intervals there are smooth square roots f (u) − f (uo ) and smooth functions F, G such that p (for uo < u < +∞) F ( f (u) − f (uo )) = u Then, letting x = Z
∞
p G( f (u) − f (uo )) =
u
(for 0 < u < uo )
p f (u) − f (uo ) in each of the two intervals, so that F (x) = u and G(x) = u, respectively,
e−sf (u) g(u) du = e−sf (uo )
Z
uo
e−s (f (u)−f (uo )) g(u) du + e−sf (uo )
0
0
= e
−sf (uo )
Z
e
∞
e−s (f (u)−f (uo )) g(u) du
uo
∞ −s x2
Z
′
g(F (x)) F (x) dx +
0
Z
∞
e
−s x2
0
g(G(x)) G (x) dx ′
√ In both integrals, replacing x by x gives Watson’s-lemma integrals Z ∞ √ √ Z ∞ Z ∞ √ dx √ dx √ √ x x e−sf (u) g(u) du = e−sf (uo ) e−sx g(F ( x)) F ′ ( x) + e−sx g(G( x)) G′ ( x) 2 x 2 x 0 0 0 [7℄ This integral is almost the Bessel function
Kν (y) =
1 2
Z
∞
1
e(u+ u )y/2 uν 0
du u
The function Kν is variously called a Bessel function of imaginary argument or MacDonald’s function or modified Bessel function of third kind. Being interested mainly in the case that the parameter ν here is purely imaginary, in √ the text we replace ν by iν. The leading factor of y arises in applications, where such an integral appears as a Whittaker function.
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Paul Garrett: Asymptotics of integrals (June 16, 2007) The assumptions f ′ (uo ) = 0 and f ′′ (uo ) > 0 assure that F has a Taylor series expansion near 0, which gives an expansion ! √ √ g(F (0)) F ′ (0) x ′ √ 1/2 g(F ( x)) F ( x) = x · + ... (small x > 0) 2 2 From this, the main term of the Watson’s lemma asymptotics for the integral involving F would be Z
∞
e−sx
√
0
√ √ dx Γ( 12 ) g(F (0)) F ′ (0) 1 x g(F ( x)) F ′ ( x) ∼ ·√ 2 x 2 s
Note that F (0) = uo . Determine F ′ (0) from F (x) = u. First, F ′ (x) · p f (u) − f (uo ) = x =
r
(u − uo
the derivative is
dx = du
r
Thus, ′
F (x) =
1 dx du
)2
dx du
r
f ′′ (uo ) · + ... = 2!
= 1. Since
f ′′ (uo ) · (u − uo ) + . . . 2
f ′′ (uo ) · 1 + O(u − uo ) 2
=
s
2 · 1 + O(u − u ) o f ′′ (uo )
which allows evaluation at x = 0, namely s
F ′ (0) =
2 f ′′ (uo )
The same argument applied to G gives G′ (0) = F ′ (0). Thus, Z
0
∞
Γ( 12 ) · 2 · g(uo ) · −s f (u) −sf (uo ) √ e du ∼ e · 2 s
Returning to √ f (y) = y
Z
q
2 f ′′ (uo )
= e
∞ 1
e−(u+ u )y uiν
0
−sf (uo )
√ 2π · g(uo ) 1 · p ·√ s f ′′ (uo )
du u
we have critical point uo = 1, and f (uo ) = 2 and f ′′ (uo ) = 2. Applying the just-derived asymptotic, √ f (y) ∼ y ·
e
−2y
! √ √ 2π · 1iν 1 √ ·√ · = π · e−2y y 2
(as y −→ +∞)
Even though the exponent is plausible, it would be easy to lose track of the power of y, which might matter. Also, the leading constant does not depend upon the index ν.
[Bleistein-Handelsman 1975] N. Bleistein, R.A. Handelsman, Asmptotic expansions of integrals, Holt, Rinehart, Winston, 1975, reprinted 1986, Dover. [Laplace 1774] P.S. Laplace, Memoir on the probability of causes of events, M´emoires de Math´ematique et de Physique, Tome Sixi‘eme. (English trans. S.M. Stigler, 1986. Statist. Sci., 1 19 364-378). 8
Paul Garrett: Asymptotics of integrals (June 16, 2007) [Lebedev 1963] N. Lebedev, Special functions and their applications, translated by R. Silverman, PrenticeHall 1965, reprinted Dover, 1972. [Watson 1917] G.N. Watson, Bessel functions and Kapteyn series, Proc. London Math. Soc. (2) xvi (1917), 150-174. 277-308, 1918. [Watson 1918a] G.N. Watson, Harmonic functions associated with the parabolic cylinder, Proc. London Math. Soc. (2) 17 (1918), 116-148. [Watson 1918b] G.N. Watson, Asymptotic expansions of hypergeometric functions, Trans. Cambridge Phil. Soc. 22, 277-308, 1918. [Watson 1922] G.N. Watson, Treatise on the Theory of Bessel Functions, Cambridge Univ. Press, 1922. [Whittaker-Watson 1927] E.T. Whittaker, G.N. Watson, A Course of Modern Analysis, Cambridge University Press, 1927, 4th edition, 1952.
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