Asymptotic Expansion of Integrals

Chee Han Tan Last modiﬁed : April 16, 2017

Abstract

These notes are largely based on the last 3 weeks of Math 6720: Applied Complex Variables and Asymptotic Methods course, taught by Christel Hohenegger in Spring 2017 and Alexander Balk in Spring 2016, at the University of Utah. Additional examples/remarks/results from other sources are added as I see ﬁt, purely for my own understanding. These notes are by no means accurate or applicable, and any mistakes here are of course my own. Please report any typographical errors or mathematical fallacy to me by email [email protected]

Motivation

A solid understanding in the asymptotic theory of integrals has proven to be invaluable for applied mathematician, but why integrals? The reason is that many functions that arise frequently in math- ematics, physics and engineering are defined by (complicated) integral expressions, and in most cases one resorts to numerical techniques to study these integrals due to the difficulty in gauging its behaviour. It is precisely this reason that many powerful analytical tools are developed to extract asymptotic behaviour of these integral functions for small or large values of the parameter. We briefly mention a few applications:

1. Integral transforms. The Fourier transform of a given function fpxq is given by

fˆpξq “ fpxqe´i2πx¨ξ dx, n żR and one is interested in estimating asymptotically fˆpξq as ξ ÝÑ 8. For simple functions such as e´α|x| and e´x2 , one can use contour integration to compute their Fourier transform explicitly. For L1 functions, Riemann-Lebesgue lemma states that fˆpξq ÝÑ 0 as |ξ| ÝÑ 8. However, these approaches are not available for complicated functions such as bump functions.

2. Special functions. Examples of special functions in mathematical physics include:

1 8 t3 Airy function : Aipxq “ cos xt ` dt π 0 3 8ż ˆ ˙ Gamma function : Γpzq “ tz´1e´t dt, Repzq ą 0. ż0 Airy functions appear in optics, electromagnetism, ﬂuid dynamics and nonlinear wave propaga- tion.

1 2 1 ASYMPTOTIC NOTATION

3. Diﬀerential equations. In special cases, one might have an integral representation for solution of ODEs and PDEs. Long time behaviour of the system can be understand using asymptotic expansion techniques. Consider the following initial value problem:

xy3pxq ` 2y “ 0, # yp0q “ 0, yp8q “ 0. Its solution has an integral representation 8 x ypxq “ exp ´t ´ ? dt. t ż0 ˆ ˙ One can show that asymptotically the solution satisﬁes

π ? x 2{3 ypxq „ 3 4x exp ´3 as x ÝÑ 8. 3 2 c ˆ ´ ¯ ˙

1 Asymptotic Notation

We begin by deﬁning asymptotic notations and asymptotic expansion. These are useful in describing the limiting behaviour of a function when the argument gets closer to a particular complex number, typically 0 or 8.

Deﬁnition 1.1. Let fpzq, φpzq be functions deﬁned on Ω Ă C.

1. We say that fpzq “ Opφpzqq as z ÝÑ z0 if there exists a constant C ą 0 and a neighbourhood U of z0 such that |fpzq| ď C|φpzq| for all z P Ω X U. fpzq i.e. is bounded locally around z0. φpzq ˇ ˇ ˇ ˇ 2. We sayˇ thatˇ fpzq “ opφpzqq as z ÝÑ z if ˇ ˇ 0 fpzq lim “ 0. zÑz0 φpzq

8 3. A sequence of gauge functions tφnun“0 is an asymptotic sequence as z ÝÑ z0 if

φn`1pzq “ opφnpzqq as z ÝÑ z0 for every n “ 0, 1,....

4. The function fpzq is said to have an asymptotic representation (expansion)

N fpzq „ fN pzq “ anφnpzq as z ÝÑ z0, n“0 ÿ if for every N “ 0, 1, 2,..., we have

fpzq ´ fN pzq “ opφN pzqq as z ÝÑ z0. In other words, asymptotic representation of a function describes its asymptotic behaviour in terms of asymptotic sequence. 1 ASYMPTOTIC NOTATION 3

Remark 1.2.

1. Intuitively, an asymptotic expansion of a given function f is a ﬁnite sum which might diverges, yet it still provides an increasingly accurate description of the asymptotic behaviour of f. There is a caveat here: for a divergent asymptotic expansion, for some z, there exists an optimal N0 “ N0pzq that gives best approximation to f, i.e. adding more terms actually gives worse accuracy.

2. However, for values of z suﬃciently close to the limiting value z0, the optimal number of terms required increases, i.e. for every ε ą 0, there exists an δ and an optimal N0 “ N0pδq such that

N fpzq ´ akφkpzq ă ε for every |z ´ z0| ă δ and N ą N0. ˇ ˇ ˇ k“0 ˇ ˇ ÿ ˇ ˇ ˇ ˇ ˇ 3. One should think of fN pzq as converging for ﬁxed N in the limit as z ÝÑ z0. Observe that the deﬁnition of asymptotic expansion implies that the remainder term is “small” compared to the last term φN pzq of fN pzq.

k ` Example 1.3. The functions φkpxq “ x form an asymptotic sequence as x ÝÑ 0 and in this case the asymptotic representation is often called an asymptotic power series. The functions ´k φkpxq “ x form an asymptotic sequence as x ÝÑ 8.

Proposition 1.4. Consider ﬁnding the leading asymptotic behaviour of the integral

b Ipxq “ fpx, tq dt as x ÝÑ x0. ża

If fpx, tq „ f0ptq as x ÝÑ x0 uniformly for t P ra, bs, i.e.

fpx, tq ´ f0ptq lim “ 0 uniformly in t, xÑx0 f0ptq then the leading behaviour of Ipxq as x ÝÑ x0 is

b b Ipxq “ fpx, tq dt „ f0ptq dt as x ÝÑ x0, ża ża provided that the integral on the RHS is ﬁnite and nonzero.

Example 1.5. For instance, to determine the leading behaviour of the integral

2 1{3 Ipxq “ cos xt2 ` x2t dt as x ÝÑ 0, ż0 ` ˘ we simply set x “ 0 and obtain

2 Ipxq „ cosp0q dt “ 2 as x ÝÑ 0. ż0 4 2 SERIES EXPANSIONS AND INTEGRATION BY PARTS

2 Series Expansions and Integration By Parts

Broadly speaking, there are two ways of approximating a function:

1. A convergent expansion, or

2. A divergent asymptotic expansion.

A convergent expansion can be easily obtained by integrating term by term the power series representation of the integrand, while a divergent expansion is usually constructed using integration by parts. Depending on the limiting value, one is more favourable than the other.

Recall the Gamma function 8 Γpzq “ tz´1e´t dt, Repzq ą 0. ż0 One can show using integration by parts that the Gamma function satisﬁes the functional equation

zΓpzq “ Γpz ` 1q, which can be used to uniquely extend Γpzq to a meromorphic function on C, with simple poles at the non-positive integers z “ ..., ´2, ´1, 0. We note that:

8 ´t 1 e 2 ? Γ “ ? dt “ 2 e´u du “ π. 2 t 8 ˆ ˙ ż0 ż0 For notational convenience, deﬁne the following function:

x γpz, xq “ tz´1e´t dt (Lower incomplete Gamma function) 0 ż 8 Γpz, xq “ tz´1e´t dt (Upper incomplete Gamma function) żx

Lemma 2.1. For any n ě 3, 8 n n ` 1 e´t dt “ Γ , n ż0 ˆ ˙ where Γp¨q is the Gamma function.

Proof. We make a change of variable u “ tn, then du “ ntn´1dt “ nun´1{ndt. The integral becomes:

8 8 8 n du 1 e´t dt “ e´u “ u´pn´1q{ne´u dt nun´1{n n ż0 ż0 ˆ ˙ ż0 1 1 “ Γ n n ˆ ˙ n ` 1 “ Γ . n ˆ ˙ Example 2.2. Consider approximating the error function

x 2 2 Erfpxq “ ? e´t dt as x ÝÑ 8. π ż0 2 SERIES EXPANSIONS AND INTEGRATION BY PARTS 5

1. Convergent expansion 2 The Taylor series of the integrand e´t around x “ 0 is

4 6 2 t t e´t “ 1 ´ t2 ` ´ ` ..., 2! 3! and it converges everywhere because it has no singularity as a function of complex variable. Integrating the Taylor series term by term, we obtain:

2 x3 x5 x7 Erfpxq “ ? x ´ ` ´ ` ... , π 3 10 42 „  and this power series converges everywhere. However, the convergence is slow for large values of x and it doesn’t capture the asymptotic behaviour of Erfpxq as x ÝÑ 8. 2. Divergent expansion We ﬁrst rewrite the error function as follows: 8 8 8 2 2 2 2 2 Erfpxq “ ? e´t dt ´ e´t dt “ 1 ´ ? e´t dt, π π ˆż0 żx ˙ żx Observe that the integrand is almost negligible if t ąą x, so it contributes most to the integral when t is close to x. Using the identity

2 1 d 2 e´t “ ´ e´t 2t dt ” ı and integration by parts, we obtain for every n ě 0:

8 ´t2 8 e 1 d 2 G n : dt e´t dt p q “ n “ ´ n`1 r s x t x 2t dt ż ż ˆ 2 ˙ ˆ ˙ 2 e´t 8 8 pn ` 1qe´t “ ´ ´ dt 2tn`1 2tn`2 ˇx żx x2 ˇ e´ ˇ n ` 1 “ ´ˇ Gpn ` 2q 2xn`1 2 ˆ ˙ 1 e´x2 n 1 G n 2 “ n`1 ´ p ` q p ` q 2 ˜x ¸ Thus,

8 ´x2 2 e 1 e´t dt “ Gp0q “ ´ Gp2q 2x 2 żx e´x2 e´x2 1 3 “ ´ ` Gp4q 2x 2p2x3q 2 2 2 2 ˆ ˙ ˆ 2 ˙ e´x e´x p1qp3qe´x p1qp3qp5q “ ´ ` ´ Gp6q 2x 22x3 23x5 23 Repeating this procedure using the recurrence relation for Gpnq, we ﬁnally obtain:

2 2e´x 1 1 1 ¨ 3 1 ¨ 3 ¨ 5 Erfpxq “ 1 ´ ? ´ ` ´ ` ... π 2x 4x3 8x5 16x7 „  2 2e´x N p2n ´ 1q!! 1 “ 1 ´ ? p´1qn ` R pxq π 2n`1 x2n`1 N`1 n“0 ÿ 6 2 SERIES EXPANSIONS AND INTEGRATION BY PARTS

2 e´x N p2n ´ 1q!! 1 “ 1 ´ ? p´1qn ` R pxq, π 2n x2n`1 N`1 n“0 ÿ where p2n ´ 1q!! “ p2n ´ 1qp2n ´ 3q ... p3qp1q and RN pxq is the remainder term, having the form

p´1qN`1 p2N ` 1q!! R pxq “ ? Gp2pN ` 1qq. N`1 π 2N`1

Clearly, the series diverges for any large x. For the asymptotic sequence

e´x2 φ pxq “ as x ÝÑ 8, n x2n`1 for a ﬁxed N we have that

8 e´t2 |R pxq| “ C dt N`1 t2N`2 ˇ żx ˇ ˇ 8 ˇ ˇ 1 ˇ d 2 “ ˇC ´ ˇ re´t s dt ˇ 2t2N`ˇ 3 dt ˇ żx ˆ ˙ ˆ ˙ ˇ ˇ ´x2 ˇ ˇ e ˇ À ˇC ˇ x2N`3 Cφ pxq “ N , x2 where C is a constant depending only on N. Hence, we show that

2 e´x N p2n ´ 1q!! 1 Erfpxq „ 1 ´ ? p´1qn as x ÝÑ 8. π 2n x2n`1 n“0 ÿ To conclude, we present a table with number of terms required for both convergent and divergent expansion to approximate Erfpxq within 10´5, for diﬀerent values of x. It clearly suggests that the divergent expansion is a better approximation compare to the convergent expansion.

Convergent expansion Divergent expansion Range x ă 1 x ă 2 x ă 3 x ă 5 x ą 3 x ą 2.5 # of terms 8 16 31 75 2 3

Example 2.3. Consider approximating the exponential integral:

8 e´t E pxq “ dt as x ÝÑ 8. 1 t żx Using integration by parts,

8 e´t e´t 8 8 e´t e´x 8 e´t E pxq “ dt “ ´ ´ dt “ ´ dt 1 t t t2 x t2 żx ˇx żx żx e´x ˇ e´t 8 8 e´t e´x e´x 8 e´t “ `ˇ ` 2 dt “ ´ ` 2 dt x ˇ t2 t3 x x2 t3 ˇx żx żx . ˇ ...... ˇ ...... “ . ˇ ...... 2 SERIES EXPANSIONS AND INTEGRATION BY PARTS 7

N pn ´ 1q! 8 e´t “ e´x p´1qn`1 ` p´1qN N! dt xn tN`1 n“1 x ÿ ż RN pxq loooooooooooomoooooooooooon For the asymptotic sequence e´x φ pxq “ , as x ÝÑ 8, n xn

We claim that the ﬁrst sum is an asymptotic expansion of E1pxq as x ÝÑ 8, with respect to the ´x k asymptotic sequence φkpxq “ e {x . A coarse estimate on the remainder gives:

N! 8 N!e´x N! |R pxq| ď e´t dt “ “ φ pxq. N xN`1 xN`1 x N żx Hence, N p´1qk`1pk ´ 1q! N E pxq „ e´x “ a φ pxq as x ÝÑ 8. 1 kn k k k“1 k“1 ÿ ÿ Note that the asymptotic expansion diverges as N ÝÑ 8 for a ﬁxed x.

` We next approximate E1pxq as x ÝÑ 0 . Diﬀerentiating E1pxq using Leibniz rule gives:

´x 2 3 dE1 e 1 x x “ ´ “ ´ 1 ´ x ` ´ ` ... . dx x x 2! 3! ˆ ˙ Integrating term by term, we obtain

x2 E pxq „ C ´ ln x ` x ´ ` .... 1 4 To ﬁnd C, we take the limit as x ÝÑ 0`:

8 e´t C “ lim dt ` ln x “ ´γ « 0.57772. xÑ0` t „żx  Note that 1 n 1 ´γ “ Γ1p1q “ lim Γz ´ “ ´ lim ´ ln n ` . zÑ0` z nÑ8 k ˜ k“1 ¸ „  ÿ

Example 2.4. Consider approximating the integral

x 1 Ipxq “ t´1{2e´t dt “ γ , x as x ÝÑ `8. 2 ż0 ˆ ˙ The Taylor series of the integrand t´1{2e´t around t “ 0 is

2 1 1 t´1{2e´t “ t´1{2 ´ t1{2 ` t3{2 ´ t5{2 ` ..., 2 6 and it converges for all t ‰ 0. Integrating this term by term gives: x 2 1 1 t´1{2e´t dt “ 2x1{2 ´ x3{2 ` x5{2 ´ x7{2 ` ..., 3 5 21 ż0 8 2 SERIES EXPANSIONS AND INTEGRATION BY PARTS

1 and it doesn’t capture the asymptotic behaviour of γ 2 , x as x ÝÑ 8. On the other hand, a direct integration by parts gives: ` ˘ x x 1 x t´1{2e´t dt “ ´t´1{2e´t ´ t´3{2e´t dt, 2 ż0 ˇ0 ż0 ˇ ˇ which diverges upon evaluating the boundary term atˇ t “ 0.

1 To ﬁnd an expansion that is useful for large x, we rewrite γ 2 , x as follows:

1 1 1 ` ˘ γ , x “ Γ ´ Γ , x 2 2 2 ˆ ˙ ˆ ˙ ˆ ˙ ? 8 “ π ´ t´1{2e´t dt. żx Using integration by parts,

1 8 1 8 γ , x “ ´t´1{2e´t ´ t´3{2e´t dt 2 2 ˆ ˙ ˇx żx ˇ 1 8 “ x´1{2e´x ´ˇ t´3{2e´t dt ˇ 2 żx 1 8 1 ¨ 3 8 “ x´1{2e´x ` t´3{2e´t ` t´5{2e´t dt 2 22 ˇx żx ´”3{2 ´x ıˇ 8 ´1{2 ´x x e ˇ1 ¨ 3 ´5{2 ´t “ x e ´ `ˇ t e dt 2 22 żx ...... “ ...... e´x N p2n ´ 1q!! p2N ` 1q!! 8 “ ? 1 ` p´1qn ` p´1qN`1 t´p2N`3q{2e´t dt, x p2xqn 2N`1 n“1 x ” ÿ ı ż RN`1pxq loooooooooooooooooooooooooomoooooooooooooooooooooooooon For the asymptotic sequence e´x φ pxq “ as x ÝÑ 8, n xnx1{2 for a ﬁxed N we have that 8 C 8 |R pxq| “ C t´p2N`3q{2e´t dt ď e´t dt N`1 x´p2N`3q{2 żx żx e´x “ C xp2N`3q{2 ˆ ˙ φ pxq “ C N , x ˆ ˙ where C is a constant depending only on N. Consequently,

1 ? e´x N p2N ´ 1q!! Γ , x „ π ´ ? 1 ` p´1qN as x ÝÑ 8. 2 x p2xqN n“1 ˆ ˙ ” ÿ ı 2 SERIES EXPANSIONS AND INTEGRATION BY PARTS 9

Example 2.5. Consider approximating the integral

8 4 Ipxq “ e´t dt as x ÝÑ 0` and as x ÝÑ 8. żx For the ﬁrst limit x ÝÑ 0`, term-by-term integration of the convergent Taylor series

8 12 4 t t e´t “ 1 ´ t4 ` ´ ` ..., 2 3 gives a divergent result. In order to make use of this convergent power series, we rewrite Ipxq to “remove” the upper bound. From Lemma 2.1, we obtain:

8 x x 4 4 5 4 Ipxq “ e´t dt ´ e´t dt “ Γ ´ e´t dt 4 ż0 ż0 ˆ ˙ ż0 5 x t8 t12 “ Γ ´ 1 ´ t4 ` ´ ` ... 4 2 3 ˆ ˙ ż0 ˆ ˙ 5 x5 x9 x13 “ Γ ´ x ´ ` ´ ` ... 4 5 18 36 ˆ ˙ ˆ ˙ N 4n`1 5 n x “ Γ ´ p´1q ` RN 1pxq, 4 p4n ` 1qn! ` n“0 ˆ ˙ ÿ ` 4n`1 where RN`1pxq “ opφN pxqq as x ÝÑ 0 for the asymptotic sequence φnpxq “ x . As a result, the leading behavior of Ipxq as x ÝÑ 0` is

8 4 5 e´t dt „ Γ ´ x as x ÝÑ 0`. 4 żx ˆ ˙ Unfortunately, this power series converges slowly for large x. For large values of x, integrating by parts gives:

8 8 4 1 d 4 e´t dt “ ´ re´t s dt 4t3 dt żx żx ˆ ˙ ˆ ˙ e´t4 8 3 8 e´t4 “ ´ ´ dt 4t3 4 t4 ˇx żx x4 ˇ t4 e´ ˇ 3 8 e´ “ ´ˇ dt 4x3 4 t4 żx One can show that 8 ´x4 4 e e´t dt „ as x ÝÑ 8. 4x3 żx

Example 2.6. Consider the integral 8 2 e´xt dt, ż0 1 π which has exact value . Integrating by parts gives: 2 x c 8 8 2 1 d 2 e´xt dt “ ´ re´xt s dt 2xt dt ż0 ż0 ˆ ˙ ˆ ˙ 10 3 LAPLACE’S METHOD

e´xt2 8 8 e´xt2 “ ´ ´ dt, 2xt 2xt2 ˇ0 ż0 ˇ and the boundary term diverges at t 0. It appearsˇ that the problem originates from the limits of “ ˇ integration, in which the parameter x appears there.

3 Laplace’s Method

Consider the Laplace integral which has the form

b Ipxq “ fptqexφptq dt. (3.1) ża where we assume that f, φ are real functions. Note that (3.1) corresponds to the Laplace transform if φptq “ ´t. To investigate the asymptotic behaviour of Ipxq as x ÝÑ 8, we try integrating by parts:

b fptq d Ipxq “ rexφptqs xφ1ptq dt ża ˆ ˙ ˆ ˙ fptqexφptq b b exφptq d fptq “ ´ dt xφ1ptq x dt φptq ˇa ża ˆ ˙ ˇ ˇ Rpxq ˇ looooooooooooomooooooooooooon If Rpxq is asymptotically smaller than the boundary term as x ÝÑ 8, then we have

fptqexφptq b Ipxq „ as x ÝÑ 8. (3.2) xφ1ptq ˇa ˇ 1 ˇ In general, (3.2) is satisﬁed if f P Cra, bs, φ P C ra,ˇ bs and one of the following three conditions holds: 1. φ1ptq ‰ 0 on ra, bs and at least one of fpaq, fpbq are not zero. These conditions are suﬃcient to ensure that Rpxq exists, and one can show that it becomes negligible compared with the boundary term as x ÝÑ 8. 2. φptq ă φpbq on ra, bq and fpbq, φ1pbq ‰ 0. In this case, Rpxq fails to exist but these conditions are strong enough to ensure that

fpbqexφpbq Ipxq „ as x ÝÑ 8. xφ1pbq

3. φptq ă φpaq on ra, bs and fpaq, φ1paq ‰ 0. In this case, Rpxq fails to exist but these conditions are strong enough to ensure that

fpaqexφpaq Ipxq „ ´ as x ÝÑ 8. xφ1paq

Remark 3.1. A diﬀerent deﬁnition for asymptotic expansion is used in Orszag’s book. We say that fpzq „ gpzq z ÝÑ z0 if fpzq ´ gpzq lim “ 0. zÑz0 gpzq This is equivalent to saying fpzq gpzq lim “ 1 ðñ lim “ 1 ðñ gpzq „ fpzq as z Ñ z0. zÑz0 gpzq zÑz0 fpzq 3 LAPLACE’S METHOD 11

If we write fpzq as fpzq “ gpzq ` rfpzq ´ gpzqs, fpzq „ gpzq as z ÝÑ z0 means that the diﬀerence fpzq ´ gpzq is small compared with gpzq as z ÝÑ z0. The diﬀerence fpzq ´ gpzq is said to be subdominant as compared with fpzq or gpzq which are dominant. [Refer to Stokes phenomenon and Example 11, page 116]

Remark 3.2. Assume t, x P R in (3.1). If fptq “ fReptq ` ifImptq, then we look at the real and imaginary parts of the integral separately. If φptq “ φReptq ` iφImptq, then we rewrite the Laplace integral as: b Ipxq “ fptqeixφImptq exφReptq dt, a ż gpx,tq

2 loooooomoooooon where gpx, tq: R ÝÑ C.

Laplace’s method is a general technique for obtaining the asymptotic behaviour as x ÝÑ `8 of integrals in which the parameter x appears in an exponential. It is based on the following important observation: if the real continuous function φptq attains its maximum at c P ra, bs and fpcq ‰ 0, then only the immediate neighbourhood of t “ c that contributes to the full asymptotic expansion of Ipxq for large x. That is, we may approximate the integral Ipxq by Ipx; εq, where

c`ε fptqexφptq dt if a ă c ă b, c´ε $ ż a ε ’ ` Ipx; εq “ ’ fptqexφptq dt if c “ a, ’ ’ a & ż b fptqexφptq dt if c “ b. ’ b ε ’ ż ´ ’ %’

Example 3.3. We use Laplace’s method to investigate the asymptotic behaviour of

10 e´xt Ipxq “ dt as x ÝÑ 8. 1 ` t ż0 Since φptq “ ´t has a maximum at t “ 0 over the interval r0, 10s, we may replace Ipxq by

ε e´xt Ipx; εq “ dt. 1 ` t ż0 Next, we choose ε ą 0 suﬃciently small such that p1 ` tq´1 „ 1, the ﬁrst term in its Taylor series about t “ 0. We then have:

ε e´xt ε 1 ´ e´εx Ipxq „ Ipx; εq „ e´xt dt “ “ as x ÝÑ 8. ´x x ż0 ˇ0 ˇ ˇ Since e´εx ! 1 as x ÝÑ 8 for any ε ą 0, the leadingˇ behaviour of Ipxq as x ÝÑ 8 is 1 Ipxq „ as x ÝÑ 8. x 12 3 LAPLACE’S METHOD

Laplace’s method also gives the full asymptotic expansion of Ipxq. For sufficiently small ε ą 0, writing p1 ` tq´1 as a geometric series 1 8 “ p´1qntn, |t| ă 1, 1 ` t n“0 ÿ we obtain: ε e´xt 8 ε Ipx; εq “ dt “ p´1qn tne´xt dt 1 ` t ż0 n“0 ż0 ÿ8 8 8 “ p´1qn tne´xt dt ´ tne´xt dt n“0 0 ε ÿ ˆż ż ˙ The first integral is just the Gamma function: 8 1 8 Γpn ` 1q n! tne´xt dt “ sne´s ds “ “ , xn`1 xn`1 xn`1 ż0 ż0 where we make a change of variable s “ xt. The second integral is subdominant compared with the first integral as x ÝÑ 8: 8 tne´xt 8 8 ntn`1e´xt εne´εx tne´xt dt “ ` dt „ as x ÝÑ 8, x x x żε ˇε żε ˇ and this is exponentially smaller thanˇ the first integral as x ÝÑ 8. Hence, the full asymptotic ˇ expansion of Ipxq as x ÝÑ 8 is 8 n! Ipxq „ p´1qn as x ÝÑ 8. xn`1 n“0 ÿ

Recall that the Laplace transform of a given function fptq is 8 Lpfqpxq “ fptqe´xt dt, ż0 and one tries to understand the asymptotic behaviour of Lpfqpxq for large x. Observe that for large x, e´xt is sufficiently small except near t “ 0, i.e. for sufficiently nice functions fptq, the main contribu- tion to Lpfqpxq occurs near t “ 0. This suggests that we could determine the asymptotic behaviour of Lpfqpxq by approximate fptq with finitely many terms of its Taylor series around t “ 0. The rigorous statement of this intuition is called Watson’s lemma, which is a powerful tool that allows one to explicitly write down an asymptotic expansion of Laplace integrals with only the knowledge of the local behaviour of the integrand.

Theorem 3.4 (Watson’s Lemma (Real version)). Consider integrals of the form b Ipxq “ fptqe´xt dt, b ą 0, b ‰ 8. (3.3) ż0 Assume that fptq P Cr0, bs and fptq has the asymptotic expansion 8 α βn ` fptq „ t ant as t ÝÑ 0 , (3.4) n“0 ÿ with α ą ´1 and β ą 0 so that the integral converges at t “ 0. Then 8 Γpα ` βn ` 1q Ipxq „ a as x ÝÑ 8. (3.5) n xα`βn`1 n“0 ÿ 3 LAPLACE’S METHOD 13

Proof. We ﬁrst split Ipxq as follows: ε b Ipxq “ fptqe´xt dt ` fptqe´xt dt, ż0 żε Ipx;εq for some ε ą 0. The second integral introduceslooooooomooooooon exponentially small errors for any ε ą 0: b b e´εx ´ e´bx fptqe´xt dt ď }f} e´xt dt “}f} , 8 8 x ˇżε ˇ żε ˆ ˙ ˇ ˇ which converges to 0 exponentiallyˇ forˇ x ÝÑ 8. Next, we substitute the asymptotic expansion (3.4) ˇ ˇ into Ipx; εq to obtain:

N ε N ε α`βn ´xt α`βn ´xt Ipx; εq “ ant e dt ` Ipx; εq ´ ant e dt . n“0 0 ˜ n“0 0 ¸ ÿ ż ÿ ż In particular, we can choose ε suﬃciently small such that

N α βn α βpN`1q fptq ´ t ant ď Kt t for every t P r0, εs, ˇ ˇ ˇ n“0 ˇ ˇ ÿ ˇ for some constant K ąˇ0. Thus, ˇ ˇ ˇ N ε ε N α`βn ´xt α βn ´xt Ipx; εq ´ ant e dt ď fptq ´ t ant e dt ˇ 0 ˇ 0 ˇ ˇ ˇ n“0 ż ˇ ż ˇ n“0 ˇ ˇ ÿ ˇ ˇ ε ÿ ˇ ˇ ˇ ˇ α`βpN`1q ´xt ˇ ˇ ˇ ď K ˇ t e dtˇ 0 ż 8 ď K tα`βpN`1qe´xt dt ż0 K 8 “ sα`βpN`1qe´s ds xα`βpN`1q`1 ż0 Γpα ` βpN ` 1q ` 1q “ K xα`βpN`1q`1 ˆ ˙ 1 “ O as x ÝÑ 8. xα`βpN`1q`1 ˆ ˙ Finally, ε 8 8 α`βn ´xt α`βn ´xt α`βn ´xt ant e dt “ ant e dt ´ ant e dt ż0 ż0 żε a 8 n sα`βne´s ds O e´εx Let s xt. “ α`βn`1 ` p q “ x 0 ż as xÝÑ8 ” ı anΓpα ` βn ` 1q “ ` Ope´εxqlooomooon xα`βn`1 as xÝÑ8 Combining all the estimates leads to looomooon

N Γpα ` βn ` 1q 1 1 Ipxq ´ a „ O “ o as x ÝÑ 8. n xα`βn`1 xα`βpN`1q`1 xα`βpN`1q n“0 ÿ ˆ ˙ ˆ ˙ The desired result follows since N was arbitrary in the asymptotic representation. 14 3 LAPLACE’S METHOD

Remark 3.5. In the case where b “ 8, a far ﬁeld decay condition is needed to ensure that the integral (3.3) converges, i.e. there exists a constant C ą 0 such that fptq ! eCt as t ÝÑ 8.

Theorem 3.6 (Watson’s Lemma (Complex version)). Suppose fptq is analytic in the sector in the complex plane 0 ă |t| ă R, |argptq| ă δ ă π (with a possible branch point at the origin) and suppose

8 n ´1 fptq “ ant N for |t| ă R, (3.6) n“1 ÿ and |fptq| ď Kebt for R ď t ď T. (3.7) π for some K, b independent of t. Then in the sector |argpzq| ď δ ă we have 2 T 8 n Ipzq “ fptqe´zt dt „ a Γ z´n{N as |z| ÝÑ 8. (3.8) n N 0 n“1 ż ÿ ´ ¯ Proof. Let z “ x ` iy. Similar to the real version, we split Ipzq into two parts:

R T ´zt ´zt Ipzq “ fptqe dt ` fptqe dt “ I1pzq ` I2pzq, ż0 żR and estimate each term. We ﬁrst estimate I2pzq using (3.7):

T T ´xt pb´xqt |I2pzq| ď |fptq|e dt ď K e dt żR żR epb´xqT ´ epb´xqR “ K b ´ x “ Ope´xRq as x ÝÑ 8.

Since the power series of fptq is convergent on the real axis,

M´1 n ´1 fptq “ ant N ` RM ptq, n“1 ÿ where the remainder term satisﬁes

M ´1 RM ptq ď Ct N for all 0 ă |t| ď R.

Thus,

R M´1 R n ´1 ´zt ´zt I1pzq “ ant N e dt ` RM ptqe dt 0 ˜ n“1 ¸ 0 ż ÿ ż M´1 8 M´1 8 R n ´1 ´zt n ´1 ´zt ´zt “ an t N e dt ´ an t N e dt ` RM ptqe dt n“1 0 n“1 R 0 ÿ ˆż ˙ ÿ ˆż ˙ ż J1pzq J2pzq loooooooooooooooomoooooooooooooooon loooooooooooooooomoooooooooooooooon We can compute J1pzq and J2pzq by making a change of variable s “ zt:

M´1 8 M´1 an n ´1 ´s an n J pzq “ s N e dt “ Γ 1 zn{N zn{N N n“1 0 n“1 ÿ ż ÿ ´ ¯ 3 LAPLACE’S METHOD 15

M´1 8 M´1 an n ´1 ´s an n J pzq “ s N e dt “ Γ , zR , 2 zn{N zn{N N n“1 zR n“1 ÿ ż ÿ ´ ¯

n and the incomplete Gamma function satisﬁes Γ , zR “ Ope´xRq as x ÝÑ 8. Finally, N ´ ¯ R R ´zt M ´1 ´xt RM ptqe dt ď C t N e dt 0 0 ˇż ˇ ż 8 ˇ ˇ M ˇ ˇ N ´1 ´xt ˇ ˇ ď C t e dt 0 ż 8 C M ´1 ´s “ s N e ds xM{N ż0 CΓp M “ N M{N x ` ˘ “ Opx´M{N q as x ÝÑ 8.

Combining all these estimates yields:

M´1 an n Ipzq “ Γ ` Ope´xRq ` Opx´M{N q ` Ope´xRq zn{N N n“1 ´ ¯ Mÿ´1 an n “ Γ ` Opx´M{N q as x ÝÑ 8. zn{N N n´1 ÿ ´ ¯ The desired result follows since M was arbitrary in the asymptotic representation.

Observe that Watson’s lemma only applies to Laplace integrals (3.1) with φptq “ ´t. For suﬃ- ciently simple φptq, we may try a change of variable of the form s “ ´φptq and obtain:

b ´φpbq fptq ´φpbq Ipxq “ fptqexφptq dt “ ´ e´xs ds “ F psqe´xs ds, φ1ptq ża ż´φpaq ż´φpaq where fptq fpφ´1p´sqq F psq “ ´ “ ´ . φ1ptq φ1pφ´1p´sqq Example 3.7. Consider approximating the integral

π{2 2 Ipxq “ e´x sin ptq dt as x ÝÑ 8. ż0 with φptq “ ´ sin2ptq. Let s “ ´φptq “ sin2ptq, then

ds “ 2 sinptq cosptqdt “ 2 sinptq 1 ´ sin2ptqdt “ 2 sp1 ´ sqdt. b a Thus, Ipxq transforms into:

1 e´xs 1 1 Ipxq “ ds “ F psqe´xs ds, 0 2 sp1 ´ sq 2 0 ż ż a 16 3 LAPLACE’S METHOD where F psq “ 1{ sp1 ´ sq. From the generalised binomial theorem,

a 1 1 1 1 8 Γ n ` 1 “ ? ? “ ? 2 sn, for |s| ă 1. s 1 ´ s s n!Γ 1 sp1 ´ sq n“0 ` 2 ˘ ˆ ˙ ÿ It follows from Watson’sa lemma (with α “ ´1{2, β “ 1) that: ` ˘

1 8 Γ n ` 1 Γ ´ 1 ` n ` 1 1 8 rΓ n ` 1 s2 Ipxq „ 2 2 “ 2 as x ÝÑ 8. 2 n!Γ 1 xn`1{2 2 n!Γ 1 xn`1{2 n“0 ˜ ` 2 ˘¸ ˜ ` ˘¸ n“0 ` 2 ˘ ÿ ÿ ` ˘ ` ˘

Laplace’s Method for Integrals with Movable Maxima

Example 3.8. Consider the Laplace transform of fptq “ e´1{t:

8 8 ´1{t ´1{t ´xt ´ 1 ´xt Lpe qpxq “ e e dt “ e t dt. ż0 ż0 Observe that e´1{t vanishes exponentially fast at t “ 0, the maximum of φptq “ ´t. If we apply Watson’s lemma, we obtain the asymptotic series expansion

Lpe´1{tqpxq „ 0 as x ÝÑ 8, since the asymptotic power series of e´1{t is 0 as t ÝÑ 0`. In this case, Watson’s lemma does not determine the behaviour of Lpe´1{tqpxq since Lpe´1{tqpxq is smaller than any power of x as x ÝÑ 8.

To find the correct behaviour of Lpe´1{tqpxq, we first determine the location of the maximum of the integrand e´1{t´xt, which occurs when: d 1 1 1 0 “ ´ ´ xt “ ´ x ùñ t “ ? . dt t t2 x ˆ ˙ Such a maximum is called a movable maximum because its location depends on the parameter x. To deal with this, we make a change of variable to transform the movable maximum to a fixed maximum. s Let t “ ? , then x 8 8 ? ´1{t ´ 1 ´xt 1 ´ x s` 1 Lpe qpxq “ e t dt “ ? e p s q ds x ż0 ż0 Example 3.9. Consider approximating the Gamma function

8 Γpxq “ tx´1e´t dt as x ÝÑ 8. ż0 This is a Laplace integral, with fptq “ e´t{t and φptq “ ln t. Laplace’s method is not immediately applicable here since max φptq “ 8, tPr0,8q and the maximum occurs as t ÝÑ 8 where fptq is exponentially small. Neglecting the term 1{t which vanishes algebraically as t ÝÑ 8, we determine the location of the maximum of txe´t, which occurs when: d 0 “ ptxe´tq “ txp´e´tq ` pxtx´1qe´t ùñ tx “ xtx´1 ùñ t “ x. dt 3 LAPLACE’S METHOD 17

Since this is a movable maximum, we make a change of variable t “ sx and obtain:

8 8 Γpxq “ tx´1e´t dt “ xx sx´1e´sx ds ż0 ż0 8 ex ln se´xs “ xx ds s ż0 8 expln s´sq “ xx ds. s ż0 18 4 PROBLEMS

4 Problems

1. Show that if ´b ` fpxq „ apx ´ x0q as x ÝÑ x0 ,

then x a 1´b ` fpxq dx „ px ´ x0q as x ÝÑ x if b ă 1. 1 ´ b 0 żx0

Solution:

2. (a) Give an example of an asymptotic relation f „ g as x ÝÑ 8 that cannot be exponentiated, i.e. efpxq „ egpxq as x ÝÑ 8 is false. Solution:

(b) Show that if fpxq ´ gpxq ! 1 as x ÝÑ 8, then efpxq „ egpxq as x ÝÑ 8.

Solution:

3. Find the leading behaviour as x ÝÑ 0` of the following integrals.

1 (a) e´x{t dt; ż0 Solution:

1 (b) cospxtq dt; żx Solution:

1{x 2 (c) e´t dt; ż0 Solution:

8 cospxtq (d) dt. t ż1 Solution:

4. Consider 8 e´t Ipxq “ dt. 1 ` xet2 ż0 ? (a) Show that Ipxq ´ 1 „ ´ exp ´ ln x as x ÝÑ 0`.

Solution: ` ˘

(b) Find the full asymptotic expansion of Ipxq as x ÝÑ 0`. 4 PROBLEMS 19

Solution:

5. Use Laplace’s method to determine the leading behaviour of the following integrals.

π{2 ? 4 (a) te´x sin t dt as x ÝÑ 8; ż0 Solution:

1 ? 2 (b) tante´xt dt as x ÝÑ 8. ż0 Solution:

6. Use Watson’s lemma to obtain an asymptotic expansion of the exponential integral

8 e´t E pxq “ dt. 1 t żx Hint: Show that 8 ´xt ´x e E1pxq “ e dt. 1 ` t ż0

Solution:

7. The modiﬁed Bessel function Inpxq has the integral representation 1 π I pxq “ ex cos θ cospnθq dθ. n π ż0 ex Show that Inpxq „ ? . 2πx

Solution: