# Asymptotics of some generalised sine-integrals

R. B. Paris
Division of Computing and Mathematics,
Abertay University, Dundee DD1 1HG, UK

###### Abstract

We obtain the asymptotic expansion for large integer $n$ of a generalised
sine-integral

$\int_{0}^{\infty}\left(\frac{\sin\,x}{x}\right)^{n}dx$

by utilising the saddle-point method. This expansion is shown to agree with
recent results of J. Schlage-Puchta in Commun. Korean Math. Soc. 35 (2020)
1193–1202 who used a different approach.

An asymptotic estimate is obtained for another related sine-integral also
involving a large power $n$. Numerical results are given to illustrate the
accuracy of this approximation. We also revisit the asymptotics of Ball’s
integral involving the Bessel function $J_{\nu}(x)$, which reduces to the
above integral when $\nu=1/2$.

Mathematics subject classification (2010): 33E20, 34E05, 41A60

Keywords: Sine integrals, asymptotic expansions, saddle-point approximation

1\. Introduction

The expansion of the generalised sine-integral

$I_{n}=\int_{0}^{\infty}\biggl{(}\frac{\sin\,x}{x}\biggr{)}^{\\!n}dx$ (1.1)

for integer $n\to\infty$ has recently been considered by Schlage-Puchta [7].
However the method used seems to be unnecessarily involved and our aim here is
to present a more direct computation using the well-known saddle-point method
for Laplace-type integrals. The interest in the integral $I_{n}$ stems from
the fact that the intersection of the unit cube with a plane orthogonal to a
diagonal and passing through the midpoint has $(n-1)$-measure equal to
$2\sqrt{n}\,I_{n}/\pi$. These intersections arise naturally in certain
probabilistic problems; see the references cited in [7].

The second related sine-integral we consider is given by

$K_{n}=\int_{0}^{\infty}e^{-ax}\biggl{(}1-\frac{\sin^{2}x}{x^{2}}\biggr{)}^{\\!n}dx\qquad(a>0)$
(1.2)

for $n\to\infty$ when the parameter $a=O(1)$. An integral of this type was
communicated to the author by H. Kaiser [3]. We employ a two-term saddle-point
approximation to estimate the growth of $K_{n}$ for large $n$ and present
numerical calculations to verify the accuracy of the resulting formula. In the
final section we revisit the expansion for large $n$ of Ball’s integral
involving the Bessel function $J_{\nu}(x)$, which reduces to (1.1) when
$\nu=\mbox{${\textstyle\frac{1}{2}}$}$.

2\. The asymptotic expansion of $I_{n}$

We begin by writing the integral in (1.1) as

$I_{n}=\int_{0}^{\pi}\biggl{(}\frac{\sin\,x}{x}\biggr{)}^{\\!n}dx+R_{n}(x),\qquad
R_{n}(x)=\int_{\pi}^{\infty}\biggl{(}\frac{\sin\,x}{x}\biggr{)}^{\\!n}dx.$

It is easily seen that

$|R_{n}(x)|<\int_{\pi}^{\infty}\frac{dx}{x^{n}}=\frac{\pi^{1-n}}{n-1}.$

The remainder term $R_{n}(x)$ is therefore bounded by
O($n^{-1}\pi^{-n}$)=O($n^{-1}e^{-n}(\pi/e)^{-n}$) and so is exponentially
small as $n\to\infty$.

Let $\psi(x)=\log\,(x/\sin x)$, where $\psi(0)=0$ and $\psi(\pi)=\infty$. Then
the integral over $[0,\pi]$ becomes

${\hat{I}}_{n}=\int_{0}^{\pi}\biggl{(}\frac{\sin\,x}{x}\biggr{)}^{\\!n}dx=\int_{0}^{\pi}e^{-n\psi(x)}dx.$

This integral has a saddle point at $x=0$ and the integration path $[0,\pi]$
is the path of steepest descent through the saddle. If we now make the
standard change of variable $\psi(x)=\tau^{2}$ discussed, for example, in [2,
p. 66] we obtain

${\hat{I}}_{n}=\int_{0}^{\infty}e^{-n\tau^{2}}\,\frac{dx}{d\tau}\,d\tau.$

From the expansion

$\tau^{2}=\log\biggl{(}\frac{x}{\sin
x}\biggr{)}=\frac{1}{6}x^{2}+\frac{1}{180}x^{4}+\frac{1}{2835}x^{6}+\frac{1}{37800}x^{8}+\frac{1}{467775}x^{10}+\cdots\
$

valid for $|x|<\pi$, we find by inversion of this series using Mathematica
that

$x=\sqrt{6}\biggl{\\{}\tau-\frac{1}{10}\tau^{3}-\frac{13}{4200}\tau^{5}+\frac{9}{14000}\tau^{7}+\frac{17597}{77616000}\tau^{9}+\frac{4873}{218400000}\tau^{11}+\cdots\
\biggr{\\}}$

whence

$\frac{dx}{d\tau}=\sqrt{6}\sum_{k=0}^{\infty}b_{k}\tau^{2k}\qquad(|\tau|<\tau_{0}).$
(2.1)

The first few coefficients $b_{k}$ are

$b_{0}=1,\quad b_{1}=-\frac{3}{10},\quad b_{2}=-\frac{13}{840},\quad
b_{3}=\frac{9}{2000},\quad b_{4}=\frac{17597}{862400},$
$b_{5}=\frac{53603}{218400000},\quad
b_{6}=-\frac{124996631}{1629936000000},\quad
b_{7}=-\frac{159706933}{4366252800000},\ldots\ .$

The circle of convergence of the series (2.1) is determined by the nearest
point in the mapping $x\mapsto\tau$ where $dx/d\tau$ is singular; that is,
when $x=3\pi/2$ (since the point $x=\pi$ maps to $\infty$ in the
$\tau$-plane). This yields the value
$\tau_{0}=|\log\,\mbox{${\textstyle\frac{3}{2}}$}\pi+\pi i|^{1/2}\doteq
1.8717$. Then we have

${\hat{I}}_{n}\sim\sqrt{6}\int_{0}^{\infty}e^{-n\tau^{2}}\sum_{k=0}^{\infty}b_{k}\tau^{2k}\,d\tau=\sqrt{\frac{3}{2n}}\sum_{k=0}^{\infty}\frac{b_{k}}{n^{k}}\int_{0}^{\infty}e^{-w}w^{k-1/2}dw$
$=\sqrt{\frac{3\pi}{2n}}\sum_{k=0}^{\infty}\frac{c_{k}}{n^{k}}\qquad(n\to\infty),$
(2.2)

where the coefficients $c_{k}$ are defined by

$c_{k}:=b_{k}\frac{\Gamma(k+\mbox{${\textstyle\frac{1}{2}}$})}{\Gamma(\mbox{${\textstyle\frac{1}{2}}$})}.$

It follows that since we have extended the integration path in (2.2) beyond
the circle of convergence of (2.1) the resulting asymptotic series is
divergent.

Table 1: The coefficients $c_{k}$ for $1\leq k\leq 12$ (with $c_{0}=1$). $k$ | $c_{k}$ | $k$ | $c_{k}$
---|---|---|---
1 | $-\mbox{${\textstyle\frac{3}{20}}$}$ | 2 | $-\mbox{${\textstyle\frac{13}{1120}}$}$
3 | $+\mbox{${\textstyle\frac{27}{3200}}$}$ | 4 | $+\mbox{${\textstyle\frac{527\,91}{394\,240\,0}}$}$
5 | $+\mbox{${\textstyle\frac{482\,427}{665\,600\,00}}$}$ | 6 | $-\mbox{${\textstyle\frac{124\,996\,631}{100\,352\,000\,00}}$}$
7 | $-\mbox{${\textstyle\frac{527\,032\,878\,9}{13\,647\,872\,000\,0}}$}$ | 8 | $-\mbox{${\textstyle\frac{747\,906\,350\,616\,1}{268\,461\,670\,400\,000}}$}$
9 | $+\mbox{${\textstyle\frac{692\,197\,762\,461\,3}{565\,182\,464\,000\,00}}$}$ | 10 | $+\mbox{${\textstyle\frac{107\,035\,304\,201\,928\,877\,41}{236\,585\,379\,430\,400\,000\,00}}$}$
11 | $+\mbox{${\textstyle\frac{509\,710\,579\,537\,397\,418\,9}{205\,726\,416\,896\,000\,000\,00}}$}$ | 12 | $-\mbox{${\textstyle\frac{123\,979\,742\,078\,372\,360\,595\,39}{362\,078\,493\,736\,960\,000\,000\,0}}$}$

Thus, neglecting exponentially small terms, we have the asymptotic expansion

$I_{n}\sim\sqrt{\frac{3\pi}{2n}}\sum_{k=0}^{\infty}\frac{c_{k}}{n^{k}}\qquad(n\to\infty),$
(2.3)

where the coefficients $c_{k}$ are listed in Table 1 for $0\leq k\leq 12$.
This expansion agrees with that obtained in [7] by less direct means, except
for the value of the coefficient $c_{10}$.

An integral of a similar nature is

$J_{n}=\int_{0}^{\infty}\biggl{(}\frac{1-\cos
x}{\mbox{${\textstyle\frac{1}{2}}$}x^{2}}\biggr{)}^{\\!n}dx=\int_{0}^{\infty}\biggl{(}\frac{\sin\mbox{${\textstyle\frac{1}{2}}$}x}{\mbox{${\textstyle\frac{1}{2}}$}x}\biggr{)}^{\\!2n}dx=2I_{2n}.$

From (2.3) its asymptotic expansion is therefore (to within exponentially
small terms)

$J_{n}\sim\sqrt{\frac{3\pi}{n}}\sum_{k=0}^{\infty}\frac{c_{k}}{(2n)^{k}}\qquad(n\to\infty).$

3\. An asymptotic estimate of another sine-integral

In this section we consider the following integral

$K_{n}=\int_{0}^{\infty}e^{-ax}\biggl{(}1-\frac{\sin^{2}x}{x^{2}}\biggr{)}^{\\!n}dx\qquad(a>0)$
(3.1)

for $n\to\infty$ (not necessarily an integer) when the parameter $a=O(1)$. We
express $K_{n}$ as a Laplace-type integral in the form

$K_{n}=\int_{0}^{\infty}e^{-n\psi(x)}f(x)\,dx,$

where

$\psi(x)=-\log\biggl{(}1-\frac{\sin^{2}x}{x^{2}}\biggr{)},\qquad
f(x)=e^{-ax}.$

For large $n$ the exponential factor in the integrand consists of a series of
peaks situated at $x=k\pi$, ($k=1,2,\ldots$) of decreasing height controlled
by the decay of $f(x)$; see Fig. 1 for a typical example. This is in marked
contrast to the situation pertaining to the integral $I_{n}$ in (1.1), where
the second and successive peaks are of height O($(k\pi)^{-n})$ ($k\geq 1$) and
so are exponentially smaller than the (half) peak in $[0,\pi]$.

Figure 1: Plot of the integrand in (3.1) when $n=5000$ and $a=1/6$ with
horizontal scale $\pi x$. The dashed curve represents $e^{-\pi ax}$.

Routine calculations show that

$\psi^{\prime\prime}(k\pi)=\frac{2}{(k\pi)^{2}},\quad\psi^{\prime\prime\prime}(k\pi)=-\frac{12}{(k\pi)^{3}},\quad\psi^{iv}(k\pi)=\frac{82}{(k\pi)^{4}}-\frac{8}{(k\pi)^{2}}.$

Application of the two-term saddle-point approximation to the $k$th peak then
yields the approximate contribution [5, p. 48], [6, §1.2.3]

$2\sqrt{\frac{\pi}{2n\psi^{\prime\prime}(k\pi)}}\biggl{\\{}1+\frac{c_{2}}{n}\biggr{\\}}e^{-k\pi
a}=k\pi\sqrt{\frac{\pi}{n}}\biggl{\\{}1+\frac{c_{2}}{n}\biggr{\\}}e^{-k\pi
a},$

where

$c_{2}=\frac{1}{2\psi^{\prime\prime}}\biggl{\\{}\frac{2f^{\prime\prime}}{f}-2\frac{\psi^{\prime\prime\prime}}{\psi^{\prime\prime}}\,\frac{f^{\prime}}{f}+\frac{5\psi^{\prime\prime\prime
2}}{6\psi^{\prime\prime 2}}-\frac{\psi^{iv}}{2\psi^{\prime\prime}}\biggr{\\}}$

with all derivatives being evaluated at $x=k\pi$. This yields

$c_{2}=\frac{1}{4}\biggl{\\{}2(1+a^{2})(k\pi)^{2}-12ak\pi+9\biggr{\\}}.$

Summing over all the peaks we then obtain

$K_{n}\sim\pi\sqrt{\frac{\pi}{n}}\biggl{\\{}\sigma_{1}+\frac{1}{8n}\biggl{(}2\pi^{2}(1+a^{2})\sigma_{3}-12\pi
a\sigma_{2}+9\sigma_{1}\biggr{)}\biggr{\\}},$

where

$\sigma_{m}:=\sum_{k=1}^{\infty}k^{m}e^{-k\pi a}.$

We have

$\sigma_{1}=\frac{1}{4\sinh^{2}\mbox{${\textstyle\frac{1}{2}}$}\pi
a},\quad\sigma_{2}=\frac{\cosh\mbox{${\textstyle\frac{1}{2}}$}\pi
a}{4\sinh^{3}\mbox{${\textstyle\frac{1}{2}}$}\pi
a},\quad\sigma_{3}=\frac{2+\cosh\pi
a}{8\sinh^{4}\mbox{${\textstyle\frac{1}{2}}$}\pi a}.$

Hence we obtain our final estimate in the form

$K_{n}\sim\frac{\pi^{3/2}}{4n^{1/2}}\biggl{\\{}1+\frac{T_{1}}{8n}\biggr{\\}}\mbox{cosech}^{2}\mbox{${\textstyle\frac{1}{2}}$}\pi
a\qquad(n\to\infty),$ (3.2)

where

$T_{1}=9-12\pi a\,\mbox{coth}\mbox{${\textstyle\frac{1}{2}}$}\pi
a+\pi^{2}(1+a^{2})\frac{(2+\cosh\mbox{${\textstyle\frac{1}{2}}$}\pi
a)}{\sinh^{2}\mbox{${\textstyle\frac{1}{2}}$}\pi a}$

with $a>0$ fixed and of O(1).

In Table 2 we show computed values of $K_{n}$ compared with the asymptotic
estimate (3.2) for different values of $n$ and the parameter $a$. It is seen
that the agreement is quite good and improves with increasing $n$. However,
since $\psi^{\prime\prime}(k\pi)$ scales like $k^{-2}$, the peaks
progressively broaden as $k$ increases with the consequence that the saddle-
point approximation eventually breaks down. In addition, the parameter $a$
cannot be too small on account of the fact that the envelope of the minima of
the integrand, given by $e^{-ax}(1-1/x^{2})^{n}$, presents a maximum value at
$x\simeq(2n/a)^{1/3}$ equal to approximately
$\exp\,[-\mbox{${\textstyle\frac{3}{2}}$}(2na^{2})^{1/3}]$. We require this
last quantity to be small for the satisfactory estimation of each peak. This
results in the condition $a\gg(2n)^{-1/2}$.

Table 2: Values of $K_{n}$ compared with asymptotic estimate (3.2). | $a=1$ | $a=3/2$
---|---|---
$n$ | $K_{n}$ | Asymptotic | $K_{n}$ | Asymptotic
100 | 0.02707847 | 0.02689533 | 0.00523230 | 0.00521489
200 | 0.01884203 | 0.01880232 | 0.00364706 | 0.00364449
500 | 0.01181371 | 0.01180983 | 0.00228888 | 0.00228866
1000 | 0.00833214 | 0.00833153 | 0.00161452 | 0.00161448
2000 | 0.00588457 | 0.00588447 | 0.00114026 | 0.00114025
4000 | 0.00415855 | 0.00415854 | 0.00080580 | 0.00080580
| $a=1/2$ | $a=2$
$n$ | $K_{n}$ | Asymptotic | $K_{n}$ | Asymptotic
100 | 0.19606514 | 0.19692975 | 0.00108887 | 0.00108697
200 | 0.13567443 | 0.13484945 | 0.00075359 | 0.00075332
500 | 0.08386120 | 0.08361625 | 0.00047067 | 0.00047064
1000 | 0.05878333 | 0.05873199 | 0.00033143 | 0.00033143
2000 | 0.04139902 | 0.04139062 | 0.00023387 | 0.00023387
4000 | 0.02921970 | 0.02921838 | 0.00016520 | 0.00016520

A closely related integral is

${\hat{K}}_{n}=\int_{1}^{\infty}e^{-ax}\biggl{(}1-\frac{\cos^{2}x}{x^{2}}\biggr{)}^{\\!n}dx.$

The peaks in the graph of the integrand are similar to those indicated in Fig.
1 but now occur at $x=(k+\mbox{${\textstyle\frac{1}{2}}$})\pi$,
$k=0,1,2,\ldots\ $. The lower limit of integration is chosen and to lie in the
interval $(\delta,\mbox{${\textstyle\frac{1}{2}}$}\pi-\delta)$ (with
$\delta>0$ so as to avoid the origin and
$\mbox{${\textstyle\frac{1}{2}}$}\pi$). With
$\psi(x)=-\log\,(1-\cos^{2}x/x^{2})$ we find

$\psi^{\prime\prime}((k+\mbox{${\textstyle\frac{1}{2}}$})\pi)=\frac{2}{(k+\mbox{${\textstyle\frac{1}{2}}$})^{2}\pi^{2}}.$

Then by similar arguments we obtain the leading asymptotic approximation

${\hat{K}}_{n}\sim\pi e^{-\pi
a/2}\sqrt{\frac{\pi}{n}}\sum_{k=0}^{\infty}(k+\mbox{${\textstyle\frac{1}{2}}$})e^{-k\pi
a}=\frac{\pi^{3/2}\cosh\mbox{${\textstyle\frac{1}{2}}$}\pi
a}{4n^{1/2}\sinh^{2}\mbox{${\textstyle\frac{1}{2}}$}\pi a}\qquad(n\to\infty).$

4\. An asymptotic expansion for Ball’s integral

Ball’s integral is given by [1]

$L(\nu;n)=\int_{0}^{\infty}\biggl{(}\frac{\Gamma(1+\nu)|J_{\nu}(x)|}{(x/2)^{\nu}}\biggr{)}^{\\!n}x^{2\nu-1}dx,\qquad
n\geq 2,\ \nu\geq\mbox{${\textstyle\frac{1}{2}}$},$ (4.1)

where $J_{\nu}(x)$ is the Bessel function of the first kind and $n$ is not
necessarily an integer. In the case $\nu=\mbox{${\textstyle\frac{1}{2}}$}$ and
integer values of $n$ the integral (4.1) (without the modulus signs) reduces
to that in (1.1), since $\sqrt{\pi/(2x)}J_{1/2}(x)=\sin x/x$. The expansion of
$L(\nu;n)$ for $n\to\infty$ has been derived in [4]; here we revisit this
result by means of the transformation used in Section 1.

We divide the integration path into the intervals $[0,j_{\nu,1}]$ and
$[j_{\nu,1},\infty)$, where $j_{\nu,1}$ denotes the first zero of
$J_{\nu}(x)$. In $[0,j_{\nu,1}]$, we have $J_{\nu}(x)\geq 0$ and the modulus
signs may be dropped in this interval, where

$\sigma(x):=\frac{\Gamma(1+\nu)J_{\nu}(x)}{(x/2)^{\nu}}=\sum_{k=0}^{\infty}\frac{(-x^{2}/4)^{k}}{k!(\nu+1)_{k}}$

with $(a)_{k}=\Gamma(a+k)/\Gamma(a)$ being Pochhammer’s symbol. We set
$\psi(x)=-\log\,\sigma(x)$ in the interval $[0,j_{\nu,1}]$, so that the
integral (4.1) becomes

$L(\nu;n)=\int_{0}^{j_{\nu,1}}e^{-n\psi(x)}x^{2\nu-1}dx+R_{n}(x),$ (4.2)

where

$R_{n}(x)=\int_{j_{\nu,1}}^{\infty}\biggl{(}\frac{\Gamma(1+\nu)|J_{\nu}(x)|}{(x/2)^{\nu}}\biggr{)}^{\\!n}x^{2\nu-1}dx.$

The tail of the integral $R_{n}(x)$ satisfies the bound

$|R_{n}(x)|<(2^{\nu}\Gamma(1+\nu))^{n}\int_{j_{\nu,1}}^{\infty}\frac{dx}{x^{(n-2)\nu+1}}=\biggl{(}\frac{2^{\nu}\Gamma(1+\nu)}{j_{\nu,1}^{\nu}}\biggr{)}^{n}\,\frac{j_{\nu,1}^{2}}{(n-2)\nu}$

since $|J_{\nu}(x)|\leq|J_{\nu}(j^{\prime}_{\nu,2})|<1$ in
$[j_{\nu,1},\infty)$, where $j^{\prime}_{\nu,2}$ is the second zero of
$J_{\nu}^{\prime}(x)$. Defining the quantity

$\xi(\nu):=\frac{2^{\nu}\Gamma(1+\nu)}{j_{\nu,1}^{\nu}},$

and noting that $j_{\nu,1}=\pi$ when $\nu=\mbox{${\textstyle\frac{1}{2}}$}$,
we see that $\xi(\mbox{${\textstyle\frac{1}{2}}$})=2^{-1/2}$. Use of
Stirling’s approximation for the gamma function and the fact that [8, p. 485]
$j_{\nu,1}>\nu$, shows that $\xi(\nu)\sim(2/e)^{\nu}\sqrt{2\pi\nu}\to 0$ as
$\nu\to\infty$. A plot of $\xi(\nu)$ for
$\nu\geq\mbox{${\textstyle\frac{1}{2}}$}$ is shown in Fig. 2 where it is seen
that $\xi(\nu)$ decreases monotonically111It is found that $\xi(\nu)$ is
monotonically decreasing on $[0,\infty)$ with $\xi(\nu)<1$ for $\nu>0$. with
increasing $\nu$. Thus as $n\to\infty$, the bound on $R_{n}(x)$ when
$\nu\geq\mbox{${\textstyle\frac{1}{2}}$}$ is of $O(n^{-1}2^{-n/2})$ and so is
exponentially small.

Figure 2: Plot of $\xi(\nu)$ against
$\nu\geq\mbox{${\textstyle\frac{1}{2}}$}$.

We now deal with the integral in (4.2), where we note that $\psi(0)=0$ and
$\psi(j_{\nu,1})=\infty$. Making the substitution $\tau^{2}=\psi(x)$, we
obtain

$L(\nu;n)=\int_{0}^{\infty}e^{-n\tau^{2}}x^{2\nu-1}\frac{dx}{d\tau}\,d\tau+R_{n}(x).$
(4.3)

From the expansion

$\tau^{2}=\psi(x)=\frac{x^{2}}{4(1+\nu)}+\frac{x^{4}}{32(1+\nu)^{2}(2+\nu)}+\frac{x^{6}}{96(1+\nu)^{3}(2+\nu)(3+\nu)}+\cdots$

valid for $x<j_{\nu,1}$, we find upon inversion

$x=2(1+\nu)^{1/2}\biggl{\\{}\tau-\frac{\tau^{3}}{4(2+\nu)}-\frac{(1+11\nu)\tau^{5}}{96(2+\nu)^{2}(3+\nu)}-\frac{(17\nu^{2}-9\nu-20)\tau^{7}}{128(2+\nu)^{3}(3+\nu)(4+\nu)}+\cdots\biggr{\\}}.$

The last expansion holds in $\tau<\tau_{0}$, where
$\tau_{0}^{2}=\log\,1/\sigma(j^{\prime}_{\nu,2})$, since
$x=j^{\prime}_{\nu,2}$ is the nearest point in the mapping $x\mapsto t$ where
$dx/d\tau$ is singular (the point $x=j_{\nu,1}$ maps to $\infty$ in the
$\tau$-plane). This then yields the expansions

$x^{2\nu-1}=(2(1+\nu)^{1/2}\tau)^{2\nu-1}\biggl{\\{}1-\frac{(2\nu-1)\tau^{2}}{4(2+\nu)}+\frac{(2\nu-1)(6\nu^{2}+\nu-19)\tau^{4}}{96(2+\nu)^{2}(3+\nu)}\hskip
85.35826pt$ $\hskip
170.71652pt-\frac{(2\nu-1)^{2}(2\nu^{3}-\nu^{2}-17\nu-20)\tau^{6}}{384(2+\nu)^{3}(3+\nu)(4+\nu)}+\cdots\biggr{\\}}$

and

$\frac{dx}{d\tau}=2(1+\nu)^{1/2}\biggl{\\{}1-\frac{3\tau^{2}}{2(2+\nu)}-\frac{5(1+11\nu)\tau^{4}}{48(2+\nu)^{2}(3+\nu)}-\frac{7(17\nu^{2}-9\nu-20)\tau^{6}}{128(2+\nu)^{3}(3+\nu)(4+\nu)}+\cdots\biggr{\\}}.$

Combination of these last two expansions then produces

$x^{2\nu-1}\frac{dx}{d\tau}=2^{2\nu}(1+\nu)^{\nu}\tau^{2\nu-1}\sum_{k=0}^{\infty}(-)^{k}b_{k}\tau^{2k}\qquad(\tau<\tau_{0}),$
(4.4)

where

$b_{0}=1,\quad b_{1}=\frac{1+\nu}{2(2+\nu)},\quad
b_{2}=\frac{3\nu^{2}+2\nu-5}{24(2+\nu)(3+\nu)},\quad
b_{3}=\frac{(1+\nu)(\nu^{3}-\nu^{2}-4\nu-8)}{48(2+\nu)^{3}(4+\nu)},$
$b_{4}=\frac{15\nu^{7}+15\nu^{6}-220\nu^{5}-918\nu^{4}+763\nu^{3}+15055\nu^{2}+26898\nu+13688}{5760(2+\nu)^{3}(3+\nu)^{2}(5+\nu)},\ldots\
.$

Insertion of the expansion (4.4) into the integral in (4.3) yields

$2^{2\nu}(1+\nu)^{\nu}\int_{0}^{\infty}e^{-n\tau^{2}}\tau^{2\nu-1}\sum_{k=0}^{\infty}(-)^{k}b_{k}\tau^{2k}d\tau\sim
2^{2\nu-1}(1+\nu)^{\nu}\Gamma(\nu)\sum_{k=0}^{\infty}\frac{(-)^{k}b_{k}(\nu)_{k}}{n^{k+\nu}}.$

As in Section 1, this will be a divergent expansion as we have integrated
beyond the circle of convergence of $x^{2\nu-1}dx/d\tau$.

Thus, neglecting exponentially small terms we finally obtain the expansion

$L(\nu;n)\sim
2^{2\nu-1}(1+\nu)^{\nu}\Gamma(\nu)\sum_{k=0}^{\infty}\frac{(-)^{k}c_{k}}{n^{k+\nu}}\qquad(n\to\infty),$
(4.5)

where the first few coefficients $c_{k}:=b_{k}(\nu)_{k}$ are

$c_{0}=1,\quad c_{1}=\frac{\nu(1+\nu)\wp_{1}(\nu)}{2(2+\nu)},\quad
c_{2}=\frac{\nu(1+\nu)\wp_{2}(\nu)}{24(2+\nu)(3+\nu)},\quad
c_{3}=\frac{\nu(1+\nu)\wp_{3}(\nu)}{48(2+\nu)^{2}(4+\nu)},$
$c_{4}=\frac{\nu(1+\nu)\wp_{4}(\nu)}{5760(2+\nu)^{3}(3+\nu)(5+\nu)},\quad
c_{5}=\frac{\nu(1+\nu)\wp_{5}(\nu)}{11520(2+\nu)^{4}(3+\nu)(6+\nu)},$
$c_{6}=\frac{\nu(1+\nu)\wp_{6}(\nu)}{2903040(2+\nu)^{5}(3+\nu)^{2}(4+\nu)(7+\nu)},$

with the polynomials $\wp_{k}(\nu)$ given by

$\displaystyle\wp_{1}(\nu)\\!\\!$ $\displaystyle=$
$\displaystyle\\!\\!1,\quad\wp_{2}(\nu)=3\nu^{2}+2\nu-5,\quad\wp_{3}(\nu)=(1+\nu)(\nu^{3}-\nu^{2}-4\nu-8),$
$\displaystyle\wp_{4}(\nu)\\!\\!$ $\displaystyle=$
$\displaystyle\\!\\!15\nu^{7}+15\nu^{6}-220\nu^{5}-918\nu^{4}+763\nu^{3}+15055\nu^{2}+26898\nu+13688,$
$\displaystyle\wp_{5}(\nu)\\!\\!$ $\displaystyle=$
$\displaystyle\\!\\!3\nu^{9}\\!-\\!7\nu^{8}\\!-\\!66\nu^{7}\\!-\\!246\nu^{6}\\!+\\!2307\nu^{5}\\!+\\!6825\nu^{4}\\!-\\!43668\nu^{3}\\!-118508\nu^{2}\\!-\\!89904\nu\\!-\\!19392,$
$\displaystyle\wp_{6}(\nu)\\!\\!$ $\displaystyle=$
$\displaystyle\\!\\!63\nu^{13}\\!-\\!3276\nu^{11}\\!-\\!16856\nu^{10}\\!+\\!131726\nu^{9}\\!+\\!781856\nu^{8}\\!-\\!4685840\nu^{7}\\!-\\!14835768\nu^{6}$
$\displaystyle\\!\\!\\!\\!+104879595\nu^{5}\\!\\!+322760624\nu^{4}\\!-\\!328990364\nu^{3}\\!-\\!1748824256\nu^{2}\\!-\\!1801386304\nu\\!-\\!590749440.$

The first coefficients (with $k\leq 3$) agree with those found by Kerman et
al. [4]. It is easily verified that when
$\nu=\mbox{${\textstyle\frac{1}{2}}$}$ the above coefficients agree with those
listed in Table 1.

The estimate for the tail of the integral over the interval
$[2^{\nu}\Gamma(1+\nu),\infty)$ in [4], however, is only $O(n^{-1})$, which is
not sufficiently sharp to justify the expansion of the main integral beyond
its leading term. We have demonstrated that the tail of the integral (4.1) is
exponentially small as $n\to\infty$.

An obvious extension of (4.1) is the integral

${\cal
L}(\nu,a;n)=\int_{0}^{\infty}\biggl{(}\frac{\Gamma(1+\nu)|J_{\nu}(x)|}{(x/2)^{\nu}}\biggr{)}^{\\!n}x^{a-1}dx,\qquad
a>0,$

where $a$ is fixed and it is assumed that $n$ satisfies the condition
$n(\nu+\mbox{${\textstyle\frac{1}{2}}$})>a$ to secure convergence at infinity.
The procedure described above then produces the expansion (when exponentially
small terms are neglected)

${\cal L}(\nu,a;n)\sim
2^{a-1}(1+\nu)^{a/2}\Gamma(\mbox{${\textstyle\frac{1}{2}}$}a)\sum_{k=0}^{\infty}\frac{(-)^{k}d_{k}}{n^{k+a/2}}\qquad(n\to\infty),$
(4.6)

where the first few coefficients $d_{k}$ are given by

$d_{0}=1,\quad
d_{1}=\frac{(\mbox{${\textstyle\frac{1}{2}}$}a)_{2}}{2(2+\nu)},\quad
d_{2}=\frac{(\mbox{${\textstyle\frac{1}{2}}$}a)_{3}}{48(2+\nu)^{2}(3+\nu)}((3a-14)\nu+9a-10),$
$d_{3}=\frac{(\mbox{${\textstyle\frac{1}{2}}$}a)_{4}}{192(2+\nu)^{3}(3+\nu)(4+\nu)}\biggl{\\{}(a^{2}-14a+64)\nu^{2}+(7a^{2}-66a+32)\nu+4(a-4)(3a+2)\biggr{\\}},$
$d_{4}=\frac{(\mbox{${\textstyle\frac{1}{2}}$}a)_{5}}{46080(2+\nu)^{4}(3+\nu)^{2}(4+\nu)(5+\nu)}\biggl{\\{}(15a^{3}-420a^{2}+4820a-23824)\nu^{4}\hskip
68.28644pt$
$+(225a^{3}-5340a^{2}+42860a-65776)\nu^{3}+(1245a^{3}-23340a^{2}+103740a+100560)\nu^{2}$
$+(3015a^{3}-39300a^{2}+45940a+252784)\nu+2700a^{3}-18000a^{2}-18800a+109504\biggr{\\}}.$

When $a=2\nu$, it is seen that the $d_{k}$ reduce to the coefficients $c_{k}$
($k\leq 4$) appearing in the expansion (4.5).

Table 3: Values of the absolute relative error in the computation of ${\cal L}(\nu,a;n)$ against truncation index $k$ when $\nu=4/3$ and $n=100$. $k$ | $a=8/3$ | $a=2/3$ | $a=10/3$
---|---|---|---
0 | $4.664\times 10^{-03}$ | $6.676\times 10^{-04}$ | $6.565\times 10^{-03}$
1 | $2.738\times 10^{-06}$ | $8.987\times 10^{-07}$ | $1.047\times 10^{-05}$
2 | $3.307\times 10^{-08}$ | $6.661\times 10^{-10}$ | $6.041\times 10^{-08}$
3 | $4.006\times 10^{-10}$ | $2.405\times 10^{-11}$ | $5.961\times 10^{-10}$
4 | $2.914\times 10^{-12}$ | $3.655\times 10^{-13}$ | $2.743\times 10^{-12}$

In Table 2 we present the values of the absolute relative error in the
evaluation of ${\cal L}(\nu,a;n)$ using the expansion (4.6) for different
truncation index $k$. The first column shows the values $\nu=4/3$, $a=8/3$,
which corresponds to the integral (4.1).

## References

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