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I am trying to approximate the integral: $$ \int_0^z \left(\frac{J_1(x\,\sin\theta)}{\sin\theta}\right)^2 {\rm d}\theta $$

My very naive approach was to do the Taylor series of the integrand. However, for rather large x, I need to have increasingly more terms in the series. For $x=1$ I need just two terms, for $x=2\pi\,10^4$ the integration does not seem to converge at all. If it would help, $z$ is typically $z\approx 10^{-1}$. Any suggestion how to approximate this integral? (I assume the definite integral does not exist). Thank you.

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    $\begingroup$ Numerically I found rather (as $x\to\infty$) : $$\int_0^z \left(\frac{J_1(x \sin \theta)}{\sin\theta}\right)^2 d\theta \sim \frac{4}{3\,\pi}x$$ $\endgroup$ Jul 31, 2015 at 9:06
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    $\begingroup$ For numerics, you can use a particular integral representation of the Bessel function that can be efficiently evaluated with the trapezoidal rule. I've posted this on this site some time in the past; search around. $\endgroup$ Jul 31, 2015 at 13:28
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    $\begingroup$ Hi J.M. I @Guesswhoitis. Glad to have you back here (playing with special functions). Btw it this the link? Cheers, $\endgroup$ Jul 31, 2015 at 13:34
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    $\begingroup$ @RaymondManzoni, That's funny, I only checked the accuracy of the guess visually, and $\frac{\sqrt{\pi}}{4} \left / \frac{4}{3\pi} \right.\! \dot= 1.04406$. Kinda interesting :) $\endgroup$ Jul 31, 2015 at 14:57
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    $\begingroup$ @RaymondManzoni Playing around with mathematica i added the first term of the taylor series of $\frac{1}{\sqrt{1-(qz)^2}}$. this gave me corrections like $ \frac{\log (8 x z)+\gamma -2}{2 \pi x}$. Does this help? $\endgroup$
    – tired
    Jul 31, 2015 at 16:43

1 Answer 1

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Let's make a substitution $\sin(\theta)=y$ $$ I(x,z)=\int_{0}^{\sin(z)}\frac{J^2_{1}(x y)}{y^2\sqrt{1-y^2}}dy $$

Because $z\approx0.1$ we may that assume $\sin(z)\approx z$. Furthermore we perform another subsitution $\frac{y}{z}=q$ yielding $$ I(x,z)=z\int_{0}^{1}\frac{J^2_{1}(x z q)}{(zq)^2\sqrt{1-(zq)^2}}dq $$

As a first step to simplify the problem we may observe that for $z\approx 0.1$, $\sqrt{1-(zq)^2}\approx 1$ over the whole domain of integration. Therefore: $$ I(x,z)\approx\frac{1}{z}\underbrace{\int_{0}^{1}\frac{J^2_{1}(x z q)}{q^2}dq }_{Q(x,z)}\quad(1) $$

Essentially there are two ways to go from here on:

1.) Ask Mathematica what the result of this integral is:

$$ I(x,z)\approx\frac{1}{4} x^2 z \, _1F_2\left(\frac{1}{2};2,3;-(x z)^2\right) \quad(2) $$

Note that this result is valid for all values of $x$! only the assumption that $z$ is sufficiently smaller then 1 was made!. If you are interested in the large $xz$ you may use 15.8.2 from here which connects small arguments of $_2F_1$ to big arguments of $_2F_1$. If i have done everything correctly this matches with the answer found by @Raymond Manzoni

$$ I(x,z)\sim\frac{4}{3\pi}x $$

2.) Second approach would be to really to dig in (1) and derive the explicit results for $xz \rightarrow \infty$ using a split of integration range and asymptotic expansions of Bessel functions. I hope i can finish this approach today or tomorrow when i'm less busy. But i'm quite sure this works out fine.

$\bf{Appendix}$

To prove (2) we may employ the following representation of $J^2_1(x)$:

$$ J^2_1(x)=\sum_{m=0}^{\infty}\frac{(-1)^m}{2^{2+2m}}\frac{x^{2m+2}\Gamma(2m+3)}{m!\Gamma(m+3)\Gamma^2(m+2)} $$

Now after switching summation and integration

$$ Q(x,z)=\sum_{m=0}^{\infty}\frac{(-1)^m}{2^{2+2m}(2m+1)}\frac{(zx)^{2m+1}\Gamma(2m+3)}{m!\Gamma(m+3)\Gamma^2(m+2)} $$

By application of the duplication formula of the gamma function this may be reduced to

$$ Q(x,z)= \sum_{m=0}^{\infty}\frac{(-1)^m}{2\sqrt{\pi}}\frac{(zx)^{2m+1}\Gamma(m+\frac{1}{2})}{m!\Gamma(m+3)\Gamma(m+2)} $$

Using the definition of Pochhammer symbol $(x)_n=\frac{\Gamma(x+n)}{\Gamma(x)}$ we find that $(\frac{1}{2})_n=\sqrt{\pi}\Gamma(n+\frac{1}{2}),(2)_n=\Gamma(n+2)$ and $(3)_n=2\Gamma(n+2)$ so we can reformulate

$$ Q(x,z)= \frac{z^2x^2}{4}\sum_{m=0}^{\infty} \frac{(-z^2x^2)^{m}(\frac{1}{2})_m}{m!(2)_m (3)_m} $$

Using the series representation of $_1F_2$ we may conclude that

$$ Q(x,z)=\frac{z^2 x^2}{4}{_1F}_2\left(\frac{1}{2};2,3;-(x z)^2\right) $$

Which proves (2)

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    $\begingroup$ This paper by Stoyanov, Farrell and Bird (no need to 'sign in' just download) could help (ch 2.1) to rewrite your integral as a hypergeometric function or get Mma's answer from the comments at $z=1$. Fine continuation, $\endgroup$ Jul 31, 2015 at 13:04
  • $\begingroup$ @RaymondManzoni Thanks a million, this works! :) $\endgroup$
    – tired
    Jul 31, 2015 at 13:36
  • $\begingroup$ Glad it did @tired. Excellent conclusion! $\endgroup$ Jul 31, 2015 at 13:38
  • $\begingroup$ @RaymondManzoni I added the proof, in case u are interested :) $\endgroup$
    – tired
    Jul 31, 2015 at 16:06

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