# Bessel Function Integral with sin argument

I would like to find if possible a solution (closed form) or approximation for the following integral: $$\int_{\pi/2}^{\pi}\int_{\pi/2}^{\pi}J_{0}\left(\alpha \sin\theta_{k}\right)J_{0}\left(\alpha \sin\theta_{j}\right)\cos\left[\gamma \left(\cos\theta_{j}-\cos\theta_{k}\right)\right]\sin^{3}\left(\theta_{k}\right)\sin^{3}\left(\theta_{j}\right)d\theta_{k}d\theta_{j}$$

where $\alpha$ and $\gamma$ are positive real constants.

• Is there any particular reason you expect a closed form solution to exist? – ASKASK Apr 28 '16 at 2:03
• That integral comes from a system analyzed in a x-z plane. I have solved the same system in a plane x-y obtaining a closed expression form for a little different integral.... – Daniel Apr 28 '16 at 15:09

I am going to play with it and see if anything happens.

Turns out that this can be represented in terms of two simpler integrals. I reached a point where I can't go further. I'll enter what I have done in hopes that it might be useful to someone else.

First I'll make it easier to type, changing $\int_{\pi/2}^{\pi}\int_{\pi/2}^{\pi}J_{0}\left(\alpha \sin\theta_{k}\right)J_{0}\left(\alpha \sin\theta_{j}\right)\cos\left[\gamma \left(\cos\theta_{j}-\cos\theta_{k}\right)\right]\sin^{3}\left(\theta_{k}\right)\sin^{3}\left(\theta_{j}\right)d\theta_{k}d\theta_{j}$ into

$\int_{\pi/2}^{\pi} \int_{\pi/2}^{\pi} J_{0}(a \sin v)J_{0}(a \sin u)\cos\left[c (\cos u-\cos v)\right]\sin^{3}v\sin^{3}udvdu$

(I used MacDown to do the editing.)

Bring out the terms independent of $v$ gives

$\int_{\pi/2}^{\pi} J_{0}(a \sin u)\sin^{3}u \int_{\pi/2}^{\pi} J_{0}(a \sin v)\cos(c (\cos u-\cos v))\sin^{3}vdvdu$

Using $\cos(c (\cos u-\cos v)) =\cos(c\cos u)\cos(c\cos v)+\sin(c \cos u)\sin(c\cos v)$ this becomes

$\int_{\pi/2}^{\pi} J_{0}(a \sin u)\sin^{3}u \int_{\pi/2}^{\pi} J_{0}(a \sin v) (\cos(c\cos u)\cos(c\cos v)+\sin(c \cos u)\sin(c\cos v)) \sin^{3}vdvdu =\int_{\pi/2}^{\pi} J_{0}(a \sin u)\sin^{3}u \int_{\pi/2}^{\pi} J_{0}(a \sin v) \cos(c\cos u)\cos(c\cos v) \sin^{3}vdvdu +\int_{\pi/2}^{\pi} J_{0}(a \sin u)\sin^{3}u \int_{\pi/2}^{\pi} J_{0}(a \sin v) \sin(c \cos u)\sin(c\cos v) \sin^{3}vdvdu =I_1+I_2$

Looking at the two integrals, $I_1 =\int_{\pi/2}^{\pi} J_{0}(a \sin u)\sin^{3}u \int_{\pi/2}^{\pi} J_{0}(a \sin v) \cos(c\cos u)\cos(c\cos v) \sin^{3}vdvdu\\ =\int_{\pi/2}^{\pi} J_{0}(a \sin u)\sin^{3}u\cos(c\cos u) \int_{\pi/2}^{\pi} J_{0}(a \sin v) \cos(c\cos v) \sin^{3}vdvdu\\ =\left( \int_{\pi/2}^{\pi} J_{0}(a \sin u)\sin^{3}u\cos(c\cos u)du \right)^2\\ =I_3^2$

and

$I_2 =\int_{\pi/2}^{\pi} J_{0}(a \sin u)\sin^{3}u \int_{\pi/2}^{\pi} J_{0}(a \sin v) \sin(c \cos u)\sin(c\cos v) \sin^{3}vdvdu\\ =\int_{\pi/2}^{\pi} J_{0}(a \sin u)\sin^{3}u \sin(c \cos u) \int_{\pi/2}^{\pi} J_{0}(a \sin v) \sin(c\cos v) \sin^{3}vdvdu =\left( \int_{\pi/2}^{\pi} J_{0}(a \sin u)\sin^{3}u \sin(c \cos u)du \right)^2\\ =I_4^2$

where

$I_3 =\int_{\pi/2}^{\pi} J_{0}(a \sin u)\sin^{3}u\cos(c\cos u)du$

and

$I_4 =\int_{\pi/2}^{\pi} J_{0}(a \sin u)\sin^{3}u\sin(c\cos u)du$

Noting that $\cos(c\cos u)+i\sin(c\cos u) =e^{ic\cos u}$,

we can write

$I_3+iI_4 =\int_{\pi/2}^{\pi} J_{0}(a \sin u)\sin^{3}ue^{ic\cos u}du$.

This simplifies the original problem, but, at this point I have no idea what else to do.

So I'll leave it at this.

• Thanks Marty. Maybe any suggestion to use an approximation form... – Daniel Apr 28 '16 at 15:10
• What are typical values of a and c? You might be able to use some approximation or expansion of $J_0$. – marty cohen Apr 28 '16 at 20:00
• One possibility is a=c=pi*m where m>0 and integer. – Daniel Apr 29 '16 at 14:18
• Then you might be able to use the asymptotic forms: en.wikipedia.org/wiki/Bessel_function#Asymptotic_forms – marty cohen Apr 29 '16 at 17:31