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.
 A: 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.
