TITLE: What is the double integral of an exponential difference of sines?
QUESTION [8 upvotes]: I am keen to know if there's a way of analytically evaluating the following integral:
$$
\int_{0}^{2\pi}\int_{0}^{2\pi}e^{i\alpha|\sin(\theta)-\sin(\varphi)|}d\theta d\varphi
$$
where $\alpha$ is a real constant. It arises when considering how arrays of charged wires interact with each other.
There are Bessel function representations for similar integrals when one does not have to take the modulus. Here, I simply don't know how to deal with the modulus.
I've tried rewriting it using
$$
\sin(\theta)-\sin(\varphi)=2\sin\left(\frac{\theta-\varphi}{2}\right)\cos\left(\frac{\theta+\varphi}{2}\right)
$$
and then transforming the coordinates to this rotated version. I think it's then fine to rotate the region of integration accordingly, owing to the double periodicity of the integrand, but then I get stuck.
Thanks in advance for any help.
Update: I've used Teresa Lisbon's suggestion of expanding out the exponential to find that
$$
\int_{0}^{2\pi}\int_{0}^{2\pi}e^{i\alpha|\sin(\theta)-\sin(\varphi)|}d\theta d\varphi=4\pi^{2}J_{0}(\alpha)^{2}+32i\alpha_{p}F_{q}(1,1;\frac{3}{2},\frac{3}{2},\frac{3}{2};-\alpha^{2})
$$
Does anyone know if the hypergeometric function part can be simplified (ideally in terms of Bessel functions)? Thanks!

REPLY [5 votes]: Let's abbreviate by $I(\alpha)$ the $\alpha$-dependent integral in
question.
Following the OP, the sine difference can be replaced according to
$\sin\theta-\sin\varphi= 2\sin\left(\frac{\theta-\varphi}{2}\right)
\cos\left(\frac{\theta+\varphi}{2}\right)$ giving
$$
  I(\alpha)=\int_{0}^{2\pi}\int_{0}^{2\pi}\exp\left(2i\alpha\left|\sin\left(\frac{\theta-\varphi}{2}\right)
\cos\left(\frac{\theta+\varphi}{2}\right)\right|\right)\;
  d\theta\, d\varphi.
$$
Setting $\theta=\varphi+2u$ yields
$$
  I(\alpha)=\int_{0}^{2\pi}\int_{-\frac{\varphi}{2}}^{\pi-\frac{\varphi}{2}}\exp\left(2i\alpha\left|\sin\left(u\right)
\cos\left(\varphi+u\right)\right|\right)\;
  2\,du\, d\varphi.
$$
Using the periodicity of the integrand in $u$, the inner integration interval can be
shifted to be independent of $\varphi$, namely $[0,\pi]$. After exchanging the
integrations, the substitution $\varphi=v-u$ gives:
$$
  I(\alpha)=2\int_{0}^{\pi}\int_{u}^{2\pi+u}\exp\left(2i\alpha\left|\sin\left(u\right)
\cos\left(v\right)\right|\right)\;
  dv\, du.
$$
Using the periodicity of the integrand in $v$, the inner integration interval can be
shifted to be independent of $u$:
$$
  I(\alpha)=2\int_{0}^{\pi}\int_{0}^{2\pi}\exp\left(2i\alpha\left|\sin\left(u\right)
\cos\left(v\right)\right|\right)\;
  dv\, du.
$$
Now expanding the exponential as suggested by Teresa Lisbon
decouples the integrals
$$
  I(\alpha)=2\sum_{n=0}^\infty\frac{1}{n!}\int_{0}^{\pi}\int_{0}^{2\pi}\left(2i\alpha\left|\sin\left(u\right)
\cos\left(v\right)\right|\right)^n\;
  dv\, du.
$$
The integrals are known to be
\begin{eqnarray*}
\int_{0}^{\pi}\left|\sin\left(u\right)\right|^ndu&=&2\int_{0}^{\frac{\pi}{2}}\left(\sin\left(u\right)\right)^ndu
=\frac{\Gamma\left(\frac{n+1}{2}\right)\sqrt{\pi}}{\Gamma\left(\frac{n+2}{2}\right)},\\
\int_{0}^{2\pi}\left|\cos\left(v\right)\right|^ndv&=&4\int_{0}^{\frac{\pi}{2}}\left(\sin\left(v\right)\right)^ndv
=2\;\frac{\Gamma\left(\frac{n+1}{2}\right)\sqrt{\pi}}{\Gamma\left(\frac{n+2}{2}\right)},
\end{eqnarray*}
giving
$$
I(\alpha)=4\pi\sum_{n=0}^\infty\frac{\left(2i\alpha\right)^{n}}{n!}\left(
\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n+2}{2}\right)}\right)^2.
$$
Separating the summation by even and odd indices, $\sum_{n=0}^\infty
f(n)=\sum_{n=0}^\infty f(2n)+\sum_{n=0}^\infty f(2n+1)$, the
$\Gamma$-functions and factorials can be expressed by (rising) Pochhammer symbols,
$$
  \Gamma\left(n+a\right) = \left(a\right)^{(n)}\Gamma(a),
  \quad\left(a\in\left\{\frac{1}{2},1, \frac{3}{2}\right\}\right),
$$
$$
  (2n)!=2^{2n}\,n!\,\left(\frac{1}{2}\right)^{(n)},\qquad
  (2n+1)!=2^{2n}\,n!\,\left(\frac{3}{2}\right)^{(n)},
$$
and the
result can be expressed by generalized hypergeometric functions:
$$
I(\alpha)=4\pi^2{}_1F_2\left(\frac{1}{2};1,1;-\alpha^2\right)
+32\,i\,\alpha\,{}_2F_3\left(1,1;\frac{3}{2},\frac{3}{2},\frac{3}{2};-\alpha^2\right).
$$
The real part can be simplified to $4\pi^2\left(J_0(\alpha)\right)^2$,
either by observing that for the even powers in the Taylor series of
the exponential taking the modulus is irrelevant, and can be evaluated
as real part of
$\int_{0}^{2\pi}\int_{0}^{2\pi}\exp\left(i\alpha\left(\sin\left(\theta\right)-
\sin\left(\varphi\right)\right)\right)\, d\theta\,
d\varphi=\left(\int_{0}^{2\pi}\exp\left(i\alpha\sin\left(\theta\right)\right)\,
d\theta\right)^2=\left(2\pi J_0(\alpha)\right)^2,$ or in the defining
series of the hypergeometric function, one can use the identity
$\sum_{n=0}^k{k\choose m}^2={2k\choose k}$ to get after a short calculation
${}_1F_2\left(\frac{1}{2};1,1;-\alpha^2\right)=\left(J_0(\alpha)\right)^2$. The
additional question which was also raised in the OP, namely whether a similar simplification is also possible for the imaginary
part, is open.