An integral of an absolute value of sinusoidals Does anyone know how to obtain an expression for the following integral:
$$
\int_{0}^{2\pi}\int_{0}^{2\pi}
\mathrm{e}^{\mathrm{i}\alpha\left\vert\,{\sin\left(\,{\theta}\,\right)
- \sin\left(\,{\varphi}\,\right)}\,\right\vert}
\,\,\,\,\,\,\mathrm{d}\theta\,\mathrm{d}\varphi
$$
The best I could do was to write the exponent as $2\mathrm{i}\alpha\left\vert\sin\left(\frac{\theta\ -\ \varphi}{2}\right)\right\vert\cos\left(\frac{\theta\ +\ \varphi}{2}\right)$ but I then get stuck. This also rather messes up the limits for the region of integration.
Thanks in advance for any help.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{\int_{0}^{2\pi}\int_{0}^{2\pi}
\expo{\ic\alpha\verts{\sin\pars{\theta}
- \sin\pars{\varphi}}}
\,\,\,\,\,\mathrm{d}\theta\,\mathrm{d}\varphi}:
\ {\Large ?}}$.

\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{2\pi}
\expo{\ic\alpha\verts{\sin\pars{\theta}
- \sin\pars{\varphi}}}
\,\,\,\,\,\mathrm{d}\theta} =
\int_{-\pi}^{\pi}
\expo{\ic\alpha\verts{-\sin\pars{\theta}
- \sin\pars{\varphi}}}\,\,\,\,\,\mathrm{d}\theta
\\[5mm] = &\
\int_{0}^{\pi}
\expo{\ic\alpha\verts{\sin\pars{\theta}
+ \sin\pars{\varphi}}}\,\,\,\,\,\mathrm{d}\theta +
\int_{0}^{\pi}
\expo{\ic\alpha\verts{\sin\pars{\theta}
- \sin\pars{\varphi}}}\,\,\,\,\,\mathrm{d}\theta
\\[5mm] = &\
\int_{-\pi/2}^{\pi/2}
\expo{\ic\alpha\verts{\cos\pars{\theta}
+ \sin\pars{\varphi}}}\,\,\,\,\,\mathrm{d}\theta +
\int_{-\pi/2}^{\pi/2}
\expo{\ic\alpha\verts{\cos\pars{\theta}
- \sin\pars{\varphi}}}\,\,\,\,\,\mathrm{d}\theta
\\[5mm] = &\
2\sum_{\sigma = \pm}\ \int_{0}^{\pi/2}
\expo{\ic\alpha\verts{\cos\pars{\theta}
+ \sigma\sin\pars{\varphi}}}\,\,\,\,\,\mathrm{d}\theta
\end{align}
Similarly,
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{2\pi}\int_{0}^{2\pi}
\expo{\ic\alpha\verts{\sin\pars{\theta}
- \sin\pars{\varphi}}}
\,\,\,\,\,\mathrm{d}\theta\,\mathrm{d}\varphi}
\\[5mm] = &\
2\sum_{\sigma = \pm}\ \int_{0}^{\pi/2}
\int_{0}^{2\pi}\expo{\ic\alpha\verts{\cos\pars{\theta}
+ \sigma\sin\pars{\varphi}}}
\,\,\,\,\,\mathrm{d}\varphi\,\mathrm{d}\theta
\\[5mm] = &\
2\sum_{\sigma = \pm}\ \int_{0}^{\pi/2}
\int_{-\pi}^{\pi}\expo{\ic\alpha\verts{\cos\pars{\theta}
- \sigma\sin\pars{\varphi}}}
\,\,\,\,\,\mathrm{d}\varphi\,\mathrm{d}\theta
\\[5mm] = &\
2\sum_{\sigma = \pm}\ \int_{0}^{\pi/2}
\\[2mm] &\ \!\!\!\!\!\!\!\!\!\!\bracks{%
\int_{0}^{\pi}\expo{\ic\alpha\verts{\cos\pars{\theta}
- \sigma\sin\pars{\varphi}}}
\,\,\,\,\,\mathrm{d}\varphi +
\int_{0}^{\pi}\expo{\ic\alpha\verts{\cos\pars{\theta}
+ \sigma\sin\pars{\varphi}}}
\,\,\,\,\,\mathrm{d}\varphi
}\mathrm{d}\theta
\\[5mm] = &\
2\sum_{\sigma = \pm}\ \int_{0}^{\pi/2}
\\[2mm] &\ \!\!\!\!\!\!\!\!\!\!\bracks{%
\int_{-\pi/2}^{\pi/2}\expo{\ic\alpha\verts{\cos\pars{\theta}
- \sigma\cos\pars{\varphi}}}
\,\,\,\,\,\mathrm{d}\varphi +
\int_{-\pi/2}^{\pi/2}\expo{\ic\alpha\verts{\cos\pars{\theta}
+ \sigma\cos\pars{\varphi}}}
\,\,\,\,\,\mathrm{d}\varphi
}\mathrm{d}\theta
\\[5mm] = &\
4\sum_{\sigma = \pm}\sum_{\sigma' = \pm}\ \int_{0}^{\pi/2}\int_{0}^{\pi/2}
\expo{\ic\alpha\verts{\sigma'\cos\pars{\theta}
+ \sigma\cos\pars{\varphi}}}
\,\,\,\,\,\mathrm{d}\varphi\mathrm{d}\theta
\\[5mm] = &\
4\sum_{\sigma = \pm}\ \int_{0}^{\pi/2}\int_{0}^{\pi/2}
\expo{\ic\alpha\bracks{\cos\pars{\theta}
+ \cos\pars{\varphi}}}
\,\,\,\,\,\mathrm{d}\varphi\mathrm{d}\theta
\\[2mm] + &\
4\sum_{\sigma = \pm}\ \int_{0}^{\pi/2}\int_{0}^{\pi/2}
\expo{\ic\alpha\verts{\cos\pars{\theta}
- \cos\pars{\varphi}}}
\,\,\,\,\,\mathrm{d}\varphi\mathrm{d}\theta
\\[5mm] = &\
8\bracks{\int_{0}^{\pi/2}
\expo{\ic\alpha\cos\pars{\theta}}
\,\mathrm{d}\theta}^{2}
\\[2mm] + &\
8\int_{0}^{\pi/2}\int_{0}^{\pi/2}
\expo{\ic\alpha\verts{\cos\pars{\theta}
- \cos\pars{\varphi}}}
\,\,\,\,\,\mathrm{d}\varphi\mathrm{d}\theta
\end{align}
I guess you can complete the evaluation.
