Calculate $I=\int_0^{2\pi}dx \int_0^{\pi} e^{\sin y(\cos x-\sin x)}\sin y\,dy$ I'm not quite sure how to calculate this integral:
$$I=\int_0^{2\pi}dx \int_0^{\pi}  e^{\sin y(\cos x-\sin x)}\sin y\,dy$$
With Mathematica I see the result is $2\pi \sqrt 2 \sinh \sqrt 2$.
Is there any special technique for calculating this integral?
Any help will be appreciated.

Thank you for these brilliant solutions using Bessel function or complex analysis.
After looking into this integral for a long time I found another "primary" solution:
Rewrite it as
$$I=\int_0^{2\pi}d\theta \int_0^{\pi}  e^{\sin \varphi(\cos \theta-\sin \theta)}\sin \varphi\,d\varphi$$
By polar coordinates it becomes
$$I=\int_{u^2+v^2+w^2=1} e^{u-v}\,dS$$
since $dS=\sin\varphi d\theta d\varphi$. Then choose an orthogonal transform such that $z=\frac{u-v}{\sqrt 2}$, we get
$$I=\int_{x^2+y^2+z^2=1} e^{\sqrt 2 z}\,dS$$
Going back to polar coordinates it becomes
$$I=\int_0^{2\pi}d\theta \int_0^{\pi}  e^{\sqrt 2 \cos\varphi}\sin \varphi\,d\varphi$$
And finally we get
$$I=2\pi \int_{-1}^1 e^{\sqrt 2 t}\,dt$$
 A: I see a solution using the modified Bessel function (not in the result, however) $$I_0(z)=\sum_{n=0}^\infty\frac{1}{n!^2}\left(\frac{z}{2}\right)^{2n}=\frac1\pi\int_0^\pi e^{z\cos t}\,dt.$$ Since $\cos x-\sin x=\sqrt2\cos(x+\pi/4)$ being integrated over its period, we may drop the $\pi/4$: $$\int_0^{2\pi}e^{r(\cos x-\sin x)}\,dx=\int_0^{2\pi}e^{r\sqrt2\cos x}\,dx=2\pi I_0(r\sqrt2),$$ and using the power series, we can compute
\begin{align*}
\int_0^\pi I_0(a\sin y)\sin y\,dy
&=\sum_{n=0}^\infty\frac{a^{2n}}{2^{2n}n!^2}\int_0^\pi\sin^{2n+1}y\,dy
\\&=\sum_{n=0}^\infty\frac{a^{2n}}{2^{2n}n!^2}\mathrm{B}\left(\frac12,n+1\right)
\\&=\sum_{n=0}^\infty\frac{a^{2n}}{2^{2n} n!^2}\frac{\sqrt\pi\cdot n!}{2^{-n-1}(2n+1)!!\sqrt\pi}
\\&=2\sum_{n=0}^\infty\frac{a^{2n}}{(2n+1)!}=\frac2a\,\sinh a.
\end{align*}
This gives the result you've got.
A: Denoting : $ f : \mathbb{C}\rightarrow\mathbb{C},\ z\mapsto\frac{\mathrm{e}^{\frac{a\left(1+\mathrm{i}\right)z}{2}}\,\mathrm{e}^{\frac{a\left(1-\mathrm{i}\right)}{2z}}}{\mathrm{i}\,z} $, where $ a\in\mathbb{R} \cdot $
$ \left(\forall n\in\mathbb{Z}\right),\ a_{n}=\left\lbrace\begin{matrix}\frac{a^{n}\left(1+\mathrm{i}\right)^{n}}{2^{n}n!},\ \text{If }n\geq 0\\ \ \ \ \ \ 0, \ \ \ \ \ \ \text{If }n<0\end{matrix}\right. $, and $ \left(\forall n\in\mathbb{Z}\right),\ b_{n}=\left\lbrace\begin{matrix}\frac{a^{-n}\left(1-\mathrm{i}\right)^{-n}}{2^{-n}\left(-n\right)!},\ \text{If }n\leq 0\\ \ \ \ \ \ 0, \ \ \ \ \ \ \text{If }n>0\end{matrix}\right. $
$$ \oint_{\left|z\right|=1}{\frac{\mathrm{e}^{\frac{a}{2}\left(z+\frac{1}{z}\right)-\frac{a}{2\,\mathrm{i}}\left(z-\frac{1}{z}\right)}}{\mathrm{i}\,z}\,\mathrm{d}z}=\oint_{\left|z\right|=1}{\frac{\mathrm{e}^{\frac{a\left(1+\mathrm{i}\right)z}{2}}\,\mathrm{e}^{\frac{a\left(1-\mathrm{i}\right)}{2z}}}{\mathrm{i}\,z}\,\mathrm{d}z} $$
\begin{aligned} f\left(z\right)&=\frac{1}{\mathrm{i}\,z}\left(\sum_{n=0}^{+\infty}{\frac{a^{n}\left(1+\mathrm{i}\right)^{n}}{2^{n}n!}z^{n}}\right)\left(\sum_{n=0}^{+\infty}{\frac{a^{n}\left(1-\mathrm{i}\right)^{n}}{2^{n}n!}z^{-n}}\right)\\ &=\frac{1}{\mathrm{i}\,z}\left(\sum_{n=-\infty}^{+\infty}{a_{n}z^{n}}\right)\left(\sum_{n=-\infty}^{+\infty}{b_{n}z^{n}}\right)\\ f\left(z\right) &=\frac{1}{\mathrm{i}\,z}\sum_{n=-\infty}^{+\infty}{c_{n}z^{n}} \end{aligned}
Where $$ \left(\forall n\in\mathbb{Z}\right),\ c_{n}=\sum_{k=-\infty}^{+\infty}{a_{k}b_{n-k}}=\sum_{k=\max\left(n,0\right)}^{+\infty}{\frac{a^{2k-n}\left(1+\mathrm{i}\right)^{k}\left(1-\mathrm{i}\right)^{k-n}}{2^{2k-n}k!\left(k-n\right)!}} $$
Setting $ n $ to $ 0 $, we'll get : $$ \sum_{k=\max\left(n,0\right)}^{+\infty}{\frac{a^{2k-n}\left(1+\mathrm{i}\right)^{k}\left(1-\mathrm{i}\right)^{k-n}}{2^{2k-n}k!\left(k-n\right)!}}=\sum_{n=0}^{+\infty}{\frac{a^{2n}}{2^{n}\left(n!\right)^{2}}} $$
Thus, applying the residue theorem : $$ \int_{0}^{2\pi}{\mathrm{e}^{a\left(\cos{x}-\sin{x}\right)}\,\mathrm{d}x}=2\pi \sum_{n=0}^{+\infty}{\frac{a^{2n}}{2^{n}\left(n!\right)^{2}}} $$
For $ y \in\mathbb{R} $, setting $ a=\sin{y} $, then integrating with respect to $ y $, we get : $$ \int_{0}^{\pi}{\sin{y}\int_{0}^{2\pi}{\mathrm{e}^{\sin{y}\left(\cos{x}-\sin{x}\right)}\,\mathrm{d}x}\,\mathrm{d}y}=4\pi\sum_{n=0}^{+\infty}{\frac{W_{2n+1}}{2^{n}\left(n!\right)^{2}}} $$
Where $ W_{n} $ for $ n\in\mathbb{N} $, is the Wallis integral $ W_{n}=\int_{0}^{\frac{\pi}{2}}{\sin^{n}{x}\,\mathrm{d}x} \cdot $
Note that since we're integrating on a segment, switching the integral and the infinite sum, which we've just done, is in deed possible.
Since $ \left(\forall n\in\mathbb{N}\right),\ W_{2n+1}=\frac{2^{2n}\left(n!\right)^{2}}{\left(2n+1\right)!} $, we have : \begin{aligned} \int_{0}^{\pi}{\sin{y}\int_{0}^{2\pi}{\mathrm{e}^{\sin{y}\left(\cos{x}-\sin{x}\right)}\,\mathrm{d}x}\,\mathrm{d}y}&=4\pi\sum_{n=0}^{+\infty}{\frac{2^{n}}{\left(2n+1\right)!}} \\ &=2\pi\sqrt{2}\sinh{\sqrt{2}}\end{aligned}
