How to evaluate $\int_{0}^{2\pi}dx\!\int_{0}^{\pi}\sin ye^{\sin y(\cos x-\sin x)}dy=\sqrt{2}\big(e^{\sqrt{2}}-e^{-\sqrt{2}}\big)\pi $ How to eveluate this double integral below,

$$\int_{0}^{2\pi}\mathrm dx\!\int_{0}^{\pi}\sin ye^{\sin y(\cos x-\sin x)}\mathrm dy=\sqrt{2}\big(e^{\sqrt{2}}-e^{-\sqrt{2}}\big)\pi $$

I only got an answer about it , and the wolframalpha gives the numerical solution which agree with it , but I have no idea how to prove it.
 A: This answer complements the solution presented by Ron Gordon.  Note that we have
$$\begin{align}
I&=\int_0^{2\pi}\int_0^\pi \sin(\theta)e^{\sin(\theta)(\cos(\phi)-\sin(\phi))}\,d\theta\,d\phi \tag 1\\\\
&=\int_0^{2\pi}\int_0^\pi \sin(\theta)e^{\sqrt{2}\sin(\theta)\cos(\phi)}\,d\theta\,d\phi \tag 2\\\\
&=\oint_{|r|=1}e^{\sqrt{2}x}\,dS \tag 3\\\\
&=2\int_{-1}^1 e^{\sqrt{2}x}\left(\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\frac{1}{\sqrt{1-x^2-y^2}}\,dy\right)\,dx \tag 4\\\\
&=2\int_{-1}^1 e^{\sqrt{2}x}\left.\left(\arctan\left(\frac{y}{\sqrt{1-x^2-y^2}}\right)\right|_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\right)\,dx \tag 5\\\\
&=2\int_{-1}^1 e^{\sqrt{2}x}\left(\pi\right)\,dx \tag 6\\\\
&=2\pi\int_{-1}^1 e^{\sqrt{2}x}\,dx \tag 7
\end{align}$$
In going from $(1)$ to $(2)$, we used the identity $\cos(\phi)-\sin(phi)=\sqrt{2}\cos(\phi+\pi/4)$ and exploited the $2\pi$-periodicity of the integrand with respect to $\phi$.
In going from $(2)$ to $(3)$, we noted that the integral is the closed-surface integral over the unit sphere.
In arriving at $(4)$, we exploited symmetry of the integration over upper and lower hemispheres, translated to Cartesian coordinates, and used $dS =\frac{1}{\sqrt{1-x^2-y^2}}\,dx\,dy$.
In going from $(4)$ to $(5)$, we began carrying out the inner integral on $y$.
In arriving at $(6)$, we completed the evaluation of the inner integral on $y$, which yielded $\pi$.
And finally in $(7)$, we simply factor the $\pi$ from the inner integral.
A: I will evaluate this integral indirectly by recognizing it for what it is: an integral over the surface of a unit sphere.
First of all, for simplicity sake, I am going to replace your notation so it looks like we are integrating over solid angle:
$$\int_0^{2 \pi} d\phi \, \int_0^{\pi} d\theta \, \sin{\theta} \, e^{\sin{\theta} (\cos{\phi}-\sin{\phi})} $$
Now, note that we can replace the $\cos{\phi}-\sin{\phi}$ in the exponential with simply $\sqrt{2} \cos{\phi}$, as we are integrating over the whole azimuth anyway.  Thus, the integral is
$$\int_0^{2 \pi} d\phi \, \int_0^{\pi} d\theta \, \sin{\theta} \, e^{\sqrt{2} \sin{\theta} \cos{\phi}} $$
Now, note that on the unit sphere, $\sin{\theta} \cos{\phi} = x$, the $x$-coordinate.  Thus, by simple geometry, we can replace this double integral with a single integral over $x$ by noting that the function we are integrating over only depends on $x$.  The contribution to the integral from a ring a distance $x$ from the plane $x=0$ is $2 \pi \sqrt{1-x^2} e^{\sqrt{2} x} ds$, where $ds$ is an element of arc length.  Of course, over a cross-section of the sphere, $y=\pm \sqrt{1-x^2}$ and
$$ds = \sqrt{1+\left ( \frac{dy}{dx} \right )^2} dx = \frac{dx}{\sqrt{1-x^2}}$$
Thus the integral is simply
$$2 \pi \int_{-1}^1 dx \, e^{\sqrt{2} x} $$
which produces the sought-after result.
