TITLE: 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 $
QUESTION [6 upvotes]: 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.

REPLY [4 votes]: 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.