TITLE: Double integral $\iint_D\arctan e^{xy}\,dy\,dx$
QUESTION [5 upvotes]: Evaluate the double integral:
$$\iint_D\arctan e^{xy}\,dy\,dx$$
where $D:\{(x,y)\in\mathbb{R}^2: x^2+y^2\leq 4x\}$
For thie integral , since our function does not have elementary anti-derivetives with respect to both variables, I tried to solve this by polar coordinate. But then it was getting worth and I got something very ugly and I cannot simplify, I stuck. Are there any tricks to deal with this?

REPLY [8 votes]: Hint:
$$\int_0^4 \int_{-\sqrt{4x-x^2}}^\sqrt{4x-x^2} \tan^{-1}(\exp(xy))\,dy\,dx=\int_0^4 \int_{-\sqrt{4x-x^2}}^0 \tan^{-1}(\exp(xy))\,dy\,dx+ \cdots $$
$$+\int_0^4 \int_0^\sqrt{4x-x^2} \tan^{-1}(\exp(xy))\,dy\,dx=$$
$$\int_0^4 \int_0^\sqrt{4x-x^2} \tan^{-1}(\exp(xy))\,dy\,dx +\int_0^4 \int_0^\sqrt{4x-x^2} \tan^{-1}(\exp(-xy))\,dy\,dx.$$
Use the identity $\tan^{-1}(c)+\tan^{-1}(\frac{1}{c})=\frac{\pi}{2}$ for $c>0$.
By the way, I tried polar coordinates too, and even Maple couldn't do the integral.
The key is to notice that the domain of integration is symmetric with respect to the $x$-axis and find a symmetry trick.