Trying to integrate $\iint_D \frac{\text{d}x\text{d}y}{x^2+y^2}$ I'm trying to integrate this:
$$\iint_D \frac{\text{d}x\text{d}y}{x^2+y^2}$$ on $D=\{(x,y)\ |\ 1\le e^x+e^y, e^{2x}+e^{2y}\le 1\}$
After trying some substitution, I can't find a resonable way to integrate it. I know it's silly, but I don't know how to get further ? …
 A: This is a very cleverly manufactured integral so that it has a nice answer.
Following standard procedure, we solve for the constraints in terms of x and we obtain that:
$$D=\{(x,y)|y\geq y_1(x)=\ln(1-e^x)~~,~~ y\leq y_2(x)=\frac{1}{2}\ln(1-e^{2x})~~,~~x<0\}$$
One easily checks that for all $x<0$, $y_1(x)\leq y_2(x)$ and the integral can be manipulated into the form
$$I=\int\limits_{-\infty}^0 dx\int\limits_{y_1(x)}^{y_2(x)}\frac{dy}{x^2+y^2}=\int\limits_{-\infty}^0 \frac{dx}{x}\big[\arctan\frac{\ln(1-e^{2x})}{2x}-\arctan\frac{\ln(1-e^{x})}{x}\big]:=\int\limits_{-\infty}^0 \frac{f(2x)-f(x)}{x}dx$$
However, for sufficiently well behaved functions integrals of the above form can be easily evaluated as follows:
$$I(a,b)=\int\limits_{-\infty}^0 dx\frac{f(ax)-f(bx)}{x}=(f(0)-f(-\infty))\ln\big(\frac{a}{b}\big)~~,~~a,b>0$$
(This can be easily proven if one takes partial derivatives under the integral sign for both a and b and then integrating back).
Thus the integral we are interested in is a special case of the above for $a=2,b=1, f(x)=\arctan\frac{\ln(1-e^{x})}{x}$ and we finally obtain
$$I=\frac{\pi}{2}\ln2$$
