How to apply a change of variables in $\int_0^1\int_0^1 \frac{\mathrm{d}x\mathrm{d}y}{1-xy}$?

Question

Denote $\mathcal{D}=\{(x,y)\in \mathbb{R}^2: 0\le x\le 1 ;\ 0 \le y\le 1 \}$.

I have to use : $x=\cos{\alpha} - t$ and $y=\cos{\alpha}+ t$.

I noticed that $\frac{x+y}{2}=\cos{\alpha}$ and $\frac{y-x}{2}=t$. So I can take $\alpha \in [0, \pi/2]$ and $t\in[-\min\{\ \cos(\alpha) \ ; 1-\cos(\alpha)\ \}, +\min\{\ \cos(\alpha)\ ; 1-\cos(\alpha)\ \}]$.

And now I do not know how to continue. I'm maybe wrong for the $\mathcal{C}^{1}$-diffeomorphism...

Thanks in advance !

Answer 1

This doesn't directly answer your question, but I would solve the integral as follows:

$$\int_0^1\int_0^1 \frac{\mathrm{d}x\mathrm{d}y}{1-xy} = \int_0^1 \int_0^1 \sum_{n=0}^{\infty}(xy)^n~\mathrm{d}x\mathrm{d}y = \sum_{n=0}^{\infty}\int_{0}^{1}\frac{y^n}{n+1}~dy = \sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}.$$