TITLE: Show that $\int_{[0,1]\times [0,1]}\frac{dx \, dy}{1-xy} = \frac{\pi^2}{6}$
QUESTION [15 upvotes]: $$
\int_{[0,1] \times [0,1]} \frac{dx \, dy}{1-xy}=\frac{\pi^2}{6},
$$
there is a hint for substitution given but I can't seem to get anywhere with it.
$$
\begin{cases}
x = \frac{u-v}{\sqrt{2}} \\
y= \frac{u+v}{\sqrt{2}}
\end{cases}
$$
and evaluate by letting $u = \sqrt{2}\sin t$. There should also encounter the term $(1-\sin t)/\cos t$ (Which I couldn't obtain)
I have managed to compute the jacobian of transformation, which is exactly 1, and did the substitution for u and v, obtaining:
$$
\int_{[0,\sqrt{2}]\times [-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}]} \frac{2}{2-u^2 +v^2} \,du\, dv
$$
doing the subtitution for $u$, we get:
$$
\int \frac{2\sqrt{2}\cos t}{2\cos^2 t+v^2}\,dt\, dv
$$
It is then not obvious how to integrate with respect to $t$ now, so change order of integration and integrate with respect to v.
which will be equal to $\int 4\arctan \frac{1}{2\cos t} \, dt$
Now this is a rather complicated integral, how to proceed from here? or is there an alternative method which I overlooked in the previous steps?
Could I have some insight on this problem please??

REPLY [4 votes]: $$\int_0^1 \int_0^1 \frac{1}{1-xy} \, dy \, dx = \int_0^1 \frac{\ln(1-x)}{-x} \, dx$$
since $\ln(1-x)=-\sum_{n=1}^\infty \frac{x^n}{n}$, then $$\int_0^1 \frac{\ln(1-x)}{-x} \, dy \, dx=\int_0^1 \sum_{n=1}^\infty \frac{x^{n-1}}{n} \, dx = \sum_{n=1}^\infty \int_0^1 \frac{x^{n-1}}{n} \, dx = \sum_{n=1}^\infty\frac{1}{n^2}=\cdots.$$