Evaluate:
$$ \int_{0}^{1} \int_{0}^{1} \frac{1}{1 - (xy)^2}\, dx \, dy $$
Can someone provide a hint to tackle this problem? I only know how to do substitution and integration by parts for one variable. My difficulty so far is to express the term $ \dfrac{1}{1 - (xy)^2} $ in term of $ x $ treating $ y $ as a constant so that I can use substitution.
$$\iint_{(0,1)^2}\frac{dx\,dy}{1-(xy)^2}=\iint_{(0,1)^2}\sum_{n\geq 0}x^{2n}y^{2n}\,dx\,dy =\sum_{n\geq 0}\frac{1}{(2n+1)^2}=\color{red}{\frac{\pi^2}{8}}.$$
The section “Three times $\pi^2/6$” in Proofs from THE BOOK by Aigner & Ziegler shows how to evaluate this integral using the completely non-obvious substitution $$ x = \frac{\sin u}{\cos v} ,\qquad y = \frac{\sin v}{\cos u} , $$ which magically turns the integral into $$ \iint_E dudv , $$ where $E$ is the triangle with corners in $(u,v)=(0,0)$, $(\pi/2,0)$ and $(0,\pi/2)$.
This is a special case of the evaluation of $\sum_{k\ge 0} (-1)^{nk} (2k+1)^{-n}$ in a paper by Beukers, Calabi and Kolk.
(See also this question.)
A much less elegant solution.
Consider $$I=\int\frac{dx}{1 - (xy)^2} $$ and change variable $$xy=t\implies x=\frac ty\implies dx=\frac{dt}{y}$$ This make $$I=\frac 1y\int\frac{dt}{1 - t^2}= \frac {\tanh ^{-1}(t)}y=\frac {\tanh ^{-1}(xy)}y\implies \int_0^1\frac{dx}{1 - (xy)^2}=\frac {\tanh ^{-1}(y)}y$$ Now, using $$\tanh ^{-1}(y)=\sum_{n=0}^\infty \frac {y^{2n+1}}{2n+1}\implies \frac{\tanh ^{-1}(y)}y=\sum_{n=0}^\infty \frac {y^{2n}}{2n+1}$$ which makes $$J=\int \frac{\tanh ^{-1}(y)}y \,dy=\sum_{n=0}^\infty \frac {y^{2n+1}}{(2n+1)^2}$$ $$\int_0^1 \frac{\tanh ^{-1}(y)}y \,dy=\sum_{n=0}^\infty \frac {1}{(2n+1)^2}=\frac{\pi ^2}{8}$$
