Proof that $\int_{0}^{1} \int_{0}^{1}\frac{1}{1-xy} \,dx\,dy = \sum_{n=1}^{\infty} \frac{1}{n^{2}}$. 
Proof that $\int_{0}^{1} \int_{0}^{1}\frac{1}{1-xy} \,dx\,dy =
 \sum_{n=1}^{\infty} \frac{1}{n^{2}}$.

From basic knowledge:
We have $\sum_{n=1}^{\infty} \frac{1}{n^{2}} = \frac{\pi}{6}$
My wrong solution: 
I tried consider special equation 
$$u = 1-xy$$ 
$$du = -ydx$$
$$\frac{du}{-y} = dx$$
So, 
$$ \int_{0}^{1} \int_{0}^{1} \frac{1}{-uy} du dy $$ $$\int_{0}^{1} \frac{1}{-y}  dy \int_{1}^{0} \frac{1}{u} du$$
$$ \log({-y})\Bigm|_0^1 \log(u)\Bigm|_1^0$$
Result: (integral does not converge)
Where is mistake?
 A: You made a mistake when you substituted the boundaries (should be $1-y$ instead of $0$). But correcting this will not help you much because you will get an integral that is not elementary solvable. However, I think the task is to show the equality
$$\int_{0}^{1} \int_{0}^{1}\frac{1}{1-xy} \,dx\,dy = \sum_{n=1}^{\infty} \frac{1}{n^{2}}$$
and not 
$$\int_{0}^{1} \int_{0}^{1}\frac{1}{1-xy} \,dx\,dy = \frac{\pi^2}{6}.$$
Therefore I suggest that you expand $\frac{1}{1-xy}$ as a geometric series. The solution will be immediate.
A: Note that
$$\begin{align}\int_0^1 \int_0^1 \frac{1}{1 - xy} \, dx \, dy &= \int_0^1 \int_0^1 \lim_{n \to \infty} \sum_{k=0}^n (xy)^k \, dx \, dy \\ &= \lim_{n \to \infty}\int_0^1 \int_0^1\sum_{k=0}^n (xy)^k \, dx \, dy  \\ &= \lim_{n \to \infty} \sum_{k=0}^n \int_0^1 \int_0^1 x^k y^k \, dx \, dy \\ &= \sum_{k=0}^\infty \frac{1}{(k+1)^2}\end{align}$$
Switching the limit and integral to obtain the second equality is justified by the monotone convergence theorem.
A: Different approach:
\begin{align}
I&=\int_0^1\int_0^1 \frac{1}{1-xy}\ dx\ dy\overset{xy=z}{=}\int_0^1\int_0^y\frac{1}{y(1-z)}\ dz \ dy=\int_0^1\frac1{1-z}\left(\int_z^1\frac1y\ dy\right)\ dz\\
&=\int_0^1\frac{-\ln z}{1-z}\ dz=\int_0^1\frac{-\ln(1-z)}{z}\ dz=\operatorname{Li}_2(y)|_0^1=\operatorname{Li}_2(1)=\zeta(2)
\end{align}
A: \begin{align}
I&=\int_0^1\left(\int_0^1 \frac{1}{1-xy}\ dx\right)\ dy=\int_0^1 \frac{-\ln(1-y)}{y}\ dy=\operatorname{Li}_2(y)|_0^1=\operatorname{Li}_2(1)=\zeta(2)
\end{align}
