1.) Expanding the integrand as a geometric series, we may write:
$$\frac{1}{1-xy}=\sum_{n=0}^{\infty}(xy)^n$$
Hence the integral becomes:
$$\int_0^1\int_0^1 \sum_{n=0}^{\infty}(xy)^n\,dx\,dy=\int_0^1\left[\sum_{n=0}^{\infty}\frac{x^{n+1}y^n}{n+1} \right]_0^1\,dy=$$
$$\int_0^{1}\sum_{n=1}^{\infty}\frac{y^{n-1}}{n}\,dy=\left[\sum_{n=1}^{\infty}\frac{y^{n}}{n^2} \right]_0^1=\sum_{n=1}^{\infty}\frac{1}{n^2}$$
2.) Using the change of variables:
$$(x,y)=(u-v,u+v)$$
we obtain:
$$\sum_{n=1}^{\infty}\frac{1}{n^2}=\iint_{R}\frac{1}{1-u^2+v^2}\left|\frac{\partial (x,y)}{\partial (u,v)} \right|\,du\,dv$$
Calculating the Jacobian matrix, we find:
$$\frac{\partial (x,y)}{\partial (u,v)}=\begin{vmatrix}\frac{\partial x}{\partial u}&\frac{\partial x}{\partial v}\\\frac{\partial y}{\partial u}&\frac{\partial y}{\partial v}\\\end{vmatrix}=\begin{vmatrix}1&-1\\1&1\\\end{vmatrix}=1(1)-(-1)(1)=2$$
Thus, we have:
$$\sum_{n=1}^{\infty}\frac{1}{n^2}=2\iint_{R}\frac{1}{1-u^2+v^2}\,du\,dv$$
Remapping the boundaries in terms of the new variables, we find $R$ is a square in the $uv$-plane with vertices:
$$(0,0),\,\left(\frac{1}{2},-\frac{1}{2} \right),\,\left(\frac{1}{2},\frac{1}{2} \right),\,(1,0)$$
Reversing the order of integration and using the symmetry of the square, we obtain:
$$\sum_{n=1}^{\infty}\frac{1}{n^2}=4\left(\int_0^{ \frac{1}{2}}\int_0^u \frac{dv\,du}{1-u^2+v^2}+\int_{ \frac{1}{2}}^1\int_0^{1-u} \frac{dv\,du}{1-u^2+v^2} \right)$$
Next, we may compute:
$$\int_0^u\frac{dv}{1-u^2+v^2}=\left[\frac{1}{\sqrt{1-u^2}}\tan^{-1}\left(\frac{v}{\sqrt{1-u^2}} \right) \right]_0^u=\frac{1}{\sqrt{1-u^2}}\tan^{-1}\left(\frac{u}{\sqrt{1-u^2}} \right)=\frac{1}{\sqrt{1-u^2}}\sin^{-1}(u)$$
$$\int_0^{1-u}\frac{dv}{1-u^2+v^2}=\left[\frac{1}{\sqrt{1-u^2}}\tan^{-1}\left(\frac{v}{\sqrt{1-u^2}} \right) \right]_0^{1-u}=\frac{1}{\sqrt{1-u^2}}\tan^{-1}\left(\frac{1-u}{\sqrt{1-u^2}} \right)=\frac{1}{\sqrt{1-u^2}}\tan^{-1}\left(\sqrt{\frac{1-u}{1+u}} \right)$$
If we let:
$$\tan(\theta)=\sqrt{\frac{1-u}{1+u}}$$
Squaring, we obtain:
$$\tan^2(\theta)=\frac{1-u}{1+u}$$
Add through by 1:
$$\tan^2(\theta)+1=\frac{1-u}{1+u}+1$$
Apply a Pythagorean identity on the left and combine terms on the right:
$$\sec^2(\theta)=\frac{2}{1+u}$$
Invert both sides:
$$\cos^2(\theta)=\frac{u+1}{2}$$
Solving for $u$, we find:
$$u=2\cos^2(\theta)-1=\cos(2\theta)$$
Hence, we find:
$$\tan^{-1}\left(\sqrt{\frac{1-u}{1+u}} \right)=\theta=\frac{1}{2}\cos^{-1}(u)$$
Using the identity $$\sin^{-1}(u)+\cos^{-1}(u)=\frac{\pi}{2}$$ we finally have:
$$\tan^{-1}\left(\sqrt{\frac{1-u}{1+u}} \right)=\frac{1}{2}\left(\frac{\pi}{2}-\sin^{-1}(u) \right)$$
Utilizing these results, we now may state:
$$\sum_{n=1}^{\infty}\frac{1}{n^2}=4\left(\int_0^{ \frac{1}{2}}\frac{1}{\sqrt{1-u^2}}\sin^{-1}(u)\,du+\frac{1}{2}\int_{ \frac{1}{2}}^1 \frac{1}{\sqrt{1-u^2}}\left(\frac{\pi}{2}-\sin^{-1}(u) \right)\,du \right)$$
Now, let's look at the first integral:
$$\int_0^{ \frac{1}{2}}\frac{1}{\sqrt{1-u^2}}\sin^{-1}(u)\,du$$
Using the substitution:
$$\alpha=\sin^{-1}(u)\,\therefore\,d\alpha=\frac{1}{\sqrt{1-u^2}}\,du$$
we now have:
$$\int_0^{\frac{\pi}{6}}\alpha\,d\alpha=\frac{1}{2} \left[\alpha^2 \right]_0^{\frac{\pi}{6}}= \frac{1}{2}\left(\frac{\pi}{6} \right)^2=\frac{\pi^2}{72}$$
Next, let's break the second integral into two parts:
i) $$\frac{\pi}{4}\int_{\frac{1}{2}}^1 \frac{1}{\sqrt{1-u^2}}\,du=\frac{\pi}{4}\left[\sin^{-1}(u) \right]_{\frac{1}{2}}^1=\frac{\pi}{4}\left(\frac{\pi}{2}-\frac{\pi}{6} \right)=\frac{\pi^2}{12}$$
ii) $$-\frac{1}{2}\int_{\frac{1}{2}}^1 \frac{1}{\sqrt{1-u^2}}\sin^{-1}(u)\,du$$
Using the substitution:
$$\alpha=\sin^{-1}(u)\,\therefore\,d\alpha=\frac{1}{\sqrt{1-u^2}}\,du$$
we now have:
$$-\frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \alpha\,d\alpha=-\frac{1}{4}\left[\alpha^2 \right]_{\frac{\pi}{6}}^{\frac{\pi}{2}}=-\frac{1}{4}\left(\left(\frac{\pi}{2} \right)^2-\left(\frac{\pi}{6} \right)^2 \right)=-\frac{\pi^2}{18}$$
Thus, putting these results together, there results:
$$\sum_{n=1}^{\infty}\frac{1}{n^2}=4\left(\frac{\pi^2}{72}+\frac{\pi^2}{12}-\frac{\pi^2}{18} \right)=4\left(\frac{\pi^2}{24} \right)$$
And, we may then state:
$$\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$$
Shown as desired.
