I've read somewhere on this site that if you consider:
$$\int_0^1 \int_0^1 \frac{1}{1-xy} \,dy\,dx$$
Then using the power series, we have this is equal to $\sum_{n=1}^{\infty} \frac{1}{n^2}$ which I decided to try and was able to show.
Apparently we can show this is equal to $\frac{\pi^2}{6}$, and using what little I know about double integrals from a few khan academy videos (I haven't taken multivariable calculus yet), I tried to evaluate this double integral by techniques of single variable calculus $u=xy$..and I got this:
$$-\int_{0}^{1} \frac{\ln (1-x)}{x}dx$$
The usual way I would evaluate this is with a Taylor series, but that just that just leads us in circles.
So I want to know how can I evaluate this, so we can prove $\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6}$.
Here, we present an approach the relies on integration methodologies only, including transformation of coordinates. To that end we proceed.
We can transform coordinates by setting $x=s+t$ and $y=s-t$. Then, $dx\,dy\to 2\,ds\,dt$ and the transformed integral domain is the square-shaped region with vertices in the $s-t$ plane at $(0,0)$, $(1/2,1/2)$, $(1/2,-1/2)$, and $(1,0)$. Then, we can write
$$\begin{align} \int_0^1\int_0^1 \frac{1}{1-xy}\,dx\,dy&=\int_0^{1/2}\int_{-s}^{s}\frac{2}{(1-s^2)+t^2}\,dt\,ds+\int_{1/2}^{1}\int_{s-1}^{1-s}\frac{2}{(1-s^2)+t^2}\,dt\,ds\\\\ &=\int_0^{1/2}\frac{4}{\sqrt{1-s^2}}\arctan\left(\frac{s}{\sqrt{1-s^2}}\right)\,ds\\\\&+\int_{1/2}^{1}\frac{4}{\sqrt{1-s^2}}\arctan\left(\sqrt{\frac{1-s}{1+s}}\right)\,ds\\\\ &=4\int_0^{1/2}\frac{\arcsin(s)}{\sqrt{1-s^2}}\,ds+4\int_{1/2}^1\frac{\arccos(s)}{2\sqrt{1-s^2}}\,ds\\\\ &=2\arcsin^2(1/2)+\arccos^2(1/2)\\\\ &=2\left(\frac{\pi}{6}\right)^2+\left(\frac{\pi}{3}\right)^2\\\\ &=\frac{\pi^2}{6} \end{align}$$
as expected!