Ways to evaluate $\int_0^1 \int_0^1 \frac{1}{1-xy}dxdy = \frac{\pi^2}{6}$ A classmate told me this, but he didn't tell me how to evaluate the integral.
$$\int_0^1 \int_0^1 \frac{1}{1-xy}dxdy=\int_0^1 \int_0^1\sum_{n=0}^\infty (xy)^n dxdy=\sum_{n=1}^\infty \frac{1}{n^2}=\zeta(2)$$
So if you can evaluate that integral this might be an easy way to solve the Basel problem.
I tried substituting $\frac{1}{x}=v$ and $\frac{1}{y}=t$ to get $$\zeta(2)=\int_1^\infty\int_1^\infty\frac{1}{vt-1}-\frac{1}{vt}dvdt=\lim_{N\to\infty}\left(\int_1^N\int_1^N\frac{1}{vt-1}dvdt-\ln(N)^2\right)$$
The areas of the cross sections of $\frac{1}{vt-1}$ are going to be infinite near $(1,1)$ so this seems like a dead end.
 A: I don't see where is the problem..
Since the range of integration is between $0$ and $1$ you can make use of the geometric Series, which general form is
$$\frac{1}{1 - z} = \sum_{k = 0}^{+\infty} z^k ~~~~~~~ |z| < 1$$
Hence in your case $z = xy$ hence
$$\int_0^1 \int_0^1 \sum_{k = 0}^{+\infty} (xy)^k \ \text{d}x\ \text{d}y$$
Split with Fubini
$$\sum_{k = 0}^{+\infty} \int_0^1 y^k\ \text{d}y \int_0^1 x^k\ \text{d}x$$
Integrations are trivial
$$\int_0^1 x^k \ \text{d}x = \frac{1}{k+1}$$
The same for $y^k$
And finally you get
$$\sum_{k = 0}^{+\infty}\frac{1}{(k+1)^2}$$
Which can be rewritten as
$$\sum_{k = 1}^{+\infty}\frac{1}{k^2} = \zeta(2)$$
Which is the famous Riemann Zeta Function.
$$\zeta(2) = \frac{\pi^2}{6}$$
A: Directly (justify steps, in particular interchanging sum and integral)
$$\int_0^1\left(\int_0^1\frac{dx}{1-yx}\right)dy=\left.\int_0^1\left(-\frac1y\log(1-xy)\right)\right|_0^1dy=$$
$$\int_0^1-\frac1y\log(1-y)dy\stackrel{\text{Power series}}=\int_0^1\frac1y\left(\sum_{n=1}^\infty\frac{y^n}n\right)dy=$$
$$\int_0^1\sum_{n=1}^\infty\frac{y^{n-1}}n\;dy=\sum_{n=1}^\infty\frac1n\int_0^1y^{n-1}dy=\sum_{n=1}^\infty\frac1{n^2}$$
