<s>[INST] How to derive $\int_0^1 \int_0^1 \frac{1}{1-xy} \,dy\,dx = \sum_{n=1}^{\infty}\frac{1}{n^{2}}$ Looking at an proof of $\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{\pi^{2}}{6}$ that argues that $\int_0^1 \int_0^1 \frac{1}{1-xy} \,dy\,dx = \sum_{n=1}^{\infty}\frac{1}{n^{2}}$ but I can't see how this is accomplished since the integral is generalized (it is a limit it the endpoint) and also if using an geometric sum to derive this it is divergent in this endpoint (value $=1$). How is this solved in a rigorous proof?
 [/INST] So there is a post in @Robert Z's link in the comments that proves
$$ \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} $$ 
So I won't restate what has already been done, but will just show that 
$$ \int_0^{1} \int_0^1 \frac{1}{1 - xy} dx dy  = \sum_{n=1}^{\infty} \frac{1}{n^2} $$ 
Observe that 
$$ 1 + x + x^2 + x^3 ... = \frac{1}{1-x}$$
Thus 
$$ \int_0^x \frac{1}{1-x'} dx'  =  x + \frac{1}{2} x^2 + \frac{1}{3}x^3 ... $$
Thus:
$$ - \ln(1 - x) = x  +\frac{1}{2}x^2 + \frac{1}{3}x^3 ... $$
Moving another step, observe that if we divide by $x$ and integrate AGAIN,
Our terms of the form $$\frac{1}{n} x^n \rightarrow_{divide} \frac{1}{n} x^{n-1} \rightarrow_{integrate} \frac{1}{n} \frac{1}{n} x^n = \frac{1}{n^2} x^n $$
So it then follows 
$$ \int_{0}^{x} - \frac{\ln(1-x')}{x'} dx' = x + \frac{1}{4}x^2 + \frac{1}{9} x^3 + ...  = \sum_{n=1}^{\infty} \frac{x^n}{n^2} $$
So now suppose we integrate to $x=1$ we then yield:
$$ \int_{0}^{1} - \frac{\ln(1 - x')}{x'} dx' = 1 + \frac{1}{4} + \frac{1}{9} ... = \sum_{n=1}^{\infty} \frac{1}{n^2} $$
So now the final punch. Observe (treating $x$ constant)
$$ \int_0^1 \int_0^1 \frac{1}{1 - xy} dx dy = -\int_0^1 \frac{\ln(1-x)}{x} dx $$ 
So we now have the desired equivalence.
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