One step in the proof that the series of $1/{n^2}$ equals ${\pi^2}/6$ that I don't understand I know questions about the specific series
$$\sum_{n=1}^{\infty} {1 \over {n^2}} = {{\pi^2} \over 6}$$
have been posted before, but there is one specific step in the proof that I don't understand I have yet to find a sufficient answer.
I had already started working out the exercise and was getting the same results as in the following link:
http://empslocal.ex.ac.uk/people/staff/rjchapma/etc/zeta2.pdf
We want to first show
$$\int_{(0,1)^2} {1 \over {1-xy}} d{\lambda^2}(x,y) = \sum_{n=1}^{\infty} {1 \over {n^2}}$$
by using a geometric series.
I defined
$${f_m}(x,y) :\doteq \lim_{m\to\infty} \sum_{n=0}^{m} {1 \over {(xy)^2}} = \lim_{m\to\infty} {{1 - {(xy)^{m+1}}} \over {1 - xy}}$$
This is a monotone increasing sequence of nonnegative functions that converges to $1 \over {1 - xy}$.
Thus, we can use the monotone convergence theorem to exchange the sum and integral; further, by using an appropriate index shift, Tonelli's theorem, and the linearity of integrals, we get:
$$\int_{(0,1)^2} {1 \over {1-xy}} d{\lambda^2}(x,y) = \lim_{m\to\infty} \sum_{n=1}^{m+1} \int_{0}^{1} \int_{0}^{1} {1 \over {(xy)^{n-1}}} dx dy$$
I can easily show that if $n = 1$ then the above double integral is equal to $1$, and if $n \geq 3$, the double integral is equal to $1 \over {n^2}$.
My problem arises by $n=2$. In this case we have
$$\int_{0}^{1} \int_{0}^{1} {1 \over xy} dx dy$$
which clearly does not exist. Am I just seeing incorrectly? Is there something obvious that I'm missing?  How can I justify this?
I appreciate any help!
 A: This is not one step proof but you may be like it
Since $\int_0^1 \frac{dx}{1+x^2}=\frac{\pi}{4}$, we have
$$\frac{\pi^2}{16}=\int_0^1\int_0^1\frac{dydx}{(1+x^2)(1+y^2)}\overset{t=xy}{=}\int_0^1\int_0^x\frac{dtdx}{x(1+x^2)(1+t^2/x^2)}$$
$$=\frac12\int_0^1\int_t^1\frac{dxdt}{x(1+x^2)(1+t^2/x^2)}\overset{x^2\to x}{=}\frac12\int_0^1\left(\int_{t^2}^1\frac{dx}{(1+x)(x+t^2)}\right)dt$$
$$=-\frac12\int_0^1\frac{\ln\left(\frac{4t^2}{(1+t^2)^2}\right)}{1-t^2}dt\overset{t=\frac{1-x}{1+x}}{=}-\frac12\int_0^1\frac{\ln\left(\frac{1-x^2}{1+x^2}\right)}{x}dx$$
$$\overset{x^2\to x}{=}-\frac14\int_0^1\frac{\ln\left(\frac{1-x}{1+x}\right)}{x}dx=-\frac14\int_0^1\frac{\ln\left(\frac{(1-x)^2}{1-x^2}\right)}{x}dx$$
$$=-\frac12\int_0^1\frac{\ln(1-x)}{x}dx+\frac14\underbrace{\int_0^1\frac{\ln(1-x^2)}{x}dx}_{x^2\to x}$$
$$=-\frac38\int_0^1\frac{\ln(1-x)}{x}dx\Longrightarrow \int_0^1\frac{-\ln(1-x)}{x}dx=\frac{\pi^2}{6}$$

Remark:
This solution can be considered a proof that $\zeta(2)=\frac{\pi^2}{6}$ as we have $\int_0^1\frac{-\ln(1-x)}{x}dx=\text{Li}_2(x)|_0^1=\text{Li}_2(1)=\zeta(2)$
