Tom Mike Apostol was an American analytic number theorist and professor at the California Institute of Technology.  Note that $\int_{0}^{1} x^n dx= \left[ \frac{x^{n+1}}{n+1}\right]_{0}^1=\frac{1}{n+1} $ With this change of variables we change the region and limits of integration  and so $$I= \int \int \frac{1}{1-xy} dxdy = \int \int \frac{2}{2-u^2+v^2} dudv $$ If we look at the integral in uv-space it doesn't change if $u \rightarrow -u $ or $v \rightarrow -v $, and so the integral in the region above the u-axis will be equal to the integral below the u-axis.  $$ I = 2I_{A} = 2 \left( \int_0^{1/\sqrt{2}} \int_{0}^{u} \frac{2}{2-u^2+v^2}dvdu + \\ \int_{1/\sqrt{2}}^{\sqrt{2}} \int_{0}^{\sqrt{2}-u} \frac{2}{2-u^2+v^2}dvdu \right) $$ If you want to learn more about Apéry's proof click [here](https://someclassicalmaths.wordpress.com/2011/10/14/roger-aperys-proof-that-zeta3-is-irrational/) and for Beuker's original proof click [here](http://www.math.wvu.edu/~mays/545/Bull.%20London%20Math.%20Soc.-1979-Beukers-268-72.pdf). Alternatively we could note that $$ \frac{d}{du} \arctan \frac{u}{\sqrt{2-u^2}} = \frac{1}{\sqrt{2-u^2}} \\ \frac{d}{du} \arctan \frac{\sqrt{2}- u}{\sqrt{2-u^2}} =-\frac{1}{2}\frac{1}{\sqrt{2-u^2}} $$ and so the integral becomes $$ I= 4 \int_{0}^{1/\sqrt{2}}f(u)f'(u) du + 4 \int^{\sqrt{2}}_{1/\sqrt{2}}g(u)(-2)g'(u) du \\ = 4\left(\left[f(u)^2/2\right]_{0}^{1/\sqrt{2}}-2\left[g(u)^2/2\right]^{\sqrt{2}}_{1/\sqrt{2}}\right ) $$ Since $f(0)=g(\sqrt2)=0$ and $f(1/\sqrt{2})=g(1/\sqrt{2})=\pi/6$ $$ I = 2\pi^2/6^2+4 \pi^2/6^2= \pi^2/6 $$ The problem of determining the value of $\zeta (2)$ also known as the Basel problem was first posed by [Pietro Mengoli](https://en.wikipedia.org/wiki/Pietro_Mengoli) in 1644. It consisted of determining the value of the following series $$ \zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2} $$ A lot of famous mathematicians tried to attack this problem but was Euler at 28 in 1735 who proved for the first time that $\zeta(2) = \frac{\pi^2}{6}$ (the problem is named after Basel, hometown of Euler as well as of the Bernoulli family who unsuccessfully attacked the problem). To prove it Euler used the Taylor expansion of the sine function and also expressed it as a product of the roots of an infinite polynomial (which at the time was still not known to be valid!). In this paper Tom Apostol gives an alternative proof based on solving a double integral which avoids Euler's expansions of the sine function. The Basel problem and Euler's ideas were later used by Riemann in his seminal 1859 paper "On the Number of Primes Less Than a Given Magnitude", in which he defined his zeta function $$ \zeta (s) = \sum_{n=1}^{\infty} \frac{1}{n^s} $$ one of the most important functions in mathematics. Here the author uses the formula for the sum of a geometric series $$ s = \sum_{n=0}^{\infty} x^n = 1 + x +x^2 + ... $$ if we multiply both sides by $x$ we have $$ sx = x+ x^2 + x^3+ ... \\ sx = s -1 \\ s = \frac{1}{1-x}, \text{ for $|x|<1$} $$ And so $\sum_{n=0}^{\infty} (xy)^n= \frac{1}{1-xy}$