Basel Problem

Theorem
Here's a proof of the well known infinite series $$\zeta(2) = \sum_{n\,=\,1}^{\infty }{\frac{1}{n^{2}}}=\frac{\pi ^{2}}{6}.$$

Proof
Note that $$\frac{1}{n^{2}}=\int_{0}^{1}\!{\int_{0}^{1}{(xy)^{n-1}\,dx}\,dy}.$$ Hence $$\sum_{n\,=\,1}^{\infty }{\frac{1}{n^{2}}}=\sum_{n\,=\,1}^{\infty }{\int_{0}^{1}\!{\int_{0}^{1}{(xy)^{n-1}\,dx}\,dy}}.$$

By the Monotone Convergence Theorem we have $$\sum_{n\,=\,1}^{\infty }{\frac{1}{n^{2}}}=\int_{0}^{1}\!{\int_{0}^{1}{\left\{ \sum_{n\,=\,1}^{\infty }{(xy)^{n-1}} \right\}\,dx}\,dy}=\int_{0}^{1}\!{\int_{0}^{1}{\frac{dx\,dy}{1-xy}}}.$$

Let $$(u,v)=\left( \frac{x+y}{2},\frac{y-x}{2} \right)$$ so that $$(x,y)=(u-v,\,u+v),$$ therefore $$\zeta (2)=2\iint_{S}{\frac{du\,dv}{1-u^{2}+v^{2}}},$$ where $$S$$ is the square defined by the coordinates $$(0,0),\,\left( \frac{1}{2},-\frac{1}{2} \right),\,(1,0),\,\left( \frac{1}{2},\frac{1}{2} \right).$$

Exploiting the symmetry of the square we have: $$\zeta (2)=4\left( \int_{0}^{\frac{1}{2}}\!{\int_{0}^{u}{\frac{dv\,du}{1-u^{2}+v^{2}}}}+\int_{\frac{1}{2}}^{1}\!{\int_{0}^{1-u}{\frac{dv\,du}{1-u^{2}+v^{2}}}} \right),$$ and after some straightforward calculations we get

$$\zeta (2)=4\left\{ \int_{0}^{\frac{1}{2}}{\frac{\arcsin u}{\sqrt{1-u^{2}}}\,du}+\int_{\frac{1}{2}}^{1}{\frac{1}{\sqrt{1-u^{2}}}\left( \frac{\pi }{4}-\frac{\arcsin u}{2} \right)\,du} \right\}=4\left\{ \frac{\pi ^{2}}{72}+\frac{\pi ^{2}}{36} \right\}=\frac{\pi ^{2}}{6}.$$

Q.E.D.