# The Basel Problem, Double Integral Method II

One another way that can evaluate $\zeta(2)$ by double integral is to write $\zeta(2)$ as follows

$\displaystyle\zeta(2)=\frac{4}{3}\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}$

This is true, since we have

$\displaystyle{\zeta(2)=\sum_{n=1}^{\infty}\frac{1}{n^2}=\sum_{n=1}^{\infty}\frac{1}{(2n)^2}+\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}=\frac{1}{4}\sum_{n=1}^{\infty}\frac{1}{n^2}+\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}}.$

Now integrating by parts yields

$\displaystyle{\int_0^1(-x^{2n}\ln(x))dx=\left[-\frac{x^{2n+1}}{2n+1}\ln(x)\right]_0^1+\int_0^1\frac{x^{2n}}{2n+1}dx=\frac{1}{(2n+1)^2}}$

and by using the monotone convergence theorem, I can write

$\displaystyle{\zeta(2)=\frac{4}{3}\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}=\frac{4}{3}\sum_{n=0}^{\infty}\int_0^1(-x^{2n}\ln(x))dx=\frac{4}{3}\int_0^1\frac{\ln(x)}{x^2-1}dx}$

Hence,

$\displaystyle\zeta(2)=\frac{4}{3}\int_0^1\frac{\ln(x)}{x^2-1}dx$

$\displaystyle=\frac{4}{3}\times\frac{1}{2}\left(\int_0^1\frac{\ln(u)}{u^2-1}du+\int_0^1\frac{\ln(v)}{v^2-1}dv\right)$

$\displaystyle=\frac{2}{3}\left(\int_0^1\frac{\ln(u)}{u^2-1}du+\int_1^{\infty}\frac{\ln(v)}{v^2-1}dv\right)$

$\displaystyle=\frac{2}{3}\left(\int_0^1\frac{\ln(u)}{u^2-1}du+\int_0^{\infty}\frac{\ln(u)}{u^2-1}du-\int_0^{1}\frac{\ln(u)}{u^2-1}du\right)$

$\displaystyle=\frac{2}{3}\int_0^{\infty}\frac{\ln(u)}{u^2-1}du=\frac{2}{3}\times\frac{1}{2}\int_0^{\infty}\frac{\ln(u^2)}{u^2-1}du$

$\displaystyle=\frac{1}{3}\int_0^{\infty}\frac{1}{1-u^2}\ln(\frac{1}{u^2})du=\frac{1}{3}\int_0^{\infty}\frac{1}{1-u^2}\left[\ln\left(\frac{1+v}{1+u^2v}\right)\right]_{v=0}^{v=\infty}du$

$\displaystyle=\frac{1}{3}\int_0^{\infty}\frac{1}{1-u^2}\left(\int_0^{\infty}\left(\frac{1}{1+v}-\frac{u^2}{1+u^2v}\right)dv\right)du$

$\displaystyle{=\frac{1}{3}\int_0^{\infty}\int_0^{\infty}\frac{1}{(1+v)(1+u^2v)}dv\,du=\frac{1}{3}\int_0^{\infty}\frac{1}{1+v}\int_0^{\infty}\frac{1}{1+u^2v}du\,dv}$

$\displaystyle{=\frac{1}{3}\int_0^{\infty}\left(\frac{1}{1+v}\left[\frac{\arctan(\sqrt{v}u)}{\sqrt{v}}\right]_{u=0}^{u=\infty}\right)dv=\frac{1}{3}\times\frac{\pi}{2}\int_0^{\infty}\frac{1}{\sqrt{v}(1+v)}dv}$

$\displaystyle{=\frac{\pi}{3}\int_0^{\infty}\frac{1}{1+w^2}dw=\frac{\pi}{3}\left[\arctan(w)\right]_0^{\infty}=\frac{\pi}{3}\times\frac{\pi}{2}=}$ $\displaystyle{\frac{\pi^2}{6}}.$