# 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,

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

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

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

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

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

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

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

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

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

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

# The Basel Problem, Double Integral Method I

In 1644, the Italian mathematician Pietro Mengoli (1625-1686) posed the question: What’s the value of the sum

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

First time, Leonhard Euler (1707-1783) in 1735 proved that above series converges to $\displaystyle\frac{\pi^2}{6}$. In this post you can see an easy proof by using double integral that published by Tom M. Apostol in 1983 in Mathematical Intelligencer. Apostol’s Proof: Note that

$\displaystyle{\frac{1}{n^2}}=\displaystyle\int_0^1\displaystyle\int_0^1x^{n-1}y^{n-1}dx\,dy$

and by the monotone convergence theorem we get

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}=\displaystyle\int_0^1\displaystyle\int_0^1\left(\sum_{n=1}^{\infty}(xy)^{n-1}\right)dx\,dy$

$=\displaystyle\int_0^1\displaystyle\int_0^1\frac{1}{1-xy}dx\,dy$

We change variables in this by putting $(u,v)=\left(\frac{x+y}{2},\frac{y-x}{2}\right)$, so that $(x,y)=(u-v,u+v)$. Hence

$\zeta$$(2)$ $=2\displaystyle\iint_S\frac{du\,dv}{1-u^2+v^2}$

where $S$ is the square with vertices $(0,0), \left(\frac{1}{2},\frac{-1}{2}\right), (1,0)$ and $(\frac{1}{2},\frac{1}{2})$.

Exploiting the symmetry of the square we get

$\zeta$$(2)$ $=4\displaystyle\int_0^{\frac{1}{2}}\displaystyle\int_0^u\frac{dv\,du}{1-u^2+v^2}+4\displaystyle\int_{\frac{1}{2}}^1\displaystyle\int_0^{1-u}\frac{dv\,du}{1-u^2+v^2}$

$=4\displaystyle\int_0^{\frac{1}{2}}\frac{\tan^{-1}\left(\frac{u}{\sqrt{1-u^2}}\right)}{\sqrt{1-u^2}}du+4\displaystyle\int_{\frac{1}{2}}^1\frac{\tan^{-1}\left(\frac{1-u}{\sqrt{1-u^2}}\right)}{\sqrt{1-u^2}}du$

Now $\tan^{-1}\left(\frac{u}{\sqrt{1-u^2}}\right)=\sin^{-1}(u)$, and if $\alpha=\tan^{-1}\left(\frac{1-u}{\sqrt{1-u^2}}\right)$ then $\tan^2(\alpha)=\frac{1-u}{1+u}$ and $\sec^2(\alpha)=\frac{2}{1+u}$.

It follows that $u=2\cos^2(\alpha)-1=\cos(2\alpha)$ and so $\alpha=\frac{1}{2}\cos^{-1}(u)=\frac{\pi}{4}-\frac{\sin^{-1}(u)}{2}$. Hence

$\zeta$$(2)$ $=4\displaystyle\int_0^{\frac{1}{2}}\frac{\sin^{-1}(u)}{\sqrt{1-u^2}}du+4\displaystyle\int_{\frac{1}{2}}^1\frac{\frac{\pi}{4}-\frac{\sin^{-1}(u)}{2}}{\sqrt{1-u^2}}du$

$\displaystyle=\left[2(\sin^{-1}(u))^2\right]_0^{\frac{1}{2}}+\left[\pi\sin^{-1}(u)-(\sin^{-1}(u))^2\right]_{\frac{1}{2}}^1$

$\displaystyle=\frac{\pi^2}{18}+\frac{\pi^2}{2}-\frac{\pi^2}{4}-\frac{\pi^2}{6}+\frac{\pi^2}{36}=$ $\displaystyle\frac{\pi^2}{6}.$