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 II

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. apostol2013Apostol’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}.

The Basel Problem, Double Integral Method I