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

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

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

# A Beautiful Convergent Series

In this post I want to find the value of the sum

$\displaystyle\sum_{n=1}^{\infty}\frac{3^n-1}{4^n} \zeta \left(n+1 \right)$

Note that for $s>1$, $\zeta(s)=\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^s}$.

I use following formula that is called Gregory-Leibniz-Madhava’s series

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{4n-3}-\frac{1}{4n-1}=\frac{\pi}{4}$

If we define $f$ function as follows:

$\displaystyle{f\left(z\right)=\frac{z}{4-3z}-\frac{z}{4-z}}$

then

$\displaystyle{f(z)=\frac{\frac{z}{4}}{1-\frac{3z}{4}}-\frac{\frac{z}{4}}{1-\frac{z}{4}}=\frac{z}{4}\sum_{n=0}^{\infty}\left(\frac{3z}{4}\right)^n-\left(\frac{z}{4}\right)^n=\frac{z}{4}\sum_{n=1}^{\infty}\left(\frac{3z}{4}\right)^n-\left(\frac{z}{4}\right)^n}$

$\displaystyle{=\frac{z}{4}\sum_{n=2}^{\infty}\left(\frac{3z}{4}\right)^{n-1}-\left(\frac{z}{4}\right)^{n-1}=\frac{1}{4}\sum_{n=2}^{\infty}\left[\left(\frac{3}{4}\right)^{n-1}-\left(\frac{1}{4}\right)^{n-1}\right]z^n=\frac{1}{4}\sum_{n=2}^{\infty}\frac{3^{n-1}-1}{4^{n-1}}z^n}.$

Now I use this theorem

THEOREM (FlajoletVardi): If $f\left(z \right)=\displaystyle\sum_{n=2}^{\infty}a_{n}z^n$ and $\displaystyle\sum_{n=2}^{\infty}|a_n|$ converges then,

$\displaystyle\sum_{n=1}^{\infty}f\left(\frac{1}{n}\right)=\displaystyle\sum_{n=2}^{\infty}a_n$$\zeta$$\left(n\right)$.

PROOF:

Because $\displaystyle\sum_{m=2}^{\infty}|a_m|<\infty$,

$\displaystyle\sum_{n=1}^{\infty}\displaystyle\sum_{m=2}^{\infty}|a_m|\frac{1}{n^m}\leq\displaystyle\sum_{n=1}^{\infty}\displaystyle\sum_{m=2}^{\infty}|a_m|\frac{1}{n^2}<\infty$

Hence, by Cauchy’s double series theorem, we can switch the order of summation:

$\displaystyle{\sum_{n=1}^{\infty}f\left(\frac{1}{n}\right)=\sum_{n=1}^{\infty}\sum_{m=2}^{\infty}a_m\frac{1}{n^m}=\sum_{m=2}^{\infty}a_m\sum_{n=1}^{\infty}\frac{1}{n^m}=\sum_{n=2}^{\infty}a_n\zeta(n)}$

This theorem implies that

$\displaystyle{\frac{\pi}{4}=\sum_{n=1}^{\infty}f\left(\frac{1}{n}\right)=\frac{1}{4}\sum_{n=2}^{\infty}\frac{3^{n-1}-1}{4^{n-1}}\zeta(n)=\frac{1}{4}\sum_{n=1}^{\infty}\frac{3^{n}-1}{4^{n}}\zeta(n+1)}$

and

$\boxed{\displaystyle\sum_{n=1}^{\infty}\frac{3^n-1}{4^n} \zeta \left(n+1 \right)=\pi}$

# Two Properties in the Elementary Set Theory

My first post is about two properties in the elementary set theory. You can see an amazing proof for them from me in the following.

$(A\cup C)\cap(B\cup C^c)\subseteq A\cup B$

$A\cap B\subseteq(A\cap C)\cup(B\cap C^c)$

In proof I used the following property in logic

$(P\Rightarrow Q)\land(R\Rightarrow S)$ $\vdash$ $(P\land R)\Rightarrow( Q\land S)$

that is called “constructive dilemma”.

First Property:

Let $x\in(A\cup C)\cap(B\cup C^c)$, then I can write

$(x\in A\lor x\in C)\land( x\in B\lor x\in C^c)$

$(x\in A^c\Rightarrow x\in C)\land(x\in B^c\Rightarrow x\in C^c)$

$(x\in A^c\land x\in B^c)\Rightarrow( x\in C\land x\in C^c)$

$\lnot(x\in A^c\land x\in B^c)\lor(x\in C\land x\in C^c)$

$(x\in A\lor x\in B)\lor(x\in C\land x\in C^c)$

which implies that $x\in A\cup B$.

Second Property:

Let $x\in A\cap B$, then

$(x\in A\land x\in B)\land\lnot(x\in C^c\land x\in C)$

$\lnot$ $\left((x\in A\land x\in B)\Rightarrow( x\in C^c\land x\in C)\right)$

$\lnot$ $\left((x\in A\Rightarrow x\in C^c)\land(x\in B\Rightarrow x\in C)\right)$

$\lnot\left((x\in A^c\lor x\in C^c)\land(x\in B^c\lor x\in C)\right)$

$(x\in A\land x\in C)\lor(x\in B\land x\in C^c)$

which implies that $x\in (A\cap C)\cup(B\cap C^c)$.