A silly little derivation of ζ(2)

(This is a cleaned-up and somewhat expanded version of this Twitter thread.)

What follows is a silly little proof that

$$\zeta (2)=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}=\frac{{\pi }^2}{6}$$

where $\zeta$ is the Riemann zeta function.

Consider the integral

$$I:={\int }_0^1\frac{\log (1-x+x^2)}{x(x-1)}dx.$$

We have, by using partial fractions and performing some other algebraic manipulations,

$$\begin{array}{rl}I& =-{\int }_0^1\frac{\log (1-x+x^2)}{x}dx-{\int }_0^1\frac{\log (1-x+x^2)}{1-x}dx\\ & =-2{\int }_0^1\frac{\log (1-x+x^2)}{x}& (x\mapsto 1-x)\\ & =2({\int }_0^1\frac{\log (1+x)}{x}dx-{\int }_0^1\frac{\log (1+x^3)}{x}dx)\\ & =\frac{4}{3}{\int }_0^1\frac{\log (1+x)}{x}dx& (x\mapsto x^{1/3}).\end{array}$$

To evaluate this integral, we take the Maclaurin series:

$${\int }_0^1\frac{\log (1+x)}{x}dx={\int }_0^1\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^n{x}^{n-1}}{n}dx$$

Since for all positive integers $N$,

$$\left|\sum _{n=1}^N\frac{(-1{)}^n{x}^{n-1}}{n}\right|\le \sum _{n=1}^N\frac{x^{n-1}}{n}\le \sum _{n=1}^{\mathrm{\infty }}\frac{x^{n-1}}{n}=-\frac{\log (1-x)}{x}$$

on $[0,1)$ and

$$-{\int }_0^1\frac{\log (1-x)}{x}dx<-{\int }_0^{1/2}\frac{\log (1-x)}{x}dx-2{\int }_{1/2}^1\log (1-x)dx<\mathrm{\infty },$$

we can invoke the dominated convergence theorem to switch summation and limit processes. We then have

$$\begin{array}{rl}{\int }_0^1\frac{\log (1+x)}{x}dx& =\sum _{n=1}^{\mathrm{\infty }}{\int }_0^1\frac{(-x{)}^{n-1}}{n}dx\\ & =\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n-1}}{n^2}=\eta (2)\end{array}$$

where $\eta (s)$ is the Dirichlet eta function.

Now, we have $\eta (s)=2^{1-s}\zeta (s)$; for $s>1$, this can be seen by separating and rearranging all even terms in the summation. Thus, we have

$$I=\frac{4}{3}\eta (2)=\frac{2}{3}\zeta (2).$$

We look at another way to evaluate $I$. Noticing that $1-x+x^2=1-x(1-x)$, we can write the integrand as a power series in $x(1-x)$:

$$\begin{array}{rl}I& =-{\int }_0^1\frac{\log (1-x(1-x))}{x(1-x)}dx\\ & ={\int }_0^1\sum _{n=1}^{\mathrm{\infty }}\frac{x^{n-1}(1-x{)}^{n-1}}{n}dx\\ & =\sum _{n=1}^{\mathrm{\infty }}{\int }_0^1\frac{x^{n-1}(1-x{)}^{n-1}}{n}\end{array}$$

where we justify the switch of integral and sum by the monotone convergence theorem (since the summand is nonnegative on $[0,1]$). Recall the Euler Beta function given by

$$\mathrm{B}(m,n)={\int }_0^1{x}^{m-1}(1-x{)}^{n-1}dx=\frac{\mathrm{\Gamma }(m)\mathrm{\Gamma }(n)}{\mathrm{\Gamma }(m+n)}.$$

In the case that $m$ and $n$ are integers, we get

$$\mathrm{B}(m,n)=\frac{(m-1)!(n-1)!}{(m+n-1)!}$$

and in particular

$${\int }_0^1\frac{x^{n-1}(1-x{)}^{n-1}}{n}dx=\frac{\mathrm{B}(n,n)}{n}=\frac{((n-1)!{)}^2}{n(2n-1)!}=\frac{2}{n^2(\genfrac{}{}{0ex}{}{2n}{n})}.$$

Thus,

$$I=2\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2(\genfrac{}{}{0ex}{}{2n}{n})}.$$

The next part may strike you as something I pulled out of nowhere. We invoke the identity

$${\arcsin }^{2}x=\frac{1}{2}\sum _{n=0}^{\mathrm{\infty }}\frac{(2x{)}^{2n}}{n^2(\genfrac{}{}{0ex}{}{2n}{n})},x\in [-1,1],$$

a nice proof of which can be found by following the links starting here. (Okay, I'll admit it's not a very commonly-taught series, and the only reason I recognized it is that I used to spend too much time on AoPS. As such, I feel bad about blackboxing it like this. But it's cute!) We deduce that

$$I=4{\arcsin }^2\left(\frac{1}{2}\right)=\frac{{\pi }^2}{9}.$$

Thus, $(2/3)\zeta (2)={\pi }^2/9$, and so $\zeta (2)={\pi }^2/6$. QED.