A Proof of Basel's Problem

MSE를 뒤적이다가 슬쩍 본 계산인데, 예뻐서 가져와봅니다.

 

We evaluate the integral

$$I={\int }_0^1{\int }_0^1\frac{x}{1-(1-y^2)x^2}\mathrm{d}x\mathrm{d}y$$

in two ways. Integrating with respect to $x$ first, we get

$$\begin{array}{rl}I& ={\int }_0^1\frac{\log (1/y)}{1-y^2}\mathrm{d}y=\sum _{n=0}^{\mathrm{\infty }}{\int }_0^1{y}^{2n}\log (1/y)\mathrm{d}y=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(2n+1{)}^2}.\end{array}$$

On the other hand, integraing with respect to $y$ first,

$$\begin{array}{rl}I& ={\int }_0^1\frac{\arcsin x}{\sqrt{1-x^2}}\mathrm{d}x={[\frac{1}{2}{\arcsin }^{2}x]}_0^1=\frac{{\pi }^2}{8}.\end{array}$$

So we conclude that

$$(1-2^{-2})\zeta (2)=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(2n+1{)}^2}=\frac{{\pi }^2}{8},$$

proving the identity $\zeta (2)=\frac{{\pi }^2}{6}$ as desired.