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{align*} I &= \int_{0}^{1} \frac{\log(1/y)}{1 - y^2} \, \mathrm{d}y = \sum_{n=0}^{\infty} \int_{0}^{1} y^{2n} \log(1/y) \, \mathrm{d}y = \sum_{n=0}^{\infty} \frac{1}{(2n+1)^2}. \end{align*}

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

\begin{align*} I &= \int_{0}^{1} \frac{\arcsin x}{\sqrt{1-x^2}} \, \mathrm{d}x = \left[ \frac{1}{2}\arcsin^2 x \right]_{0}^{1} = \frac{\pi^2}{8}. \end{align*}

So we conclude that

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

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