ζ(2)の値

\begin{align*} \sum_{k=1}^{\infty}\frac{1}{k^{2}} & =\sum_{k=1}^{\infty}\frac{1}{(2k)^{2}}+\sum_{k=1}^{\infty}\frac{1}{(2k-1)^{2}}\\ & =\frac{1}{4}\sum_{k=1}^{\infty}\frac{1}{k^{2}}+\sum_{k=1}^{\infty}\frac{1}{(2k-1)^{2}}\\ & =\frac{4}{3}\sum_{k=1}^{\infty}\frac{1}{(2k-1)^{2}}\\ & =\frac{4}{3}\sum_{k=0}^{\infty}\frac{1}{(2k+1)^{2}}\\ & =\frac{4}{3}\sum_{k=0}^{\infty}\frac{1}{(2k+1)}\int_{0}^{1}x^{2k}dx\\ & =\frac{4}{3}\sum_{k=0}^{\infty}\frac{1}{(2k+1)}\left(\left[x^{2k+1}\log x\right]_{0}^{1}-\left(2k+1\right)\int_{0}^{1}x^{2k}\log xdx\right)\\ & =-\frac{4}{3}\sum_{k=0}^{\infty}\int_{0}^{1}x^{2k}\log xdx\\ & =-\frac{4}{3}\int_{0}^{1}\frac{1}{1-x^{2}}\log xdx\\ & =-\frac{2}{3}\int_{0}^{1}\frac{1}{1-x^{2}}\log x^{2}dx\\ & =-\frac{2}{3}\int_{0}^{1}\frac{1}{1-x^{2}}\left[\log\frac{1+x^{2}y^{2}}{1+y^{2}}\right]_{y=0}^{y=\infty}dx\\ & =-\frac{2}{3}\int_{0}^{1}\frac{1}{1-x^{2}}\int_{0}^{\infty}\frac{\partial}{\partial y}\left(\log\frac{1+x^{2}y^{2}}{1+y^{2}}\right)dydx\\ & =-\frac{4}{3}\int_{0}^{1}\frac{1}{1-x^{2}}\int_{0}^{\infty}\left(\frac{x^{2}y}{1+x^{2}y^{2}}-\frac{y}{1+y^{2}}\right)dydx\\ & =-\frac{4}{3}\int_{0}^{1}\frac{1}{1-x^{2}}\int_{0}^{\infty}\frac{-y(1-x^{2})}{\left(1+x^{2}y^{2}\right)\left(1+y^{2}\right)}dydx\\ & =\frac{4}{3}\int_{0}^{1}\int_{0}^{\infty}\frac{y}{\left(1+x^{2}y^{2}\right)\left(1+y^{2}\right)}dydx\\ & =\frac{4}{3}\int_{0}^{\infty}\frac{y}{1+y^{2}}\int_{0}^{1}\frac{1}{1+x^{2}y^{2}}dxdy\\ & =\frac{4}{3}\int_{0}^{\infty}\frac{y}{1+y^{2}}\left[\frac{1}{y}\tan^{\bullet}\left(xy\right)\right]_{x=0}^{x=1}dy\\ & =\frac{4}{3}\int_{0}^{\infty}\frac{\tan^{\bullet}y}{1+y^{2}}dy\\ & =\frac{4}{3}\int_{0}^{\infty}\tan^{\bullet}yd\tan^{\bullet}y\\ & =\frac{2}{3}\left[\tan^{\bullet,2}y\right]_{0}^{\infty}\\ & =\frac{2}{3}\left(\frac{\pi}{2}\right)^{2}\\ & =\frac{\pi^{2}}{6} \end{align*}