積分問題

\begin{align*} \int_{0}^{\infty}\frac{x^{s}}{\cosh^{2}x}dx & =4\int_{0}^{\infty}\frac{x^{s}}{(e^{x}+e^{-x})^{2}}dx\\ & =4\int_{0}^{\infty}\frac{x^{s}e^{-2x}}{(1+e^{-2x})^{2}}dx\\ & =4\int_{0}^{\infty}x^{s}e^{-2x}\sum_{u=0}^{\infty}\left(-e^{-2x}\right)^{u}\sum_{v=0}^{\infty}\left(-e^{-2x}\right)^{v}dx\\ & =4\sum_{u=0}^{\infty}\sum_{v=0}^{\infty}(-1)^{u+v}\int_{0}^{\infty}x^{s}e^{-2(u+v+1)x}dx\\ & =4\sum_{u=0}^{\infty}\sum_{v=0}^{\infty}(-1)^{u+v}\frac{1}{2^{s+1}(u+v+1)^{s+1}}\int_{0}^{\infty}y^{s}e^{-y}dy\qquad,\qquad y=2(u+v+1)x\\ & =4\sum_{u=0}^{\infty}\sum_{v=0}^{\infty}(-1)^{u+v}\frac{1}{2^{s+1}(u+v+1)^{s+1}}\Gamma(s+1)\\ & =\frac{\Gamma(s+1)}{2^{s-1}}\sum_{u=0}^{\infty}\sum_{v=0}^{\infty}\frac{(-1)^{u+v}}{(u+v+1)^{s+1}}\\ & =\frac{\Gamma(s+1)}{2^{s-1}}\sum_{j=0}^{\infty}\sum_{k=0}^{j}\frac{(-1)^{j}}{(j+1)^{s+1}}\\ & =\frac{\Gamma(s+1)}{2^{s-1}}\sum_{j=0}^{\infty}\frac{(-1)^{j}}{(j+1)^{s}}\\ & =\frac{\Gamma(s+1)}{2^{s-1}}\eta(s) \end{align*}