積分解説16

2020/11/17に 白茶 さんが出題した問題です。

https://mathlog.info/articles/721

[解説]

$\begin{array}{rcl}& & {\int }_0^{\mathrm{\infty }}\frac{1-\cos x}{x(e^x-1)}dx\\ & =& -{\int }_0^{\mathrm{\infty }}\frac{1}{x(e^x-1)}{\int }_0^1\frac{\partial }{\partial x}\cos txdtdx\\ & =& {\int }_0^1{\int }_0^{\mathrm{\infty }}\frac{\sin tx}{e^x-1}dxdt\\ & =& {\int }_0^1{\int }_0^{\mathrm{\infty }}\sin tx\sum _{k=1}^{\mathrm{\infty }}{e}^{-kx}dxdt\\ & =& \mathrm{Im}\sum _{k=1}^{\mathrm{\infty }}{\int }_0^1{\int }_0^{\mathrm{\infty }}{e}^{-(k-it)x}dxdt\\ & =& \mathrm{Im}\sum _{k=1}^{\mathrm{\infty }}{\int }_0^1{[-\frac{1}{k-it}{e}^{-(k-it)x}]}_0^{\mathrm{\infty }}dt\\ & =& \mathrm{Im}\sum _{k=1}^{\mathrm{\infty }}{\int }_0^1\frac{1}{k-it}dt\\ & =& \mathrm{Im}\sum _{k=1}^{\mathrm{\infty }}{\int }_0^1\frac{k+it}{k^2+t^2}dt\\ & =& \sum _{k=1}^{\mathrm{\infty }}{\int }_0^1\frac{t}{k^2+t^2}dt\\ & =& \frac{1}{2}\sum _{k=1}^{\mathrm{\infty }}{[\log (k^2+t^2)]}_0^1\\ & =& \frac{1}{2}\sum _{k=1}^{\mathrm{\infty }}\log (1+\frac{1}{k^2})\\ & =& \frac{1}{2}\log \prod _{k=1}^{\mathrm{\infty }}(1+\frac{1}{k^2})\\ & =& \frac{1}{2}\log \frac{\sinh \pi }{\pi }\end{array}$

よって、この問題の解答は${\displaystyle \frac{1}{2}\log \frac{\sinh \pi }{\pi }}$となります。