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
$$ \displaystyle \int_0^\infty \frac{1-\cos x}{x(e^x-1)}dx $$
[解説]
$ \begin{eqnarray*} && \int_0^\infty \frac{1-\cos x}{x(e^x-1)}dx\\ &=&-\int_0^\infty\frac1{x(e^x-1)}\int_0^1\frac{\partial}{\partial x}\cos txdtdx\\ &=&\int_0^1 \int_0^\infty\frac{\sin tx}{e^x-1}dxdt\\ &=&\int_0^1\int_0^\infty \sin tx\sum_{k=1}^\infty e^{-kx}dxdt\\ &=&\Im\sum_{k=1}^\infty \int_0^1 \int_0^\infty e^{-(k-it)x}dxdt\\ &=&\Im\sum_{k=1}^\infty\int_0^1\left[-\frac1{k-it}e^{-(k-it)x} \right]_0^\infty dt\\ &=&\Im\sum_{k=1}^\infty\int_0^1\frac1{k-it}dt\\ &=&\Im\sum_{k=1}^\infty \int_0^1 \frac{k+it}{k^2+t^2}dt\\ &=&\sum_{k=1}^\infty \int_0^1 \frac t{k^2+t^2}dt\\ &=&\frac12\sum_{k=1}^\infty\left[\log(k^2+t^2)\right]_0^1\\ &=&\frac12\sum_{k=1}^\infty \log\l1+\frac1{k^2} \r\\ &=&\frac12\log \prod_{k=1}^\infty \l1+\frac1{k^2} \r \\ &=&\frac12\log\frac{\sinh\pi}\pi \end{eqnarray*} $
よって、この問題の解答は$\displaystyle \frac12\log\frac{\sinh\pi}\pi$となります。