イータ関数の導関数がでてきます

\begin{align*} \int_{0}^{\infty}\frac{\log x}{1+e^{x}}dx & =\int_{0}^{\infty}\frac{e^{-x}\log x}{1+e^{-x}}dx\\ & =\int_{0}^{\infty}e^{-x}\log x\sum_{k=0}^{\infty}\left(-e^{-x}\right)^{k}dx\\ & =\sum_{k=0}^{\infty}\left(-1\right)^{k}\int_{0}^{\infty}e^{-\left(k+1\right)x}\log xdx\\ & =\sum_{k=0}^{\infty}\left(-1\right)^{k}\mathcal{L}_{x}\left[H\left(x\right)\log x\right]\left(k+1\right)\\ & =\sum_{k=0}^{\infty}\left(-1\right)^{k}\frac{-1}{k+1}\left(\log\left(k+1\right)+\gamma\right)\\ & =-\gamma\sum_{k=0}^{\infty}\frac{\left(-1\right)^{k}}{k+1}-\sum_{k=0}^{\infty}\frac{\left(-1\right)^{k}\log\left(k+1\right)}{k+1}\\ & =-\gamma\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k-1}}{k}-\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k-1}\log k}{k}\\ & =-\gamma\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k-1}}{k}-\left[\frac{d}{dt}\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k-1}k^{t}}{k}\right]_{t=0}\\ & =-\gamma\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k-1}}{k}-\left[\frac{d}{dt}\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k-1}}{k^{1-t}}\right]_{t=0}\\ & =-\gamma\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k-1}}{k}-\left[\frac{d}{dt}\eta\left(1-t\right)\right]_{t=0}\\ & =-\gamma\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k-1}}{k}+\eta'\left(1\right)\\ & =-\gamma\log2+\left(-\frac{1}{2}\log^{2}2+\gamma\log2\right)\cmt{\because\eta'\left(1\right)=-\frac{1}{2}\log^{2}2+\gamma\log2}\\ & =-\frac{1}{2}\log^{2}2 \end{align*}