# Integrate $$\int_{0}^{\infty} \frac{x}{\mathrm{e}^{x}+1} \mathrm{d} x$$

How to evaluate

\int\limits_{0}^{\infty} \frac{x}{\mathrm{e}^{x}+1} \mathrm{d} x

was a question on Mathematics Stack Exchange. The original post asked for a contour integral solution and there is a good one among the answers. However, this integral can be evaluated trivially by recognizing an integral definition of the Dirichlet eta function or use of the Mellin transform.

We have

\eta(s) = \frac{1}{\Gamma(s)}\int\limits_{0}^{\infty} \frac{x^{s-1}}{\mathrm{e}^{x}+1} \mathrm{d} x

thus our integral is

\int\limits_{0}^{\infty} \frac{x}{\mathrm{e}^{x}+1} \mathrm{d} x = \eta(2)\Gamma(2) = \frac{\pi^{2}}{12}

Also
\begin{align}
\mathcal{M}\left[(\mathrm{e}^{ax}+1)^{-1}\right](s) & = \int\limits_{0}^{\infty} \frac{x^{s-1}}{\mathrm{e}^{ax}+1} \mathrm{d} x \\
& = a^{-s}\Gamma(s)(1-2^{1-s})\zeta(s) \\
& = a^{-s}\Gamma(s)\eta(s)
\end{align}
With $$a=1$$ and $$s=2$$ we arrive at our result.