Evaluate $\int_{0}^{\infty} \frac{x}{1+e^x}dx$

Question

$$\int_{0}^{\infty} \frac{x}{1+e^x}dx=\int_{0}^{\infty} \frac{x(e^x+1-e^x)}{1+e^x}dx=\int_{0}^{\infty} x\cdot dx - \int_{0}^{\infty} \frac{x\cdot e^x}{1+e^x}$$

I can't think of a way to proceed further

Answer 1

Note that

$${x\over1+e^x}={xe^{-x}\over1+e^{-x}}=xe^{-x}(1-e^{-x}+e^{-2x}-e^{-3x}+\cdots)=xe^{-x}-xe^{-2x}+xe^{-3x}-xe^{-4x}+\cdots$$

It's easy to see that

$$\int_0^\infty xe^{-nx}\,dx={1\over n^2}$$

Thus

$$\begin{align} \int_0^\infty{x\over1+e^x}\,dx &=1-{1\over4}+{1\over9}-{1\over16}+\cdots\\ &=\left(1+{1\over4}+{1\over9}+{1\over16}+\cdots\right)-2\left({1\over4}+{1\over16}+\cdots \right)\\ &=\left(1+{1\over4}+{1\over9}+{1\over16}+\cdots\right)-{1\over2}\left(1+{1\over4}+\cdots\right)\\ &={1\over2}\left(1+{1\over4}+{1\over9}+{1\over16}+\cdots\right)\\ &={\pi^2\over12} \end{align}$$