Finding the integral of $\frac{x}{e^x + 1}$

Question

I've having some difficulty with finding this integral:

$$ \int_0 ^{\infty} \frac{x}{e^x + 1}$$

Now usually I would use the monotone convergence theorem to write (using geometric series):

$$f_n (x) = \sum_0 ^n (-1)^k x e^{-(k+1)x},$$

but this isn't a sequence of positive terms, so how do we justify moving the integral inside?

Thanks.

Answer 1

$$ \begin{align} \int_0^\infty\frac{x}{e^x+1}\mathrm{d}x &=\int_0^\infty x\sum_{k=1}^\infty(-1)^{k-1}e^{-kx}\mathrm{d}x\\ &=\sum_{k=1}^\infty(-1)^{k-1}\int_0^\infty xe^{-kx}\mathrm{d}x\\ &=\sum_{k=1}^\infty(-1)^{k-1}\frac1{k^2}\int_0^\infty xe^{-x}\mathrm{d}x\\ &=\sum_{k=1}^\infty(-1)^{k-1}\frac1{k^2}\\ &=\frac{\pi^2}{12} \end{align} $$ Although the sum is not positive, the integrand is dominated by $\frac{x}{e^x-1}$ which has the same series, but without the alternation. We can then use dominated convergence to justify the exchange of integration and summation.

Answer 2

Note that using the replacement $u=e^x+1, du = (u-1)dx$, leads to an integral involving the definition of the second order polylogarithm, $\text{Li}_2(x)$.

Answer 3

Let us generalize the problem, we will evaluate $$ \int_0 ^{\infty} \frac{x^{s-1}}{e^x + 1}\ dx. $$ Rewrite the integral above as \begin{align} \int_0 ^{\infty} x^{s-1}\cdot\frac{e^{-x}}{1+e^{-x}}\ dx&=\int_0 ^{\infty} x^{s-1}\cdot\sum_{n=1}^\infty(-1)^{n-1}e^{-nx} \ dx\\ &=\sum_{n=1}^\infty(-1)^{n-1}\int_{x=0}^{\infty} x^{s-1} e^{-nx} \ dx\quad;\quad\text{let }u=nx\\ &=\color{blue}{\sum_{n=1}^\infty(-1)^{n-1}\frac1{n^s}}\color{red}{\int_{u=0}^{\infty} u^{s-1} e^{-u} \ du}\\ &=\color{blue}{\eta(s)}\color{red}{\Gamma(s)}, \end{align} where $\color{blue}{\eta(s)}$ is the Dirichlet eta function and $\color{red}{\Gamma(s)}$ is the gamma function.

In our case, we have $s=2$. Hence $$ \int_0 ^{\infty} \frac{x}{e^x + 1}\ dx=\eta(2)\Gamma(2)=\large\color{blue}{\frac{\pi^2}{12}}. $$