Computing $\int_0^\infty\mathrm{d} x\frac{x}{e^x+1}$ with contour integration

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\large\tt\mbox{Just another one !!!}$

In the first step we integrate by parts: \begin{align} &\color{#00f}{\large\int_{0}^{\infty}{x \over \expo{x} + 1}\,\dd x} =\int_{0}^{\infty}\ln\pars{1 + \expo{-x}}\,\dd x= \int_{0}^{\infty}\sum_{\ell = 1}^{\infty}{\pars{-1}^{\ell + 1} \over \ell}\, \expo{-\ell x}\,\dd x \\[3mm]&=\sum_{\ell = 1}^{\infty}{\pars{-1}^{\ell + 1} \over \ell}\ \overbrace{\int_{0}^{\infty}\expo{-\ell x}\,\dd x}^{\ds{1 \over \ell}} =\sum_{\ell = 1}^{\infty}{\pars{-1}^{\ell + 1} \over \ell^{2}} =\sum_{\ell = 0}^{\infty} \bracks{{1 \over \pars{2\ell + 1}^{2}} - {1 \over \pars{2\ell + 2}^{2}}} \\[3mm]&=\bracks{\sum_{\ell = 1}^{\infty}{1 \over \ell^{2}} -\sum_{\ell = 1}^{\infty}{1 \over \pars{2\ell}^{2}}} -{1 \over 4}\sum_{\ell = 1}^{\infty}{1 \over \ell^{2}} =\half\ \underbrace{\sum_{\ell = 1}^{\infty}{1 \over \ell^{2}}} _{\ds{\zeta\pars{2} = {\pi^{2} \over 6}}} =\ \color{#00f}{\large{\pi^{2} \over 12}} \end{align}

$\zeta\pars{z}$ is the Riemann Zeta Function .

Consider the contour integral

$$\oint_C dz \frac{z^2}{e^z+1}$$

where $C$ is the rectangle with vertices at $0$, $R$, $R+i 2 \pi$, and $i 2 \pi$, in that order, with a small semicircular indentation of radius $\epsilon$ at $z=i \pi$ into the rectangle. Thus, the contour integral is equal to

$$\int_0^R dx \frac{x^2-(x+i 2 \pi)^2}{e^x+1} + i \int_0^{2 \pi} dy \frac{(R+i y)^2}{e^{R+i y}+1} \\+i PV \int_0^{2 \pi} dy \frac{y^2}{e^{i y}+1} + i \epsilon \int_{\pi/2}^{-\pi/2} d\phi \, e^{i \phi} \frac{(i \pi+\epsilon e^{i \phi})^2}{-e^{\epsilon e^{i \phi}}+1}$$

where $PV$ denotes the Cauchy principal value. As $R \to \infty$, the second integral vanishes. As $\epsilon \to 0$, the fourth integral approaches $-i \pi^3$. By Cauchy's theorem, the contour integral is zero. Thus, we have

$$-i 4 \pi \int_0^{\infty} dx \frac{x}{e^x+1} + 4 \pi^2 \int_0^{\infty} \frac{dx}{e^x+1}\\+i PV \int_0^{2 \pi} dy \frac{y^2}{e^{i y}+1} - i \pi^3 = 0$$

Now,

$$i PV \int_0^{2 \pi} dy \frac{y^2}{e^{i y}+1} = \frac12 PV \int_0^{2 \pi} dy \, y^2 \, \tan{\frac{y}{2}} + i \frac12 \int_0^{2 \pi} dy \, y^2 $$

Equating imaginary parts, we get that

$$-4 \pi \int_0^{\infty} dx \frac{x}{e^x+1} = \pi^3 - \frac{4 \pi^3}{3}$$

or

$$\int_0^{\infty} dx \frac{x}{e^x+1} = \frac{\pi^2}{12}$$