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

Question

Let's set: $$ \int_0^\infty\mathrm{d}x\frac{x}{e^x+1}=I. $$ I would like to compute it using, presumably, the methods of complex analysis and contour integration.

Answer 1

We start from the contour integral evaluated through Cauchy's theorem: $$ \oint_{\Gamma_{\epsilon,R}}\mathrm{d}z\frac{z^2}{e^z+1} = 0 $$ where $\Gamma_{\epsilon, R}$ is the rectangle of vertices $0,\ R,\ R+i2\pi,\ i2\pi$, indented at the singularity $i\pi$ with a semicircle of radius $\epsilon$.

Splitting it into its natural branches yields: $$ \int_0^R\mathrm{d}x\frac{x^2}{e^x+1}+ \int_0^{2\pi}i\mathrm{d}y \frac{(iy+R)^2}{e^{iy+R}+1}+ \int_R^0\mathrm{d}x\frac{(x+i2\pi)^2}{e^{x+i2\pi}+1}+ \left(\int_{2\pi}^{\pi+\epsilon}+ \int_{\pi-\epsilon}^0\right)i\mathrm{d}y\frac{(iy)^2}{e^{iy}+1}+ \int_{\pi/2}^{-\pi/2}i\epsilon e^{i\theta}\mathrm{d}\theta\frac{(\epsilon e^{i\theta}+i\pi)^2}{e^{i\pi + \epsilon e^{i\theta}}+1}=0.$$

Now we observe that the first integral cancels out with the first term of the expansion of the third integral, furthermore the second one $\to0$ as $R\to \infty$. Our desired integral appears as the rectangular term of said expansion.

We are now left with the following calculations: $$ -\int_{-\pi/2}^{\pi/2}i\epsilon e^{i\theta}\mathrm{d}\theta\frac{(\epsilon e^{i\theta}+i\pi)^2}{e^{i\pi + \epsilon e^{i\theta}}+1}=i\pi^2\int_{-\pi/2}^{\pi/2}\mathrm{d}\theta(-1+O(\epsilon)) \to_{\epsilon\to0}-i\pi^3 $$ and, using Euler's formula and splitting into real and imaginary part: $$ i\int_0^{2\pi}\mathrm{d}y\frac{y^2}{\cos y+1+i\sin y}=i\int_0^{2\pi}\mathrm{d}y\ \frac{y^2}{2}- \int_0^{2\pi}\mathrm{d}y\frac{y^2\sin y}{2(\cos y +1}. $$ Therefore, using the easy-to-verify $\int\mathrm{d}x\frac{1}{e^x+1}=\ln 2$: $$ 0-i4\pi I + 4\pi^2\ln2+i\int_0^{2\pi}\mathrm{d}y\ \frac{y^2}{2}-\int_0^{2\pi}\mathrm{d}y\frac{y^2\sin y}{2(\cos y +1}-i\pi^3=0. $$ By comparison of real and imaginary part: $$ I=\frac{\pi^2}{12} $$ $$ \int_0^{2\pi}\mathrm{d}y\frac{y^2\sin y}{2(\cos y +1}=4\pi^2\ln 2 $$

Answer 2

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

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

Answer 3

By way of diversity and since you admit complex variable methods note that what we have here is a special value of a Mellin transform: $$\mathfrak{M}\left(\frac{1}{e^x+1}; s\right) = \int_0^\infty \frac{1}{e^x+1} x^{s-1} dx.$$ To compute this transform we may proceed in an admittedly somewhat unorthodox manner by expanding the function being transformed into a series: $$\int_0^\infty \frac{1}{e^x+1} x^{s-1} dx = \int_0^\infty \frac{e^{-x}}{1+e^{-x}} x^{s-1} dx = \int_0^\infty \sum_{q\ge 1} (-1)^{q+1} e^{-qx} x^{s-1} dx \\ = \sum_{q\ge 1} (-1)^{q+1} \int_0^\infty e^{-qx} x^{s-1} dx = \Gamma(s) \sum_{q\ge 1} \frac{(-1)^{q+1}}{q^s} = \Gamma(s) \left(1-\frac{2}{2^s}\right) \zeta(s).$$ Put $s=2$ to obtain $$\Gamma(2) \times \frac{1}{2} \times \zeta(2) = \frac{\pi^2}{12}.$$ Here we have used the integral representation of the gamma function which I suppose is admissible in a complex variable method type answer.

Answer 4

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

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 .