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.


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

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

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


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

Equating imaginary parts, we get that

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


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

  • $\begingroup$ Is there also an answer when the original integral has an integer exponent 2,3,4,... on the $x$? Would that change the power on the pi or is it another big can of worms? $\endgroup$ – imranfat Mar 5 '14 at 22:13
  • $\begingroup$ @imranfat: what happens is that you get integrals over the power of $x$ you want, plus lower powers of $x$. In effect, you would end up solving an upper-diagonal linear system of equations [$(n-1) \times (n-1)$ I believe] to get the integral you actually want. $\endgroup$ – Ron Gordon Mar 5 '14 at 22:27
  • $\begingroup$ Well, that's a bit more difficult than I thought but I can see where those lower powers come from since you work out the brackets in the numerator. I thought maybe we can simply adjust the exponent on the pi and get done with it. How silly... $\endgroup$ – imranfat Mar 6 '14 at 15:11

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.

  • $\begingroup$ Nice! However, as far as I know, using analytic functions such as Zeta and Gamma somewhat "hides" the real problem of integration itself. Is is correct? $\endgroup$ – Brightsun Mar 5 '14 at 21:43
  • $\begingroup$ Consider for example the integral representation of the beta function which is quite popular on this site. By using beta function identities you can sometimes get impressive results quickly. Of course the real work was in proving those identities in the first place. I think both approaches have their advantages. Once you have proved an integral representation like the one for the gamma function (which is often taken as the definition) it makes no sense not to use it, quite the contrary. $\endgroup$ – Marko Riedel Mar 5 '14 at 21:51
  • 1
    $\begingroup$ Actually, what I think is a cool problem is to prove that $\zeta(2)$ takes on its well-known value by working this problem both ways. $\endgroup$ – Ron Gordon Mar 5 '14 at 21:55

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: $$ I=\frac{\pi^2}{12} $$ $$ \int_0^{2\pi}\mathrm{d}y\frac{y^2\sin y}{2(\cos y +1}=4\pi^2\ln 2 $$

  • $\begingroup$ Yes, Ron you are totally right, as usual. I wanted to make the notation a little lighter by specifying the $\epsilon$ in the first passage. $\endgroup$ – Brightsun Mar 5 '14 at 18:57

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