I have been self-studying CA and find it very interesting. So, working through problems in a book I have, I ran across

$$\int_{0}^{\infty}\frac{\sin(ax)}{e^{2\pi x}-1}dx=\frac{1}{4}\coth(a/2)-\frac{1}{2a}$$

and $$\int_{0}^{\infty}\frac{\sin(ax)}{e^{x}+1}dx=\frac{1}{2a}-\frac{\pi}{2\sinh(\pi a)}$$

For the former, I wrote it as $\frac{e^{aiz}}{e^{2\pi z}-1}$ and used a rectangular contour with vertices $0, \;\ R. \;\ R+i, \;\ i$

$e^{2\pi z}-1$ has poles at $ni$. Of these, I think $i$ only lies within the contour.

Unless I am in error, I then calculated the residue at $i$ to be $\frac{e^{-a}}{2\pi}$

So, $2\pi i(\frac{e^{-a}}{2\pi})=ie^{-a}$

Now, where I get hung up is setting up the integrals around the contour. Two of which should tend to 0 as $R\to \infty$.

Here is what I done.

$$\int_{0}^{R}\frac{e^{iax}}{e^{2\pi x}-1}dx+\int_{0}^{\infty}\frac{e^{ai(R+iy)}}{e^{2\pi (R+iy)}-1}idy+\int_{R}^{0}\frac{e^{ai(x+i)}}{e^{2\pi (x+i)}-1}dx+\int_{\infty}^{0}\frac{e^{ai(iy)}}{e^{2\pi iy}-1}dy=-\sinh(a)$$

I am unsure of the limits on the second and fourth integrals.

I am not so sure this is correct. The second and fourth ones, which represent the vertical sides, should tend to 0 as $R\to\infty$. I hope :).

This than gave me $(1-e^{-a})\int\frac{e^{iax}}{e^{2\pi x}-1}dx=-\sinh(a)$.

Which does not look correct. I did manage to solve this using series and $\pi csc(\pi z)$, but the contour I am unsure of.

Can someone lend a hand here?. Any advice on either would be appreciated.

For the other one, I only posted it because it looks similar, but I believe is actually more involved and 'tougher' if you will. If I can get one, perhaps I can manage to evaluate the other.

With that one, I think the same rectangle with the same vertices can be used, but $\pi i$ would have to be avoided. Perhaps a Principal Value in there somewhere. But, I was told to use vertices $0, \;\ R, \;\ R+2\pi i, \;\ 2\pi i$

Since I am relatively new to CA, setting up those integrals is the confusing part.

I can do easier types of contour integrals, but want to learn more about these more challenging ones.

Thanks for any assistance.

  • $\begingroup$ Wow... talk about a race condition! $\endgroup$
    – Asaf Karagila
    May 5, 2012 at 18:33
  • $\begingroup$ I think your contour may be goingthrough a pole. $\endgroup$
    – Argon
    May 5, 2012 at 18:47
  • $\begingroup$ Yes, I see that. There is a pole at 'i' which lies on the rectangle. Perhaps vertices of $0, \;\ R, \;\ R+2\pi i, \;\ 2\pi i$ would be better. The book I got the problem from gave a hint that suggested using a rectangle with vertices $0, \;\ R, \;\ r+i, \;\ i$. It is a miscellaneous problem from Schaum's Complex Variables. $\endgroup$
    – Cody
    May 5, 2012 at 20:30
  • $\begingroup$ @Cody I see, its question 7.81-2. Perhaps we can approach the singularity in the limit? $\endgroup$
    – Argon
    May 5, 2012 at 21:30
  • $\begingroup$ Yeah, maybe we can. I am certainly open for suggestions :). If the rectangle had corners $0,R,2\pi i, R+2\pi i$, maybe we could set it up as $$\int_{0}^{R}\frac{e^{aiz}}{e^{2\pi x}-1}dx+\int_{0}^{2\pi}\frac{e^{ai(R+iy)}}{e^{2\pi(R+iy)}-1}idy+\int_{R}^{0}\frac{e^{ai(x+2\pi i)}}{e^{2\pi(x+2\pi i)}-1}dx+\int_{2\pi}^{0}\frac{e^{ai(iy)}}{e^{2\pi(iy)}-1}dy=\frac{e^{-a}}{2\pi}$$. Just an idea. I really don't know what I'm doing at this point. :) The problem here though, is the third integral. $\endgroup$
    – Cody
    May 5, 2012 at 22:24

5 Answers 5


$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Abel-Plana Formula:

\begin{align} \sum_{n = 0}^{\infty}\expo{-n\verts{a}} & = \int_{0}^{\infty}\expo{-\verts{a}x}\,\dd x + \bracks{{1 \over 2}\,\expo{-\verts{a}x}}_{\ x\ =\ 0} - 2\,\Im\int_{0}^{\infty}{\expo{-\verts{a}x\ic} \over \expo{2\pi x} - 1}\,\dd x \end{align}

\begin{align} {1 \over 1 - \expo{-\verts{a}}} & = {1 \over \verts{a}} + {1 \over 2} + 2\,\mrm{sgn}\pars{a}\int_{0}^{\infty} {\sin\pars{ax} \over \expo{2\pi x} - 1}\,\dd x \end{align}
\begin{align} &2\,\mrm{sgn}\pars{a}\int_{0}^{\infty} {\sin\pars{ax} \over \expo{2\pi x} - 1}\,\dd x = {1 \over 1 - \expo{-\verts{a}}} - {1 \over 2} - {1 \over \verts{a}} = {1 + \expo{-\verts{a}} \over 1 - \expo{-\verts{a}}}\,{1 \over 2} - {1 \over \verts{a}} \end{align}
\begin{align} &\int_{0}^{\infty} {\sin\pars{ax} \over \expo{2\pi x} - 1}\,\dd x = \bbx{\ds{{1 \over 4}\,\coth\pars{a \over 2} - {1 \over 2a}}} \end{align}

  • $\begingroup$ The Abel-Plana formula is obtained by a change of contour and the residue theorem, did you check the growth condition ? $\endgroup$
    – reuns
    Mar 27, 2017 at 17:55
  • $\begingroup$ @user1952009 $\left\vert\exp\left(-\left\vert a\right\vert z\right)\right\vert < 1/\left\vert az\right\vert$ as $\left\vert z\right\vert \to \infty$. $\endgroup$ Mar 27, 2017 at 19:41

The given integral is the imaginary part of $$ \int_{0}^{+\infty}\sum_{n\geq 1}\exp\left((ia-2\pi n)x\right)\,dx $$ that equals $$ \color{red}{\frac{-2 + a \coth(a/2)}{4 a}} $$ by the Eisenstein series of the cotangent function. In particular, the given integral is bounded between $-\frac{1}{4}$ and $\frac{1}{4}$ for any $a\in\mathbb{R}$ and it is an increasing function of the $a$ variable.

  • 1
    $\begingroup$ Eisenstein series is not the best reference, see people.reed.edu/~jerry/311/cotan.pdf (due to Euler when he proved the Basel problem ?) $\endgroup$
    – reuns
    Mar 26, 2017 at 23:31
  • $\begingroup$ @user1952009: of course, the value of $\sum_{n\geq 1}\frac{1}{n^2+a^2}$ can be computed from Fourier series or from the logarithmic derivative of a suitable Weierstrass product. $\endgroup$ Mar 27, 2017 at 0:19

I found $$\int_{C}\frac{e^{iaz}}{e^{2\pi z}-1}dz$$ in "Applied Complex Analysis with PDE" by Nakhle Asmar.

If anyone is interested, he uses a contour with a line segment and quarter circle with $I_{1}, ...., I_{6}$

$\int_{C}\frac{e^{iaz}}{e^{2\pi z}-1}dz=I_{1}+I_{2}+I_{3}+I_{4}+I_{5}+I_{6}$

starting with a line segment $[\epsilon,R]$ and moving counterclockwise. As $\epsilon\to 0^{+}$ and $R\to \infty$, $I_{1}\to I$.

i.e. $I_{2}, \;\ z=R+iy$, where y varies from 0 to 1.

$I_{3}, \;\ z=x+i$, where x varies from $R$ to $\epsilon$

$I_{4}$ is done over the quarter circle from $(\epsilon,\epsilon+i)$ to $0,i-i\epsilon)$

$I_{5}, \;\ z=iy$ where y varies from $i-\epsilon$ to $\epsilon$

$I_{6}$ is done over the quarter circle from $(0,i\epsilon)$ to $(\epsilon,0)$

For instance, $I_{6}$ gives $$\lim_{\epsilon\to 0}\int\frac{e^{iaz}}{e^{2\pi z}-1}dz=\frac{-\pi}{2}\text{Res}\left(\frac{e^{iaz}}{e^{2\pi z}-1},0\right)$$

$$=\frac{-\pi i}{2}\cdot \frac{e^{iaz}}{2\pi e^{2\pi (i)}}=-i\cdot \frac{e^{-a}}{4}$$

It is rather lengthy and ultimately results in $\frac{-1}{2a}+\frac{1}{4}\frac{e^{a}+1}{e^{a}-1}=\frac{-1}{2a}+\frac{1}{4}\coth(\frac{a}{2})$


Contour integration: We consider the integral:

$$I = \oint_{C\left(\epsilon,R\right)} \frac{\exp(i a z)dz}{\exp(2\pi z) - 1}$$

with $C\left(\epsilon,R\right)$ a rectangle from $\epsilon$ to $R$ on the real axis, from there to $R + i$ parallel to the imaginary axis, from there to $\epsilon + i$ parallel to the real axis, then a clockwise quarter turn with radius $\epsilon$ and center $i$ to the point $i - i\epsilon$, from there we move on the imaginary axis to the point $i\epsilon$ and then we take a clockwise quarter turn with radius $\epsilon$ and center the origin to move back to the starting point at $\epsilon$.

The integrand is analytic inside the contour, therefore $I = 0$. The sum of the two parts parallel to the real axis is:

$$I_r = \left[1-\exp(-a)\right]\int_\epsilon^R \frac{\exp(i a x)dx}{\exp(2\pi x) - 1}$$

So, the desired integral will follow from the imaginary part of $I_r$. The part of the contour integral from $R$ to $R + i$ tends to zero in the limit of $R\to\infty$ so this can be disregarded. The part of the contour integral along the imaginary axis can be written as:

$$I_i = -\int_{\epsilon}^{1-\epsilon}\frac{\exp(-a y)\exp(-\pi i y)}{2\sin(\pi y)}dy$$

We then see that

$$\lim_{\epsilon\to 0}\operatorname{Im}I_i = \frac{1-\exp(-a)}{2a}$$

The two quarter circles can be evaluated, we can borrow from the derivation of the residue theorem that each of them in the limit $\epsilon\to 0$ can be evaluated as $-\frac{\pi}{2} i$ times the residue at the poles that are at the centers of the quarter circles (unlike the case of a complete contour, this is only valid in the limit $\epsilon\to 0$). The residue at the pole at $z = 0$ is $\frac{1}{2\pi}$, while the residue at $z = i$ is $\frac{\exp(-a)}{2\pi}$. Having completed the evaluation of all parts of the contour integral, we can now equate the imaginary part of the contour integral to zero, and take the limit of $\epsilon\to 0$ and $R\to \infty$. This yields:

$$\int_0^{\infty}\frac{\sin(a x)dx}{\exp(2\pi x)-1} = \frac{1}{4}\coth\left(\frac{a}{2}\right) - \frac{1}{2 a}$$


For any $b>0$, $$ \left|\int_{b}^{\infty}\frac{\sin(ax)}{e^{2\pi x}-1}dx\right|\leq\int_{b}^{\infty}\frac{1}{e^{2\pi x}-1}dx=b-\frac{\ln(e^{2\pi b}-1)}{2\pi}. $$ Therefore, if you would like to compute your integral within an $\epsilon$ tolerance, you can compute numerically the integral $$ \int_{0}^{b^{*}}\frac{\sin(ax)}{e^{2\pi x}-1}dx $$ where $b^{*}$ is a positive solution of $$ b^{*}-\frac{\ln(e^{2\pi b^{*}}-1)}{2\pi}\leq \epsilon. $$


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