A curious integral

Question

The following integral has been on my mind for a while $$\int_0^\infty \frac{\sin(x)}{e^x-1}\,\mathrm d x \tag{$\dagger$}$$

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Let us indicate the integrand as $f(x)=\frac{\sin(x)}{e^x-1}$. The following are a couple of observations.

1. The integrand can be extended by continuity in $0$ since $$\lim_{x\to 0}f(x)=\lim_{x\to 0}\frac{\frac{\sin(x)}{x}}{\frac{e^x-1}{x}}=1$$ This is the original reason I started playing around with this integral.

2. Mathematica yields the result $$\int_0^\infty \frac{\sin(x)}{e^x-1}\,\mathrm d x=\frac{\pi}{2}\textrm{Coth}(\pi)-\frac 12 \approx 1.076674047$$ which, following numerical evidence, seems correct.

3. Complex analysis may be useful here, since the integrand is a holomorphic function on $\mathbb C$. I tried writing $\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$, but I was not able to find an appropriate integration contour to solve the problem.

4. The $-1$ in the denominator breaks the simmetry of the expression. This made most of my substitutions useless.

Can this integral be evaluated correctly, preferrably through complex analytic methods?

Answer 1

Without resorting to the Abel-Plana formula, you may simply notice that

$$ I=\int_{0}^{+\infty}\frac{\sin x}{e^x-1}\,dx = \sum_{n\geq 1}\int_{0}^{+\infty}\sin(x)e^{-nx}\,dx = \sum_{n\geq 1}\frac{1}{n^2+1} \tag{1}$$ and since $\frac{\sin(\pi x)}{\pi x}=\prod_{n\geq 1}\left(1-\frac{x^2}{n^2}\right)$, by applying $\frac{d}{dx}\log(\cdot)$ to both sides we get: $$ -\frac{1}{x}+\pi\cot(\pi x) = \sum_{n\geq 1}\frac{2x}{x^2-n^2}\tag{2} $$ as well as: $$ \sum_{n\geq 1}\frac{1}{n^2+z^2} = \frac{-1+\pi z \coth(\pi z)}{2 z^2}\tag{3}$$ from which $I=\frac{-1+\pi\coth(\pi)}{2}$ clearly follows.
Have a look at this thread for further proofs.

Answer 2

The Abel-Plana theorem states $$\sum _{n=0} ^{\infty} f(n) - \int_0^{\infty} f(s) \mathrm{d}s = \frac{1}{2} f(0) +i \int_0^{\infty} \frac{f(iy) - f(-iy)}{e^{2 \pi y} - 1} $$

Expressing the $sin$ using complex exponential your integral reads $$ \frac{1}{2i}\int_0^{\infty} \frac{e^{iz} - e^{-iz}}{e^z -1} \mathrm{d}z$$ by scaling the variable of integration, $$ \frac{2 \pi}{2i}\int_0^{\infty} \frac{e^{2 \pi iy} - e^{-2 \pi iy}}{e^{2 \pi y} -1} \mathrm{d}z = -\pi i\int_0^{\infty} \frac{e^{2 \pi iy} - e^{-2 \pi iy}}{e^{2 \pi y} -1} \mathrm{d}z$$

By stating $$ f(x) = - e^{-2 \pi x} $$ one can compute the series and the integral on the l.h.s of the Theorem

$$ \sum _{n=0} ^{\infty} f(n) - \int_0^{\infty} f(s) \mathrm{d}s = -\frac{e^{2 \pi i}}{ e^{2 \pi } -1} + \frac {1}{2 \pi} $$

Putting it all together, the sought integral equals

$$ - \pi [-\frac{e^{2 \pi i}}{ e^{2 \pi } -1} + \frac {1}{2 \pi} +\frac{1}{2}] = \frac{\pi}{2}[-1 +\frac{2 e^{2 \pi}}{e^{2 \pi}-1}]-\frac{1}{2}$$ which coincides with the given solution as $$ coth (x) = \frac{e^{2x} + 1}{e^{2x} - 1}$$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{\infty}{\sin\pars{x} \over \expo{x} - 1}\,\dd x & = {1 \over 2\ic}\int_{0}^{\infty}{\expo{-\ic x} - \expo{\ic x} \over 1 - \expo{-x}}\,\pars{-\expo{-x}}\,\dd x \,\,\,\stackrel{t\ =\ \expo{\large -x}}{=}\,\,\, -\,{1 \over 2\ic}\int_{0}^{1}{t^{\large\ic} - t^{\large -\ic} \over 1 - t} \,\dd t \\[5mm] & = -\,{1 \over 2\ic}\pars{H_{-\ic} - H_{\ic}}\qquad\qquad\qquad \pars{~H_{z}:\ Harmonic Number~} \\[5mm] & = -\,{1 \over 2\ic}\bracks{H_{-\ic} - \pars{H_{\ic - 1} + {1 \over \ic}}} \qquad\qquad\qquad\pars{~H_{z}\ Recurrence~} \\[5mm] & = -\,{1 \over 2} - {1 \over 2\ic}\bracks{H_{-\ic} - H_{\ic - 1}} = -\,{1 \over 2}-\,{1 \over 2\ic}\bracks{\pi\cot\pars{\pi\ic}} \pars{\substack{Euler \\[1mm] Reflection\ Formula}} \\[5mm] & = -\,{1 \over 2}-\,{\pi \over 2\ic}\bracks{-\ic\coth\pars{\pi}} = \bbx{{\pi \over 2}\,\coth\pars{\pi} - {1 \over 2}} \approx 1.077 \end{align}

Answer 4

Contour integration solution: We consider the integral

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

with $C\left(\epsilon,R\right)$ a rectangle from $\epsilon$ to $R$ on the real axis, from there to $R + 2\pi i$ parallel to the imaginary axis, from there to $\epsilon + 2\pi i$ parallel to the real axis, then a clockwise quarter turn with radius $\epsilon$ and center $2\pi i$ to the point $2\pi i - i\epsilon$, 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:

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

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:

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

We then see that

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

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 $1$, while the residue at $z = 2\pi i$ is $\exp(-2\pi)$. 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:

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