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I know the following result thanks to the technique "Integral Milking":

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

So I have a proof (I might list it here later, if it turns out this question seems very hard to solve) of the result, but I wouldn't be able to solve it if I would start with the integral. I tried a few things, e.g. expanding and substitution, but I didn't come anywhere. WolframAlpha doesn't have the closed-form, but you can check numerically if you want.

How would you solve the integral without knowing the result?

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4 Answers 4

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Divide the numerator and denominator by $e^{2\pi x}$: $$I=-\int_0^{\infty} \frac{e^{-2\pi x} \sin{(2x)}}{1-e^{-2\pi x}} \; dx$$ $$I=-\int_0^{\infty} \sum_{n=1}^{\infty} e^{-2\pi x n} \sin{(2x)} \; dx$$ Due to Fubini theorem we can interchange the summation and integral: $$I=-\sum_{n=1}^{\infty} \int_0^{\infty} e^{-2\pi x n} \sin{(2x)} \; dx$$ Then, use integration by parts: $$I=-\sum_{n=1}^{\infty} \frac{1}{2 \pi^2 n^2+2}$$ $$I=-\frac{1}{4} \left( \coth{1}-1\right)$$ $$I=\frac{1}{2-2e^2}$$

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HINT:

Expand the denominator as

$$\frac{1}{1-e^{2\pi x}}=-\sum_{n=0}^{\infty}e^{-2(n+1)\pi x}$$

Then note that this leaves a series

$$-\frac1{2}\,\sum_{n=1}^{\infty} \frac1{\pi^2 n^2+1}$$

The series can be found in closed form using for example contour integration or Fourier series and Parseval's theorem. See This Answer as an example.

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  • $\begingroup$ Your answer is really great, I might mark Ty.'s as the answer since he has the "full solution", or what do you think? (I'm not so experienced on this site yet) $\endgroup$ Jun 14, 2020 at 16:46
  • $\begingroup$ I've added a sentence with a link with examples of how to evaluate the series. That should be usefu in understanding how the closed form can be derived. $\endgroup$
    – Mark Viola
    Jun 14, 2020 at 16:52
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ This is an interesting application of the Abel-Plana Formula: \begin{align} {1 \over 1 - \expo{-2}} & = \sum_{n = 0}^{\infty}\expo{-2n} \\ & = \overbrace{\int_{0}^{\infty}\expo{-2n}\dd n} ^{\ds{1 \over 2}}\ +\ \overbrace{\left.{1 \over 2}\expo{-2n} \right\vert_{\ n\ =\ 0}}^{\ds{1 \over 2}}\ -\ 2\,\int_{0}^{\infty}{\Im\pars{\expo{-2\ic x}} \over \expo{2\pi x} - 1} \,\dd x \\[5mm] {1 \over 1 - \expo{-2}} & = {1 \over 2} + {1 \over 2} + 2\,\int_{0}^{\infty}{\sin\pars{2x} \over \expo{2\pi x} - 1}\,\dd x \\[5mm] \int_{0}^{\infty}{\sin\pars{2x} \over 1 - \expo{2\pi x}}\,\dd x & = {1 \over 2}\pars{1 - {1 \over 1 - \expo{-2}}} = \bbox[15px,#ffd,border:1px solid navy]{1 \over 2 - 2\expo{2}}\ \approx\ -0.0783 \\ & \end{align}


This integral was first evaluated by Legendre.

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For the computation of $$I=\int_0^{\infty} e^{-2\pi x n} \sin{(ax)} \; dx$$ you even do not need integration by parts. Write it as $$I=\Im\left(\int_0^{\infty} e^{-2\pi x n} e^{iax} \; dx \right)=\Im\left(\int_0^{\infty} e^{-(2\pi n-ia)x} \; dx \right)=\Im\left(\frac{1}{2 \pi n-i a}\right)=\frac{a}{4 \pi ^2 n^2+a^2}$$

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