How do you prove that $\int_0^\infty \frac{\sin(2x)}{1-e^{2\pi x}} dx = \frac{1}{2-2e^2}$? 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?
 A: 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:
$$I=-\sum_{n=1}^{\infty} \frac{1}{2 \pi^2 n^2+2}$$
$$I=-\frac{1}{4} \left( \coth{1}-1\right)$$
$$I=\frac{1}{2-2e^2}$$
A: 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.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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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.
A: 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
$$I=\Im\left(\int_0^{\infty}  e^{-2\pi x n} e^{iax} \; dx \right)=\Im\left(\int_0^{\infty}  e^{-(2\pi  n-ia)x}  \; dx \right)=\Im\left(\frac{1}{2 \pi  n-i a}\right)=\frac{a}{4 \pi ^2 n^2+a^2}$$
