# 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?

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: $$I=-\sum_{n=1}^{\infty} \frac{1}{2 \pi^2 n^2+2}$$ $$I=-\frac{1}{4} \left( \coth{1}-1\right)$$ $$I=\frac{1}{2-2e^2}$$

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.

• 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) Jun 14, 2020 at 16:46
• 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. Jun 14, 2020 at 16:52


This integral was first evaluated by Legendre.

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 $$I=\Im\left(\int_0^{\infty} e^{-2\pi x n} e^{iax} \; dx \right)=\Im\left(\int_0^{\infty} e^{-(2\pi n-ia)x} \; dx \right)=\Im\left(\frac{1}{2 \pi n-i a}\right)=\frac{a}{4 \pi ^2 n^2+a^2}$$