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I would like to get a hint how to integrate following function:

$$\int_{0}^{\infty} \frac{\sin(t)}{e^t-1} dt$$

I guess i have to multiply or add some constant t somewhere, then differentiate the "inner" term for d/dt and then smartly substitute to get an answer which should (i guess again) have the form of sin and exp. And in the last step I would have to get the right value for t, to get back to the original integral.

The following task would be to show the equality of $ \int_{0}^{\infty} \frac{\sin(t)}{e^t-1} dt = \sum_{n=1}^{\infty} \frac{1}{1+n^2}$. That's why I think the solution is some representation of sin and exp. With a representation of the integral in sin and exp terms, this part would get easy.

Greetings

Simply Beautiful Art
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Dedechild
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    You can expand $(e^t -1)^{-1} = -\sum_{k=0}^{\infty}e^{kt}$ and then do the resulting integral (by parts twice should do the trick) per each term in series. This will prove the result. – fGDu94 Nov 28 '19 at 02:51

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