Let $a>0$ we want to show that
$$\int_{0}^{\infty}\dfrac{\sin(t)}{e^{at}-1}=\sum_{0}^{\infty}\dfrac{1}{a^2n^2+1}$$
I assume that we want to find the power series expansion of $\dfrac{\sin(t)}{e^{at}-1}$ and then interchange the integral and the sum, but I was not able to find a closed formulafor the power series expansion, and I was only able to write the few first terms of the expansion which does not seem to have a clear pattern..
The second question is to find an equivalent to $\int_{0}^{\infty}\dfrac{\sin(t)}{e^{at}-1}$ when $a\rightarrow \infty$.
My try $\dfrac{1}{n^2a^2+1}\sim \dfrac{1}{a^2n^2}$ hence
$$\int_{0}^{\infty}\dfrac{\sin(t)}{e^{at}-1}\sim \sum_{0}^{\infty}\dfrac{1}{a^2n^2}=\dfrac{\pi^2}{6a^2}$$ Is this correct?
Thank you for your help!!
You might try using this. $${1\over e^{at} - 1} = {e^{-at} \over 1 - e^{-at}} = \sum_{n=1}^\infty e^{-ant}.$$