Show that $\int_0^\infty \frac{\sin(xt)}{e^t-1}dt = \sum_{n=1}^\infty \frac{x}{n^2+x^2}$

Question

Show that $\int_0^\infty \frac{\sin(xt)}{e^t-1}dt = \sum_{n=1}^\infty \frac{x}{n^2+x^2}$.

From an earlier question I have $ \int_o^\infty \frac{x^p}{e^x-1}dx = p!\sum_{n=1}^\infty \frac{1}{n^{p+1}}$

I'm pretty sure I need to do something with the Taylor Expansion of $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} $ ... $ = \sum_{n=0}^\infty (-1)^{(n-1)}\frac{x^2n-1}{(2n-1)!} = \sum_{n=1}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)!}$, but I'm not sure what to do with it as $\sin(xt)$

Alernatively I was thinking that if I set $F(x) = \int_0^\infty \frac{\sin(xt)}{e^t-1}dt$, then $F(0) = 0$ and $|F(x)| \leq p!\sum_{n=1}^\infty \frac{1}{n^{p+1}}$ if $x^p \geq 1$, but again I get stuck here.

Answer 1

$$I=\int_0^\infty\frac{\sin(xt)}{e^t-1}dt=\Im\int_0^\infty\frac{e^{ixt}}{e^t-1}dt$$ and if $v=e^t-1$ $$I=\Im\int_0^\infty\frac{(v+1)^{ix-1}}{v}dv$$ then you could use the binomial expansion of $(1+y)^n=1+ny...$

Answer 2

Expand $\frac1{e^t-1}=\sum_{n\ge 1} e^{-nt }$ instead to obtain $$\int_0^\infty \frac{\sin xt}{e^t-1}dt = \sum_{n \ge1}\int_0^\infty \sin xt \ e^{-nt}dx=\sum_{n\ge 1}\frac{x}{n^2+x^2}$$