Help with $\int_0^\infty {\frac{{\sin t}}{{{e^t} - x}}dt} = \sum\limits_{n = 1}^\infty {\frac{{{x^{n - 1}}}}{{{n^2} + 1}}} $ The question is to show

$\int_0^\infty  {\frac{{\sin t}}{{{e^t} - x}}dt}  = \sum\limits_{n = 1}^\infty  {\frac{{{x^{n - 1}}}}{{{n^2} + 1}}} $ for $-1<x<1$.

The integration is a Lebesgue integration. I have worked on this for half an hour but I still don't know how to solve it except for I can brutally show the Taylor expansion of the left-hand side is consistent with the right-hand side.
Hope someone can help on this. Thank you!
 A: The solution method for these kinds of problems is almost always to expand the troublesome part (here $\frac{1}{e^t-x}$) of the integrand in some series and integrating term by term. In some cases (usually if the integrand has a periodic factor) one can split the integration over $[0,\infty)$ into pieces (e.g. $[0,2\pi]$,$[2\pi,4\pi]$,$\ldots$) and then performing some substitution to simplify each integral before solving integral-by-integral.

For this problem the former method seems to be the way to go. The term $\frac{1}{e^t-x}$ begs to be expanded in a power-series $e^{-t}\frac{1}{1-e^{-t}x} = \sum_{n=1}^\infty x^{n-1} e^{-nt}$. This series converges if $|x| e^{-t} < 1$ and since $t\in[0,\infty]$ this is satisfied for all $|x|<1$.
The tricky part is usually to justify interchanging the integration and summation. The conditions that has to be satisfied for this to be allowed is explained in this answer and for this problem we can safely do this if we can prove that
$$\sum_{n=1}^\infty\int_0^\infty |x^{n-1}\sin(t)e^{-nt}|\, {\rm d}t < \infty$$
By using $|\sin(t)| \leq 1$ we get that the integral above is bounded by $$\sum_{n=1}^\infty|x|^{n-1}\int_0^\infty e^{-nt}\, {\rm d}t = \sum_{n=1}^\infty \frac{|x|^{n-1}}{n}=-\frac{\log(1-|x|)}{|x|} < \infty$$ for all $|x|<1$. 

Finally we need to evaluate $\int_0^\infty \sin(t)e^{-nt}\, {\rm d}t$. This can be done in many ways. If you know contour integration then the integral can be written and evaluated as $\Im \int_0^\infty e^{(i-n)t}\, {\rm d}t = \Im\left(\frac{1}{i-n}\right) = \frac{1}{1+n^2}$. 
Another method one can use is integration by parts (two times)
$$\int_0^\infty \sin(t)e^{-nt}\, {\rm d}t = 1 - n\int_0^\infty \cos(t)e^{-nt}\, {\rm d}t$$
and
$$\int_0^\infty \cos(t)e^{-nt}\, {\rm d}t = n\int_0^\infty \sin(t)e^{-nt}\, {\rm d}t$$
which gives $\int_0^\infty \sin(t)e^{-nt}\, {\rm d}t = \frac{1}{1+n^2}$.

Putting all pieces togeather gives us the desired result
$$\int_0^\infty \frac{\sin(t)}{e^t-x}\,{\rm d}t = \sum_{n=1}^\infty \frac{x^{n-1}}{1+n^2}$$
for all $|x|<1$. The formula above can also be shown to hold for $x=\pm 1$ where it has closed form solutions. For example for $x=1$ the sum evaluates to $\frac{\pi\coth(\pi)-1}{2}$, see e.g. this answer.
A: Factor, 
$$ \frac{\sin t}{e^t - x} = e^{-t} \frac{\sin t}{1 - \frac{x}{e^t}} = e^{-t}\sin t \sum_{n=0}^{\infty} \left( \frac{x}{e^t} \right)^n   $$
Now integrate, 
$$ \int_0^{\infty} e^{-t(n+1)}\sin t ~ dt = \frac{1}{n^2 + 2n + 2} = \frac{1}{(n+1)^2+1} $$ 
Thus, we are left with, 
$$ \sum_{n=0}^{\infty} \frac{x^n}{(n+1)^2+1} = \sum_{n=1}^{\infty} \frac{x^{n-1}}{n^2+1} $$
A: HINT:
$$\frac{\sin t}{e^t-x}=\frac{e^{-t}\sin t}{1-xe^{-t}}=e^{-t}\sin t\sum_{k=0}^{\infty}e^{-kt}x^k$$
Then integrate term by term.
