I have $\frac{1}{2 \pi i} \int_{\infty-iT}^{\infty+iT} \frac{e^{-s(1-t)}-e^{st}}{-s+e^{-s}-1} ds$ and I am trying to solve it using a contour. So I could have t>0, or t<0. I have a pole at 0. For t<0 I got that is equal to 0 because I have a contour with no residue in it. For t>0 I have a contour with a vertical line on the positive side with a semicircle extending to the left to close it. So because the pole is inside, I expect the solution to be $2\pi i \sum residues$ where my residue is at 0.
So I have $2\pi i \frac{1}{2 \pi i} \frac{e^{-s(1-t)}-e^{st}}{-s+e^{-s}-1} (s-0)ds$. Which leaves me with $\frac{e^{-s(1-t)}-e^{st}}{-s+e^{-s}-1} s$ in the limit that s approaches 0. When I solve it I get 0. This doesn't seem right to me, so I am wondering if I am misunderstanding a part of it. Any help or direction is welcomed.
The problem is that the denominator is messy. Rewrite your integrand as
$$\frac{1-e^{-s}}{1+s-e^{-s}} e^{s t}$$
There really is not a pole at $s=0$ because the singularity is removed by the numerator. On the other hand, you may have zeroes in the complex plane - these are related to Lambert's W-function and may take some work to track down. You will also need to worry about selecting a branch for those zeroes. Here is a table of some zeroes that correspond to a particular branch of the W-function.