Calculating the residues What are the residues at $s=0,\pm 1$ for the following function
$$I=\frac{1}{e^{2\pi is}-1}\frac{e^{isu}}{s^2(s^2-1)}?$$
Now it appears to me that, two pieces, $e^{2\pi is}$ and $s^2(s^2-1)$, diverge in the denominator. What are the residues then?
 A: Since you asked only for the residues and not how to get them:
$$\text{res}\left(I,0\right)=-\frac{i \left(3 u^2-6 \pi  u+2 \pi ^2-6\right)}{12 \pi }$$
$$\text{res}\left(I,1\right)=\frac{e^{i u} (2 u+5 i-2 \pi )}{8 \pi }$$
$$\text{res}\left(I,-1\right)=\frac{e^{-i u} (-2 u+5 i+2 \pi )}{8 \pi }$$

To get the first residue, we note that $s=0$ is a $4$th order pole of the function $I(s)$, then we must compute
$$\text{res} (I,0)={\frac {1}{3!}\lim _{s\to 0}{\frac {d^3}{ds^3}}}\left((s-0)^{4}I(s)\right)$$
which is quite "time consuming" :) (I did it with Mathematica)
$$\frac{1}{3!}\underset{s\to 0}{\text{lim}}\left(\frac{\left(-\frac{i e^{i s u} u^3}{-1+e^{2 i \pi  s}}+\frac{6 e^{i u s+2 i \pi  s} i \pi  u^2}{\left(-1+e^{2 i \pi  s}\right)^2}+3 e^{i s u} \left(\frac{4 e^{2 i \pi  s} \pi ^2}{\left(-1+e^{2 i \pi  s}\right)^2}-\frac{8 e^{4 i \pi  s} \pi ^2}{\left(-1+e^{2 i \pi  s}\right)^3}\right) i u+e^{i s u} \left(\frac{48 i e^{6 i \pi  s} \pi ^3}{\left(-1+e^{2 i \pi  s}\right)^4}-\frac{48 i e^{4 i \pi  s} \pi ^3}{\left(-1+e^{2 i \pi  s}\right)^3}+\frac{8 i e^{2 i \pi  s} \pi ^3}{\left(-1+e^{2 i \pi  s}\right)^2}\right)\right) s^2}{s^2-1}+3 \left(-\frac{e^{i s u} u^2}{-1+e^{2 i \pi  s}}+\frac{4 e^{i u s+2 i \pi  s} \pi  u}{\left(-1+e^{2 i \pi  s}\right)^2}+e^{i s u} \left(\frac{4 e^{2 i \pi  s} \pi ^2}{\left(-1+e^{2 i \pi  s}\right)^2}-\frac{8 e^{4 i \pi  s} \pi ^2}{\left(-1+e^{2 i \pi  s}\right)^3}\right)\right) \left(\frac{2 s}{s^2-1}-\frac{2 s^3}{\left(s^2-1\right)^2}\right)+\frac{e^{i s u} \left(\left(\frac{24 s}{\left(s^2-1\right)^3}-\frac{48 s^3}{\left(s^2-1\right)^4}\right) s^2+6 \left(\frac{8 s^2}{\left(s^2-1\right)^3}-\frac{2}{\left(s^2-1\right)^2}\right) s-\frac{12 s}{\left(s^2-1\right)^2}\right)}{-1+e^{2 i \pi  s}}+3 \left(-\frac{2 i e^{i u s+2 i \pi  s} \pi }{\left(-1+e^{2 i \pi  s}\right)^2}+\frac{i e^{i s u} u}{-1+e^{2 i \pi  s}}\right) \left(\left(\frac{8 s^2}{\left(s^2-1\right)^3}-\frac{2}{\left(s^2-1\right)^2}\right) s^2-\frac{8 s^2}{\left(s^2-1\right)^2}+\frac{2}{s^2-1}\right)\right)=\frac{i \left(-3 u^2+6 \pi  u-2 \pi ^2+6\right)}{12 \pi }$$
$s=1;\;s=-1$ are a double poles
$$\text{res}\left(I,1\right)=\underset{s\to 1}{\text{lim}}\frac{dI}{ds}(s-1)^2$$
and
$$\text{res}\left(I,-1\right)=\underset{s\to 1}{\text{lim}}\frac{dI}{ds}(s+1)^2$$
It is useful L'Hopital rule in the last two limits
