How would you calculate the residue of:
$$\frac{\sinh(p^{1/2}(1-x))}{p\sinh(p^{1/2})}$$
Where the variable of interest is $p$.
I think the poles are:
$p = 0 $ and $p = -(n\pi)^2$
I'm not sure how to go about even determining the order. I'm guessing the pole at $0$ is of order $2$ and all the other ones are of order $1$?
Thanks in advance!
The poles are simple and their residues can be evaluated accordingly. For the residues at $p=-n^2\pi^2$ are given by
$$\begin{align} \lim_{p\to -n^2\pi^2}(p+n^2\pi^2)\frac{\sinh((1-x)p^{1/2})}{p\sinh(p^{1/2})}&=\lim_{p\to -n^2\pi^2}\frac{\sinh((1-x)p^{1/2})}{\sinh(p^{1/2})+\frac12p^{1/2}\cosh(p^{1/2})}\\\\ &=\frac{\sinh((1-x)in\pi)}{\frac12 \cosh(in\pi)/(in\pi)}\\\\ &=-2\frac{\sin(n\pi x)}{n\pi} \end{align}$$
The residue at $p=0$ is given by
$$\begin{align} \lim_{p\to 0}p\frac{\sinh((1-x)p^{1/2})}{p\sinh(p^{1/2})}&=\lim_{p\to 0}\frac{\sinh((1-x)p^{1/2})}{\sinh(p^{1/2})}\\\\ &=1-x \end{align}$$