how to find residues of $\frac{e^{st}}{\cosh(a\sqrt{s})}$? Can someone give me a hint on how to find residues of  $\displaystyle\frac{e^{st}}{\cosh(a\sqrt{s})}$, where $a\in\mathbb{R}$?
I am trying to solve an integral using residue method (actually inverse Laplace transform).
I know the zeros of $\cosh(x)$ are $(\pm\frac{-i\pi}{2})+2i\pi n$ for $n\in\mathbb{Z}$, but I do not know how to write the denominator in factor form in order to cancel a zero and take the limit as $s\to\mathrm{pole}$. How do I use the above information on zeros $\cosh(x)$ in order to find the residue?
 A: The residue of a function $f(z)$ at $z_0$  (assuming it has a simple pole  -- do yours?) is equal to $\lim_{z\rightarrow z_0} (z-z_0) f(z).$ Now, use L'Hopital for your particular function.
A: How about this - Assuming $\sqrt{z}$ can be defined uniquely in the domain;
$\dfrac{e^{zt}}{cosh(a\sqrt{z})} = \dfrac{2e^{zt}}{e^{a\sqrt{z}}+e^{-a\sqrt{z}}}$
$=\dfrac{2e^{zt+a\sqrt{z}}}{1+ e^{2a\sqrt{z}}} $. This function will have poles at all the points where $ e^{2a\sqrt{z}} = -1 = e^{i(2k+1)\pi}$. This means $\sqrt{z} = i(2k+1)\pi/(2a)$.
Now $\sqrt{z} = e^{1/2\log{z}} = e^{1/2\ln{|z|}}e^{i/2 \arg{z}}$. From the last result we know  $\sqrt{z}$ is purely imaginary. Hence assuming $ 1/2 \arg{z} = \theta$ we can write $\cos{\theta} = 0$ which means $\arg{z} = n\pi$.
Also $\sqrt{|z|} = \dfrac{(2k+1)\pi}{2a}$ which gives us $|z| =\dfrac{(4m+1)\pi^2}{4a^2}$.
Thus we know the exact locations of the poles. We need to check if they are simple poles.
If $f(z) = 1+ e^{2a\sqrt{z}}$,
then  $f'(z) = \dfrac{ae^{2a\sqrt{z}}}{\sqrt{z}}$. Since we initially assumed that $\sqrt{z}$ was well defined, that is $\log{z}$ was defined, hence $z=0$ cannot be in the domain and $f'(z) \neq 0$. Thus all the zeros of $f(z)$ are simple zeros and hence the poles are simple poles. 
However this will make the calculation of residues very messy unless some special tricks can be used from the contour. 
