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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?

Daniel W. Farlow
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Nasser
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2 Answers2

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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.

Igor Rivin
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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.

Sourav D
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