# Laplace transform of a product of functions

While trying to compute the Laplace transform of a certain product, part of the calculation leaves me with a Bromwich integral which has the form:

$$\mathcal{G}(s)=\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{1}{p(s-p)}sech\left(\sqrt{2p}\right) sech \left(\sqrt{2(s-p)}\right)\:dp,\:\: s,p \in \mathbb{C},\: \gamma \in \mathbb{R}.$$

It is easy to see the simple poles for this are $p=0$, $p=s$, and writing $sech(\sqrt{2p})$ as $1/\cos\left(i\sqrt{2p}\right)$, there are poles for $n \in \mathbb{Z}$:

$$p=-\frac{\pi^{2}(n+1/2)^{2}}{2}.$$

Similarly, for $sech(\sqrt{2(s-p)})$ poles are located at:

$$p=s+\frac{\pi^{2}(n+1/2)^{2}}{2}.$$

From the bits I have read about Laplace inversion, $\gamma$ is supposed to be chosen so that the real part of any poles (say in some set $L$) are less than $\gamma$: or $$\gamma= \max_{w \in L} \text{Real}(w).$$

My plan was to define a suitable Bromwich contour and make use of residues - although with the poles extending to $\pm \infty$ on the real axis, I'm uncertain how to construct a suitable contour...has anyone does this before, or know how to proceed? Thanks

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I'm really not so sure that you should be including poles of $\text{sech}{(\sqrt{2 p})}$ inside your contour, as these are branch points. – Ron Gordon Apr 2 '13 at 12:50
Yes, thank you. That makes sense. Will give that a try. – LaplaceKis Apr 8 '13 at 11:44