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Consider the equation $$L[u(x,t)] = \tilde u(s,t) = \frac{e^{-t\sqrt{s^2-1}}}{s-2}$$ I want to find $u(x,t)$ in the form of an integral. I first need to find the poles and singularities of the function. I think it is right to say there is a pole at $s=2$, but I am wondering if the square root in the exp also causes branch points at $s=\pm 1$, if someone could clarify (and maybe explain why) that would be great.

I ultimately want to put $u(x,t)$ in the form of an integral around some contour and then find the leading asympotitic behavior for large $x$, which would be given by expanding $\tilde u(s,t)$ about the singularity with the largest real part, which if I am correct above would be $s=2$. Any help on this expansion too would be great. Here is what I have though: Rewrite $$\frac{e^{-t\sqrt{s^2-1}}}{s-2} = e^{-t\sqrt{s^2-1}} \cdot \left[ -\frac{1}{2} (1- \frac{s}{2})^{-1}\right] \approx e^{-t\sqrt{s^2-1}} \cdot \left[ -\frac{1}{2} (1 + \frac{s}{2})\right]\,\,\,\text{valid for} |s| < 2$$ So it is not valid since I want to expand about $s=2$, but I am not sure how to progress.

Many thanks

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