bromwich inverse laplace of $\frac{1}{\sqrt{s+1}}$

Question

I want to use the Bromwich integral to evaluate the inverse laplace of $\frac{1}{\sqrt{s+1}}$.

The complex function $\frac{e^{st}}{\sqrt{s+1}}$ has a pole and branch point in -1. I cannot find a good contour to evaluate. Am I right when I say the countour should enclose the poles but not cross any branch lines?

Answer 1

One evaluates the ILT by considering the contour integral

$$\oint_C dz \frac{e^{z t}}{\sqrt{z+1}} $$

where $C$ is a Bromwich contour that is deformed to avoid the branch point at $z=-1$. The deformation includes going up and back above and below, respectivly, along the negative real axis up to $z=-1$ (i.e., $z \in [-1,\infty)$ is a branch cut). The Bronwich contour includes arcs of radius $R$ in the left-half plane; these vanish in the limit as $R \to \infty$. (You would need to show this.) Thus, the contour integral is equal to

$$\int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{\sqrt{s+1}} + e^{i \pi} \int_{\infty}^1 dx \frac{e^{-x t}}{e^{i \pi/2} \sqrt{x-1}}+ e^{-i \pi} \int_1^{\infty} dx \frac{e^{-x t}}{e^{-i \pi/2} \sqrt{x-1}} $$

By Cauchy's theorem, the contour integral is zero. Thus, we have that the ILT is

$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{\sqrt{s+1}} = \frac1{\pi} \int_1^{\infty} dx \frac{e^{-x t}}{\sqrt{x-1}} = \frac{2}{\pi} e^{-t} \int_0^{\infty} dx \, e^{-t x^2}$$

or

$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{\sqrt{s+1}} = \frac{e^{-t}}{\sqrt{\pi t}} $$