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

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

A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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\begin{align}
&\bbox[15px,#ffd]{\int_{\pars{-1}^{\large +} - \infty\ic}
^{\pars{-1}^{\large +} + \infty\ic}{1 \over \root{s + 1}}\,\expo{ts}{\dd s \over 2\pi\ic}} =
\expo{-t}\int_{0^{\large +} - \infty\ic}
^{0^{\large +} + \infty\ic}s^{-1/2}\expo{ts}{\dd s \over 2\pi\ic}
\\[5mm] = &\
-\expo{-t}\int_{-\infty}^{0}
\bracks{\pars{-s}\expo{\ic\pi}}^{-1/2}\expo{ts}
{\dd s \over 2\pi\ic} -\expo{-t}\int_{0}^{-\infty}
\bracks{\pars{-s}\expo{-\ic\pi}}^{-1/2}\expo{ts}
{\dd s \over 2\pi\ic}
\\[5mm] = &\
\ic\expo{-t}\int_{0}^{\infty}s^{-1/2}\expo{-ts}
{\dd s \over 2\pi\ic} +
\ic\expo{-t}\int_{0}^{\infty}s^{-1/2}\expo{-ts}
{\dd s \over 2\pi\ic}
\\[5mm] = &\
{\expo{-t} \over \pi}\int_{0}^{\infty}s^{-1/2}\expo{-ts}\dd s =
{\expo{-t}t^{-1/2} \over \pi}\int_{0}^{\infty}s^{-1/2}\expo{-s}\dd s =
{\expo{-t}t^{-1/2} \over \pi}\
\overbrace{\Gamma\pars{1 \over 2}}^{\ds{\root{\pi}}}
\\[5mm] = &\ \bbx{\expo{-t} \over \root{\pi t}}
\end{align}
