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?

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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}
