Complex integral with non integer pole multiplicity I'm trying to calculate the following complex integral:
$$
\int\limits_{\sigma-i\infty}^{\sigma+i\infty}\frac{1}{s^\alpha}e^{as}\,ds
$$
where $a>0$, $\sigma\in\mathbb{R}$ and $\alpha\in (0,1)$. So far I have tried to use the residue theorem, but as $\alpha$ is not an integer it does not work. I have tried also to do a change of variables to get only integer exponents (without success so far).
Maybe someone can give some advice on this.
Thank you for reading.
 A: You are looking for the inverse Laplace transform of $\frac{1}{s^\alpha}$, and it is useful to recall that
$$\forall\beta>-1,\qquad (\mathcal{L}x^\beta)(s) = \int_{0}^{+\infty} x^{\beta} e^{-sx}\,dx = \frac{\Gamma(\beta+1)}{s^{\beta+1}} \tag{1}$$
from which
$$\forall \alpha\in(0,1),\qquad  \left(\mathcal{L}^{-1}\frac{1}{s^\alpha}\right)(a) = \frac{a^{\alpha-1}}{\Gamma(\alpha)}\tag{2} $$
nice and easy.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Hereafter, I'll show an explicit evaluation:

$\ds{\left.z^{-\alpha}\right\vert_{\ z\ \not=\ 0} = \verts{z}^{-\alpha}\exp\pars{-\alpha\,\mrm{arg}\pars{z}\ic}\,,\qquad -\pi < \,\mrm{arg}\pars{z} < \pi}$.

With $\ds{a > 0}$, $\ds{\sigma \in \mathbb{R}}$ and $\ds{\alpha \in \pars{0,1}}$, it's clear that the following integral vanishes out whenever $\ds{\sigma < 0}$. Then,
\begin{align}
&\left.\int_{\sigma - \infty\ic}^{\sigma + \infty\ic}{1 \over s^{\alpha}}\,\expo{as}\,\dd s\,\right\vert_{\ \sigma\ >\ 0}
\\[5mm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim}\,\,\,&\
-\int_{-\infty}^{-\epsilon}\pars{-s}^{-\alpha}\expo{-\ic\pi\alpha}\expo{as}
\,\dd s -
\int_{\pi}^{-\pi}\epsilon^{-\alpha}\expo{-\ic\alpha\theta}
\epsilon\expo{\ic\theta}\ic\,\dd\theta -
\int_{-\epsilon}^{-\infty}\pars{-s}^{-\alpha}\expo{\ic\pi\alpha}\expo{as}\,\dd s
\\[5mm]= &\
-\expo{-\ic\pi\alpha}\int_{\epsilon}^{\infty}s^{-\alpha}\expo{-as}
\,\dd s -
2\ic\,\epsilon^{1 - \alpha}\,{\sin\pars{\pi\alpha} \over \alpha - 1} +
\expo{\ic\pi\alpha}\int_{\epsilon}^{\infty}s^{-\alpha}\expo{-as}\,\dd s
\\[5mm] & =
2\ic\sin\pars{\pi\alpha}\int_{\epsilon}^{\infty}s^{-\alpha}\expo{-as} \,\dd s -
2\ic\,\epsilon^{1 - \alpha}\,{\sin\pars{\pi\alpha} \over \alpha - 1}
\\[5mm] & =
2\ic\sin\pars{\pi\alpha}\bracks{%
-\,{\epsilon^{-\alpha + 1}\expo{-a\epsilon} \over -\alpha + 1} - \int_{\epsilon}^{\infty}{s^{-\alpha + 1} \over -\alpha + 1}\,\expo{-as}\pars{-a} \,\dd s} -
2\ic\,\epsilon^{1 - \alpha}\,{\sin\pars{\pi\alpha} \over \alpha - 1}
\\[5mm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\to}\,\,\,&\
-2\ic\sin\pars{\pi\alpha}\,{a \over \alpha - 1}
\int_{0}^{\infty}s^{-\alpha + 1}\,\expo{-as}\,\dd s =
-2\ic\sin\pars{\pi\alpha}\,{a \over \alpha - 1}\,{1 \over a^{2 - \alpha}}
\int_{0}^{\infty}s^{-\alpha + 1}\,\expo{-s}\,\dd s
\\[5mm] & =
-2\ic\sin\pars{\pi\alpha}\,{1 \over \alpha - 1}\,{1 \over a^{1 - \alpha}}\,
\Gamma\pars{2 - \alpha} =
2\ic\sin\pars{\pi\alpha}\,{1 \over a^{1 - \alpha}}\,\Gamma\pars{1 - \alpha}
\\[5mm] = &\
{2\ic \over a^{1 - \alpha}}\,\Gamma\pars{1 - \alpha}\sin\pars{\pi\alpha} =
{2\pi\ic \over a^{1 - \alpha}\,\Gamma\pars{\alpha}}
\end{align}

$$\bbox[15px,#ffe,border:1px dotted navy]{\ds{%
\left.\int_{\sigma - \infty\ic}^{\sigma + \infty\ic}
{1 \over s^{\alpha}}\,\expo{as}\,\dd s\,
\right\vert_{\large{\alpha\ \in\ \pars{0,1} \atop a\ >\ 0}} =
\left\{\begin{array}{lcl}
\ds{0} & \mbox{if} & \ds{\sigma < 0}
\\[2mm]
\ds{2\pi\ic\,{a^{\alpha - 1}\, \over \Gamma\pars{\alpha}}} & \mbox{if} & \ds{\sigma > 0}
\end{array}\right.}}
$$
