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.
On
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ 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}
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.