I want to solve the following integral when $L$ is odd:
$$ f(x) = \int_{-\infty}^{\infty} \left[ \frac{1}{1 + \sigma^{4} t^{2}} \right]^{\frac{L}{2}} e^{-jtx} dt $$ which can be simplified to: $$ f(x) = \int_{-\infty}^{\infty} \frac{1}{ \left(1 + \sigma^{4} t^{2} \right)^{\frac{L}{2}} } e^{jt|x|} dt $$
So we can simplify the denominator:
$$ \left(1 + \sigma^{4} t^{2}\right)^{\frac{L}{2}} = \left(\sigma^{4}(t^{2} + \sigma^{-4}) \right)^{\frac{L}{2}} = \sigma^{4 \cdot \frac{L}{2}}(t^{2} + \sigma^{-4})^{\frac{L}{2}} $$
$$ \left(t^{2} + \sigma^{-4}\right)^{\frac{L}{2}} = \left[(t-j\sigma^{-2}) \cdot (t+j\sigma^{-2})\right]^{\frac{L}{2}} = (t-j\sigma^{-2})^{\frac{L}{2}} \cdot (t+j\sigma^{-2})^{\frac{L}{2}} $$
So our integral becomes:
$$ f(x) = \sigma^{-4 \cdot \frac{L}{2}} \int_{-\infty}^{\infty} \frac{e^{jt|x|}}{(t-j\sigma^{-2})^{\frac{L}{2}} \cdot (t+j\sigma^{-2})^{\frac{L}{2}} } dt $$
Here is where it gets fuzzy for me. Now when $L = 2M$ $$ f(x) = \sigma^{-4M} \int_{-\infty}^{\infty} \frac{e^{jt|x|}}{(t-j\sigma^{-2})^{M} \cdot (t+j\sigma^{-2})^{M} } dt $$ We can solve this by creating a contour $\Gamma$ in the upper half plane that encloses the pole $z = js$, where $s = \sigma^{-2}$ for ease of notation. It turns out that:
$$ f(x) = j s^{2M} \text{Res}\left\{ g(z) \right\}\Big|_{z = js} $$
where: $$ g(z) = \frac{e^{jz|x|}}{(z-js)^{M} \cdot (z+js)^{M} } $$
So now we need to evaluate the residue. We know that if $g(z)$ has a pole of order $M$ at $z=z_{0}$, then:
$$ \text{Res}\left\{ g(z) \right\}\Big|_{z = z_{0}} = \underset{z \rightarrow z_{0}}{\text{lim}} \; \frac{1}{(M-1)!} \frac{d^{M-1}}{dz^{M-1}} \left[ (z-z_{0})^{M} g(z) \right] $$
Now we can solve this via the Generalized Leibniz Rule and we get some answer.
Question:
What happens when $L = 2K+1$? In this case, we would have a fractional derivative, ie $M = K + \frac{1}{2}$. How do we evaluate the residue here? Is this even the right thing to do anymore?
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \mbox{Note that}\quad\on{f}\pars{x} & \equiv \bbox[5px,#ffd]{ \int_{-\infty}^{\infty} \pars{1 \over 1 + \sigma^{4} t^{2}}^{L/2} \expo{-\ic tx}\,\dd t} \\[2mm] & = {1 \over \sigma^{2}}\int_{-\infty}^{\infty} {\expo{-\ic\verts{\xi}t} \over \pars{1 + t^{2}}^{L/2}} \,\dd t\,,\qquad \xi \equiv {x \over \sigma^{2}} \end{align}
With $\ds{\xi \in \mathbb{R}}$ and $\ds{L \in \mathbb{R}\setminus\left[2,\infty\right)}$, the integration is switched to an integration in the "left complex plane" around a key-hole contour which "takes care" of the $\ds{\root{1 + z}}$-principal branch-cut ( the arc contribution vanishes out as its radius $\ds{\to \infty}$ ). Namely, \begin{align} &\bbox[5px,#ffd]{\int_{-\infty}^{\infty} {\expo{-\ic\verts{\xi}t} \over \pars{1 + t^{2}}^{L/2}} \,\dd t} \\[5mm] \stackrel{t\ \mapsto\ \ic\,t}{=}\,\,\,& -\ic\int_{-\infty\ic}^{\infty\ic} {\expo{\verts{\xi}t} \over \pars{1 - t}^{L/2}\ \pars{1 + t}^{L/2}}\,\dd t \\[5mm] = &\ \ic\int_{-\infty}^{-1} {\expo{\verts{\xi}t} \over \pars{1 - t}^{L/2}\ \pars{-1 - t}^{L/2}\expo{\ic\pi L/2}}\,\dd t \\[2mm] + &\ \ic\int_{-1}^{-\infty} {\expo{\verts{\xi}t} \over \pars{1 - t}^{L/2}\ \pars{-1 - t}^{L/2}\expo{-\ic\pi L/2}}\,\dd t \\[5mm] = &\ \ic\expo{-\ic\pi L/2}\int_{1}^{\infty} {\expo{-\verts{\xi}t} \over \pars{1 + t}^{L/2}\ \pars{-1 + t}^{L/2}}\,\dd t \\[2mm] - &\ \ic\expo{\ic\pi L/2}\int_{1}^{\infty} {\expo{-\verts{\xi}t} \over \pars{1 + t}^{L/2}\ \pars{-1 + t}^{L/2}}\,\dd t \\[5mm] = &\ 2\sin\pars{\pi L \over 2}\int_{1}^{\infty} \expo{-\verts{\xi}t}\pars{t^{2} - 1}^{-L/2}\ \dd t \\[5mm] = &\ {2^{3/2 - L/2} \over \root{\pi}}\, \sin\pars{\pi L \over 2}\,\,\, \verts{\xi}^{L/2 - 1/2}\,\,\, \on{K}_{\pars{1 - L}/2}\ \pars{\verts{\xi}} \\[2mm] &\ \Gamma\pars{1 - {L \over 2}} \label{1}\tag{1} \\[5mm] = &\ \root{\pi}\,{2^{3/2 - L/2} \over \Gamma\pars{L/2}} \,\,\,\verts{\xi}^{L/2 - 1/2}\,\,\, \on{K}_{\pars{1 - L}/2}\ \pars{\verts{\xi}} \label{2}\tag{2} \end{align}