$\int_{-\infty}^{\infty}\frac{e^{\mathrm{i}tx}} {(\,t - \mathrm{i})^{n/2}}dt$, Fourier transform via contour integral with branch cut I am trying to evaluate the following Fourier transform 
$$F(x,n) =
\int_{-\infty}^{\infty}\frac{\mathrm{e}^{\mathrm{i}tx}}
{(\,t - \mathrm{i}\,)^{n/2}}\,\mathrm{d}t\,,\quad
\forall\ x \in \mathbb{R}^{+}\,,\ n\in \mathbb{N}.
$$

For even $n$, we can use the contour integral on the complex plane $t$ where the contour consists of the interval $\left[-a, a\right]$ on the real axis and a hemisphere on the upper half plane centered at the origin with a radius of $a$ for $a &gt; 0$. We obtain the result using Cauchy's differentiation formula. 

For odd natural number $n$, there is a branch cut for
$\displaystyle(t - \mathrm{i})^{\frac n 2}$. Now if we make a slit from
$t = \mathrm{i}$ and integrate
$\displaystyle\frac{e^{\mathrm{i}tx}}{\left(\,t -\mathrm{i}\,\right)^{\frac n 2}}$ starting from $t = \mathrm{i}$ we will have a divergent integral for $n \geq 3$. What could be done&nbsp;?
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
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  $\ds{\mrm{F}\pars{x,n} \equiv
\int_{-\infty}^{\infty}{\expo{\ic tx} \over \pars{t - \ic}^{n/2}}\,\dd t\,,\quad
\forall\ x \in \mathbb{R}\setminus\braces{0}\,,\ n\in \mathbb{N}_{\color{#f00}{\ \geq\ 1}}}$.



\begin{equation}\bbox[15px,#ffe,border:1px dotted navy]{\ds{%
\begin{array}{l}
\mbox{I'll consider a 'more general' case: Namely;}
\ds{\,\,\, x \in \mathbb{R}\setminus\braces{0}\,,\ n \in \mathbb{R}^{+}}.
\\[5mm]
\mbox{Lets consider,}
\\\ds{\quad
\left.\vphantom{\Large A}\mc{F}\pars{x,\alpha}
\right\vert_{~\substack{x\ \in\ \mathbb{R}\setminus{0}\\[0.5mm] \alpha\ \in\ \mathbb{R}^{\large+}}} \equiv
\int_{-\infty}^{\infty}{\expo{\ic tx} \over \pars{t - \ic}^{\alpha/2}}\,\dd t}
\quad\mbox{where}\ z^{\alpha/2}\ \mbox{is its}\ Principal\ Branch.
\end{array}}}\label{1}\tag{1}
\end{equation}

\begin{align}
\mc{F}\pars{x,\alpha} &amp; =
\expo{-x}\int_{-\infty - \ic}^{\infty - \ic}
{\expo{\ic xt} \over t^{\alpha/2}}\,\dd t
\,\,\,\stackrel{\substack{t\ =\ -\ic s\\ s\ =\ \phantom{-}\ic t\\[0.5mm] \mbox{}}}{=}\,\,\,
\expo{-x}\int_{\ic\pars{-\infty - \ic}}^{\ic\pars{\infty - \ic}}
{\expo{xs} \over \pars{-\ic s}^{\alpha/2}}\,\pars{-\ic}\,\dd s
\\[5mm] &amp; =
-\bracks{x &gt; 0}\ic\expo{\alpha\pi\ic/4}\expo{-x}\int_{1 - \infty\ic}^{1 + \infty\ic}
{\expo{xs} \over s^{\alpha/2}}\,\dd s\label{2}\tag{2}
\end{align}
The remaining integral is evaluated along a key-hole contour which
'takes care' of the $\ds{z^{\alpha/2}\mbox{-Branch-Cut}}$:
\begin{align}
\left.\int_{1 - \infty\ic}^{1 + \infty\ic}{\expo{xs} \over s^{\alpha/2}}\,\dd s
\,\right\vert_{\ x\ &gt;\ 0} &amp;
\,\,\,\stackrel{\mrm{as}\ \epsilon\ \to\ 0^{\large +}}{\sim}\,\,\,
-\int_{-\infty}^{-\epsilon}\pars{-s}^{-\alpha/2}\expo{-\alpha\pi\ic/2}\expo{xs}
\,\dd s
\\[2mm] &amp;
\phantom{\,\,\,\stackrel{\mrm{as}\ \epsilon\ \to\ 0^{\large +}}{\sim}\,\,\,}
-\int_{\pi}^{-\pi}\epsilon^{-\alpha/2}
\expo{-\ic\alpha\theta/2}\epsilon\expo{\ic\theta}\ic\,\dd\theta
\\[2mm] &amp;
\phantom{\,\,\,\stackrel{\mrm{as}\ \epsilon\ \to\ 0^{\large +}}{\sim}\,\,\,}
-\int_{-\epsilon}^{-\infty}\pars{-s}
^{-\alpha/2}\expo{\alpha\pi\ic/2}\expo{xs}\,\dd s
\\[5mm] &amp; =
-\expo{-\alpha\pi\ic/2}\int_{\epsilon}^{\infty}s^{-\alpha/2}\expo{-xs}\,\dd s +
{2\sin\pars{\alpha\pi/2}\,\epsilon^{1 - \alpha/2} \over 1 - \alpha/2}\,\ic
\\[2mm] &amp; +
\expo{\alpha\pi\ic/2}\int_{\epsilon}^{\infty}s^{-\alpha/2}\expo{-xs}\,\dd s
\\[5mm] &amp; =
2\ic\sin\pars{\alpha\pi \over 2}
\int_{\epsilon}^{\infty}s^{-\alpha/2}\expo{-xs}\,\dd s +
{2\sin\pars{\alpha\pi/2}\epsilon^{1 - \alpha/2} \over 1 - \alpha/2}\,\ic
\end{align}


  Note that I'll can consider the case where $\ds{2 \not\mid \alpha}$ such that the case $\ds{2 \mid \alpha}$ is a 'limiting one'.


Therefore,
\begin{align}
\left.\int_{1 - \infty\ic}^{1 + \infty\ic}{\expo{xs} \over s^{\alpha/2}}\,\dd s
\,\right\vert_{\ x\ &gt;\ 0} &amp;
\,\,\,\stackrel{\mrm{as}\ \epsilon\ \to\ 0^{\large +}}{\sim}\,\,\,
2\ic\sin\pars{\alpha\pi \over 2}\bracks{%
{1 \over x^{1 - \alpha/2}}\int_{x\epsilon}^{\infty}s^{-\alpha/2}\expo{-s}\,\dd s + {\epsilon^{1 - \alpha/2} \over 1 - \alpha/2}}
\\[5mm] &amp; =
2\ic\sin\pars{\alpha\pi \over 2}\bracks{%
x^{\alpha/2 - 1}\Gamma\pars{1 - {\alpha \over 2},x\epsilon} +
{\epsilon^{1 - \alpha/2} \over 1 - \alpha/2}}\label{3}\tag{3}
\end{align}


  $\ds{\Gamma\pars{1 - {\alpha \over 2},x\epsilon}}$ is the
  Incomplete Gamma Function.


In the present case $\large\left(\right.$&nbsp;see $\ds{\mathbf{\color{#000}{8.7.3}}}$ in DLMF&nbsp;${\large\left.\right)}$:&nbsp;$\ds{\pars{~1 - {\alpha \over 2} \not= 0,-1,-2,\ldots\ \mbox{or/and}\ \alpha \not= 2,4,6,\ldots~}}$ 
\begin{align}
&amp;x^{\alpha/2 - 1}\,\Gamma\pars{1 - {\alpha \over 2},x\epsilon}
\,\,\,\stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim}\,\,\,
x^{\alpha/2 - 1}\,\Gamma\pars{1 - {\alpha \over 2}}- {\epsilon^{1 - \alpha/2}\expo{-x\epsilon} \over 1 - \alpha/2}\label{3.a}\tag{3.a}
\\[5mm] &amp;
\bbx{\mbox{Note that the cases}\ \alpha = 2,4,6,\ldots\
\mbox{are 'limiting cases' of the final result.}}
\end{align}
\eqref{3} becomes, in the $\ds{\epsilon \to 0^{+}}$ limit,
\begin{align}
\left.\int_{1 - \infty\ic}^{1 + \infty\ic}{\expo{xs} \over s^{\alpha/2}}\,\dd s
\,\right\vert_{\ x\ &gt;\ 0} &amp; =
2\ic\sin\pars{\alpha\pi \over 2}\,x^{\alpha/2 - 1}
\,\Gamma\pars{1 - {\alpha \over 2}}
\\[5mm] &amp; =
2\ic\sin\pars{\alpha\pi \over 2}\,x^{\alpha/2 - 1}
\,{\pi \over \Gamma\pars{\alpha/2}\sin\pars{\alpha\pi/2}} =
{2\pi\ic \over \Gamma\pars{\alpha/2}}\,x^{\alpha/2 - 1}\label{4}\tag{4}
\end{align}

Finally, with \eqref{2} and \eqref{4}:
$$\bbox[15px,#ffe,border:1px dotted navy]{\ds{
\left.\vphantom{\Large A}\int_{-\infty}^{\infty}
{\expo{\ic tx} \over \pars{t - \ic}^{\alpha/2}}\,\dd t
\,\right\vert_{\ \substack{x\ \in\ \mathbb{R}\setminus\braces{0}\\[0.5mm]
\alpha\ \in\ \mathbb{R}^{+}}} =
\bracks{x &gt; 0}2\pi\,{\expo{\alpha\pi\ic/4} \over \Gamma\pars{\alpha/2}}\,\,
x^{\alpha/2 - 1}\expo{-x}}}
$$
It stays related to $\Gamma(s)$ and its properties. For $x &gt; 0$ and $n &gt; 0$
$$F(x)  =\int_{-\infty}^\infty \frac{e^{itx}}{(t-i)^{n/2}} dt=e^{-x}\int_{-\infty-i}^{\infty-i} \frac{e^{itx}}{t^{n/2}} dt=e^{-x}x^{n/2-1}\int_{-\infty-ix}^{\infty-ix} \frac{e^{i\tau}}{\tau^{n/2}} d\tau$$
$$ =F(1) e^{1-x}x^{n/2-1}$$
For $x &lt; 0$ I'm not sure.