Inverse Fourier transform of $(z-i/2)^{-N/2}$ How can I compute the integral corresponding to the inverse Fourier transform of $(z-i/2)^{-N/2}$:
\begin{equation}
I \equiv \frac{1}{2\pi}\int_{-\infty}^{+\infty}dz\frac{e^{itz}}{\left(z-\frac{i}{2}\right)^{\frac{N}{2}}} ?
\end{equation}
I reached this integral while studying the $\chi^2$ distribution, for which the characteristic function is $(1-2ik)^{-1/2}$.
My ideia was to consider the order of the pole as being $\frac{N}{2}$ and applying the residues theorem, with the residue of order $n$ being 
\begin{equation}
\mathrm{Res}^{(n)}\left(f(z),c\right)=\frac{1}{(n-1)!}\displaystyle\lim_{z\rightarrow c}\left\{ \frac{d^{n-1}}{dz^{n-1}}\left[(z-c)^{n}f(z)\right]\right\}
\end{equation}
In our case, $n=\frac{N}{2}$, thus
\begin{equation}
\mathrm{Res}^{\left(\frac{N}{2}\right)}\left(f(z),c\right)=\frac{1}{\Gamma\left( \frac{N}{2}\right)}\displaystyle\lim_{z\rightarrow c}\left\{ \frac{d^{\frac{N}{2}-1}}{dz^{\frac{N}{2}-1}}\left[\left(z-\frac{i}{2}\right)^{\frac{N}{2}}\frac{e^{itz}}{\left(z-\frac{i}{2}\right)^{\frac{N}{2}}}\right]\right\}
\end{equation}
\begin{equation}
\Rightarrow \mathrm{Res}^{\left(\frac{N}{2}\right)}\left(f(z),c\right)= \frac{1}{\Gamma\left( \frac{N}{2}\right)} (it)^{\frac{N}{2}-1} e^{-\frac{t}{2}},
\end{equation}
where I have used a result from fractional calculus to compute the derivative of the exponential function: $D^\alpha e^{\lambda x} = \lambda^\alpha e^{\lambda x}$. Therefore,
\begin{equation}
I=2\pi i \mathrm{Res}^{\left(\frac{N}{2}\right)}\left(f(z),c\right) = \frac{1}{\Gamma\left( \frac{N}{2}\right)} i^{\frac{N}{2}} t^{\frac{N}{2}-1} e^{-\frac{t}{2}}.
\end{equation}
I am under the impression that this is not the way to do it, because I've never heard about such thing as a non-integer pole. I need some opinion on this.
Edit based on @paul garrett 's answer
We use the identity
\begin{equation}
\int_{0} ^{+\infty} \frac{dt}{t} t^s e^{-tx} = x^{-s}\Gamma(s)
\end{equation}
and make the changes $t\rightarrow it$ and $k\rightarrow \left(k - \frac{i}{2}\right)$, now with $s=\frac{N}{2}$:
\begin{equation}
i^{\frac{N}{2}} \int_{0}^{+\infty} dt \hspace{2pt} t^{\frac{N}{2}-1} e^{-itk} e^{-\frac{t}{2}} = \left(k - \frac{i}{2}\right)^{-\frac{N}{2}} \Gamma\left(\frac{N}{2}\right).
\end{equation}
Taking the inverse Fourier transform, we get for $t>0$,
\begin{equation}
i^{\frac{N}{2}} \frac{1}{2\pi} \int_{-\infty} ^{+\infty} dk \hspace{2pt} e^{itk} \int_{0}^{+i\infty} dt' \hspace{2pt} {t'}^{\frac{N}{2}-1} e^{-it'k} e^{-\frac{t'}{2}} \\ = \frac{1}{2\pi} \int_{-\infty} ^{+\infty} dk \hspace{2pt} e^{itk} \left(k - \frac{i}{2}\right)^{-\frac{N}{2}} \Gamma\left(\frac{N}{2}\right).
\end{equation}
\begin{equation}
\Rightarrow i^{\frac{N}{2}} \int_{0}^{+\infty} dt' \hspace{2pt} \delta(t-t') \hspace{2pt} {t'}^{\frac{N}{2}-1} e^{-\frac{t'}{2}} = \frac{1}{2\pi} \int_{-\infty} ^{+\infty} dk e^{itk} \left(k - \frac{i}{2}\right)^{-\frac{N}{2}} \Gamma\left(\frac{N}{2}\right).
\end{equation}
\begin{equation}
\Rightarrow \frac{1}{\Gamma\left(\frac{N}{2}\right)} i^{\frac{N}{2}} t^{\frac{N}{2}-1} e^{-\frac{t}{2}} = \frac{1}{2\pi} \int_{-\infty} ^{+\infty} dk \frac{e^{itk}}{ \left(k - \frac{i}{2}\right)^{\frac{N}{2}} }, \hspace{5pt} t>0
\end{equation}
which is our initially defined integral $I(t>0)$. The case $t<0$ corresponds to $t\neq t'$, for which $\delta(t-t')$ is null; thus, $I(t<0)=0$.
 A: You have two problems: $N$ even and $N$ odd. When $N$ is even, it's easier to find the residue by using the expansion of
$\exp\left({\rm i}\left\lbrack t - {\rm i}/2\right\rbrack\right)$. When $N$ is odd, you have some branch cut as related to the denominator.
A: One approach, important in many applications of all kinds, is to start with a change-of-variables identity
$$
\int_0^\infty t^s\, e^{-tx}\;{dt\over t}
\;=\; x^{-s}\cdot \int_0^\infty t^s\,e^{-t}\;{dt\over t}
\;=\; x^{-s}\cdot \Gamma(s)
$$
at least for $\Re(s)>0$ and real $x>0$. By the identity principle,
$$
\int_0^\infty t^s\,e^{-t(x+iy)}\;{dt\over t} \;=\; (x+iy)^{-s}\Gamma(s)
$$
for all real $y$. Let $f_x(t)$ be $0$ for $t<0$ and $t^{s-1}\,e^{-tx}$ for $t>0$. Then the previous identity can be construed as asserting that (up to normalizing constant...) the Fourier transform of $f_x$ is $(x+iy)^{-s}\cdot \Gamma(s)$. By Fourier inversion, we learn the Fourier transform of $y\to (x+iy)^{-s}$.
(This device is at least 100 years old, if not 200.)
Edit: that is, explicitly,
$$
{1\over 2\pi} \int_{-\infty}^\infty {e^{itx}\;dx\over x+iy)^s} \;=\; \left\{
\matrix{
\Gamma(s)^{-1} t^{s-1}\,e^{-t} & \hbox{(for $t>0$)}
\cr
\cr
0 & \hbox{(for $t<0$)}
}\right.
$$
