Fourier transform of $(x-c)^{-p/q}$, $p$ and $q$ co-prime. I am trying to do 
$$\int_{-\infty}^\infty e^{iax} (x-c)^{-p/q} dx$$
for co-prime $p,q>0$ and $c\in \mathbb{C}$, $\mathrm{Re}(c)=0$. The case $q=1$ is in the complex analysis textbooks ($c$ is a pole of order $p$ then). $q=2$ and $p=1$ is slightly harder, but still not bad: a branch cut along the $i$-axis, two large quarter-circles, a small circle around $c$, and Jordan Lemma do the job (look here for even more general 2 roots in the denominator). I could not find anything on the general case, even for $q=2$ and $p>2$.
The case $q=1$ uses 
$$\oint_C \frac{e^{iax}}{(x-c)^p} dx=\frac{2\pi i}{(p-1)!} 
\left(\frac{d^{p-1}}{d x^{p-1}} e^{iax}\right)\bigg|_{x=c}$$ 
for a contour $C$ around $c$. This brought me to the somewhat esoteric "Fractional calculus" which more or less takes the above as the definition of a generalized derivative. But I am not sure if there is anything like "Complex fractional calculus" and how does that help me with the integral. At this point, I have the feeling that my lack of basic understanding of complex analysis might have taken me too far afield... 
It seems to me that $\Gamma$ function, Riemann surface etc will have to appear in the solution, but I am not sure how.
 A: Here is what I worked up for $q=2$. For $x>0$
$
\DeclareMathOperator{\Res}{Res}
$
\begin{gather*}
f(x)&\stackrel{def}=\frac{1}{2\pi}\int_{-\infty}^{\infty} \frac{e^{itx}}{(1+it)^{\frac{n}{2}}}dt=\frac{ x^{\frac{n}{2}-1} e^{-x}}{\Gamma\left(\frac{n}{2}\right)}
\end{gather*}
This is not surprising, because I was actually trying to get the $\chi^2_n$-distribution by inverting its characteristic function. In this case the denominator looks like $1+2it$ which results in a $2^{-\frac{n}{2}}$ factor in the result.
Proof:
\begin{align*}
f(x)\stackrel{i(\frac{z}{x}+1)}{=} \frac{i x^{\frac{n}{2}-1}e^{-x}}{2 \pi (-1)^\frac{n}{2}}\int_{-x+i\infty}^{-x-i\infty} z^{-\frac{n}{2}}e^{-z}dz
\end{align*}
I needed the following two.
More or less Jordans' Lemma:
\begin{align*}
\int_{-\pi/2}^{\pi/2} e^{-x \cos\theta}d\theta&=\int_0^{\pi} e^{-x \sin\theta}d\theta= 2\int_0^{\pi/2} e^{-x\sin\theta}d\theta\leq 2\int_0^{\pi/2} e^{- 2 x\theta/\pi}d\theta\\
&\stackrel{-\pi\theta/{2x}}{=}\frac{\pi}{2 x}\int_{-x}^0 e^{\theta}d\theta=O(x^{-1})
\end{align*}
This is kind of trivial:
\begin{align*}
\int_{0}^{2 \pi} e^{i a \theta}e^{-b e^{i \theta}}d\theta
&=\frac{1}{ia} \int_{0}^{2 \pi} e^{-b e^{i \theta}}d e^{i a \theta}
=-\frac{1}{ia} \int_{0}^{2 \pi} e^{i a\theta }d e^{-b e^{i \theta}}\\
&=\frac{b}{a} \int_{0}^{2\pi} e^{i (a+1) \theta} e^{-b e^{i \theta}} d\theta
\end{align*}
\begin{gather*}
g(z)\stackrel{def}{=}z^{-\frac{n}{2}}e^{-z}
\end{gather*}
Case $n=2m$.
\begin{gather*}
C_1=[R, -R]i-x,\  
C_2=Re^{i [-\pi/2, \pi/2]}-x
\end{gather*}
\begin{align*}
\Res(g(z), 0)=\frac{1}{\Gamma(m)}\frac{ d^{m-1} }{dz^{m-1}} e^{-z} \bigg|_{0}=\frac{(-1)^{m-1}}{\Gamma(m)}
=\frac{1}{2\pi i} \left( \int_{C_1} g(z) dz +  \int_{C_2} g(z) dz \right)
\end{align*}
\begin{align*}
\left|\int_{C_2} g(z) dz\right|&\leq  \frac{R e^x}{(R-x)^m} \int_{-\pi/2}^{\pi/2} e^{-R \cos\theta} d\theta=O(R^{-m})
\end{align*}
\begin{gather*}
f(x)= \frac{i x^{\frac{n}{2}-1} e^{-x}}{2 \pi (-1)^m}  \cdot 2\pi i\cdot \frac{(-1)^{m-1}}{\Gamma(m)}
= \frac{x^{\frac{n}{2}-1}e^{-x}}{\Gamma(\frac{n}{2})}
\end{gather*}
Case $n=2m+1$.
\begin{gather*}
C_1=[R, -R]i-x, 
C_2=Re^{i [-\pi/2, 0]}-x, 
C_3=[R-x,\varepsilon]\\
C_4=\varepsilon e^{i[2\pi, 0]}, 
C_5=[\varepsilon, R-x],
C_6=Re^{i [0, \pi/2]}-x
\end{gather*}
\begin{align*}
\left|\int_{C_2} g(z) dz\right|&\leq \frac{R e^x}{(R-x)^\frac{n}{2}}\int_{-\pi/2}^0 e^{-R\cos\theta}d\theta=O(R^{-m})
\end{align*}
Similar for $C_6$.
\begin{align*}
\left|\int_{C_4} g(z) dz\right|
&=\varepsilon^{1-n/2} \left|\int_{2\pi}^{0} e^{i\theta}e^{-\varepsilon e^{i\theta}} e^{-n i\theta/2}d\theta\right|
=\varepsilon^{1/2-m} \left|\int_0^{2\pi} e^{i(1/2-m)\theta}e^{-\varepsilon e^{i\theta}} d\theta\right|\\
&=\frac{\varepsilon^{1/2}}{(-1/2)_m} \left|\int_{0}^{2\pi} e^{i\theta/2}e^{-\varepsilon e^{i\theta}} d\theta\right|
=O(\varepsilon^{1/2})
\end{align*}
\begin{align*}
\lim \int_{C_5} g(z) dz&=\Gamma\left(1-\frac{n}{2}\right)=\frac{\pi }{\Gamma\left(\frac{n}{2}\right) \sin {\frac{\pi n}{2}}}
=\frac{(-1)^{\frac{n-1}{2}} \pi }{\Gamma\left(\frac{n}{2}\right)}
=\frac{ \pi (-1)^{\frac{n}{2}}}{i \Gamma\left(\frac{n}{2}\right)}
\end{align*}
\begin{align*}
\lim \int_{C_3} g(z) dz&= \int_\infty^0 (-1)^n t^{-\frac{n}{2}}e^{-t} dt = (-1)^{n-1} \int_0^\infty t^{-\frac{n}{2}}e^{-t} dt=\lim \int_{C_5} g(z) dz
\end{align*}
\begin{gather*}
f(x)= \frac{i x^{\frac{n}{2}-1} e^{-x}}{2 \pi (-1)^{\frac{n}{2}}}\cdot 2\cdot \frac{ \pi (-1)^{\frac{n}{2}} }{i \Gamma\left(\frac{n}{2}\right)}
=\frac{ x^{\frac{n}{2}-1} e^{-x}}{\Gamma\left(\frac{n}{2}\right)}
\end{gather*}
A: Your answer heads in the right direction, but perhaps working harder than turns out to be necessary. There is a by-now-standard computation of the Fourier transform of $(x-w)^\alpha$ for $w\not\in \mathbb R$ for any $\alpha\in\mathbb R$, at first for $\Re(\alpha)<-1$ for literal integrability, and then for arbitrary real $\alpha$, by meromorphic continuation, etc.
The computation in the range of convergence invokes a widely useful trick involving $\Gamma(s)$: for $\Re(s)>0$ and $y>0$, $\int_0^\infty t^s\,e^{-ty}\;dt/t= y^{-s}\Gamma(s)$, by changing variables. By the Identity Principle from Complex Analysis, the same holds for $y$ complex with $\Re(y)>0$. Thus, replacing $y$ by $y-2\pi i x$,
$$
\int_0^\infty t^{s-1}\,e^{-ty}\cdot e^{-2\pi ixt}\;dt
\;=\; (y-2\pi ix)^{-s}\,\Gamma(s)
$$
That is, letting $f_y(t)$ be the function that is $t^{s-1}e^{-ty}$ for $t>0$, and $0$ for $t<0$, the Fourier transform of $f_y$ at $x$ is 
$(y-2\pi ix)^{-s}\,\Gamma(s)$.
Then Fourier inversion gives the sort of result you want...
