Proving Fourier transform of $\int_0^\infty e^{-x}x^{a-1}dx = \Gamma(a)(1+i\omega)^{-a}$ 
Given   $a > 0$, 
  $$f(x) = \begin{cases} e^{-x}x^{a-1}, &\mbox{if } x > 0 \\ 
0, & \mbox{if } x \leq 0 \end{cases}$$
Prove the the Fourier transform of $f$, $\hat{f}(\omega)$ is equal to $\Gamma(a)(1+i\omega)^{-a}$.

I know the Gamma function is defined as $\int^{\infty}_0e^{-x}x^{a-1} dx$
Then, $$\hat{f}(\omega) = \int^{\infty}_0(e^{-x}x^{a-1})(e^{-i\omega x})dx$$
But from here Im a little lost as to what to do, do I just integrate by parts then it works out? Any help/tips is welcome!
 A: Gamma function for integers is basically a factorial $\Gamma(a)=(a-1)!, a\in \mathbb {N}$ .
Use iterative integration by parts as in (very end of the pdf) http://www.willamette.edu/~mjaneba/courses/ma256/integrationbyparts.pdf
A: I think you need a little Complex Analysis. The Fourier transform is
$$
         \frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}e^{-i\omega x}e^{-x}x^{a-1}dx
    = \frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}e^{-z(1+i\omega)}z^{a-1}dz,
$$
which can be viewed as a line integral along the positive real axis. Assume a branch cut for $z^{a-1}$ along the negative real axis. You can trade this line integral for one along the ray $\{ r(1-i\omega) : r \ge 0 \}$, which can be written using the parameterization
$$
       z(r)=\frac{r(1-i\omega)}{|1-i\omega|^2}
$$
as
$$
       \frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}e^{-r}\frac{r^{a-1}(1-i\omega)^{a-1}}{|1-i\omega|^{2a-2}}\,\frac{(1-i\omega)}{|1-i\omega|^2}dr \\
  = \frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}e^{-r}r^{a-1}dr\;\frac{(1-i\omega)^{a}}{|1-i\omega|^{2a}} \\
  = \Gamma(a)\frac{(1-i\omega)^{a}}{|1-i\omega|^{2a}}
$$
Because $|1-i\omega|^{2a}=(1-i\omega)^{a}(1+i\omega)^{a}$ for $a > 0$ and real $\omega$, then the above becomes $\Gamma(a)/(1+i\omega)^{a}$.
