Complex integral, Cauchy's theorem I am currently doing some complex analysis and stumbled upon the following problem:
I want to calculate the following integral:
$$\frac{1}{2\pi i}\int_{-\infty}^{\infty}e^{ix}(ix)^{c-1}dx$$
where $ 0< c < 1$
I tried using the pole in $x=0$ and using Cauchy's theorem, but I find it hard to calculate the residue.
Could you tell me if this approach is correct? And if it is, how I can calculate the residue?
If I am using the wrong approach, how can I solve this?
Thanks in advance
 A: The integrand actually has a branch point at $x=0$, not a pole. The idea is to fold the contour around the branch point to give two integrals that you can evaluate.
In particular, the integral is equal to
$$ \frac{1}{2\pi i}\int_{\gamma} e^{z} z^{c-1} \, dz, $$
where $\gamma$ is composed of a line just below the negative real axis starting at $-\infty$ and ending at $-\epsilon$, a(n almost-)circle of radius $\epsilon$, and a line from $-\epsilon$ to $-\infty$ just above the negative real axis. (Jordan's lemma shows that the integrals over the large semicircles we use to deform the contour in this way tend to zero.)
The integral over the small circle tends to zero because the singularity at $z=0$ is weaker than $1/z$. This leaves the two lines. A simple change of variables ($t=e^{\pm i\pi}x$) will give you
$$ \frac{1}{2\pi i}\int_{\gamma} e^{z}z^{c-1} \, dz = \frac{2i\sin{c\pi}}{2\pi i} \int_0^{\infty} t^{c-1}e^{-t} \, dt = \frac{\sin{c\pi}}{\pi} \, \Gamma(c). $$
The reflection formula then implies this is $1/\Gamma(1-c)$.
