A tricky residue theorem problem I thought I was fairly well-versed with using the residue theorem to evaluate improper integrals, but one problem has been giving me grief.
How does one compute the integral
$$\int_{-\infty}^{\infty}\frac{e^{\alpha+ix}}{(\alpha+ix)^\beta} dx$$
for real numbers $\alpha>1$, $\beta>0$?
Note that $\beta$ is not necessarily an integer, which restricts the contours that can be used (ie. one needs to use contours on which a branch of logarithm can be defined).
 A: This solution doesn't involve the residue theorem, only Cauchy theorem (to manipulate the integration contour).
Let $t=\alpha+ix$, so that your integral is
$$I(\beta)=-i\int^{\alpha+i\infty}_{\alpha-i\infty}t^{-\beta}e^tdt$$
(integration over a vertical line - I hope the notation is clear).
By Cauchy theorem the integral is equal to 
$$-i\int_Ct^{-\beta}e^tdt$$
where $C$ comes from $-\infty$ below the real axis, makes a small circle around $0$, and returns to $-\infty$ above the real axis.
Suppose that $\beta<1$; the integral over the circle part of $C$ goes to $0$ as the radius of the circle goes to zero. We can pass to the limit and forget about the circle. The integral then becomes ($r=-t$)
$$-i(e^{\pi i\beta}-e^{-\pi i\beta})\int_0^\infty r^{-\beta}e^{-r}dr=2\sin(\pi\beta)\Gamma(1-\beta)$$
(using $t^\beta=e^{\pm\pi i\beta}r^\beta$, the sign given by whether we go above or below the real axis).
You can use the functional equation 
$$\Gamma(1-\beta)\Gamma(\beta)\sin(\pi\beta)=\pi$$
to simplify the result to
$$I(\beta)=2\pi/\Gamma(\beta).$$
We supposed that $\beta<1$. Notice however that $I(\beta)$ is an analytic function of $\beta$ for $\Re\beta>0$ (it converges uniformly), so the result is true for every $\beta$ with $\Re\beta>0$. You can try to use the residue theorem to check this result for $\beta\in\mathbb{N}$ (I hope I didn't make too many mistakes, so perhaps you'll get it :)
