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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).

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1 Answer 1

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 :)

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Do you not mean "where $C$ comes from $−\infty$ below the real axis, makes a semi-circle around $0$, and goes to $+\infty$ above the real axis." ? –  pbs Oct 2 '12 at 14:37
@pbs: no, I really meant to return to $-\infty$ –  user8268 Oct 2 '12 at 14:39
But how can you return to $-\infty$ above the real axis? I may be getting my wires crossed. The first integral is over $\alpha-i\infty$ to $\alpha+i\infty$. So isn't it an integral from $-i\infty$ to $+i\infty$ on the imaginary axis? –  pbs Oct 2 '12 at 14:39
@pbs: you replace the original path $|$ with $\supset$; you need to use Cauchy thm. to prove that the integrals are the same –  user8268 Oct 2 '12 at 14:52
by Cauchy's theorem you mean ? –  pbs Oct 2 '12 at 15:26

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