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I ran into this problem when I was doing some residue computations.

For real $a\neq0$, compute,

$$I=\int_{-\infty}^{+\infty} \frac{e^{iax}}{(x+i)^3} $$

Be sure to treat both cases when $a<0, a>0$

I know how to deal with this when $a<0$. One proceeds as usual and end up having to use the so called "Jordan's lemma". (essentially you use the fact that $\sin \theta \geq 2\theta/\pi$, for $0\leq \theta \leq \pi/2$)

But my question is how about when $a>0$. I know that the integral must be zero. Is there an easy way to see this or am I completely not seeing something here?. For example is there a way to obtain both results using the same contour or should I proceed differently?

I am sure somebody ran into this problem before.

Any comments, hints?.

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Close the contour in the other (upper) half-plane. –  Fabian Jan 9 '13 at 9:44
@Fabian:Thanks. think I get what you mean. If one uses the upper half plane semicircular contour then you have no residues inside. So by cauchy's theorem the integral must be zero. But I still need to show that the contribution from the semicircular part goes to zero as well. Is this easy to see?. –  Jack Dawkins Jan 9 '13 at 9:46
It is the same as in the lower contour. Use Jordan's lemma. –  Fabian Jan 9 '13 at 9:49

1 Answer 1

up vote 2 down vote accepted

For $a<0$, you close the contour in the lower half plane. You then pick up the contributions from the pole at $x=-i$ and you obtain $$I = - 2 \pi i \, \text{Res}_{x=-i} \frac{e^{ia x}}{(x+i)^3} = i \pi a^2 e^{a} .$$ For $a> 0$, you have to close the contour in the upper half plane. There is no pole in this contour and thus $$I=0.$$

For $a=0$ you can use either contour.

In both cases the vanishing of the integral along the half circle is garanteed by Jordan's lemma.

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Got it thanks. I was getting confused over nothing. :) –  Jack Dawkins Jan 9 '13 at 9:57

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