Suppose I have to calculate the following integral using residue calculus - $$\int_ {-\infty}^{\infty} \! \frac{e^{-ix}}{x^2 + 1 } \, \mathrm{d}x. $$
Now my approach is to construct a contour in the complex plane. So I make semicircles in upper half plane and lower half plane each of radius $R$ . Now I need to calculate $$\lim_{R \rightarrow \infty} \int_ {-R}^{+R} \! \frac{e^{-iz}}{z^2 + 1 } \, \mathrm{d}z.$$
There are two poles $i$ and $-i$ of the above integrand. One is in the upper half plane and the other is in the lower half plane. Now I am supposed to calculate the residue but I am not sure whether to calculate it at $i$ or $-i$. Can someone please explain this to me?
In general, What is the idea behind choosing the points at which residue needs to be calculated in case there are more than one poles?
Answer
Either one will work, and will give the same result (provided you're careful about the orientation). Frequently, it feels a bit more natural to work in the upper half plane because it's consistent with counterclockwise orientation.
Here, though, the difficulty is not the same between the two choices. Note that
$$|e^{-iz}| =e^y$$
In the upper half plane, this is unbounded, whereas it's quite bounded in the lower half plane. As such, it's easier to study the integral using the lower half.
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