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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?

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    $\begingroup$ You can pick either, just keep track of your orientation, since you're going from $-\infty$ to $\infty$, I recommend the upper-half plane choice, i.e. $i$. $\endgroup$ Apr 14, 2016 at 1:23
  • $\begingroup$ Considering T. Bongers answer, it seems $i$ lands us in some difficulty due to the unboundedness condition. $\endgroup$ Apr 14, 2016 at 1:35

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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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  • $\begingroup$ Their is also $1+ z^2$ in the denominator. Still the integrand diverges? $\endgroup$ Apr 14, 2016 at 1:34
  • $\begingroup$ @Dark_Knight The $|z|^2$ decay is not enough to kill of the exponential growth in the numerator. $\endgroup$
    – user296602
    Apr 14, 2016 at 1:35
  • $\begingroup$ Ok, so beside boundedness, are there any other criterions which affects the choice of choosing pole? $\endgroup$ Apr 14, 2016 at 1:39
  • $\begingroup$ Whatever is simplest, depending on how hard it is to compute residues or whether one region contains more poles, I suppose. But being able to control the bounds is probably the hardest part, usually. $\endgroup$
    – user296602
    Apr 14, 2016 at 1:44

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