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How to prove that when solving improper integrals with residue theorem, we should only include residues in the upper complex plane? I've tried working my way through with Jordan's lemma, but I got stuck.

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closed as unclear what you're asking by Ron Gordon, Claude Leibovici, Cesareo, user99914, Shailesh Jun 26 '18 at 15:33

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Consider real integrals of the form $\int_{-\infty}^\infty f(x)\,dx$.

In order to apply contour techniques this integral must be transformed into a complex contour integral.

Take a symmetric interval, $[-R,R]$ and adding in a upper semi-circle of radius $R$ (centered at the origin). Then as $R \to \infty$ the semi-circle part should tend to zero so that in the limit one gets the (PV) integral over $(-\infty,\infty)$

As the half circle goes bigger and bigger it gathers more and more of the upper half plane, so in the limit you have essentially captured the entire upper half plane. Thus when you go to compute the contour integral using residues, you should only consider poles in the upper half plane.

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You cannot prove that because it is false. For instance, if we compute$$\int_{-\infty}^{+\infty}e^{iax}f(x)\,\mathrm dx$$with $a\in(-\infty,0)$ and where $f$ is an analytic function such that $\lim_{z\to\infty}f(z)=0$, then we use the residues of the lower halfplane.

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  • $\begingroup$ But why does the theorem state that we take only residues in the lower half plane? $\endgroup$ – Lazar Šćekić Jun 26 '18 at 11:51
  • $\begingroup$ @LazarŠćekić Which theorem? If you don't state the theorem, nobody can help you. $\endgroup$ – José Carlos Santos Jun 26 '18 at 11:52
  • $\begingroup$ Well I don't think the theorem has its real name, since only the appliance of Cauchy residue theorem to improper integrals of rational functions from negative infinity to positive infinity $\endgroup$ – Lazar Šćekić Jun 26 '18 at 11:56
  • $\begingroup$ @LazarŠćekić And for some of those integrals we use the residues from the upper halfplane, whereas for some other integrals we use the residues from the lower halsplane. $\endgroup$ – José Carlos Santos Jun 26 '18 at 12:03

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