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When giving an example of applications of the residue theorem, my professor wrote the following integral: $$\int_{-\infty}^\infty\frac{z^3}{z-i}\mathrm{d}z=\oint_\Gamma\frac{z^3}{z-i}\mathrm{d}z=2\pi i \mathrm{Res}(f,i)$$ Where $\Gamma$ is the (counterclockwise) contour made up by taking the limit $R\to\infty$ of the two pieces $[-R,R]$ and a semicircle parameterized by $Re^{i\theta}$, for $\theta\in[0,\pi]$. My professor claims that the integral over the semicircle vanishes as $R\to\infty$ by Jordan's Lemma. It seems to me, naively, that since the integrand scales as $R^2$ and the length of the semicircular arc scales as $R$, the integral over the semicircle should scale as $R^3$, and hence would not vanish. This leads me to also to conclude that $\int_{-\infty}^\infty\frac{z^3}{z-i}\mathrm{d}z$ might not converge.

Why does the integral over the semicircle of $\frac{z^3}{z-i}$ vanish?

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    $\begingroup$ You're right, the integral over the semicircle does not vanish in the limit—indeed the original function is definitely not integrable on $(-\infty,\infty)$ by the same observation that the integrand grows like $z^2$ on both ends. $\endgroup$ Feb 15, 2023 at 7:25
  • $\begingroup$ @GregMartin My professor has replied that the integral over the semicircular contour vanishes by Jordan's lemma. I can't see how to apply Jordan's lemma here. Does it apply, and if so, how? $\endgroup$
    – DanDan面
    Feb 16, 2023 at 0:56
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    $\begingroup$ The integral you are considering is divergent. You cannot evaluate it by any means. Tell your professor that the improper integral does not converge. There is nothing to solve here. $\endgroup$
    – Gary
    Feb 16, 2023 at 1:04
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    $\begingroup$ I don't believe Jordan's lemma can be applied (there is no factor of the form $e^{iaz}$ with $a>0$), and even if it were applied it wouldn't imply that the integral over the contour vanishes. It sounds like your professor is wrong, but are you 100% sure that you've copied the integrand correctly? $\endgroup$ Feb 16, 2023 at 16:50

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