# Residue theorem to find $\int_{-\infty}^\infty\frac{z^3}{z-i}\mathrm{d}z$

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?

• 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. Feb 15, 2023 at 7:25
• @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? Feb 16, 2023 at 0:56
• 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.
– Gary
Feb 16, 2023 at 1:04
• 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? Feb 16, 2023 at 16:50