This integral made me wonder, what should be used:

$ \underset{|z-3 \pi|=4}{\int} \frac{1}{z \sin{z}} dz$

Here $0$ is not a relevant pol since it's not in the circle. so the 3 relevant pols are:

$z_0 = -4 \pi \qquad z_1 = -3 \pi \qquad z_2 = -2 \pi $

I solved with the residue theorem and got $0$

  1. Could I have seen from the beginning its $0$ without calculation using Cauchy's contour rule? What do I look for in order to use that?

  2. I know residue theorem and Cauchy's formula for first order pols are sometimes interchangeable, but when exactly? what do I look for to rule residue theorem out immediately?

  3. Is there an isolated singularity that prevents the use of residue theorem? I know it has a version for pols but what of removable and essential singularities?


The residue theorem is a direct generalization of the Cauchy integral formula, so everything that can be solved using the Cauchy integral formula can also be solved using the residue theorem. The residue theorem also holds for removable and essential singularities, although for removable singularities the residue is always $0$, of course.

  • $\begingroup$ Is a removable singularity the reason for the 0? which one is the removable? $\endgroup$
    – Toma
    Jul 30 '20 at 16:21
  • 1
    $\begingroup$ No, none of these are removable. By the way, the relevant poles should be $2\pi$, $3\pi$ and $4\pi$ and the corresponding residues are $\frac{1}{2\pi}$, $-\frac{1}{3\pi}$ and $\frac{1}{4\pi}$. Not sure how you got $0$. $\endgroup$
    – Klaus
    Jul 30 '20 at 16:29
  • $\begingroup$ Oh you're right I forgot that the middle is not $-3 \pi$ but $3 \pi $ haha $\endgroup$
    – Toma
    Jul 30 '20 at 20:32

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