In my textbook(H.Silverman - Complex variables), the residue theorem required that there are finitely many isolated singularities inside the contour. Similarly, Residue theorem at infinity (viewing as a point in Riemann sphere), it stated only 'finitely many' sense.

However, in some solution about the following integral $$\int_{\vert{z}\vert=1}\frac{z}{\sin\bar{z}}\,dz,$$ using the substitution $z\mapsto\tfrac{1}{w}$, applied residue theorem (as mentioned first) to get the answer.

Is it possible?

Not only the integrand have infinitely many singularities inside the contour $\vert{z}\vert=1$ but also it has a limit point of the set of its singularities inside the contour $\vert{z}\vert=1$.

What is the difference between 'the substitution method' and 'Residue at infinity' for calculating the contour integrals?

Anyone give some comment or related reference, please. Thank you!

  • 2
    $\begingroup$ Furthermore, $z\mapsto\sin\left(\overline z\right)$ is not an analytic function. $\endgroup$ – José Carlos Santos Jul 1 at 20:33
  • 1
    $\begingroup$ But $\overline{z} = 1/z$ on $|z|=1$. So the substitution $z = 1/w$ makes this into $$ \oint_{|w|=1}\frac{ w}{\sin(w)}\; dw$$ and this has only one singularity inside the contour. $\endgroup$ – Robert Israel Jul 1 at 21:39

To apply the residue theorem to

$$\int_{|z|=1} \frac{z}{\sin 1/z}dz $$ you need to show that $$\lim_{r \to 0}\int_{|z|=r} \frac{z}{\sin 1/z}dz=0$$ obtaining

$$\int_{|z|=1} \frac{z}{\sin 1/z}dz = \lim_{N \to \infty} \int_{|z|=1} \frac{z}{\sin 1/z}dz-\int_{|z|=2\pi/(N+1/2)} \frac{z}{\sin 1/z}dz \\ = \lim_{N \to \infty} 2i\pi\sum_{n=7}^N \frac{2\pi/n}{ \frac{-1}{(2\pi /n)^2}\cos(\frac{1}{2\pi /n})}+\frac{-2\pi/n}{ \frac{-1}{(-2\pi/n)^2}\cos(\frac{1}{-2\pi /n})}$$

Here the obtained series converges absolutely but in general you need to keep the order of summation according to the annulus where each residue comes from, that $\lim_{r \to 0}\int_{|z|=r} \frac{z}{\sin 1/z}dz=0$ implies the series converges.

  • $\begingroup$ thanks for comment. I thought the substitution itself was a problem, but is the substitution itself has no problem? $\endgroup$ – Primavera Jul 2 at 6:43
  • $\begingroup$ Given my answer, what do you mean exactly with "substitution" and "substitution method" and "residue at $\infty$" ? $\endgroup$ – reuns Jul 2 at 18:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.