# Infinitely many singularities inside the contour with residue theorem

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!

• Furthermore, $z\mapsto\sin\left(\overline z\right)$ is not an analytic function. – José Carlos Santos Jul 1 at 20:33
• 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. – Robert Israel Jul 1 at 21:39

## 1 Answer

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.

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