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!
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.