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!

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    $\begingroup$ Furthermore, $z\mapsto\sin\left(\overline z\right)$ is not an analytic function. $\endgroup$ Commented Jul 1, 2019 at 20:33
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    $\begingroup$ 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. $\endgroup$ Commented Jul 1, 2019 at 21:39

1 Answer 1


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? $\endgroup$
    – AnonyMath
    Commented Jul 2, 2019 at 6:43
  • $\begingroup$ Given my answer, what do you mean exactly with "substitution" and "substitution method" and "residue at $\infty$" ? $\endgroup$
    – reuns
    Commented Jul 2, 2019 at 18:54

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