Let $C$ be a circle $\gamma=\partial B (0,2)$ oriented positively.

I have to calculate $$\int_\gamma \frac{-\cos(1/z)}{\sin(1/z)z^2}dz$$

My attempt:

Notice that $\sin(1/z)$ is meromorphic inside and on $\gamma$ and doesnt have any poles on $\gamma$.

Moreover, $[\sin(1/z)]'=\frac{-\cos(1/z)}{z^2}$ (in neighbourhood of $\gamma$).

I will use now the Argument Principle. $f(z)=\sin(1/z)$ has only one pole (at $0$) inside $\gamma$ [obviously its not a pole, but essential singularity].

On the other hand, $f(z)$ has countably many zeros inside $\gamma$, at points $z=\frac{1}{k\pi}, k\in\mathbb{Z}$. I'm stuck now.

So integral $\int_\gamma \frac{-\cos(1/z)}{\sin(1/z)z^2}dz$ is not finite?

Edit: I want to solve this using argument principle on "raw" integral.

  • 1
    $\begingroup$ Hint: $u=1/z$, the entire integral is screaming for this substitution. $\endgroup$ – orion Jul 4 '15 at 12:58
  • $\begingroup$ I know, but I wanted to use argument principle in this case. I know I can substitute $\endgroup$ – luka5z Jul 4 '15 at 13:00
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    $\begingroup$ $z=0$ is not a pole but an essential singularity for $f(z)=\sin(1/z)$. $\endgroup$ – Jack D'Aurizio Jul 4 '15 at 13:02
  • $\begingroup$ My bad. You are right $\endgroup$ – luka5z Jul 4 '15 at 13:03

$$\int_{\gamma}\frac{-\cos(1/z)\,dz}{z^2\sin(1/z)} = \int_{|w|=\frac{1}{2}}\cot w\,dw = 2\pi i\cdot\text{Res}\left(\cot w,w=0\right)=\color{red}{2\pi i}.$$

  • $\begingroup$ Nice, but could you explain why my attempt using Argument Principle is incorrect? Thanks $\endgroup$ – luka5z Jul 4 '15 at 13:02
  • $\begingroup$ @luka5z: I did that: $z=0$ is an essential singularity for $\sin(1/z)$, not a pole. $\endgroup$ – Jack D'Aurizio Jul 4 '15 at 13:03
  • $\begingroup$ Sorry, you are obviously right. I dont know what I did :) $\endgroup$ – luka5z Jul 4 '15 at 13:04

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