Computing $\int_{|z|=1}z^3\sin(\frac{1}{z})dz$ and $\int_{|z|=\frac{2}{3}}\sin(\frac{1}{z^2}+e^{z^2}\cos(z))dz $

Question

Compute the following integrals using residues: $$\begin{align}\int_{|z|=1}z^3\sin(\frac{1}{z})dz\\\int_{|z|=\frac{2}{3}}\sin(\frac{1}{z^2}+e^{z^2}\cos(z))dz \end{align}$$

In the first integral I could not find any singular points, but removable ones. therefore to check out I was doing it right I derived the Laurent series:

$\sin(\frac{1}{z})=z^2-\frac{1}{3!}+\frac{1}{5!z^2}-….$

So there is no residue at infinity.

Regarding the second integral

$\int_{|z|=\frac{2}{3}}\sin(\frac{1}{z^2}+e^{z^2}\cos(z))dz$

I found no singularities but removable ones. However I was not able to compute the Laurent series to check that out.

So the value of both integrals would equal $0$.

Question:

Is my answer right?

Thanks in advance!

Answer 1

In the case of the first integral, you have a clear non-removable singularity at $0$. But the residue at $0$ is $0$ (it's an even function), and so the integral is equal to $0$. The same argument applies to the second integral.