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

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:

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.