How can I evaluate this complex integral $\int_{|z|=1}e^{\frac{1}{z}}\cos{\frac{1}{z}}dz$?

Question

I'm trying to evaluate the following complex integral using the residue method. $$\int_{|z|=1}e^{\frac{1}{z}}\cos{\frac{1}{z}}dz$$

The point $z_0=0$ seems to be a singularity. I'm not sure but I think it's also a non-removable one. I tried using the Taylor expansion of $e^x$ and $\cos{x}$ as that usually helps.

$$e^{\frac{1}{z}}\cos{\frac{1}{z}}=(1+\frac{1}z+\frac{1}{2!z^2}+\frac{1}{3!z^3}+...)(1-\frac{1}{2!z^2}+\frac{1}{4!z^4}-...)\\=>e^{\frac{1}{z}}\cos{\frac{1}{z}}=(1+\frac{1}{z}+...)$$

It seems like the negative power terms are infinite showing that $z_0=0$ is no pole. If I'm correct, the coefficient of $1/z$, which is $1$, is the residue of the singularity and this leads to the result:$$ \int_{|z|=1}e^{\frac{1}{z}}\cos{\frac{1}{z}}dz=2\pi i$$

I don't think I've evaluated other integrals with non-removable singularities and I'm not sure about the whole process..

Answer 1

Assuming that $|z|=1$ is positively oriented (which is the convention), then your working looks fine to me. The residue theorem does not care about the type of singularities within the simply-connected domain around which you're integrating, just that there are finitely many such singularities. Furthermore, the method you used to compute the required residue does not require the implied singularity to be of a certain type; it may be taken as the definition of the residue.

Answer 2

You are right. It is an essential singularity with residue $1$. You can find a primitive (defined away from zero) for $$ e^{\frac{1}{z}}\cos{\frac{1}{z}} - \frac{1}{z} $$ just using that $\displaystyle\frac{z^{1-n}}{1-n}$ is a primitive for $z^{-n}$, with $n>1$. Hence the $$ \int_{|z|=1}e^{\frac{1}{z}}\cos{\frac{1}{z} - \frac{1}{z}}dz= 0 $$ And the only term of $e^{\frac{1}{z}}\cos{\frac{1}{z}}$ that contributes to the integral is $\frac{1}{z}$. This type of argument appears in the proof of the Residue Theorem.