Using Residue Theorem to calculate the integral

Question

for $$I=\int_{|z|=1}{z^m \cos\left(\frac{1}{z}\right)}\,dz$$ where $m=0,1,2...$

Is the singularity $z=0$ or there are some other singularities?

if it is $z=0$, what's order of pole?

Answer 1

Note that $$ I=\int_{|z|=1}{z^m \cos\left(\frac{1}{z}\right)}\,dz =\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}\int_{\lvert z\rvert=1}z^{m-2n}\,dz, $$ and that $$ \int_{\lvert z\rvert=1}z^{m-2n}\,dz=\left\{ \begin{array}{ccc} 2\pi i & \text{if} & m-2n=-1,\\ 0 & \text{if} & m-2n\ne-1. \end{array}\right. $$ So, if $m$ is even, then $I=0$, while if $m=2k-1$, then $$ I=\frac{2\pi i(-1)^k}{(2k)!}. $$

Answer 2

$z=0$ is the only singularity, and it is not a pole.

Answer 3

Let $z=1/\zeta$; then the integral is equal to

$$-\int_{|\zeta|=1} d\zeta \frac{\cos{\zeta}}{\zeta^{m+2}} $$

which, by Cauchy's theorem, is

$$\frac{i 2 \pi}{(m+1)!} \left [\frac{d^{m+1}}{d\zeta^{m+1}} \cos{\zeta}\right ]_{\zeta=0} = \begin{cases} 0 & m \; \text{even} \\ \displaystyle \frac{i 2 \pi (-1)^{\frac{m+1}{2}}}{(m+1)!} & m \; \text{odd}\end{cases}$$

Note that, in substituting, we traverse the circle in the opposite direction; thus we multiply by $-i 2 \pi$.