Let $\gamma=\{z\in \Bbb C: \lvert z \rvert=2\}$ in anti-clockwise orientation. Then $I=\frac {1}{2\pi i} \int_{\gamma} z^7 \cos \frac 1{z^2} dz$=?

Question

$$I=\frac {1}{2\pi i} \int_{\gamma} z^7 \cos \frac 1{z^2} dz=?$$

The function $\cos \frac 1{z^2}$ is neither analytic at $z=0$ and nor it has a pole at $z=0$. By Cauchy Integral Formula can I get the right answer? Or there is some other method?

Since $\cos \frac 1{z^2}$ does not have pole at $z=0$, can I say that the integral is zero by Cauchy Integral Theorem?

Answer 1

Hint : Do the change $w = \frac{1}{z}.$

You get $$\frac{1}{2i\pi}\int_{\gamma} z^7 \cos \frac 1{z^2} dz = -\frac{1}{2i\pi} \int_{\gamma'} \frac{cos(w^2)}{w^9}dw,$$ where $\gamma' = \{z : |z| = 1/2\}.$

Now Cauchy Integral formula can help you.

Answer 2

You can just extract the coefficient of the $\frac1z$-term of the series expansion of the integrand. Specifically, $$ z^7\cos(1/z^2) =z^7\left(1 - \frac{z^{-4}}{2} + \frac{z^{-8}}{24} - \cdots\right) = z^7 - \frac{z^3}{2} + \frac{z^{-1}}{24} - \cdots $$ which means that we have $$ I = \frac{1}{2\pi i}\int_\gamma z^7\cos\frac{1}{z^2} dz =\frac1{2\pi i}\int_\gamma \frac{1}{24}\cdot \frac{1}{z}dz = \frac{1}{2\pi i}\cdot\frac{2\pi i}{24}= \frac{1}{24}$$