I'm been banging my head against the wall trying to solve the following question which ask to solve the following integral using the Cauchy integral formula, and hence evaluating the corresponding real integrals.
$\int_{\gamma} e^zz^n dz$ where $\gamma$ is the unit circle {$e^{i\theta}: -\pi \leq \theta \leq \pi$} and $n\in \mathbf{Z}$.
To solve the question, I'm attempting to use the generalised form of the Cauchy integral formula. Although to use it, the $z^n$ is normally in the denominator not the numerator.
Any help will be much appreciated, thanks!!
Best Answer
That integral is equal to $0$ if $n\geqslant0$. In fact,\begin{align}\int_\gamma e^zz^n\,\mathrm dz&=\int_\gamma\frac{e^zz^{n+1}}z\,\mathrm dz\\&=2\pi ie^00^{n+1}\\&=0.\end{align}On the other hand, if $n<0$, then\begin{align}\int_\gamma e^zz^n\,\mathrm dz&=\int_\gamma\frac{e^z}{z^{-n}}\,\mathrm dz\\&=\frac{2\pi i}{(-n-1)!}\exp^{(-n-1)}(0)\\&=\frac{2\pi i}{(-n-1)!}.\end{align}