First, Cauchy's Integral Theorem:
If $f$ is a continuous function on $U$ admitting a holomorphic primitive $g$, and $\gamma$ is a closed path in $U$, then
\begin{equation}
\int_\gamma f = 0
\end{equation}
When applying this theorem to $z^n$ over some circle $C_R$ of radius $R$, we have
\begin{equation}
\int_{C_R} z^n dz = \cases{0, & $n \ne -1$ \\ 2\pi i, & $n=-1$}
\end{equation}
(Note, we are allowed to apply Cauchy's Integral Theorem to $f(z)=z^n$ since a primitive $g$, namely $g(z)=\frac{z^{n+1}}{n+1}$, exists).
My question is, why isn't $\int_{C_R} z^n \ dz$ identically zero? Why doesn't Cauchy's Theorem hold for the case $n=-1$?
Thanks!
Best Answer
Parametrize $C_R$ by $z=R e^{i \phi}$ for $\phi \in [0, 2 \pi)$. Then
$$\oint_{C_R} dz \, z^{-1} = i R \int_0^{2 \pi} d\phi \, e^{i \phi} R^{-1} e^{-i \phi} = i 2 \pi$$