Applying Cauchy's Integral Theorem to $\int_{C_R} z^n \ dz$

Question

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!

Answer 1

Parametrize $C_R$ by $z=R e^{i \phi}$ for $\phi \in [0, 2 \pi)$. Then

$$\oint_{C_R} dz \, z^{-1} = i R \int_0^{2 \pi} d\phi \, e^{i \phi} R^{-1} e^{-i \phi} = i 2 \pi$$

Answer 2

I see the problem now.

Note that when $n=-1$, the primitive $g(z)=\frac{z^{n+1}}{n+1}$ is undefined, and thus the theorem doesn't hold. I wish I had seen this faster...

Answer 3

Actually there is a homology version of Cauchy's Theorem.

$\textbf{Theorem.}$ Suppose $V$ is an open subset of $\mathbf{C}$, $\Gamma$ is a formal sum of closed curves. Suppose that $\text{Ind}(\Gamma,a)=0$ for all $a\in \mathbf{C}\backslash V$. Suppose that $f\in H(V)$. Then $$ \int_{\Gamma}f(w)dw=0. $$

$\textbf{Remark.}$ It is OK to apply this theorem for $n\ge 0$. When $n=-1$, $f(z)=1/z$ has a singularity at $z=0$. We can not apply this theorem. We could either do computation directly, or apply the Residue Theorem.