# Closing the contour over circular path ## Direct calculation by parameterizing Integrating over a circle in either a _positive orientation_ (on $S_{+}$) or a _negative orientation_ (on $S_{-}$) yields: $\oint_{S\pm} dz(z-z_0)^n = \pm2\pi i \delta_{n,-1}$ To show this consider the following integration over an arc from the first example in [Examples](Contour%20integrals.md#Examples%20of%20contour%20integrals) of contour integrals:  meaning that $\int_{S_{\phi}} dz(z-z_0)^n = \int_0^{\phi}dt \frac{dz(t)}{dt}(z(t)-z_0)^n= \int_0^{\phi}dt(iRe^{it})(Re^{it})^n$ $= \int_0^{\phi}dt i(R)^{n+1}e^{it(n+1)} = \begin{cases} \frac{R^{n+1}}{n+1}\big(e^{i(n+1)\phi}-1\big), & n\neq -1 \\ i\phi, & n=-1 \end{cases}$ at $\phi =2\pi$, $e^{2\pi i(n+1)}=1$, turning the arc into a closed circle, and thus the cases reduce to $\begin{cases} 0, & n \neq -1 \\ i\phi, & n=-1\end{cases}$. We can then quickly check to see that the sign changes when we switch the order of the limits (and thus the direction of the integration), letting us conclude that $\oint_{S\pm} dz(z-z_0)^n = \pm2\pi i \delta_{n,-1}$. First, notice that the case where the integral doesn't vanish corresponds to the case where the integrand _isn't holomorphic_ since $(z-z_0)^{-1}$ blows up to infinity at $z=z_0$. The integral disappears when the integrand isn't holomorphic. This hints at there maybe being a more general conclusion we can make about integration of holomorphic functions over closed contours on $\mathbb{C}$, which is elaborated on by [[Cauchy's theorem]], which shows that such integrals over holomorphic domains are 0. #MathematicalFoundations/Analysis/ComplexAnalysis/Integrals