# Understanding Cauchy's integral formula

Let $f$ be an analytic on a simply connected domain $D$, and $z_0\in D$

then $f(z_0)=\frac{1}{2\pi i}\int_C{\frac{f(z)}{z-z_0}}$ for a closed path C which is all in C, and surrounds $z_0$...

Is it really necessary for $f$ to be analytic on every point in $C$? or continuity of $f$ on $C$ is sufficient?

• Shouldn't it be $\frac{f(z)}{z-z_0} dz$ ? – C. Dubussy Apr 13 '16 at 19:32
• yes it is really necessary for $f$ to be analytic (holomorphic) on an open connected set containing the contour $C$ and $z_0$ – reuns Apr 13 '16 at 19:34
• You are right, edited :) – Daniel Apr 13 '16 at 19:34
• read a complex analysis course for proving the Cauchy integral theorem that $\int_C g(z) dz = 0$ whenever $g(z)$ is holomorphic on $U$ containing $C$. then use that if $f(z)$ is holomorphic on $U$ then $g(z) = \frac{f(z)-f(z_0)}{z-z_0}$ is holomorphic on $U$ hence $\int_C \frac{f(z)-f(z_0)}{z-z_0} dz = 0$ and $\int_C \frac{f(z)}{z-z_0} dz = \int_C \frac{f(z_0)}{z-z_0} dz = 2 i \pi f(z_0)$ (if $z_0$ is in the interior of the contour) – reuns Apr 13 '16 at 19:45

Based on your question, let's assume that $f$ is continuous on $C$. If $f$ is not analytic at some point $z_1 \in C$, but continuous at $z_1$, we may take the analytic continuation of $f$ to the entire boundary $C$, and then it reduces to the case when $f$ is analytic on $C$.
• In general, there is no reason to believe that a holomorphic function continuous up to boundary admits an analytic continuation (not even to a single point on $C$). – mrf Apr 13 '16 at 20:24
• We already know that $C$ is contained in the $D^\circ$, so if $f$ is holomorphic in $D-\{z_1\}$, and continuous on $D$ (or bounded on $D$), then we may take the analytic continuation. – Yuxin Wang Apr 13 '16 at 20:38
• It's not clear from the statement that $C$ is contained in $D$. If it is, the OP is already assuming that $f$ is analytic on $D$. I suspect he wants to ask about when $C = \partial D$ (but your intepretation may very well be the intended one). – mrf Apr 13 '16 at 20:41
It's known (but not easy to prove in full generality) that if $\Omega$ is a simply connected domain such that $\partial\Omega$ is a rectifiable Jordan curve, and $f$ is a continuous function on $\bar\Omega$, holomorphic on $\Omega$, it is true that $$\int_{\partial\Omega} f(z)\,dz = 0.$$ See this question for details. Consequently, under the same assumptions, $$\int_{\partial\Omega} \frac{f(z)}{z-z_0}\,dz = 2\pi i f(z_0)$$ for $z_0 \in \Omega$.