Let $f:\Omega\to\mathbb C$ be holomorphic, where $\Omega$ is a simply connected domain. Suppose we know the values of $f$ on a simple closed curve $\gamma$ contained in $\Omega$.

The Cauchy Integral Formula tells us how to calculate $f(a)$ for $a$ inside $\gamma$. But $f(a)$ is also uniquely determined for $a\in\Omega$ outside $\gamma$ (by the Identity Theorem). For such $a$, is there an explicit Cauchy-like formula for $f(a)$ in terms of the values of $f$ on $\gamma$?


I don't think there are easy ways to do this. It will require at least several steps for all points outside $\gamma$ covered. Certainly, we would not expect any easy formula for $f(a_1)$ in terms of values on $\gamma$ in the figure 1 below. Let $R$ be the region inside $\gamma$.

First, you have already that inside of $\gamma$ is covered.

Consider $$ R_1=\{z\in\Omega | \exists a_0\in R, \exists \epsilon>0 \ \mathrm{such \ that} \ |z-a_0|<\epsilon \ \mathrm{ and }\ \overline{D}(a_0,\epsilon)\subseteq \Omega\}. $$ (Figure 2 below)

The function values on $R_1$ in terms of $\gamma$ is obtained as follows: $$ f(z)=\sum_{k=0}^{\infty} \left(\frac1{2\pi i} \int_{\gamma} \frac{f(w)}{(w-a_0)^{k+1}} dw\right)(z-a_0)^k $$ The function $f$ is holomorphic on $\Omega$, so the series is valid for any $z\in R_1$. The region $R_1$ properly contains $R$. We repeat this idea to obtain a larger region $R_2$ by using the boundary of $R_1$ as a new curve $\gamma_1$.

For any $z\in\Omega$, the above process is repeated finitely many times to reach $z$ from the inside of $\gamma$. (See figure 3).

Alternative approach

I present an alternative approach using Riemann mapping theorem. Fix $a_0$ inside $\gamma$. Let $\phi:\Omega \rightarrow \mathbb{D}$ such that $\phi(a_0)=0$ and $\phi$ is bijective holomorphic. See Figure 1.

Practically, this is almost as equally difficult as my previous approach, since $\phi$ encodes how to reach from $a_0$ to any point in $\Omega$. The presentation is more elegant than the previous approach.

Consider for $z\in \mathbb{D}$, the function $$ F(z)=f(\phi^{-1}(z)). $$ This is a holomorphic function on $\mathbb{D}$ and $\phi(\gamma)$ encloses $0\in\mathbb{D}$. Thus, we are able to write $F$ as a power series about $0$ using Cauchy integral formula as before. $$ F(z)=\sum_{k=0}^{\infty} \left(\frac1{2\pi i} \int_{\phi(\gamma)} \frac{F(w)}{w^{k+1}} dw\right) z^k. $$

Since $F$ is holomorphic on $\mathbb{D}$, the power series is valid for any $z\in \mathbb{D}$.

Then for any $z\in\Omega$, we have $$ f(z)=\sum_{k=0}^{\infty} \left(\frac1{2\pi i} \int_{\phi(\gamma)} \frac{F(w)}{w^{k+1}} dw\right) (\phi(z))^k. $$ By a change of variable $\phi^{-1}(w)$->$w$, we have $$ f(z)=\sum_{k=0}^{\infty} \left(\frac1{2\pi i} \int_{\gamma} \frac{f(w)\phi'(w)}{(\phi(w))^{k+1}} dw\right) (\phi(z))^k. $$

Figure 1enter image description here

Figure 2enter image description here

Figure 3enter image description here


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