(Cauchy’s formula for “exterior” region) Let \(\gamma\) be piecewise smooth simple closed curve with interior \(\Omega_1\) and exterior \(\Omega_2\). Assume \(f'(z)\) exists in an open set containing \(\gamma\) and \(\Omega_2\) and \(\lim_{z \rightarrow \infty } f(z) = A\). Show that \[\frac{1}{2 \pi i} \int_\gamma \frac{f(\xi)}{\xi - z} \, d \xi = \begin{cases} A, & \text{if\ $z \in \Omega_1$}, \\ -f (z) + A, & \text{if\ $z \in \Omega_2$} \end{cases}\]
Solution
Goal: Let \(\gamma\) be a piecewise smooth simple closed curve with interior \(\Omega_1\) and exterior \(\Omega_2\); assume \(f'\) exists on an open set containing \(\gamma \cup \Omega_2\) and \(\lim_{z\to\infty} f(z) = A\). Show \(\frac{1}{2\pi i}\int_\gamma \frac{f(\xi)}{\xi - z}\, d\xi = A\) for \(z \in \Omega_1\) and \(= -f(z) + A\) for \(z \in \Omega_2\).
1. Fix \(z\) and \(R\) large enough that \(\gamma \subseteq \overline{D_R(0)}\) and \(\abs{z} < R\); let \(\Gamma_R = \theset{\abs{\xi} = R}\) oriented counterclockwise, and consider the region \(\Omega_2 \cap D_R(0)\) whose positively oriented boundary is \(\Gamma_R\) together with \(\gamma\) traversed clockwise.
The region between \(\gamma\) and the large circle is \(\Omega_2 \cap D_R(0)\); its boundary is \(\Gamma_R\) plus \(\gamma\) with the opposite of its (counterclockwise) orientation, since \(\Omega_2\) lies outside \(\gamma\).
2. \(\int_{\Gamma_R} \frac{f(\xi)}{\xi - z}\, d\xi \to 2\pi i A\) as \(R \to \infty\), uniformly in \(z\) with \(\abs{z}\) bounded.
Write \(f = A + (f - A)\). Then \(\int_{\Gamma_R} \frac{A}{\xi - z}\, d\xi = 2\pi i A\) for \(R > \abs{z}\) by the Cauchy integral formula applied to the constant \(A\). For the remainder, \(\abs{\int_{\Gamma_R} \frac{f(\xi) - A}{\xi - z}\, d\xi} \leq 2\pi R \cdot \frac{\sup_{\abs{\xi} = R}\abs{f(\xi) - A}}{R - \abs{z}} \to 0\), since \(f(\xi) \to A\) as \(\xi \to \infty\).
3. If \(z \in \Omega_2\), then \(\frac{1}{2\pi i}\int_\gamma \frac{f(\xi)}{\xi - z}\, d\xi = -f(z) + A\).
On the region of 1, the function \(\xi \mapsto \frac{f(\xi)}{\xi - z}\) is holomorphic except for the simple pole at \(\xi = z\) with residue \(f(z)\). By the residue theorem, \(\int_{\Gamma_R} \frac{f(\xi)}{\xi - z}\, d\xi - \int_\gamma \frac{f(\xi)}{\xi - z}\, d\xi = 2\pi i f(z)\) (the boundary of the region, with \(\gamma\) clockwise, contributes \(\int_{\Gamma_R} - \int_\gamma\)). Rearranging and taking \(R \to \infty\) with 2 gives \(\frac{1}{2\pi i}\int_\gamma \frac{f(\xi)}{\xi - z}\, d\xi = A - f(z)\).
4. If \(z \in \Omega_1\), then \(\frac{1}{2\pi i}\int_\gamma \frac{f(\xi)}{\xi - z}\, d\xi = A\).
For \(z \in \Omega_1\) the function \(\xi \mapsto \frac{f(\xi)}{\xi - z}\) is holomorphic on the whole region of 1 (its only possible pole, at \(\xi = z\), is outside), so by the Cauchy integral theorem \(\int_{\Gamma_R} \frac{f(\xi)}{\xi - z}\, d\xi - \int_\gamma \frac{f(\xi)}{\xi - z}\, d\xi = 0\). Taking \(R \to \infty\) and using 2 yields \(\frac{1}{2\pi i}\int_\gamma \frac{f(\xi)}{\xi - z}\, d\xi = A\).
5. Q.E.D.
3 proves the claim for \(z \in \Omega_2\) and 4 for \(z \in \Omega_1\), which is the full statement.