Let \(\gamma\) be a positively oriented (counterclockwise) piecewise smooth simple closed curve in \(\mathbb{C}\) with interior \(\Omega_1\) and exterior \(\Omega_2\). Assume \(f\) is holomorphic on an open set containing \(\gamma \cup \Omega_2\), with \(\lim_{z \to \infty} f(z) = A\).
Prove that the Cauchy integral \[ F(z) = \frac{1}{2\pi i} \int_{\gamma} \frac{f(\xi)}{\xi - z} \, d\xi = \begin{cases} A & \text{if } z \in \Omega_1, \\ -f(z) + A & \text{if } z \in \Omega_2. \end{cases} \]
Solution
Goal: Compute \(F(z)\) for \(z \in \Omega_1\) and \(z \in \Omega_2\) by applying Cauchy’s theorem on the annular domain between \(\gamma\) and a large circle \(C_R = \{|\xi| = R\}\), and taking the limit as \(R \to \infty\).
1. Asymptotic limit of the integral along a large circle \(C_R\):
1.1. Let \(C_R\) denote the circle \(\{|\xi| = R\}\), oriented counterclockwise, for \(R > 0\) sufficiently large such that \(\gamma \subset \mathbb{D}_R\) and (if \(z \in \Omega_2\)) \(|z| < R\).
1.2. Split the integral along \(C_R\):\[\frac{1}{2\pi i} \int_{C_R} \frac{f(\xi)}{\xi - z} \, d\xi = \frac{A}{2\pi i} \int_{C_R} \frac{d\xi}{\xi - z} + \frac{1}{2\pi i} \int_{C_R} \frac{f(\xi) - A}{\xi - z} \, d\xi.\]
1.3. Since \(|z| < R\), the winding number of \(C_R\) around \(z\) is 1:\[\frac{1}{2\pi i} \int_{C_R} \frac{d\xi}{\xi - z} = 1.\]
1.4. Bound the error term:- Let \(\varepsilon > 0\). Since \(\lim_{\xi \to \infty} f(\xi) = A\), there exists \(R_0 > 0\) such that \(|f(\xi) - A| < \varepsilon\) for all \(|\xi| \ge R_0\).- For \(R \ge R_0\) with \(R > |z|\):\[\left| \frac{1}{2\pi i} \int_{C_R} \frac{f(\xi) - A}{\xi - z} \, d\xi \right| \le \frac{1}{2\pi} \cdot \frac{\varepsilon}{R - |z|} \cdot 2\pi R = \varepsilon \frac{R}{R - |z|}.\]- Taking \(R \to \infty\), the upper bound approaches \(\varepsilon\). Since \(\varepsilon > 0\) was arbitrary:\[\lim_{R \to \infty} \frac{1}{2\pi i} \int_{C_R} \frac{f(\xi) - A}{\xi - z} \, d\xi = 0.\]
1.5. Therefore:\[\lim_{R \to \infty} \frac{1}{2\pi i} \int_{C_R} \frac{f(\xi)}{\xi - z} \, d\xi = A \cdot 1 + 0 = A.\]
2. Case \(z \in \Omega_1\):
2.1. Let \(z \in \Omega_1\).
2.2. Choose \(R > 0\) large enough that \(\gamma \subset \mathbb{D}_R\).
2.3. Consider the bounded region \(U_R = \mathbb{D}_R \cap \Omega_2\) between \(\gamma\) and \(C_R\).
2.4. The boundary is \(\partial U_R = C_R - \gamma\) (where \(-\gamma\) indicates that the inner boundary is oriented clockwise relative to \(U_R\)).
2.5. Since \(z \in \Omega_1\), \(z \notin U_R\). Thus the function \(\xi \mapsto \frac{f(\xi)}{\xi - z}\) is holomorphic on an open neighborhood of \(\overline{U_R}\).
2.6. By Cauchy’s Integral Theorem on \(U_R\):\[\frac{1}{2\pi i} \int_{C_R} \frac{f(\xi)}{\xi - z} \, d\xi - \frac{1}{2\pi i} \int_\gamma \frac{f(\xi)}{\xi - z} \, d\xi = 0.\]
2.7. Thus:\[F(z) = \frac{1}{2\pi i} \int_\gamma \frac{f(\xi)}{\xi - z} \, d\xi = \frac{1}{2\pi i} \int_{C_R} \frac{f(\xi)}{\xi - z} \, d\xi.\]
2.8. Since the left-hand side \(F(z)\) is independent of \(R\), take \(R \to \infty\) and apply 1:\[F(z) = A.\]
3. Case \(z \in \Omega_2\):
3.1. Let \(z \in \Omega_2\).
3.2. Choose \(R > |z|\) large enough that \(\gamma \subset \mathbb{D}_R\).
3.3. Now the point \(z\) lies in the interior of the bounded region \(U_R = \mathbb{D}_R \cap \Omega_2\).
3.4. The function \(\xi \mapsto \frac{f(\xi)}{\xi - z}\) is holomorphic on \(\overline{U_R} \setminus \{z\}\) with a simple pole at \(\xi = z\).
3.5. The residue at \(\xi = z\) is:\[\operatorname{Res}_{\xi = z} \frac{f(\xi)}{\xi - z} = f(z).\]
3.6. By the Residue Theorem / Cauchy’s Integral Formula on \(U_R\):\[\frac{1}{2\pi i} \int_{C_R} \frac{f(\xi)}{\xi - z} \, d\xi - \frac{1}{2\pi i} \int_\gamma \frac{f(\xi)}{\xi - z} \, d\xi = f(z).\]
3.7. Rearranging for \(F(z)\):\[F(z) = \frac{1}{2\pi i} \int_\gamma \frac{f(\xi)}{\xi - z} \, d\xi = -f(z) + \frac{1}{2\pi i} \int_{C_R} \frac{f(\xi)}{\xi - z} \, d\xi.\]
3.8. Taking \(R \to \infty\) and applying 1:\[F(z) = -f(z) + A.\]
4. Conclusion:
$F(z) = A$ for $z \in \Omega_1$ and $F(z) = -f(z) + A$ for $z \in \Omega_2$.