Let \(\gamma\) be a piecewise smooth simple closed curve with interior \(\Omega_1\) and exterior \(\Omega_2\). Assume \(f'\) exists in an open set containing \(\gamma\) and \(\Omega_2\) with \(\lim_{z\to \infty} f(z) = A\). Show that \[ \frac{1}{2 \pi i} \int_{\gamma} \frac{f(\xi)}{\xi-z} d \xi=\left\{\begin{array}{ll} A, & \text { if } z \in \Omega_{1} \\ -f(z)+A, & \text { if } z \in \Omega_{2} \end{array}\right. .\]
Solution
Let \[ I(z)=\frac1{2\pi i}\int_\gamma\frac{f(\xi)}{\xi-z}\,d\xi. \] Choose a large positively oriented circle \(C_R\) containing \(\gamma\) and \(z\). Since \(f(\xi)\to A\) as \(\xi\to\infty\), the function \[ h(\xi)=\frac{f(\xi)-A}{\xi-z} \] is holomorphic in the exterior of \(\gamma\) and satisfies \(h(\xi)=o(1/|\xi|)\) at infinity. Hence \[ \int_{C_R}h(\xi)\,d\xi\to0. \] Also \[ \frac1{2\pi i}\int_{C_R}\frac{A}{\xi-z}\,d\xi=A. \]
If \(z\in\Omega_1\), then \(f(\xi)/(\xi-z)\) is holomorphic in the annulus between \(\gamma\) and \(C_R\), so the two contour integrals agree. Letting \(R\to\infty\) gives \[ I(z)=A. \]
If \(z\in\Omega_2\), choose \(R\) so large that \(z\) lies between \(\gamma\) and \(C_R\). Cauchy’s theorem on that annular region punctured at \(z\) gives \[ \int_{C_R}\frac{f(\xi)}{\xi-z}\,d\xi -\int_\gamma\frac{f(\xi)}{\xi-z}\,d\xi =2\pi i f(z). \] Passing to the limit yields \[ A-I(z)=f(z), \] so \[ I(z)=A-f(z). \] Thus \[ \boxed{ I(z)= \begin{cases} A,&z\in\Omega_1,\\ A-f(z),&z\in\Omega_2. \end{cases}} \]