Let \(0\leq \alpha \leq 2\pi\) be a fixed angle. Suppose \(f\) is continuous on the region \(\Omega = \ts{\abs{z} \geq R, \Arg(z) \in [0, \alpha]}\) and \(\lim_{z\to \infty} zf(z) = A\). Show that \[ \lim_{z\to \infty} \int_{\gamma_R} f(z) \dz = iA\alpha ,\] where \(\gamma_R \da \ts{ \abs{z} = R, \Arg(z) \in [0, \alpha]}\) is an arc.
Solution
Key observation: \[ iA\alpha = \int_\gamma {A\over z}\dz .\] Why this is true: \[ \int_\gamma {A\over z}\dz = \int_0^\alpha {1\over Re^{it}} iRe^{it}dt = \int_0^\alpha iA \dt = iA\alpha .\]
Now estimate the difference:
\[ \abs{ \int_\gamma f(z) \dz - iA\alpha } &= \abs{ \int_\gamma f(z) \dz - \int_\gamma {A\over z} \dz}\\ &= \abs{\int_\gamma f(z) - {A\over z} \dz} \\ &= \abs{\int_\gamma{zf(z) - A \over z} \dz} \\ &\leq \int_\gamma \abs{zf(z) - A \over z} \dz \\ &= \int_\gamma { \abs{zf(z) - A} \over R} \dz \\ &\leq {1\over R } \int_\gamma \norm{zf(z) - A}_{\infty, \gamma} \dz \\ &= {\eps\over R}\cdot \length(\gamma) \\ &= {\eps \over R} \cdot R\alpha \\ &= \eps \alpha \\ &\convergesto{R\to\infty}0 .\]