Assume \(f\) is continuous in the region \[ R_0\le |z-a|<\infty, \qquad 0\le \arg(z-a)\le \beta_0, \qquad 0<\beta_0\le 2\pi, \] and that the limit \[ \lim_{z\to\infty}(z-a)f(z)=A \] exists. Show that \[ \lim_{r\to+\infty}\int_{\gamma_r} f(z)\,dz=iA\beta_0, \] where \[ \gamma_r:=\{z:z=a+re^{it},\ 0\le t\le\beta_0\}. \]
Solution
Parametrize the arc by \[ z=a+re^{it},\qquad 0\le t\le \beta_0, \] so that \(dz=ire^{it}\,dt\) and \(z-a=re^{it}\). Then \[ \int_{\gamma_r}f(z)\,dz =i\int_0^{\beta_0}(z-a)f(z)\,dt. \] Because \((z-a)f(z)\to A\) as \(z\to\infty\) within the stated sector, the convergence is uniform on the arc parameter \(0\le t\le\beta_0\): for every \(\varepsilon>0\), once \(r\) is sufficiently large, \[ \left|(a+re^{it}-a)f(a+re^{it})-A\right|<\varepsilon \] for every such \(t\). Hence \[ \left| \int_{\gamma_r}f(z)\,dz-iA\beta_0 \right| \le \int_0^{\beta_0}\left|(z-a)f(z)-A\right|dt \le \beta_0\varepsilon. \] Therefore \[ \boxed{\displaystyle \lim_{r\to+\infty}\int_{\gamma_r}f(z)\,dz=iA\beta_0}. \]