Assume \(f\) is continuous in the region \(\theset{x+iy \suchthat x\geq x_0, ~ 0\leq y \leq b}\), and the following limit exists uniformly with respect to \(y\) (and is independent of \(y\)): \[ \lim_{x\to +\infty}f(x+iy) = A .\]
Show that if \(\gamma_x \definedas \theset{z = x+it \suchthat 0 \leq t \leq b}\), then \[ \lim_{x\to +\infty} \int_{\gamma_x} f(z) \,dz = iAb .\]
Solution
Parametrize the vertical segment by \[ z=x+it, \qquad 0\le t\le b, \qquad dz=i\,dt. \] Then \[ \int_{\gamma_x}f(z)\,dz =i\int_0^b f(x+it)\,dt. \] Since \(f(x+iy)\to A\) uniformly for \(0\le y\le b\), \[ \sup_{0\le t\le b}|f(x+it)-A|\longrightarrow0. \] Therefore \[ \begin{aligned} \left|\int_{\gamma_x}f(z)\,dz-iAb\right| &=\left|i\int_0^b\bigl(f(x+it)-A\bigr)\,dt\right|\\ &\le b\sup_{0\le t\le b}|f(x+it)-A|\longrightarrow0. \end{aligned} \] Hence \[ \lim_{x\to\infty}\int_{\gamma_x}f(z)\,dz=iAb. \]