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 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
The convergence \(f(x+iy)\to A\) “independent of \(y\)” means uniform convergence in \(y\in[0,b]\): \[\sup_{0\le y\le b}\abs{f(x+iy)-A}\longrightarrow0\qquad(x\to+\infty).\] Parametrize \(\gamma_x\) by \(z=x+it\), \(0\le t\le b\), so \(dz=i\,dt\) and \(\length(\gamma_x)=b\).
1. \(\displaystyle\int_{\gamma_x} A \,dz = iAb\) for every \(x\ge x_0\).
\(\int_{\gamma_x} A\,dz = \int_0^b A\, i\,dt = iAb\).
2. For every \(x\ge x_0\),\[\abs{\int_{\gamma_x} f(z)\,dz - iAb} \le b\sup_{0\le y\le b}\abs{f(x+iy)-A}.\]
By step 1, \[\abs{\int_{\gamma_x} f(z) \,dz - iAb} = \abs{\int_{\gamma_x} \bigl(f(z) - A\bigr) \,dz} \le \length(\gamma_x)\sup_{z\in\gamma_x}\abs{f(z)-A}.\] The supremum is finite because \(f\) is continuous on the compact segment \(\gamma_x\).
3. Q.E.D.
By uniform convergence, the right-hand side of step 2 tends to \(0\) as \(x\to+\infty\), so \(\lim_{x\to +\infty} \int_{\gamma_x} f(z) \,dz = \boxed{iAb}\).