Assume \(f\) is continuous in the region: \(x \geq x_0, \; 0 \leq y \leq b\) and the limit \[\displaystyle \lim_{x \rightarrow + \infty} f(x + iy) = A\] exists uniformly with respect to \(y\) (independent of \(y\)). Show that \[\lim_{x \rightarrow + \infty} \int_{\gamma_x} f(z) dz = iA b \; , \; \;\] where \(\gamma_x : = \{ z \; | \; z = x + it, \; 0 \leq t \leq b\}.\)
Solution
Goal: If \(f\) is continuous on \(\{x \ge x_0, 0 \le y \le b\}\) and \(\lim_{x \to +\infty} f(x + iy) = A\) uniformly in \(y \in [0,b]\), show \(\lim_{x \to +\infty} \int_{\gamma_x} f(z)\,dz = iAb\), where \(\gamma_x\) is the vertical segment from \(x\) to \(x + ib\).
1. Parametrize the segment: \(\gamma_x: z(t) = x + it\), \(t \in [0, b]\); then \(\int_{\gamma_x} f = i\int_0^b f(x + it)\,dt\).
\(\int_{\gamma_x} f(z)\,dz = \int_0^b f(x + it) \cdot i\,dt = i\int_0^b f(x + it)\,dt\).
2. By uniform convergence, \(\lim_{x \to +\infty} f(x + it) = A\) uniformly for \(t \in [0, b]\).
Given \(\eps > 0\), uniform convergence gives \(X\) with \(\abs{f(x + it) - A} < \eps\) for all \(x > X\) and all \(t \in [0,b]\).
3. \(\lim_{x \to +\infty} \int_0^b f(x + it)\,dt = \int_0^b A\,dt = Ab\).
For \(x > X\) (from <1>2), \(\abs{\int_0^b f(x+it)\,dt - Ab} = \abs{\int_0^b (f(x+it) - A)\,dt} \le \int_0^b \abs{f(x+it) - A}\,dt < b\eps\).
4. Q.E.D.
1 and 3: \(\int_{\gamma_x} f = i\int_0^b f(x + it)\,dt \to i\cdot Ab = iAb\).