For \(R>0\) let \(C_R\) be the upper semicircle \(t\mapsto Re^{it}\), \(t\in[0,\pi]\).
Let \(\alpha>0\), let \(g\) be continuous on \(C_R\), and let \(M_R\coloneqq\sup_{z\in C_R}\abs{g(z)}\). Then \[ \abs{\int_{C_R}e^{i\alpha z}g(z)\dz}\leq\frac{\pi}{\alpha}M_R. \] In particular, if \(g\) is continuous on \(\{z : \operatorname{Im}z\geq0,\ \abs{z}\geq R_0\}\) for some \(R_0\) and \(M_R\to0\) as \(R\to\infty\), then \(\int_{C_R}e^{i\alpha z}g(z)\dz\to0\).
For \(\alpha<0\) the same statements hold with \(C_R\) replaced by the lower semicircle \(t\mapsto Re^{-it}\), \(t\in[0,\pi]\), and \(\pi/\alpha\) replaced by \(\pi/\abs{\alpha}\).
For \(\alpha>0\) and \(z=Re^{it}\), \(\abs{e^{i\alpha z}}=e^{-\alpha R\sin t}\) and \(\abs{\dz}=R\,dt\), so \[ \abs{\int_{C_R}e^{i\alpha z}g(z)\dz}\leq M_R\int_0^\pi e^{-\alpha R\sin t}R\,dt=2M_R\int_0^{\pi/2}e^{-\alpha R\sin t}R\,dt. \] Since \(\sin t\geq 2t/\pi\) for \(t\in[0,\pi/2]\), \[ \int_0^{\pi/2}e^{-\alpha R\sin t}R\,dt\leq\int_0^{\pi/2}e^{-2\alpha Rt/\pi}R\,dt\leq\frac{\pi}{2\alpha}, \] which gives the bound. For \(\alpha<0\) and \(z=Re^{-it}\) with \(t\in[0,\pi]\), \(\abs{e^{i\alpha z}}=e^{\alpha R\sin t}=e^{-\abs{\alpha}R\sin t}\), and the same estimate applies with \(\abs{\alpha}\) in place of \(\alpha\).