Let \(C(R)\) be the circular arc defined by \(|z|=R\) and \(\operatorname{Im}z>0\). Prove that \[ \int_{C(R)}\frac{e^{iz}}{z}\,dz\longrightarrow0 \qquad\text{as }R\to\infty. \]
Solution
Parametrize the upper semicircle by \[ z=Re^{i\theta},\qquad 0\le\theta\le\pi. \] Then \[ \left|\int_{C(R)}\frac{e^{iz}}z\,dz\right| \le \frac{\pi}{R}, \] so the integral tends to \(0\) as \(R\to\infty\).
1. Under the parametrization \(z=Re^{i\theta}\),\[ \int_{C(R)}\frac{e^{iz}}z\,dz =i\int_0^\pi e^{iRe^{i\theta}}\,d\theta. \]
Since \[ dz=iRe^{i\theta}\,d\theta=iz\,d\theta, \] we have \(dz/z=i\,d\theta\), which gives the displayed identity.
2. Consequently,\[ \left|\int_{C(R)}\frac{e^{iz}}z\,dz\right| \le\int_0^\pi e^{-R\sin\theta}\,d\theta. \]
For \(z=Re^{i\theta}\), \[ i z=iR\cos\theta-R\sin\theta, \] so \[ |e^{iz}|=e^{-R\sin\theta}. \] Apply the triangle inequality to 1.
3. The last integral is at most \(\pi/R\).
On \(0\le\theta\le\pi/2\), concavity of \(\sin\theta\) gives \[ \sin\theta\ge\frac{2\theta}{\pi}. \] By symmetry, on \(\pi/2\le\theta\le\pi\), \[ \sin\theta\ge\frac{2(\pi-\theta)}{\pi}. \] Hence \[ \begin{aligned} \int_0^\pi e^{-R\sin\theta}\,d\theta &\le2\int_0^{\pi/2}e^{-2R\theta/\pi}\,d\theta\\ &=\frac{\pi}{R}(1-e^{-R})\\ &\le\frac{\pi}{R}. \end{aligned} \]
4. Therefore\[ \int_{C(R)}\frac{e^{iz}}z\,dz\longrightarrow0. \]
By 2 and 3, \[ 0\le \left|\int_{C(R)}\frac{e^{iz}}z\,dz\right| \le\frac{\pi}{R}\longrightarrow0. \]