Calculate $\lim_{R\to \infty} \int_{C_R} e^{iz}/z dz$

Question

Let $C_R$ be the lower semicircle of radius $R$. Calculate $$\lim_{R\to \infty} \int_{C_R} \frac{e^{iz}}{z} dz$$

Does the following make sense? Expand $e^{iz}$ as its taylor series and observe that

$$\lim_{R\to \infty} \int_{C_R} \frac{e^{iz}}{z} dz =\lim_{R\to \infty} \sum_{n=0}^\infty \int_{C_R} \frac{{iz}^{n-1}}{n!} $$

All of the terms besides $n=1$ have a primitive and so we need only calculate:

$$\lim_{R\to \infty} \int_{C_R} \frac{1}{z} dz$$

We will make a brach cut of $\log z$ on the imaginary axis so that if $z$ is in this region, then $\log z = \log |z| + i\arg z$ where $\arg z \in (\pi/2,5\pi/2)$. Thus,

$$\lim_{R\to \infty} \int_{C_R} \frac{1}{z} dz = \lim_{R\to \infty} \log z|_{Re^{i\pi}}^{Re^{2\pi i}} = \lim_{R\to \infty} \log R + 2\pi i - (\log R + \pi i) = \pi i $$

Any comments or suggestions would be appreciated.

Answer 1

Let $\phi(t) = Re^{it}$. Then $$\begin{align}\int_{C_R} \dfrac {e^{iz}}z\,dz &= \int_\pi^{2\pi}\dfrac{e^{i\phi(t)}}{\phi(t)} \phi'(t)\,dt \\&=\int_\pi^{2\pi} \dfrac{e^{iRe^{it}}}{Re^{it}} iRe^{it}\,dt \\&=\sum_{n=0}^\infty\dfrac{i^{n+1}R^n}{n!}\int_\pi^{2\pi}e^{iRnt}\,dt \end{align}$$