TITLE: Calculating $\int\frac{e^{iz}}{z}\, dz$ on the semi-circle given by $re^{i\theta}$ where $\theta:\,0\to\pi$
QUESTION [5 upvotes]: As a part of an exercise I need to calculate 

$$ \lim_{r\to0}\int_{\sigma_{r}}\frac{e^{iz}}{z}\, dz $$
Where  $$ \sigma_{r}:\,[0,\pi]\to\mathbb{C} $$
$$ \sigma_{r}(t)=re^{it} $$

I know how to calculate the integral on the full circle ($|z|=r$)
using Cauchy integral formula: $$2\pi ie^{iz}|_{z=0}=2\pi i$$
I have checked if 
$$
(\overline{\frac{e^{iz}}{z})}=\frac{e^{iz}}{z}
$$
so that the integral I want to calculate is $$\frac{1}{2}\cdot2\pi i=\pi i$$
but I got that that the equality above does not hold.
Can someone please suggest a way of calculating this integral ?

REPLY [3 votes]: Note that $\dfrac{e^{iz}-1}{z}$ is bounded near $z=0$. Since the length of $\sigma_r$ tends to $0$, we get
$$
\lim_{r\to0^+}\int_{\large\sigma_r}\frac{e^{iz}-1}{z}\,\mathrm{d}z=0
$$
Thus,
$$
\begin{align}
\lim_{r\to0^+}\int_{\large\sigma_r}\frac{e^{iz}}{z}\,\mathrm{d}z
&=\lim_{r\to0^+}\int_{\large\sigma_r}\frac{e^{iz}-1}{z}\,\mathrm{d}z
+\lim_{r\to0^+}\int_{\large\sigma_r}\frac{1}{z}\,\mathrm{d}z\\
&=0+\lim_{r\to0^+}\int_0^\pi\frac{1}{re^{it}}\,\mathrm{d}re^{it}\\
&=\lim_{r\to0^+}\int_0^\pi i\,\mathrm{d}t\\[9pt]
&=\pi i
\end{align}
$$