Calculating $\int\frac{e^{iz}}{z}\, dz$ on the semi-circle given by $re^{i\theta}$ where $\theta:\,0\to\pi$ 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 ?
 A: For $k \neq -1$, the function $z \mapsto z^k$ has a primitive on $\mathbb{C}\setminus \{0\}$, namely $z \mapsto \frac{1}{k+1}z^{k+1}$.
Thus, for $f$ holomorphic in a punctured neighbourhood of $0$, with the Laurent expansion
$$f(z) = \sum_{-\infty}^\infty a_k z^k,$$
you can semi-explicitly evaluate the integral over $\sigma_r$ (for small enough $r$) as
$$\begin{align}
\int_{\sigma_r} f(z)\,dz &= a_{-1}\int_{\sigma_r}\frac{dz}{z} + \sum_{k\neq -1} a_k \int_{\sigma_r}z^k\\
&= \pi i a_{-1} + \sum_{k \neq -1} \frac{a_k r^{k+1}}{k+1} \left((-1)^{k+1} - 1\right)\\
&= \pi i a_{-1} -2 \sum_{-\infty}^\infty \frac{a_{2k}r^{2k+1}}{2k+1}.
\end{align}$$
From that, you can read off that in general the integral depends on $r$, and it converges to $\pi i \operatorname{res}_0 f$ when $f$ has a simple pole in $0$ (generally, if $f$ has an odd principal part).
Your integrand
$$\frac{e^{iz}}{z} = \sum_{k=0}^\infty \frac{i^k}{k!}z^{k-1} = \frac1z + i -\frac{z}{2} -i\frac{z^2}{6} + \frac{z^3}{24} + \dotsb$$
has $0$ as a simple pole, hence the above applies. Since not all even coefficients of the Laurent expansion vanish, the integral depends on $r$, but the limit is $\pi i$.
A: 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,
$$
\begin{align}
\lim_{r\to0^+}\int_{\large\sigma_r}\frac{e^{iz}}{z}\,\mathrm{d}z
&=\lim_{r\to0^+}\int_{\large\sigma_r}\frac{e^{iz}-1}{z}\,\mathrm{d}z
+\lim_{r\to0^+}\int_{\large\sigma_r}\frac{1}{z}\,\mathrm{d}z\\
&=0+\lim_{r\to0^+}\int_0^\pi\frac{1}{re^{it}}\,\mathrm{d}re^{it}\\
&=\lim_{r\to0^+}\int_0^\pi i\,\mathrm{d}t\\[9pt]
&=\pi i
\end{align}
$$
