I got the function $f(x) = \frac{\exp(ix)}{x}$
and I want to integrate over a halfcircle from $-R$ to $R$ where I also put another half circle around 0 from $-r$ to $r$ to not include 0.
Since I got no poles and f is holomorphic I know from Cauchys integral theorem that the integral over my path should be zero.
I parameterize my path the following way:
$$\begin{eqnarray} \gamma_1(t) &=& t &t \in [-R,-r] \\
\gamma_2(t) &=& r \cdot \exp(i (\pi-t)) & t \in [0, \pi] \\
\gamma_3(t) &=& t & t \in [r,R] \\
\gamma_4(t) &=& R \cdot \exp(i t) & t \in [0, \pi]
\end{eqnarray}
$$
Now my problem is, if I try to estimate that the integral over $\gamma_4$ has to be zero, I end up with an integral like
$$
\int_{0}^\pi \exp(i R \exp(it)) i ~ \operatorname{d}t
$$
where I can find the right inequality that this integral go to zero for $R$ going to $\infty$.
Moreover I don't know how I could calculate that the integral over $\gamma_2$ has to be $\pi i$.
HINT 1: The vanishing of the integral $\gamma_4$ follows pretty directly from Jordan's lemma, with $g(z) = 1/z$.
HINT 2: The integrand over the path $\gamma_2$ can be found by writing down an integral similar to your integral for $\gamma_4$, and then considering the limit as $r \to 0$. For full rigor, note that $|e^{i r e^{i t}}| = |e^{i r \cos t - r \sin t}| = e^{-r \sin t}$, and then use the dominated convergence theorem.