Limit of this complex integral I'm trying to compute $\lim_{r\to\infty}I(r)$, where $I(r)$ is given as follows:

Let $\gamma (t)=re^{it},\ 0\le t\le \pi$ and
  $$I(r)=\int_{\gamma}\frac{e^{iz}}{z}dz.$$

I found this integral very hard to calculate. I've already tried to find some Cauchy theorems to use, but without success.
Am I missing something?
 A: We have
$$I(r)=\int_{\gamma} \frac{e^{iz}}{z} dz = i \int_0^{\pi} e^{i r e^{it}} dt$$
Note that 
$$|e^{ire^{it}}| = e^{\mathrm{Re}(i r e^{it})}=e^{-r\sin(t)}.$$
Thus,
$$|I(r)|\le \int_0^\pi e^{-r\sin(t)} dt.$$
Since $|e^{-r\sin(t)}|\le 1$ and $e^{-r\sin(t)}\to 0$ for $t\in(0,\pi)$, we have by Lebesgue's dominated convergence theorem that
$$\lim_{r\to\infty} \int_0^\pi e^{-r\sin(t)} dt = 0$$
Therefore $\lim_{r\to\infty} |I(r)|$ exists and equals $0$.
Edit: Here is an alternative proof that $\lim_{r\to\infty} \int_0^\pi e^{-r\sin(t)} dt = 0$:
Choose some small $\epsilon>0$ and write
$$\int_0^\pi e^{-r\sin(t)} dt = \Big(\int_0^\epsilon e^{-r\sin(t)} dt + \int_{\pi-\epsilon}^\pi e^{-r\sin(t)} dt\Big) + \int_\epsilon^{\pi-\epsilon} e^{-r\sin(t)} dt.$$
Now estimate
$$\Big|\int_0^\epsilon e^{-r\sin(t)} dt + \int_{\pi-\epsilon}^\pi e^{-r\sin(t)} dt\Big| \le \int_0^\epsilon dt + \int_{\pi-\epsilon}^\epsilon dt = 2\epsilon$$
and
$$\int_\epsilon^{\pi-\epsilon} e^{-r\sin(t)} dt \le\pi e^{-r\sin(\epsilon)}.$$
Thus,
$$|I(r)|\le \int_0^\pi e^{-r\sin(t)}dt\le 2\epsilon + \pi e^{-r\sin(\epsilon)}$$
Taking $\lim_{n\to\infty}$ on both sides (the limit $\lim_{r\to\infty} |I(r)|$ exists because $|I(r)|$ is bounded and decreasing) and noting that $\sin(\epsilon)>0$:
$$\lim_{r\to\infty} |I(r)|\le 2\epsilon$$
Since this holds for all $\epsilon>0$, we have $\lim_{r\to\infty} |I(r)|=0$.
