Under what conditions does $\int_C \frac{e^{iz}}{z} dz$ converge? $C$ is a arc on the complex plane having 0 as its center, and $R$ as its radius. It starts on angle $\theta_0$ and ends on $\theta_1$ ($0 \leq \theta_0 \leq \theta_1 \leq 2\pi$. My instructor's claim is that the following always holds, no matter the value of $\theta_0$ and $\theta_1$. 
$$ \lim_{R\to +\infty} \int_C \frac{e^{iz}}{z} dz = 0$$
Is this claim true? If so, why does this integral converge for $\theta_0 = \frac{5\pi}{4},\theta_1 = \frac{7\pi}{4}$ ?
 A: You do not give a precise definition of $C$, but it seems you mean that the arc is part of the circle with radius $R$ and center $0$. A parametrization is given by $z : [\theta_0,\theta_1] \to \mathbb C, z(t) = R e^{it}$. Since $z'(t) = iz(t)$ and $iz(t) = iR\cos t - R\sin t$, you get
$$I(R,\theta_0,\theta_1) = \int_C \frac{e^{iz}}{z}dz = \int_{\theta_0}^{\theta_1}\frac{e^{iz(t)}}{z(t)}z'(t)dt = i \int_{\theta_0}^{\theta_1}e^{iz(t)}dt = i \int_{\theta_0}^{\theta_1} \frac{e^{iR\cos t}}{e^{R\sin t}}dt .$$
Hence the claim is true for $\theta_1 \le \pi$ because
$$\lvert I(R,\theta_0,\theta_1) \rvert \le \int_{\theta_0}^{\theta_1} \left\lvert \frac{e^{iR\cos t}}{e^{R\sin t}}\right \rvert dt = \int_{\theta_0}^{\theta_1} \frac{1}{e^{R\sin t}} dt \le \int_{0}^{\pi} \frac{1}{e^{R\sin t}} dt .$$
Let $\varepsilon > 0$. Choose $\delta > 0$ such that $\int_{0}^{\delta} \frac{1}{e^{R\sin t}} dt , \int_{\pi- \delta}^{\pi} \frac{1}{e^{R\sin t}} dt < \varepsilon /4$. Let $\mu = \min \{\sin t \mid t \in [\delta, \pi-\delta] \}$. Then
$$\int_{\delta}^{\pi-\delta} \frac{1}{e^{R\sin t}} dt \le  \int_{\delta}^{\pi-\delta} \frac{1}{e^{R \mu}} dt$$
which is $< \varepsilon/2$ for $R \ge R_0$.
For $\theta_0 = 0$ and $\theta_1 = 2\pi$ the claim ist not true. By Cauchy's integral formula we have
$$\frac{1}{2\pi i}\int_C \frac{e^{iz}}{z - 0}dz = e^{i0} = 1, $$
thus
$$I(R,0,2\pi) = 2 \pi i .$$
We conclude that the claim is also false when $0 \le \theta_0 \le \pi$ and $\theta_1 = 2\pi$. In fact
$$I(R,\theta_0,2\pi) = I(R,0,2\pi) - I(R,0,\theta_0) \to 2\pi i .$$
