Show that $\lim_{R \to{+}\infty}{I_{R}}= 0$. 
Consider $$\displaystyle I_{R}=\int_{C_{R}}^{} \frac{e^{iz}}{z^{2}}\, dz,$$ where $C_{R}$ is the semicircle with radius R in the upper half plane with endpoints $(-R,0)$ and $(R,0)$ $(C_{R}$ is open, it does not include the x axis). Show that $$\displaystyle \lim_{R \to{+}\infty}{I_{R}}= 0.$$

Could someone help me through this problem?
 A: On the semicircle $C_R$, we have $z = R e^{i \theta}$. Now try to bound the integrand $\displaystyle \frac{e^{iz}}{z^2}$ using a function decaying faster than $\displaystyle \frac1R$ on this circle. Then note that $$\displaystyle \left \lvert \int_{C_R} \frac{e^{iz}}{z^2} dz \right \rvert \leq \left \lvert \frac{e^{iz}}{z^2} \right \rvert_{\max \text{ on } C_R} \times \text{length of }C_R$$ and let $R \rightarrow \infty$. Move your cursor over the gray region below for detailed answer. 

 On the semicircle $C_R$, we have $z = R e^{i \theta}$. Hence, the integrand is$$\displaystyle \frac{e^{iz}}{z^2} = \frac{e^{iR(\cos(\theta) + i \sin(\theta))}}{R^2 e^{2i \theta}}.$$
 Hence, $$\displaystyle \left \lvert \frac{e^{iz}}{z^2} \right \rvert = \left \lvert \frac{e^{iR(\cos(\theta) + i \sin(\theta))}}{R^2 e^{2i \theta}} \right \rvert = \left \lvert \frac{e^{-R \sin(\theta) + iR\cos(\theta)}}{R^2 e^{2i \theta}} \right \rvert = \frac{e^{-R \sin(\theta)}}{R^2}.$$ Note that $R \sin(\theta) > 0$ since $\theta \in \left (0,\pi \right )$. Hence, we get that $\displaystyle \frac{e^{-R \sin(\theta)}}{R^2} < \frac1{R^2}$.
Now we can bound the integral as shown below.$$\displaystyle \left \lvert I_R \right \rvert = \left \lvert \int_{C_R} \frac{e^{iz}}{z^2} dz \right \rvert \leq \left \lvert \frac{e^{iz}}{z^2} \right \rvert_{\max \text{ on } C_R} \times \text{length of }C_R < \frac{2 \pi}{R}.$$ Now take the limit as $R \rightarrow \infty$ to get that $I_R \rightarrow 0$.

