I'm having trouble showing that $\displaystyle \lim_{R \to \infty} \int_{C_{R}} \frac{e^{iz^{2}}}{1+z^{4}} \ \mathrm{d}z = 0 $ where $C_{R}$ is the portion of the circle $|z|= R$ in the first quadrant.
$$ \begin{align} \left|\int_{C_{R}} \frac{e^{iz^{2}}}{1+z^{4}} \ \mathrm{d}z \ \right| &\le \int_{C_{R}} \left|\frac{e^{iz^{2}}}{1+z^{4}} \ \mathrm{d}z \right| \\&= \int_{0}^{\frac{\pi}{2}} \left| \frac{e^{iR^{2}e^{2it}}}{1+R^{4}e^{4it}} Re^{it}\right| \ \mathrm{d}t \\ &\le \int_{0}^{\frac{\pi}{2}} \frac{e^{-R^{2} \sin 2t}}{R^{4}-1} R \ \mathrm{d}t \end{align}$$
Now I want to use the fact that $\displaystyle \sin t \ge \frac{2 t}{\pi}$ for $ \displaystyle 0 \le t \le \frac{\pi}{2}$, but the limits of integration seem to make that not possible.
Hint:
When $t\in\left[0,\frac{\pi}{2}\right],$ $\sin(2t)\geq0,$ so $\displaystyle e^{-R^{2}\sin(2t)}\leq1$. It follows that $$\int_{0}^{\frac{\pi}{2}}\frac{e^{-R^{2}\sin(2t)}}{R^{4}-1}\, \mathrm{d}t\leq\int_{0}^{\frac{\pi}{2}}\frac{1}{R^{4}-1}\, \mathrm{d}t=\frac{\pi}{2}\frac{1}{R^{4}-1}.$$