inequality in absolute value with exponential

Question

Can you help me with this problem please ??

If $r>0$ . Show that $\left |{\displaystyle\int_{\gamma}e^{iz^{2}}dz}\right |\leq{\displaystyle\frac{\pi(1-e^{-r^{2}})}{4r}}$ where $\gamma(t)=re^{it}$, $0\leq{t}\leq{pi/4}$

Thanks

Answer 1

$$\left|\int e^{iz^2}dz \right| \leq \int_{0}^{\frac{\pi}{4}}|e^{i (re^{it})^2} r e^{it} |dt = \int_{0}^{\frac{\pi}{4}}|re^{ir^2\cos(2t) - r^2 \sin(2t)}| dt=\int_{0}^{\frac{\pi}{4}} re^{-r^2\sin(2t)} dt$$

Now $\sin(2t) > t$ with $0 \leq t \leq \pi/4$ and $e^{-t}$ is decreasing function then $e^{-r^2\sin(2t)}<e^{-r^2t}$, now,

$$\left|\int e^{iz^2}dz \right| \leq \int_{0}^{\frac{\pi}{4}} re^{-r^2 t} dt \leq \frac{\pi}{4r^2}r (1-e^{-r^2}) = \frac{\pi}{4r}(1-e^{-r^2}) $$

Check it ! Because I could have a mistake !