On a contour defined by sector of radius $R$ making an angle of $\frac{\pi}{4}$, I applied the ML inequality on the curvy(angular) part ofcontour and this is what i found $|\int_{\gamma} e^{iaz^{2}}dz|\leq\frac{\pi R}{4}$ since $|f(z)|\leq 1$ and $L=\frac{\pi R}{4}$
But this doesn't seem right as $\lim_{R\to\infty}\frac{\pi R}{4}=\infty$,even though I know that it should go to zero.
I am evaluating the fresnel integrals using this.
Note that
$$ \left\vert \int_{\gamma} e^{iaz^2} dz \right\vert \le \int_{\gamma} \left\vert e^{iaz^2} \right\vert \left\vert dz \right\vert = \int_0^{\pi/4} e^{-aR^2\sin 2\theta} R \ d\theta $$
You can show geometrically that on $\theta \in [0,\pi/4]$, the function $g(\theta) = \sin 2\theta$ lies completely above its secant line, i.e.
$$ \sin 2\theta \ge \frac{4}{\pi}\theta \implies e^{-aR^2\sin2\theta} \le e^{-4aR^2 \theta /\pi} $$
Hence, the integral is bounded above by
$$ \int_0^{\pi/4} e^{-4aR^2 \theta /\pi} R\ d\theta = \frac{\pi}{4aR}\left(1 - e^{-aR^2}\right) $$
which tends to $0$ as $R \to \infty$