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Let $r > 0$ and $\gamma : [0,\pi/4] \to \Bbb C, \gamma(t)=re^{it}$ a path. Show that $$\left|\int_{\gamma} e^{iz^2} \ dz \ \right| \le \frac{\pi(1-e^{-r^2})}{4r}$$

This question has been answered here already, but I would like to approach it using the $ML$-lemma. I think that the length of the path is $L=\frac14r$. The trouble I'm facing is approximating $|e^{iz^2}|$. If I write $z=r\cos \alpha +ir\sin \alpha$, then $z^2 =r^2\cos \left(2α\right)+r^2\sin \left(2α\right)i$ and so $$|e^{iz^2}|=|e^{i(r^2\cos \left(2α\right)+r^2\sin \left(2α\right)i)}| =|e^{-r^2\sin \left(2α\right)+r^2\cos \left(2α\right)i}| =e^{-r^2 \sin(2\alpha)} \le e^{-r^2}$$ but I'll just end up with $$\left|\int_{\gamma} e^{iz^2} \ dz \ \right| \le \frac{re^{-r^2}}{4}$$ which is not quite what I needed to show. What can I do here?

Werner
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  • I doubt that the ML lemma gives an estimate as strong as the one to be proved, since the maximum of a nonnegative continuous function $f$ on a closed interval $[a,b]$ is generally greater than the average value of $f$ over $[a,b]$. The result $\pi(1 - e^{-r^2})/(4r)$ is obtained from integrating $re^{-4\alpha r^2/\pi}$ from $\alpha = 0$ to $\alpha = \pi/4$. – kobe May 08 '22 at 18:47
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    Note that $e^{-r^2 \sin(2\alpha)} \le e^{-r^2}$ only holds when $\sin(2\alpha)=1$. – John B May 08 '22 at 19:07

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We have $$ \left|\int_{\gamma} e^{iz^2} \ dz \ \right| \le\int_0^{\pi/4}|e^{i\gamma(t)^2}|\cdot|\gamma'(t)|\,dt= \int_0^{\pi/4}e^{-r^2 \sin(2t)}r\,dt. $$ Since $\dfrac{\sin\alpha}\alpha\ge\dfrac2\pi$, we get $$ \left|\int_{\gamma} e^{iz^2} \ dz \ \right| \le \int_0^{\pi/4}e^{-r^2 4t/\pi}r\,dt=\frac{\pi(1-e^{-r^2})}{4r}. $$

PS: For a proof of the inequality $\dfrac{\sin\alpha}\alpha\ge\dfrac2\pi$ see Proving $\frac2\pi x \le \sin x \le x$ for $x\in [0,\frac {\pi} 2]$.

John B
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