$$\left|\int_{\gamma (t)} e^{iz^2} \ dz \right| ≤ \frac{\pi (e-1)}{4e}$$, Where $$\gamma (t)=e^{it},t\in[0,\pi/4]$$
This is question in Complex Analysis by Ponnusamy Page_128 ,I tried it but unable to get this My Attempt: $L=\pi/4$ and $M=\max_{t\in[0,\pi/4]}|e^{i{\gamma (t)}^2}|=1$
Which gives $\pi/4$ instead of $\frac{\pi (e-1)}{4e}$
Because $\sin x$ is a concave function for $0\leq x\leq \frac{\pi}{2}$, we get the inequality $$\sin x\geq \frac{2}{\pi}x,\quad 0\leq x\leq \frac{\pi}{2}.$$ We're going to use this inequality applied to $x=2t$ in the following calculation: $$\begin{align*} \left|\int_{\gamma(t)}e^{iz^2}\; dz\right|&\leq \int_{\gamma(t)}\left|e^{iz^2}\right|\; |dz|\qquad\text{(by the $M$-$L$ inequality)}\\ &=\int_0^{\pi/4}e^{-\sin 2t}\; dt\\ &\leq\int_0^{\pi/4}e^{-(4/\pi)t}\; dt\\ &=\frac{\pi}{4}\left(1-e^{-(4/\pi)(\pi/4)}\right)\\ &=\frac{\pi(e-1)}{4e}. \end{align*}$$