A different approach

Consider the contour given by the sector of the circle with radius R and an angle $\frac{\pi}{4}$ .Then we have

$0=\large\int_\zeta e^{i z^2} dz = \large\int_{\gamma_1}\ e^{i z^2} dz+\large\int_{\gamma_2}\ e^{i z^2} dz+\large\int_{\gamma_3}\ e^{i z^2} dz$ where $\gamma_1$ is the segment along the real axis, $\gamma_2$ is the arc of the sector and $\gamma_3$ is the segment joining $Re^{\frac{i\pi}{4}}$ with 0.Now,calculating with the natural parameterization , we have $\large\int_{0}^{R} e^{i z^2} dz+ \large\int_{0}^{\pi/4} e^{i(Re^{i \theta})^2} d\theta + \large\int_{0}^{1} e^{i[Re^{i\pi/4}(1-t)]^2}(-Re^{i\pi/4}) dt.$ Obsereve that $\large |\large\int_{0}^{\pi/4} e^{i(Re^{i \theta})^2} d\theta\large| \leq \large\int_{0}^{\pi/4} Re^{-R^2 \sin\theta} d\theta$ $\leq\large\int_{0}^{\pi/4} R e^{-R^2 (2 /\pi \cdot\theta d\theta)}$ $= \dfrac{-\pi e^{-R^2/2}}{R}+\dfrac{\pi}{R}$ . Which clearly goes to zero as R tends to infinity.Thus,we have $\large\lim_{R\rightarrow \infty}\large \int_0^R e^{iz^2} dz = \large\lim_{R\rightarrow \infty} R e^{i \pi/4} \large\int_0^1 e^{i(Re^{i \pi/4}(1-t))^2} dt. \quad\quad\quad (*)$ Now, if we do u-substitution,using $u =R(1-t)$ , the right hand side becomes $\begin{aligned}\lim_{R \rightarrow \infty} R e^{i\pi/4}\large \int_R^0 e^{-u^2} \dfrac{-1}{R} du &= \lim_{R\rightarrow \infty} R e^{i\pi/4} \large\int_0^R e^{-u^2} \dfrac{1}{R} du\\ &= \lim_{R \rightarrow \infty} e^{i \pi/4}\large \int_0^R e^{-u^2} du\\ &= e^{i\pi/4}\large \int_0^\infty e^{-u^2} du\\&=e^{i\pi/4}\dfrac{\sqrt{\pi}}{2},\end{aligned}$ where the last equality follows from the fact that $e^{-u^2}$ is even and so $\large\int_0^\infty e^{-u^2} du = \dfrac{1}{2}\large\int_{-\infty}^\infty e^{-u^2} du = \dfrac{1}{2} \sqrt{\pi}.$ Combining this result with $(*)$ , we have $\large\int_0^\infty e^{iz^2} dz = e^{i\pi/4}\dfrac{\sqrt{\pi}}{2} = \dfrac{\sqrt{2\pi}}{4}+ i \dfrac{\sqrt{2\pi}}{4}.$ So as we are required the integral of $\cos(z^2)$ we take $\Re \large\int_0^\infty e^{iz^2} dz$ i.e $\dfrac{\sqrt{2\pi}}{4}$ .

Fresnel Integrals!!!!