I have been doing some research and found out that the most used function to contour integrate when it comes to fresnel integrals is $e^{iz^2}$ on the boundary of $|z| \leq R$ and $0 \leq \arg(z) \leq \frac{\pi}{4}$. For this one I understand the splitting process into $\cos$ and $\sin$ thanks to $e^{iz}$ identity.
Recently, I also found out that, sometimes, it is used the function $e^{-z^2}$ on the same contour, achieving the same result for both integrals of $\cos(x^2)$ and $\sin(x^2)$
My question is: which of those 2 functions simplify the proof and calculations? How can you use the exponential identity with the last one if there is no $i$ in its exponent?
thanks in advance
To answer your second question, "How can you use the exponential identity with the last one if there is no $i$ in its exponent?," we proceed as follows.
Let $C_R$ denote the closed contour comprised of $(1)$ the straight line segment from $0$ to $R$, $(2)$ the circular arc from $R$ to $Re^{i\pi/4}$, and $(3)$ the straight line segment from $Re^{i\pi/4}$ to $0$.
Inasmuch as $e^{-z^2}$ is analytic, Cauchy's integral theorem guarantees that
$$\begin{align} 0&=\oint_{C_R} e^{-z^2}\,dz\\\\ &=\int_0^R e^{-x^2}\,dx +\int_0^{\pi/4}e^{-R^2 e^{i2\phi}}\,ie^{i\phi}\,d\phi+\int_{R}^0e^{-t^2 e^{-i\pi/2}}e^{-i\pi/4}\,dt\\\\&=\int_0^R e^{-x^2}\,dx-e^{-i\pi/4}\int_0^R e^{it^2}\,dt +\int_0^{\pi/4}e^{-R^2 e^{i2\phi}}\,ie^{i\phi}\,d\phi\tag1 \end{align}$$
As $R\to \infty$, the thrid integral on the right-hand side of $(1)$ vanishes and we find that
$$\begin{align} \int_0^\infty e^{it^2}\,dt&=e^{i\pi/4}\int_0^\infty e^{-x^2}\,dx\\\\ &=\frac{\sqrt \pi}{4}(1+i) \end{align}$$
And we are done!