All right, let's do some complex analysis! Let's integrate $e^{iz^2}$ over the closed contour defined in three pieces (the arrows indicating the direction of contour integration)
$$
\begin{cases}
\Gamma_1: & |z|:0\rightarrow R, & \theta=0 \\
\Gamma_2: & |z| = R, & \theta: 0\rightarrow \pi/4\\
\Gamma_3: & |z|:R\rightarrow0, & \theta=\pi/4
\end{cases}
$$
which we will eventually want to take the limit $R\rightarrow\infty$.
It can be seen that
$$
\int_0^\infty e^{ix^2}dx=\lim\limits_{R\rightarrow\infty}\int_{\Gamma_1}e^{iz^2}dz
$$
and since $e^{ix^2}$ is an even function
$$
\int_{-\infty}^{\infty}e^{ix^2}dx=2\int_0^\infty e^{ix^2}dx
$$
so we're heading in the right direction.
Now Cauchy's Theorem states that
$$
\oint_D f(z)dz =0
$$
for $f(z)$ analytic in $D$. Our function, $e^{iz^2}$, has no singularities and is defined on the entire complex plane, so it is considered an entire function, and Cauchy's Theorem holds for our closed contour:
$$
\int_0^R e^{ix^2}dx+\int_{\Gamma_2}e^{iz^2}dz+\int_{\Gamma_3}e^{iz^2}dz=0
$$
For our second integral above, we show that it vanishes as $R\rightarrow\infty$ using the ML test given by
$$
\left|\int_\Gamma f(z)dz\right|\leq ML
$$
where $M$ is a finite upper bound of $f(z)$ and $L$ is the length of the contour $\Gamma$. Of course, we need to assume that $f(z)$ is bounded and analytic on $\Gamma$ for this.
In order to apply the ML test, we substitute into our integrand $z=re^{i\theta}$ so that
$$
z^2 = r^2e^{2i\theta} = r^2cos(2\theta)+ir^2sin(2\theta)
$$
$$
|e^{iz^2}|=|e^{ir^2cos(2\theta)-r^2sin(2\theta)}|\leq e^{-R^2}=M
$$
because $r=R$ on this contour and $sin(2\theta)\leq1$. While,
$$
L=\frac{\pi R}{4}
$$
since we are looking at $1/8$th of the perimeter of the circle with radius $R$. By the ML test
$$
\left|\int_{\Gamma_2} e^{iz^2}dz\right| \leq e^{-R^2}\frac{\pi R}{4}
$$
which goes to $0$ as $R\rightarrow\infty$.
Now we want to deal with the 3rd contour integral $\Gamma_3$. Fortunately, the contour we picked allows us to easily parameterize this integral, as $y=x$. We will also need $z^2=(x+iy)^2=x^2-y^2+2ixy$. Recalling that $dz=dx+idy$ the integral becomes
$$
\int_{\Gamma_3} e^{i(x^2-y^2)-2xy}(dx+idy)
=\int_{R}^{0} e^{-2x^2}dx+i\int_{R}^{0} e^{-2y^2}dy
\rightarrow-\sqrt{\frac{\pi}{8}}(1+i)\ \text{as}\ R\rightarrow0
$$
from our real Gaussian integral identities.
Taking $R\rightarrow\infty$, our results for the contour integrals in our Cauchy's Theorem equation imply that
$$
\int_0^\infty e^{ix^2}dx = \sqrt{\frac{\pi}{8}}(1+i)
$$
The integral from $-\infty$ to $\infty$ is just twice this. So boom.
If you want, you can rewrite $e^{ix^2}=cos(x^2)+isin(x^2)$ and equate the real and imaginary parts in the last equation and you will get the limiting values of the Fresnel Integrals.
Boom.
Also, since
$$
(1+i)=\sqrt{2}e^{i\pi/4}=\sqrt{2e^{i\pi/2}}=\sqrt{2i}
$$
we have
$$
\int_0^\infty e^{ix^2}dx = \sqrt{\frac{i\pi}{4}} = \frac{1}{2}\sqrt{-\frac{\pi}{i}}
$$
which exactly matches the well-known Gaussian integral identity
$$
\int_0^\infty e^{-\alpha x^2}dx = \frac{1}{2}\sqrt{\frac{\pi}{\alpha}}
$$
with $\alpha=-i$. Boom. Thus, this suggests that this identity can work for imaginary $\alpha$, and possibly certain complex $\alpha$ with the right combination of real and imaginary parts as well as choice of contours that do not make our integrals blow up.
