= Solution
Integrate the entire function $e^{iz^2}$ around the positively oriented sector boundary: the real segment $0\to R$, the arc $Re^{i\theta}$ for $0\leq\theta\leq\pi/4$, and the radial segment back to zero. <Cauchy integral theorem> gives
$$
\int_0^Re^{ix^2}\,dx+\int_{\mathrm{arc}}e^{iz^2}\,dz
-e^{i\pi/4}\int_0^Re^{-r^2}\,dr=0.
$$
The last term follows from $z=re^{i\pi/4}$ and $iz^2=-r^2$; its minus sign is the return orientation.
On the arc, $|e^{iz^2}|=e^{-R^2\sin(2\theta)}$. The supplied sine bound gives $\sin(2\theta)\geq4\theta/\pi$ throughout this sector, so
$$
\left|\int_{\mathrm{arc}}e^{iz^2}\,dz\right|
\leq R\int_0^{\pi/4}e^{-4R^2\theta/\pi}\,d\theta
=\frac\pi{4R}(1-e^{-R^2})\longrightarrow0.
$$
The Gaussian integral then yields $\int_0^\infty e^{ix^2}\,dx=e^{i\pi/4}\sqrt\pi/2$. Taking real and imaginary parts proves \b[both <Fresnel integrals>]:
$$
\boxed{\int_0^\infty\cos x^2\,dx=\int_0^\infty\sin x^2\,dx=\sqrt{\frac\pi8}.}
$$
The contour estimate proves existence of the improper oscillatory integrals as well as their values; absolute convergence is neither asserted nor required.
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