Why $\lim_{R\to\infty}\int_{0}^{\pi}\sin(R^{2}e^{2i\theta})iRe^{i\theta}\:\mathrm{d}\theta = -\sqrt{\frac{\pi}{2}}$ This is a short question, but I'm simply not sure where to start, I know by Jordan's Lemma that the integral is not $0$, but I only know the below result due to Mathematica.

$$\lim_{R\to\infty}\int_{0}^{\pi}\sin(R^{2}e^{2i\theta})iRe^{i\theta}\:\mathrm{d}\theta=-\sqrt{\frac{\pi}{2}}$$

I need the result in order to proceed with evaluating a real integral using a contour integral.
Does anyone have any advice on how to approach the integral.

Just in order to add something to the other answers, in order to compute the Fresnel integrals you may also use the Laplace transform. Since:
$$\mathcal{L}^{-1}\left(\frac{1}{\sqrt{x}}\right) = \sqrt{\frac{1}{\pi s}},\qquad\mathcal{L}(\sin x)=\frac{1}{1+s^2},\qquad\mathcal{L}(\cos x)=\frac{s}{1+s^2} $$
we have:
$$ \int_{0}^{+\infty}\sin(x^2)\,dx = \frac{1}{2}\int_{0}^{+\infty}\frac{\sin x}{\sqrt{x}}\,dx = \frac{1}{2}\int_{0}^{+\infty}\frac{ds}{\sqrt{\pi s}(1+s^2)}=\frac{1}{\sqrt{\pi}}\int_{0}^{+\infty}\frac{du}{1+u^4}$$
as well as:
$$ \int_{0}^{+\infty}\cos(x^2)\,dx = \frac{1}{2}\int_{0}^{+\infty}\frac{\cos x}{\sqrt{x}}\,dx = \frac{1}{2}\int_{0}^{+\infty}\frac{s\,ds}{\sqrt{\pi s}(1+s^2)}=\frac{1}{\sqrt{\pi}}\int_{0}^{+\infty}\frac{u^2\,du}{1+u^4}.$$
Through the substitution $u\to\frac{1}{u}$ we can see that the last two integrals are the same.
