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.

You're probably better off using a different contour. Presumably you're interested in
$$ \int_{0}^{\infty} \sin{(x^2)} \, dx. $$
Then this is the imaginary part of
$$ \int_0^{\infty} e^{ix^2} \, dx. $$
We know the value of $\int_0^{\infty} e^{-x^2} \, dx$, so let's try to rotate the contour to get to this. Look at the sector of the circle of radius $R$, between the real axis and the line $\Re(z)=\Im(z)$. Then the integral along the real axis tends to the Fresnel integral, the bit along the other straight part is
$$ \int_R^0 e^{-y^2} e^{i\pi/4} \, dy = -\frac{1+i}{\sqrt{2}} \int_0^R e^{-y^2} \, dy \to -\frac{(1+i)\sqrt{\pi}}{\sqrt{2}}. $$
There are no singularities in the contour, so we now just have to check the last bit tends to zero. This is
$$ \int_0^{\pi/4} e^{i R e^{i\theta}} iRe^{i\theta} \, d\theta $$
Essentially the trick now is that
$$ \lvert e^{i R^2 e^{2i\theta}} \rvert = e^{-R^2\sin{2\theta}} < e^{-4R^2 \theta/\pi}, $$
and so its integand tends to zero as $R \to \infty$.
