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.
 A: Let $u=Re^{i\theta}$, so that $du=iRe^{i\theta}\,d\theta$.  Thus
$$\lim_{R\to \infty}\int_0^{\pi}\sin(R^2e^{i2\theta})iRe^{i\theta}\,d\theta=-\int_{-\infty}^{\infty}\sin(u^2)\,du=-\sqrt{\frac{\pi}{2}}$$
where we used the well-know results of the Fresnel Integral.

NOTE:
To evaluate the Fresnel integral, let's analyze the following complex integral.
$$\begin{align}
\oint e^{iz^2}\,dz&=\int_0^Re^{ix^2}\,dx+\int_0^{\pi/4}e^{iR^2e^{i2\phi}}iRe^{i\phi}\,d\phi+\int_R^0e^{i(1+i)^2t^2}(1+i)\,dt\\\\
&=\int_0^Re^{ix^2}\,dx+\int_0^{\pi/4}e^{iR^2e^{i2\phi}}iRe^{i\phi}\,d\phi+(1+i)\int_R^0e^{-2t^2}\,dt
\end{align}$$
Now, as $R\to \infty$, the first integral becomes
$$\lim_{R\to \infty}\int_0^Re^{ix^2}\,dx=\int_0^{\infty}\cos (x^2)\, dx+i\int_0^{\infty}\sin (x^2)\, dx$$
the second integral goes to zero since
$$\begin{align}
\left|\int_0^{\pi/4}e^{iR^2e^{i2\phi}}iRe^{i\phi}d\phi\right|&\le R\int_0^{\pi/4}e^{-R^2\sin 2\phi}\,d\phi\\\\
&\le R\int_0^{\pi/4}e^{-4R^2\phi/\pi}\,d\phi\\\\
&=\frac{\pi}{4}\frac{1-e^{-R^2}}{R}\to 0
\end{align}$$
and the third integral becomes the Gaussian Integral
$$(1+i)\int_{\infty}^0e^{-2t^2}\,dt=-(1+i)\sqrt{\frac{\pi}{8}}$$
Since the contour integral is zero, we have

$$\bbox[5px,border:2px solid #C0A000]{\int_0^{\infty}\cos (x^2)\, dx=\int_0^{\infty}\sin (x^2)\, dx=\sqrt{\frac{\pi}{8}}}$$

A: 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.
A: 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 $R \to \infty$.
