How can we compute $\int^{\infty}_0\sin(x^2)dx$ using Fourier transform? 
How can we compute $\int^{\infty}_0\sin(x^2)dx$ using Fourier transform?

I had an idea in my mind. To use the $\text{sinc}$ function and take its inverse Fourier Transform or something like that. I also tried using the inverse Fourier Transform of $\frac1{x^2}$, but I cannot use the fact to this improper integral.
Can someone point me out how can we use the Fourier Transform here?
Please only use Fourier Transform, nothing else.
 A: Only a partial answer, but maybe useful. Assuming $\int_0^\infty\sin(x^2)dx=\int_0^\infty\cos(x^2)dx$ (which is numerically verified and possibly easy to prove), and employing the Fourier transform
$$
\hat{f}(\xi)=\int_{-\infty}^\infty dx \frac{1}{\sqrt{|x|}}e^{-2\pi\mathrm{i}x\xi}=2\mathrm{Re}\int_0^\infty dx \frac{1}{\sqrt{x}}e^{-2\pi\mathrm{i}x\xi}=\frac{1}{\sqrt{|\xi|}}\ ,
$$
we have easily 
$$
I=\int_0^\infty dx\cos(x^2)=\mathrm{Re}\int_0^\infty dx\ e^{\mathrm{i}x^2}=\frac{1}{2}\mathrm{Re}\int_0^\infty dy\ \frac{e^{\mathrm{i}y}}{\sqrt{y}}=\frac{1}{4}\hat{f}\left(-\frac{1}{2\pi}\right)=\sqrt{\frac{\pi}{8}}\ .
$$
A: I hope the Laplace transform is allowed too, since it is just a "rotated variant" of the Fourier transform. We have:
$$ \mathcal{L}(\sin x) = \frac{1}{1+s^2},\qquad \mathcal{L}^{-1}\left(\frac{1}{\sqrt{x}}\right) = \frac{1}{\sqrt{\pi s}}\tag{1}$$
hence:
$$ \begin{eqnarray*}\int_{0}^{+\infty}\sin(x^2)\,dx &=& \frac{1}{2}\int_{0}^{+\infty}\frac{\sin x}{\sqrt{x}}\,dx\\&=&\frac{1}{2\sqrt{\pi}}\int_{0}^{+\infty}\frac{ds}{(1+s^2)\sqrt{s}}\\&=&\frac{1}{\sqrt{\pi}}\int_{0}^{+\infty}\frac{ds}{1+s^4}\tag{2}\end{eqnarray*}$$
and the last integral can be computed in many ways, for instance through the residue theorem and/or through partial fraction decomposition ($1+4s^4 = (1-2s+2s^2)(1+2s+2s^2)$).
That leads to:

$$ \int_{0}^{+\infty}\sin(x^2)\,dx = \frac{1}{\sqrt{\pi}}\cdot\frac{\pi}{\sqrt{8}} = \color{red}{\sqrt{\frac{\pi}{8}}.}\tag{3}$$

