Prove that $\int_{0}^{\infty} \sin(t^2)\,dt$ equals $\sqrt{\pi/8}$ Why is $$\int_{0}^{\infty} \sin(t^2)\,dt=\sqrt{\frac{\pi}{8}}$$ I am kinda new to this kind of calculus and I only know that $$\int_{-1}^{1}\sqrt{1-x^2}dx=\frac{\pi}{2}$$
The problem in which I have to use it is:
$$\lim_{x\to\infty}\frac{\int_{0}^{x} \sin(t^2)\,dt}{x^3}$$

Hint. By setting, for any real number $t$ such that $t>0$,
$$
I_t:=\int_0^\infty e^{-tx^2}\cos (x^2)\:dx,\quad J_t:=\int_0^\infty e^{-tx^2}\sin (x^2)\:dx. \tag1
$$ One may prove that
$$
\begin{align}
\left(I_t\right)^2-\left(J_t\right)^2 &=\left(\int_0^\infty e^{-tx^2}\cos (x^2)\:dx\right)^2-\left(\int_0^\infty e^{-tx^2}\sin (x^2)\:dx\right)^2
\\\\&=\int_0^\infty\!\int_0^\infty e^{-t(x^2+y^2)}\left(\cos (x^2)\cos (y^2)-\sin (x^2)\sin (y^2)\right)dxdy
\\\\&=\int_0^\infty\!\int_0^\infty e^{-t(x^2+y^2)}\cos (x^2+y^2)\:dxdy
\\\\&=\int_0^{\pi/2}d\theta\int_0^\infty r\:e^{-tr^2}\cos(r^2)\:dr
\\\\&=\frac{\pi}4\:\frac t{1+t^2}. \tag2
\end{align}
$$ Similarly
$$
\begin{align}
2\:I_tJ_t &=\int_0^\infty\!\int_0^\infty e^{-t(x^2+y^2)}\sin (x^2+y^2)\:dxdy
\\\\&=\int_0^{\pi/2}d\theta\int_0^\infty r\:e^{-tr^2}\sin (r^2)\:dr
\\\\&=\frac{\pi}4\:\frac1{1+t^2}. \tag3
\end{align}
$$ From $(2)$ and $(3)$, one easily gets
$$
I_t=\sqrt{\frac{\pi}8}\:\sqrt{\frac{\sqrt{1+t^2}+t}{1+t^2}},\quad J_t=\sqrt{\frac{\pi}8}\:\sqrt{\frac{\sqrt{1+t^2}-t}{1+t^2}} \tag4
$$ As $t\to 0$, this leads to

$$
\int_0^\infty \cos (x^2)\:dx=\int_0^\infty \sin (x^2)\:dx=\sqrt{\frac{\pi}8}. \tag5
$$

