Solving $\left|\int_0^\infty{\sin(x^2)}dx\right|$

$$Let\:I=\int_0^\infty{\sin(x^2)dx}$$

Using \(\sin(x)=-\mathfrak{Im}(e^{-ix})\), we get,

$$I = -\mathfrak{Im}\left(\int_0^\infty{e^{-ix^2}}dx\right)$$

Substituting \(\sqrt{i}x = t\) and using this result,

$$I = -\mathfrak{Im}\left(\int_0^\infty{e^{-t^2}}\frac{dx}{\sqrt{i}}\right) =-\mathfrak{Im}\left(\frac{\sqrt{\pi}}{2\sqrt{i}}\right)$$

$$\sqrt{i}=e^{i\frac{\pi}{4}}=\frac{1}{\sqrt{2}}(1+i)$$

Therefore, we have,

$$I = -\mathfrak{Im}\left(\frac{\sqrt{\pi}}{2}\frac{\sqrt{2}}{(i+1)}\right)=-\mathfrak{Im}\left(\frac{\sqrt{\pi}}{2}\frac{\sqrt{2}(1-i)}{(i^2-1)}\right)$$

$$\boxed{\left|I\right| = \frac{\sqrt{\pi}}{2\sqrt{2}}}$$