Fresnel integral $\int\limits_0^\infty\sin(x^2) dx$ calculation

Well, if one puts $v=\frac{1}{u}$ then: $$I=\int_0^\infty\frac{v^2}{1+v^4} dv =\int_0^\infty\frac{1}{1+u^4} du$$ So summing up the two integrals from above gives: $$2I=\int_0^\infty\frac{1+u^2}{1+u^4} du$$

There's a neat trick to evaluate the integral $$S_n(t)=\int_0^\infty \sin(tx^n)dx.$$ First, take the Laplace transform: $$\begin{align} \mathcal{L}\{S_n(t)\}(s)&=\int_0^\infty e^{-st}S_n(t)dt\\ &=\int_0^\infty\int_0^\infty \sin(tx^n)e^{-st}dxdt\\ &=\int_0^\infty\int_0^\infty \sin(tx^n)e^{-st}dtdx\\ &=\int_0^\infty \frac{x^n}{x^{2n}+s^2}dx\tag1\\ &=s^{1/n-1}\int_0^\infty \frac{x^n}{x^{2n}+1}dx\\ &=\frac{s^{1/n-1}}{n}\int_0^\infty \frac{x^{1/n}}{x^2+1}dx\\ &=\frac{s^{1/n-1}}{n}\int_0^{\pi/2} \tan(x)^{1/n}dx\\ &=\frac{s^{1/n-1}}{n}\int_0^{\pi/2} \sin(x)^{1/n}\cos(x)^{-1/n}dx\\ &=\frac{s^{1/n-1}}{2n}\Gamma\left(\frac12(1+1/n)\right)\Gamma\left(\frac12(1-1/n)\right)\tag2\\ &=\frac{s^{1/n-1}\pi}{2n\cos\frac{\pi}{2n}}\tag3\\ &=\frac{\pi \sec\frac{\pi}{2n}}{2n\Gamma(1-1/n)}\mathcal{L}\{t^{-1/n}\}(s). \end{align}$$ Thus, taking the inverse Laplace transform on both sides, $$S_n(t)=\frac{\pi \sec\frac{\pi}{2n}}{2nt^{1/n}\Gamma(1-1/n)}.$$ Choose $n=2$ and $t=1$ to get your integral: $$S_2(1)=\frac12\sqrt{\frac\pi2}\ .$$

Explanation:

$(1)$: for real $q$ and $s$, $$\begin{align} \int_0^\infty \sin(qt)e^{-st}dt&= \text{Im}\int_0^\infty e^{iqt}e^{-st}dt\\ &=\text{Im}\int_0^\infty e^{-(s-iq)t}dt\\ &=\text{Im}\left[\frac{1}{s-iq}\right]\\ &=\frac{1}{s^2+q^2}\text{Im}\left[s+iq\right]\\ &=\frac{q}{s^2+q^2} \end{align}$$

$(2)$: See here.

$(3)$: See here.