real analysis - Generalized Fresnel integral $int_0^infty sin x^p , {rm d}x$

I am stuck at this question. Find a closed form (that may actually contain the Gamma function) of the integral

$$\int_0^\infty \sin (x^p)\, {\rm d}x$$

I am interested in a Laplace approach, double integral etc. For some weird reason I cannot get it to work.

I am confident that a closed form may actually exist since for the integral:

$$\int_0^\infty \cos x^a \, {\rm d}x = \frac{\pi \csc \frac{\pi}{2a}}{2a \Gamma(1-a)}$$

there exists a closed form with $\Gamma$ and can be actually be reduced further down till it no contains no $\Gamma$. But trying to apply the method of Laplace transform that I have seen for this one , I cannot get it to work for the above integral that I am interested in.

May I have a help?

Answer

If $p>1$,
$$ I(p)=\int_{0}^{+\infty}\sin(x^p)\,dx = \frac{1}{p}\int_{0}^{+\infty}x^{\frac{1}{p}-1}\sin(x)\,dx \tag{1}$$
but since $\mathcal{L}(\sin(x))=\frac{1}{s^2+1}$ and $\mathcal{L}^{-1}\left(x^{1/p-1}\right)=\frac{s^{-1/p}}{\Gamma\left(1-\frac{1}{p}\right)}$ we have:
$$ I(p)=\frac{1}{p\,\Gamma\left(1-\frac{1}{p}\right)}\int_{0}^{+\infty}\frac{s^{-1/p}}{1+s^2}\,ds = \color{red}{\frac{\pi}{2p\,\Gamma\left(1-\frac{1}{p}\right)}\,\sec\left(\frac{\pi}{2p}\right)}.\tag{2}$$