If you want to derive the result from scratch using contour integration (and justify the substitution that Lucian made), consider $f(z) = e^{iz^{p}}$ and integrate around a sector/wedge of radius $R$ that makes an angle of $\frac{\pi}{2p}$ with the positive real axis.
I'm just going to consider the case where $p$ is an integer greater than $1$.
Then
$$\int_{0}^{\infty} f(x) \ dx + \lim_{R \to \infty} \int_{0}^{\pi/(2p)} f(Re^{it}) iRe^{it}\ dt + \int^{\infty}_{0} f(te^{i \pi / (2p)}) e^{i \pi/(2p)} \ dt =0 $$
But
$$\Big| \int_{0}^{\pi/(2p)} f(Re^{it}) iRe^{it}\ dt \Big| =\int_{0}^{\pi/(2p)} |e^{iR^{p}e^{ipt}} | R \ dt = \int_{0}^{\pi/(2p)} e^{-R^{p} \sin pt} R \ dt$$
$$ \le \int_{0}^{\pi /(2p)} e^{-R^{p} \frac{2}{\pi} pt} R \ dt \ \ (\text{Jordan's inequality})$$
$$ = R \frac{\pi}{2} \frac{1}{pR^{p}} \Big(1-e^{-R^{p}} \Big) \to 0 \ \text{as} \ R \to \infty \ \text{since p >1}$$
And
$$\int_{\infty}^{0} f(te^{i \pi / (2p)}) e^{i \pi/(2p)} \ dt = - \int_{0}^{\infty} e^{it^{p}e^{i \pi/2}}e^{i \pi/(2p)} \ dt = -e^{i \pi/(2p)}\int_{0}^{\infty} e^{-t^{p}} \ dt $$
So we have
$$ \int_{0}^{\infty} e^{ix^{p}} \ dx = e^{i \pi /(2p)} \int_{0}^{\infty} e^{-t^{p}} \ dt$$
Or
$$ \int_{0}^{\infty} \Big( \cos(x^{p}) + i \sin(x^{p}) \Big) \ dx = \left[ \cos \left(\frac{\pi}{2p} \right) + i\sin \left( \frac{\pi}{2p} \right) \right]\frac{1}{p}\Gamma\left( \frac{1}{p} \right)$$
If you want to extend the result to all real values of $p$ greater than $1$, indent the contour around the origin and show that it's contribution is vanishing small as the radius of the indentation goes to $0$. The computation is very similar to the one showing that the integral around the big wedge vanishes.
