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?

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$\ds{\bbox[5px,#ffd]{}}$

With
$\ds{\pars{~x \equiv t^{1/\pars{2p}} \implies t = x^{2p}~}}$:
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}\sin\pars{x^{p}}\,\dd x} =
{1 \over 2p}\int_{0}^{\infty}
t^{\bracks{\color{red}{1/\pars{2p} + 1/2}} - 1}
\,\,{\sin\pars{\root{t}} \over \root{t}}\,\dd y
\end{align}
Note that
$\ds{{\sin\pars{\root{t}} \over \root{t}} =
\sum_{k = 0}^{\infty}
{\pars{-1}^{k} \over \pars{2k + 1}!}\,t^{k} =
\sum_{k = 0}^{\infty}
\color{red}{\Gamma\pars{1 + k} \over
\Gamma\pars{2 + 2k}}\,{\pars{-t}^{k} \over k!}}$.

With Ramanujan-MT:
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}\sin\pars{x^{p}}\,\dd x}
\\[5mm] = &\
{1 \over 2p}\,
\Gamma\pars{{1 \over 2p} + {1 \over 2}}\,
{\Gamma\pars{1 - \bracks{1/\pars{2p} + 1/2}} \over
\Gamma\pars{2 - 2\bracks{1/\pars{2p} + 1/2}}}
\\[5mm] = &\
{1 \over 2p}\,{\Gamma\pars{1/\bracks{2p} + 1/2}
\Gamma\pars{1/2 - 1/\bracks{2p}} \over
\Gamma\pars{1 - 1/p}}
\\[5mm] = &\
{1 \over 2p}\,{\pi/\sin\pars{\pi\braces{1/\bracks{2p} + 1/2}}
\over \Gamma\pars{1 - 1/p}} =
\bbx{\pi\sec\pars{\pi/\bracks{2p}} \over
2p\,\Gamma\pars{1 - 1/p}} \\ &\
\end{align}
