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?
 A: 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:
$$ I(p)=\frac{1}{p\,\Gamma\left(1-\frac{1}{p}\right)}\int_{0}^{+\infty}\frac{s^{-1/p}}{1+s^2}\,ds = \color{red}{\frac{\pi}{2p\,\Gamma\left(1-\frac{1}{p}\right)}\,\sec\left(\frac{\pi}{2p}\right)}.\tag{2}$$
A: The integral evaluates to
$$
\int_0^{\infty}\sin x^a\ dx=\Gamma\left(1+\frac{1}{a}\right)\sin\frac{\pi}{2a},
$$
but the way I know uses complex analysis.

Added by request:
We will integrate the function $\exp(-x^a)$ around the circular wedge of radius $R$ and opening angle $\pi/(2a)$, for $a>1$. By the Residue Theorem,
$$
0=\int_0^R\exp(-x^a)\ dx+\int_0^{\pi/(2a)}\exp(-R^ae^{i\theta})iRe^{i\theta}\ d\theta-e^{i\pi/(2a)}\int_0^R\exp(-ix^a)\ dx.
$$
The middle integral is $O(Re^{-R^a})$, so sending $R\to\infty$ yields
$$
e^{i\pi/(2a)}\int_0^{\infty}\exp(-ix^a)\ dx=\int_0^{\infty}\exp(-x^a)\ dx.
$$
To compute the latter integral, let $t=x^a$ so that $dt/t=a\ dx/x$. This yields
$$
\int_0^{\infty}\exp(-x^a)\ dx=\frac{\Gamma(1/a)}{a}.
$$
Putting everything together,
$$
\int_0^{\infty}\exp(-ix^a)\ dx=e^{-i\pi/(2a)}\Gamma\left(1+1/a\right).
$$
Taking imaginary parts yields the result.
A: Hint: Use $~\displaystyle\int_0^\infty\exp\Big(-\sqrt[n]x\Big)~dx~=~n!~\iff~\int_0^\infty e^{-x^n}~dx~=~\Gamma\bigg(1+\frac1n\bigg)~$ in conjunction with Euler's formula.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

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}
