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?

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.
