How to integrate $\int_0^{\infty} \frac{\sin(x^{-p})}{x^2}dx$ where $p>1$? How to integrate the following integral:
$$\int_0^{\infty} \dfrac{\sin(x^{-p})}{x^2} dx, p>1 ?$$
Thank you for any help.
Attempt: I have tried simple sub: $x^{-p} =u \implies du=dx (-p)x^{-p-1}.$ 
$$\int_0^{\infty} \dfrac{\sin(x^{-p})}{x^2} dx = \dfrac{-1}{p}\int_{\infty}^{0} \sin(u)x^{p-1}du .$$
However, I can not proceed further with this sub.

From the change of variable $u=x^{-p}$, the integral becomes
$$
-\frac{1}{p}\int_{0}^\infty \sin(u)u^{1/p-1}du=-\frac{1}{p}\mathcal{M}\{\sin(u)\}(1/p),
$$
where $\mathcal{M}$ denotes the Mellin transform.
Since $0<1/p<1$, one can easily infer from http://mathworld.wolfram.com/MellinTransform.html
that
$$
-\frac{1}{p}\Gamma\left(\frac{1}{p}\right)\sin\left(\frac{\pi}{2p}\right)=-\Gamma\left(\frac{1}{p}+1\right)\sin\left(\frac{\pi}{2p}\right)
$$
is the value of the aforementioned integral.
