How to integrate $\int_0^{\infty} \frac{\sin(x^{-p})}{x^2}dx$ where $p>1$?

Question

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.

Answer 1

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.