Generalized sine integral $ \int_0^\infty \sin^m(x^n)/x^p\,\mathrm dx$ I have seen that both the integrals 

$$ 
\color{black}{ 
    \int_0^\infty \operatorname{sinc}(x^n)\,\mathrm{d}x 
                        =
    \frac{1}{n-1} \cos\left( \frac{\pi}{2n}\right)\Gamma \left(\frac{1}{n} \right)
}
$$

and 

$$ 
\color{black}{
    \int_0^\infty \operatorname{sinc}^n(x)\,\mathrm{d}x 
                           = 
    \frac{\pi}{2^n (n-1)!}  
    \sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k {n \choose k} (n-2k)^{n-1}
}
$$ 

beeing computed here. Is there a similar expression for the more general
$$ 
 \int_0^\infty \frac{\sin^m(x^n)}{x^p}\,\mathrm{d}x 
$$
sine integral? I do not know precielly the restriction on the constant. I assume that they have to be positive. From my research $m$ and $p$ does not neccecary have to be equal, for example
$$
   \int_0^\infty \frac{\sin^3(x)}{x^2}\,\mathrm{d}x = \frac{4}{3} \log 3
   \qquad \text{and} \qquad 
    \int_0^\infty \frac{\sin^3(x^2)}{x^2}\,\mathrm{d}x 
                                 = 
    \frac{1}{4} \sqrt{\frac{\pi}{2}}\left( \sqrt{3} - 3\right)
$$
but any closer restrictions on $n,m$ and $p$ I have not been able to gather. 
 A: Through the substitution $x=z^{1/m}$ the problem boils down to finding
$$ I(m,\alpha)=\int_{0}^{+\infty}\frac{\sin(x)^m}{x^\alpha}\,dx $$
and assuming $m\in\mathbb{N}$ we have that $\sin(x)^m$ can be expressed as a finite Fourier sine/cosine series.
So the problem boils down to finding, through the Laplace (inverse) transform,
$$ \int_{0}^{+\infty}\frac{\sin(x)}{x^\alpha}\,dx = \Gamma(1-\alpha)\cos\left(\frac{\pi\alpha}{2}\right),\qquad \int_{0}^{+\infty}\frac{\cos(x)}{x^\alpha}\,dx=\Gamma(1-\alpha)\sin\left(\frac{\pi\alpha}{2}\right)$$
then extend the range of validity of such identities through integration by parts.
A: One alternative approach (compared to the answer of Jack D'Aurizio) is to write $$\tag{1} I(m,\alpha) = \int_0^1 \zeta(\alpha,x) \, \sin(x)^m \, \mathrm{d} x,$$ where $\zeta(\alpha,x)$ denotes the Hurwitz zeta function. We can use the binomial theorem in the form $$\sin(x)^m = (2 i)^{-m} \sum_{k=0}^m (-1)^k \binom{m}{k} (-1)^k e^{ix(n-2k)}$$ in order to reduce the problem to determining the fourier representation of the Hurwitz zeta function.
The right-hand side of (1) can be extended analytically for any $\alpha \neq 1$. For $\mathrm{Re}(\alpha) <1$ one has (by using the well-known 'Hurwitz formula') the identity
$$ \zeta(\alpha,x)=  2 (2\pi )^{\alpha-1} \Gamma (1-\alpha) \left( \sin \big(\tfrac {\pi \alpha}{2}\big) \sum _{k=1}^{\infty} \frac{\cos(2 \pi x k)}{k^{1-\alpha}}+\cos \big(\tfrac{\pi \alpha}{2}\big) \sum_{k=1}^{\infty} \frac{\sin(2\pi x k)}{k^{1-\alpha}}  \right).$$
By analytic continuation this gives explicite formulas for (1) also for complex-values.
