Can we simplify $\int_{0}^{\infty}\frac{{\sin}^px}{x^q}dx$? We know the followings : 
$$\int_{0}^{\infty}\frac{{\sin}x}{x}dx=\int_{0}^{\infty}\frac{{\sin}^2x}{x^2}dx=\frac{\pi}{2},\int_{0}^{\infty}\frac{{\sin}^3x}{x^3}dx=\frac{3\pi}{8}.$$
Also, we can get
$$\int_{0}^{\infty}\frac{{\sin}^3x}{x^2}dx=\frac{3\log 3}{4},\int_{0}^{\infty}\frac{{\sin}^4x}{x^3}dx=\log 2.$$
Then, I got interested in their generalization. 

Question : Letting $p,q\in\mathbb N$, can we simplify the following?
  $$\int_{0}^{\infty}\frac{{\sin}^px}{x^q}dx$$

I don't have any good idea. Could you show me how to simplify this?
Remark : This question has been asked previously on math.SE without receiving any answers.
 A: Gradshteyn and Ryzhik gives the recursion relation:

(see also volume 1 of Integrals and Series by Prudnikov, Brychkov, and Marichev, page 387)
so if $m$ (your $q$) and $p$ are both even or both odd, with $p\ge m$ to assure convergence of the integral, we arrive at integrals of the form
$$\int_0^\infty \frac{\sin^{2n+1}x}{x}\,dx = \frac{(2n-1)!!}{(2n)!!}\frac{\pi}{2}$$
$$\int_0^\infty \frac{\sin^{2n}x}{x^2}\,dx = \frac{(2n-3)!!}{(2n-2)!!}\frac{\pi}{2}$$
This recursion produces the following closed form expression for $p$ and $m$ of the same parity (see Prudknikov, page 388):
$$
\int_{0}^{\infty}\frac{\sin^{p}{x}}{x^{m}}dx=\frac{\pi}{2^{p}(m-1)!{\cal S}[(m-1)\pi/2]}$$
$$\qquad\qquad\times\sum_{j=0}^{[(p-1)/2]}(-1)^{[(p-1)/2]+j}\frac{p!(p-2j)^{(m-1)}}{j!(p-j)!}
$$
with ${\cal S}(x)=\cos x$ for $p$ odd and ${\cal S}(x)=\sin x$ for $p$ even.
what remains is the case that $m$ and $p$ have different parity; the integral contains logarithms, but I haven't found a general formula (at least neither in Gradshteyn nor in Prudnikov). 
A: I'm posting an answer just to inform that the question has received an answer by Nick Strehlke on MSE.
\begin{align*}
 \int_0^\infty {\sin^p x\over x^q}\,dx & = \left\{\begin{array}{ll}
 \displaystyle{(-1)^{(p+q)/2}\pi\over 2^{p+1}(q-1)!}\sum_{k = 0}^p(-1)^k{p\choose k} |p - 2k|^{q-1} & \text{$p,q$ even,} \\[2em]
 \displaystyle {(-1)^{(p+q)/2-1}\pi\over 2^{p+1}(q-1)!}\sum_{k = 0}^p (-1)^k{p\choose k} \operatorname{sign}(p-2k) |p-2k|^{q-1} & \text{$p,q$ odd,} \\[2em]
 \displaystyle {(-1)^{(p+q+1)/2} \over 2^p (q-1)!} \sum_{k = 0\atop k\not = p/2}^p (-1)^k {p\choose k} |p-2k|^{q-1}\log{|p - 2k|} & \text{$p$ even, $q$ odd,} \\[2em]
 \displaystyle {(-1)^{(p+q-1)/2} \over 2^p (q-1)!} \sum_{k = 0\atop k\not = (p\pm1)/2}^p (-1)^k {p\choose k} \operatorname{sign}(p-2k) |p-2k|^{q-1}\log{|p - 2k|} & \text{$p$ odd, $q$ even,}
 \end{array}\right.
\end{align*}
By the way, I noticed that these formulas can be simplified a bit as the followings which might be easier to calculate : 
$$\frac{(−1)^{(p+q+1)/2}}{2^{ p−q} (q−1)! } \sum_{k=0}^{ (p−4)/2} (−1)^k\binom pk\left(\frac p2−k\right)^{ q−1} \log\left(\frac p2−k\right) $$
 for $p$  even and $q$  odd such that $3\le q\le p−1$ .
$$\frac{(−1)^{(p+q-1)/2}}{2^{ p−1} (q−1)! } \sum_{k=0}^{ (p−3)/2} (−1)^k\binom pk\left(p−2k\right)^{ q−1} \log\left(p−2k\right) $$
 for $p$  odd and $q$  even such that $2\le q\le p−1$ .
A: Suppose that $f$ is a smooth function which satisfies in the following condition
$$f(\pi+x)=f(x)\space    \space  and   \space   f(\pi-x)=f(x)$$ then, if the following integral exists 
$$\int_0^\infty\frac{\sin^2x}{x^2}f(x)$$
We have the following nice equalities
$$\int_0^\infty\frac{\sin^2x}{x^2}f(x)=\int_0^{\infty}\frac{\sin x}{x}f(x)=\int_0^{\pi/2}f(x)dx$$
I proved the following formula as a part of my Bachelor project which extend the Lobachevsky's work that if we have
 $f(\pi+x)=f(x)\space    \space  and   \space   f(\pi-x)=f(x)$ then 
$$\int_0^\infty\frac{\sin^4x}{x^4}f(x)=\int_0^{\pi/2}f(x)dx+\frac{2}{3}\int_0^{\pi/2}\sin^2 x\,f(x)dx$$
and I gave a method for finding an explicit formula for higher degree
$$\int_0^\infty\frac{\sin^{2n}x}{x^{2n}}f(x)$$
see http://arxiv.org/pdf/1004.2653.pdf
