I'm posting an answer just to inform that the question has received an answer by Nick Strehlke on [MSE][1].

\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 $2\le q\le p−1$ .

  [1]: http://math.stackexchange.com/questions/558014/can-we-simplify-int-0-infty-frac-sinpxxqdx/590849#590849