TITLE: Can we simplify $\int_{0}^{\infty}\frac{{\sin}^px}{x^q}dx$?
QUESTION [33 upvotes]: 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.

REPLY [10 votes]: 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