integration of $\int_{0}^{\infty}\frac{2\sin x-\sin2x}{x^n}dx$ How to find $$\int_{0}^{\infty}\cfrac{2\sin x-\sin2x}{x^n}dx$$? Any suggestion will be very helpful.  
 A: Given the integral: 
$$I := \int_0^{+\infty} \frac{2\,\sin x - \sin 2x}{x^n}\,\text{d}x = \int_0^{+\infty} \frac{2\,\sin x\,(1 - \cos x)}{x^n}\,\text{d}x$$
with $n \in \mathbb{Z}$, you get:
$$ I = \begin{cases} 2\left(1-2^{n-2}\right)\cos\left(\frac{\pi}{2}\,n\right)(-n)! & \text{if} \; \; n \le 0 \\ \left\{\frac{\pi}{2},\,\log 4,\,\frac{\pi}{2}\right\} & \text{if} \; \; n = \{1,\,2,\,3\} \\ +\infty & \text{if} \; \; n \ge 4 \end{cases}$$
A: For $n=3$ we have
$$\int\limits_{0}^{\infty}\cfrac{2\sin x-\sin2x}{x^n}dx=4\int\limits_{0}^{\infty}\left(\frac{\sin{x}}{x}\cdot\frac{\sin\frac{x}{2}}{x}\cdot\frac{\sin\frac{x}{2}}{x}\right)dx=4\cdot\frac{\pi}{2}\cdot1\cdot\frac{1}{2}\cdot\frac{1}{2}=\frac{\pi}{2}.$$
There is the following result.
If $a_k>0$ and $a_0>\sum\limits_{k=1}^na_k$ then
$$\int\limits_0^{+\infty}\prod_{k=0}^n\frac{\sin{a_kx}}{x}dx=\frac{\pi}{2}\prod_{k=0}^na_k$$
A: I suspect a problem with some $n$'s since, around $x=0$
$$\cfrac{2\sin x-\sin2x}{x^n}=\frac{x^3-\frac{x^5}{4}+O\left(x^7\right) }{x^n}\approx x^{3-n}$$ So, $n<4$ is a requirement.
Effectively, a CAS gives $$I_n= \int_0^{+\infty} \frac{2\,\sin x - \sin 2x}{x^n}\,\,dx =-\frac{1}{2} \left(2^n-4\right) \cos \left(\frac{\pi  n}{2}\right) \Gamma (1-n)$$ $n$ being or not an integer.
