Finding the value of improper integral given some other integral's value Calculate the value of I. $$I=\int_0^{\infty} \frac{\sin(x)}{x^p}dx$$. Where 
$0<p<1,\int_0^{\infty}\frac{x^{p-1}}{1+x}dx=\frac{\pi}{\sin(p\pi)}$. 
$$\text{Attempt}$$ No concrete progress at all to be honest. Just tried doing $I(s)=\int e^{-sx^p}\frac{\sin(x)}{x^p}dx$ and then calculating $\frac{dI(s)}{ds}$ . But didn't help much . Another idea was calculating definite integral of $\frac{e^{ix}}{x^p}$ and then extracting the value of imaginary part of this integral.

Using Laplace transform, we get
$$I=\int_0^\infty\mathscr{L}_t(\sin x)\mathscr{L}^{-1}_t(x^{-p})dt\\
=\int_0^\infty\frac{t^{p-1}}{1+t^2}\frac1{\Gamma(p)}dt$$
We can easily find $$\int_0^\infty\frac{t^{p-1}}{1+t^2}dt$$ by using substitution $u=t^2$:$$\int_0^\infty\frac{t^{p-1}}{1+t^2}dt=\frac12\int_0^\infty\frac{u^{p/2-1}}{1+u}du=\frac{\pi}2\csc\frac{p\pi}2$$
Therefore, $$I=\frac{\pi}2\csc\frac{p\pi}2\frac1{\Gamma(p)}= \Gamma (1-p)\cos\frac{p\pi}{2}$$
When $0<p<2$.
