Absolute and Conditional Convergence of the integral $\frac{\sin(x)}{x^p}$ for real values of $p$ I need to determine the values of p for which this integral converges conditionally and absolutely.
$$\int_{0}^{\infty} \dfrac{\sin(x)}{x^p} dx
$$
I think the interval for conditional convergence is $0 < x < 2$ and for absolute convergence the interval is probably $0 < x < 1$. I'm guessing that I need to somehow compare it with the $\dfrac{1}{x^p}$ integral, but I am not sure exactly where to begin and how to logically proceed. I first thought about dividing the integral into two separate improper integrals, with one of them integrating from $0$ to $1$ and the other from $1$ to infinity, but I do not know how to continue from that point on.

Hints: The following facts are useful for your problem. Close to zero the integrand behaves as

$$ \frac{\sin x}{x^p} \sim \frac{x}{x^p},  $$

and 

$$ \Bigg| \frac{\sin(x)}{x^p}\Bigg|\leq \frac{1}{x^p}. $$

