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.

(1) For $\,p=0\,$ the integral diverges.
(2) Lets prove that the integral converges for $\,p\in(0,2)$.
$$\left|\frac{\sin{x}}{x^p}\right|\leqslant \frac{1}{x^{p-1}}\quad ,\,x>0 \quad \Rightarrow\quad \int_{0}^{\lambda}{\left|\frac{\sin{x}}{x^p}\right|dx}\leqslant \int_{0}^{\lambda}{\frac{dx}{x^{p-1}}}$$
which converges for $\enspace p-1< 1\enspace \Leftrightarrow \enspace p<2$, which is true. Dirichlet's test implies that $\int_{\lambda}^{\infty}{\frac{\sin{x}}{x^p}dx}$ converges for  $\,p>0$.
(3) Lets prove that the integral diverges for $\,p\geqslant2$.
$$\lim_{x \rightarrow 0}\frac{x^{p-1}}{\sin{x}}=0\,(or\,1,\,if\,p=2)\quad\Rightarrow\quad \frac{\sin{x}}{x^p}\geqslant \frac{17}{x}\quad ,x\in(0,\varepsilon)$$
$$\Rightarrow\quad\int_{0}^{\varepsilon}{\frac{\sin{x}}{x^p}dx}\geqslant 17\int_{0}^{\varepsilon}{\frac{dx}{x}}$$
which diverges.
Therefore, the whole integral diverges for $\,p\geqslant2\enspace$because $\int_{0}^{\infty}=\int_{0}^{\varepsilon}+\int_{\varepsilon}^{\infty}$.

