What values of $\alpha$ make this improper integral convergent? I'm having trouble discussing what values of $\alpha$ make
$$\int_{0}^{+\infty}\frac{1-\cos{x}}{x^{\alpha}}dx$$
convergent. The problem explicits that $\alpha \gt 1$. 
I've seen that the integral can be written like
$$\lim_{a \to 0^+}\int_{a}^{b}\frac{1-\cos{x}}{x^{\alpha}}dx \,\,\,+ \,\,\, \lim_{c \to +\infty}\int_{b}^{c}\frac{1-\cos{x}}{x^{\alpha}}dx,$$
where b $\gt 0$. From this point I don't have a clue on how to continue. Can someone give me a hand? Thank you so much.

At infinity, $1-\cos x$ is bounded and it will be convergent (even absolutely) for any $\alpha>1$. At zero you observe that
$$
1-\cos x\sim x^2/2
$$
And then the integrand is equivalent to
$x^{2-\alpha}$ and will be convergent iff $2-\alpha>-1$ or $\alpha<3$. Finally, it is convergent iff $\alpha\in (1,3)$.
