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.

Regarding your first integral, for $x$ near zero, we know $1-\cos x \approx x^2/2$  so your integrand is close to $x^2/x^\alpha$.  This puts a condition on $\alpha$ that I will let you figure out.
In your second integral, the first term of the integrand is $1/x^\alpha$.  I think you can see that this is divergent for the "wrong" $\alpha$.  There is no hope that the second term will help cancel out the divergence because $\cos x$ oscillates around zero and will not produce a value large enough to "help".  So $\alpha$ has to be such that the integral of $1/x^\alpha$ will not diverge.
