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.
 A: Divide it to two improper integrals with one singularity each. 
$\int_{0}^{\infty}\displaystyle\frac{1-cos{x}}{x^\alpha} \,dx=\int_{0}^{1}\displaystyle\frac{1-cos{x}}{x^\alpha} \,dx+\int_{1}^{\infty}\displaystyle\frac{1-cos{x}}{x^\alpha} \,dx$
The original integral will converge $\iff$ Both those integrals will converge
$\int_{1}^{\infty}\displaystyle\frac{1-cos{x}}{x^\alpha} \,dx<\int_{1}^{\infty}\displaystyle\frac{2}{x^\alpha} \,dx$
This integral converges for all $\alpha>1$.
Now to the more interesting integral.
Using Taylor's series for $1-cos{x}$ at $x=0$ we get that:
$1-cos{x}$=$\displaystyle\frac{x^2}{2}+R_{2}(x)$
So:  $\displaystyle\frac{1-cos{x}}{x^\alpha}=\displaystyle\frac{\frac{x^2}{2}+R_{2}(x)}{x^\alpha}=\displaystyle\frac{\frac{1}{2}+\frac{R_{2}(x)}{x^2}}{x^{\alpha-2}}$
$\int_{0}^{1}\displaystyle\frac{\frac{1}{2}+\frac{R_{2}(x)}{x^2}}{x^{\alpha-2}} \,dx$ converges for $\alpha<3$.
So the final answer is $1<\alpha<3$
A: 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.
A: 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 $\alpha\in (1,3)$.
A: This is not a full answer, but maybe you can utilise the following result
$$
\cos(2x) = 1 - 2\sin^2x\implies 1 - \cos(x) = 2\sin^2 \left(\frac{x}{2}\right)
$$
we then have an integral
$$
\int_0^\infty \frac{1-\cos x}{x^\alpha}dx = \frac{2}{2^\alpha}\int_0^\infty \frac{\sin^2 \left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)^\alpha}dx 
$$
then we have
$$
\frac{1}{2^{\alpha-2}}\int_0^\infty \frac{\sin^2 \left(u\right)}{u^\alpha}du = \frac{1}{2^{\alpha-2}}\int_0^\infty \frac{\sin^2 \left(u\right)}{u^2}\frac{1}{u^{\alpha-2}}du
$$
We also know
$$
\frac{\sin^2 \left(u\right)}{u^2}\frac{1}{u^{\alpha-2}} \leq \frac{1}{u^{\alpha-2}}
$$
