Convergence of $\int^{\infty}_{0} \frac{\sin^2x}{x^a}dx$ I'm trying to prove the convergence of 
$$\int^{\infty}_{0} \frac{\sin^2x}{x^a}dx$$
First we split the integral
$$\int^{1}_{0} \frac{\sin^2x}{x^a}dx+\int^{\infty}_{1} \frac{\sin^2x}{x^a}dx$$
Since
$$\frac{\sin^2x}{x^a}\leq\frac{1}{x^a}$$
by the p-test, the term $\int^{\infty}_{1} \frac{\sin^2x}{x^a}dx$ converges if and only if $a>1$.
And for the other term we use
$$\frac{\sin^2x}{x^a}\cong\frac{x^2}{x^a}=\frac{1}{x^{a-2}}$$
So that the integral converges if and only if $a-2<1 \to a<3$.
Summing up, $\int^{\infty}_{0} \frac{\sin^2x}{x^a}dx$ converges if $a \in (1,3)$, or else diverges.
Is my reasoning correct?
Thanks in advance for any suggestion!

If you attempt to use the comparison test with $\sin^2(x)/x^a$ and $x^2/x^a$, then for $a\in(-\infty,3]$ the integral does not converge in the latter function and therefore says nothing about the convergence of the actual integral.
Instead, you could use the same approximation as the first integral and get that it converges for $a\in[1,\infty)$
