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!

Assuming that $a>0$, your considerations for the integral on $[0,1]$ are fine: as $\sin^2(x)/x^a$ is non negative you can use the asymptotic comparison test and the integral is convergent on $(0,1]$ iff $a<3$. 
On the other hand, I think that the inequality
$$\frac{\sin^2x}{x^a}\leq\frac{1}{x^a}$$
tells you only that the integral on $[1,+\infty)$ is convergent for $a>1$ but does not help to say that it is not convergent for $0<a\leq 1$ ($1/x^2\leq 1/x$ for $x\geq 1$ does not imply that $\int_1^{\infty}1/x^2\ dx=+\infty$!).
Alternatively for that part use $\sin(x)^2=(1-\cos(2x))/2$. Then your integral becomes 
$$\int^{\infty}_{1} \frac{\sin^2x}{x^a}dx=\frac{1}{2}\int^{\infty}_{1} \frac{1}{x^a}dx-\frac{1}{2}\int^{\infty}_{1} \frac{\cos(2x)}{x^a}dx.$$
On the R.H.S., the first integral is convergent iff $a>1$ and the second integral is convergent iff $a>0$ because
$$\int_{1}^{+\infty} \frac{\cos (2x)}{x^a} dx=\left[\frac{\sin (2x)}{ 2x^a}\right]_1^{+\infty}+ \frac{a}{2}\int_1^{+\infty} \frac{\sin( 2x)}{x^{a+1}} dx=\frac{\sin (2)}{ 2}+ \frac{a}{2}\int_1^{+\infty} \frac{\sin( 2x)}{x^{a+1}} dx$$
and
$$\int_1^{+\infty} \frac{|\sin( 2x)|}{x^{a+1}} dx\leq \int_1^{+\infty} \frac{1}{x^{a+1}} dx<\infty.$$
Hence your integral is convergent on $[1,+\infty)$ iff $a>1$.
