Improper integral for $x^{-a} \sin(x)$ for $1I was trying to show that the improper integral $\int\limits_0^{\infty} x^{-a} \sin(x) \, dx$ exists for $0<a<2$ (problem 6.3.2 #6 in Strichartz). I took care of the case where $0 < a \leq 1$ but now I'm running into problems for the case when $1<a<2$. It's easy to show that there aren't problems on the infinity side, so my idea was to show that it satisfied a Cauchy criterion as $x$ goes to 0:
Want to show: For every $\epsilon>0$, there exists $r$ such that $0<y \leq z< r$ implies that $\left|\int\limits_y^z x^{-a}\sin(x) \, dx\right| < \epsilon$.
Integration by parts gives $\left|\dfrac{z^{-a+1}}{-a+1}-\dfrac{y^{-a+1}}{-a+1}-\int\limits_{y}^z\dfrac{x^{-a+1}\cos(x)}{-a+1}\, dx\right|$. I ran into issues after this with getting this quantity bounded less than epsilon. Any suggestions; does this seem like a reasonable approach at least?

Hint:
$$
x^{-a}\sin x=x^{1-a}\,\frac{\sin x}{x},
$$
$0<1-a<1$ and $\sin x/x$ is bounded.
