Convergence of $\int_{0}^{1}\frac{\sin(x^a)}{x^b}$ for $a<0$ and $b\geq 1$

Question

I've been asked about the convergence of the integral $\int_{0}^{1}\frac{\sin(x^a)}{x^b} dx$ for every $a, b\in \mathbb{R}$ and $a<0$, $b\geq 1$ is the only case I couldn't figure out.

To be honest, I'm quite embarrassed I can't manage to handle such an basic-looking improper integal, but neither of the methods I've tried (Cauchy's critetion, Dirichlet's test, finding antiderivative) seems to work.

Note: Obviously the function is continuous in $(0, 1]$ so the only "problem" is that the function rapidly alters between very large (positive and negative) values.

Thank you for helping.

Answer 1

To start, try a specific case, like $a=-1.$ Then let $x = 1/u.$ We arrive at

$$\int_1^\infty\frac{\sin u}{u^{2-b}}\,du.$$

Dirichlet gives convergence if $2-b> 0.$ What happens if $2-b\le 0?$

Answer 2

Let $x = y^{1/a}, dx = \frac{1}{a}y^{1/a-1}dy$, so that we have: $$\frac{1}{a}\int_0^1 y^{(1-a)/b}\sin y \ dy$$ The power $(1-a)/b$ is larger than $0$, and therefore both factors are bounded, so that the integral is always convergent.