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

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.
