When is $\int_{0}^1|x^{a-b-1}\cos(x^{-b})|\ dx<\infty$? 
Let $a,b>0$. Consider the function
  $$
f(x)=x^{a-b-1}\cos(x^{-b}),\quad x\in(0,1]. 
$$
  When is $f$ absolutely integrable on $[0,1]$?

If $a>b>0$, it is not hard to show that $f$ is absolutely integrable on $[0,1]$. For the case $0<a\leq b$, is it true that
$$
\int_0^1|x^{a-b-1}\cos(x^{-b})|\ dx=\infty?
$$
By change of variables, one has
$$
\int_1^\infty y^{-a/b}|\cos y|\ dy\tag{1}
$$
One can show that $\int_1^\infty y^{-a/b}\ dy=\infty$ since $-a/b\geq -1$. How would you estimate (1)?

$\left|\cos y\right|$ is a continuous, bounded, $\pi$-periodic, non-negative function with mean value $\frac{2}{\pi}$. 
It follows that
$$ \int_{1}^{M} y^{-a/b}\left|\cos y\right|\,dy \approx \frac{2}{\pi}\int_{1}^{M} y^{-a/b}\,dy \tag{1} $$
and the RHS of $(1)$ converges as $M\to +\infty$ iff $\frac{a}{b}>1$, i.e.  $\color{red}{a>b}$.
In such a case, we may write 
$$\int_{1}^{+\infty}y^{-a/b}\left|\cos(y)\right|\,dy = \frac{2b}{\pi(a-b)}-\frac{4}{\pi}\sum_{k\geq 1}\frac{(-1)^k}{4k^2-1}\int_{1}^{+\infty} y^{-a/b}\cos(2ky)\,dy\tag{2} $$
by exploiting the Fourier cosine series of $\left|\cos y\right|$.
