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 $\left|\cos y\right|$.
By change of variables $y =x^{-b}$ we have, $$ \int_{0}^{1} |x^{a-b-1 \cos(x^{-b})}|dx = -\frac{1}{b}\int_{1}^{\infty}\frac{|\cos y|}{y^{\frac{-a}{b}}} dy=-\frac{1}{b}\int_{1}^{\infty}\frac{|\cos y|}{y^{\alpha}} dy$$
With $\alpha=\frac{-a}{b}$. by continuity we have the convergence of $$\int_{1}^{\frac{\pi}{2}}\frac{|\cos y|}{y^{\alpha}} dy<\infty.~~\forall~~\alpha$$
Now, suppose $\alpha>0$ Then, we have, $$\int_{\frac{\pi}{2}}^{\infty} \frac{|\cos y|}{y^{\alpha}}\,dy =\lim_{n\to \infty} \sum_{k=0}^{n}\int_{\frac{\pi}{2}+k\pi }^{\frac{\pi}{2}+(k+1)\pi}\frac{|\cos y|}{y^{\alpha}}\,dy=\sum^{\infty}_{k=0}a_k$$ Where $$ a_k = \int_{\frac{\pi}{2}+k\pi }^{\frac{\pi}{2}+(k+1)\pi}\frac{|\cos y|}{y^{\alpha}}\,dy =\int_{0 }^{\pi}\frac{|\cos (\frac{\pi}{2}+k\pi+t)|}{(\frac{\pi}{2}+k\pi+t)^{\alpha}}\,dt =\int_{0 }^{\pi}\frac{\sin t}{(\frac{\pi}{2}+k\pi+t)^{\alpha}}\,dt \tag{I}\label{I}$$ $ |\cos (\frac{\pi}{2}+k\pi+t)| = |\sin (k\pi +t)| =\sin t$ for $0\le t\le \pi$
Moreover, we also have,that $$0\le t \le \pi\implies \frac{1}{(\frac{\pi}{2}+(k+1)\pi)^{\alpha}}\le \frac{1}{(\frac{\pi}{2}+k\pi +t)^{\alpha}}\le \frac{1}{(\frac{\pi}{2}+k\pi)^{\alpha}} $$
This implies that $$\frac{2}{(\frac{\pi}{2}+k\pi)^{\alpha}}\,dt\le a_k =\int_{0 }^{\pi}\frac{\sin t}{(\frac{\pi}{2}+k\pi+t)^{\alpha}}\,dt\le \frac{2}{(\frac{\pi}{2}+(k+1)\pi)^{\alpha}}$$
since $\int_0^{\pi}\sin t dt =2$
If $\alpha \lt 0$ then, we have the reverse inequality i.e $$\frac{2}{(\frac{\pi}{2}+k\pi)^{\alpha}}\,dt\ge a_k =\int_{0 }^{\pi}\frac{\sin t}{(\frac{\pi}{2}+k\pi+t)^{\alpha}}\,dt\ge \frac{2}{(\frac{\pi}{2}+(k+1)\pi)^{\alpha}}$$
In any case we obtain that, $$a_k \sim \frac{1}{k^\alpha}$$ that is $$ \frac{C_1}{k^\alpha}\le a_k \le \frac{C_2}{k^\alpha}$$ By the wel konw Rieamnn series, we know that the series $\sum a_k$ converges iff $\alpha>1.$ hence from \eqref{I} we deduce that, $$\int_{\frac{\pi}{2}}^{\infty} \frac{|\cos y|}{y^{\alpha}}\,dy =\sum^{\infty}_{k=0}a_k <\infty \Longleftrightarrow \color{red}{\alpha = \frac{-b}{a}>1} $$
