This came up in a previous question, but was closed because the question wasn't terribly clear. I don't want to edit the other question substantially because it's not mine so I'm asking a new one and including what I did.
My book asks me to find, for each $p \in (0, \infty)$, a continuous function $f : (0, \infty) \to (0, \infty)$ such that $f^q$ is integrable if and only if $q = p$. The hint is to look at functions of the form $x^{-a} |\log x|^b$. $p$ isn't included in the problem statement directly but my guess matches the answer to the other question that a function of the form $h(x) = (x^{-a} |\log x|^b)^{1/p}$ is the way to go.
If $q = p$, then $h(x) = x^{-a} |\log x|^b$. I broke this up into two integrals due to the properties of the log function at $x = 1$: $$ \int_0^{\infty}\dfrac{|\log^b(x)|}{x^a}dx=\int_0^{1}\dfrac{(-\log(x))^b}{x^a}dx+\int_1^{\infty}\dfrac{\log^b(x)}{x^a}dx $$
My efforts to use the comparison test were stymied because I know that the integral $\displaystyle\int_1^\infty \frac{1}{x^a} dx$ converges if $a > 1$, but the integral $\displaystyle\int_0^1 \frac{1}{x^a} dx$ diverges if $a > 1$, so I don't see how $a = 1$ allows the combined integral to converge.
Did I miss something obvious, or is there a simpler way to apply the comparison test?