I am interested in finding the values of $a, b$ such that the integral
$$ \int_0^{\infty}\frac{{\left|\log x\right|}^b}{x^a} dx $$ converges.
My idea was to separate this integral: $$ \int_0^{1}\frac{\left|\log x\right|^b}{x^a} dx + \int_1^{\infty}\frac{\left|\log x\right|^b}{x^a} dx. $$ From the first part, we can see that $a$ needs to be less than $1$. Indeed, any powers of $x$ will dominate the $\log$. So $a<1$ is a necessary condition. To see that the integral converges for every $a<1$, let $\epsilon>0$ be such that $a+\epsilon <1$. Then $x^{\epsilon}{\left|\log x\right|}^{\beta}\to 0$ when $x\to \infty$. We can then rewrite the first integral as $$ \int_0^{1}\frac{x^{\epsilon}\left|\log x\right|^b}{x^{a+\epsilon}} dx, $$ which converges.
The problem comes from the second integral. It seems to me that when $a<1$, the second integral will never converge, since $\int_1^{\infty}\frac{1}{x^a} dx$ and $\int_1^{\infty}\left|\log x\right|^b dx$ both never converges.
Any suggestions would be much appreciated!