I want to find all pairs $(a,b)$ where $a,b\in \mathbb{R}$ such that $\int_{1}^{\infty}\frac{(\ln x)^{b}}{x^{a}}dx$ is finite.
I found some parts of the solution. For instance, when $b<-2$ and $a>1$ or when $b=1$ and $a>2$, this integral is finite. But I don't know how to analyze it in general. Is there any way to see all $(a,b) $ easier?
On
hint
Let $$f(x)=\frac{(\ln(x))^b}{x^a}$$ Assume $ b\ge 0$. $ f $ is continuous at $[1,+\infty)$, so it is locally integrable and positive at $ [1,+\infty)$.
If $ a>1$ then catch $\alpha \in (1,a)$.
$$\lim_{x\to+\infty}x^{\alpha}f(x)=0$$ $$\implies 0\le f(x) \le \frac{1}{x^\alpha} $$ for $ x$ great enough. thus the integral converges.
If $ a<1 $, take $ \beta\in (a,1)$
$$\lim_{x\to+\infty}x^\beta f(x)=+\infty$$ $$\implies \frac{1}{x^\beta}\le f(x)$$ for $ x $ great enough. then the integral diverges.
If $ a=1$, then for $ X>1$
$$\int_1^Xf(x)dx=\Bigl[\frac{(\ln(x))^{b+1}}{b+1}\Bigr]_1^X$$
the integral converges $ \iff b+1<0$
I let for you the cases $ a=1$ and $ b<0$.
The convergence of this integral is related to convergence of Bertrand's Series through application the integral test. But we need to be careful due to the possible singularity at $x=1^+$ arising from $\log^b(x)$ for $b<0$.
One way to see things a bit clearer is to enforce the substitution $x\mapsto e^x$. Proceeding, we obtain
$$\int_1^\infty \frac{\log^b(x)}{x^a}\,dx=\int_0^\infty x^be^{(1-a)x}\,dx\tag 1$$
And now the singularity at $x=1$ for $b<0$ of the integrand on the left-hand side of $(1)$ is more transparent by looking at the singularity at $x=0$ for $b<0$ of the integrand on the right-hand side of $(1)$.
From the right-hand side of $(1)$, it is straightforward to see that the integral diverges for $a\le 1$ and converges for $a>1$ and $b>-1$.