Help analyzing the convergence of $\int_{2}^{+\infty} \frac{1}{\ln^p(x) x^s}\,dx$

Question

so far I've been able to establish what happens when,

$p=0$ and $s=0$, diverges

$p=0$ and $s>1$, diverges

$p=0$ and $s<1$, converges

$p=0$ and $s=1$, diverges

But when $p\neq 0$ and $s\in \mathbb{R}$ I don't really know how to integrate

$$\int_{2}^{+\infty} \frac{1}{\ln^p(x)x^s}\,dx$$

Thanks,

Answer 1

The conditions of convergence are Re(s)>1 or Re(s)=1 and Re(p)>1. these conditions are obtained from the Euler-Maclaurin formula and the conditions of convergence of the Bertrand's series.

elementary explaination: for p, s real

As the primitive of $\frac1{x \ln(x)} = \ln \ln x$, which diverge, the integral diverge when $s <1$ or $p\le 1$ and for s>1, evidently the integral converge whatever p because $\ln^p(x) = o(x^\epsilon)$ for all $\epsilon>0$.

And for $s=1$ and $p \neq 1$, the primitive of $\frac1{x \ln^p(x)}$ is $\frac{\ln(x)^{1-p}}{1-p}$ and the integral converge iff $p>1$.

Answer 2

Let $u = \ln x$, then $$ \int_a^\infty \frac{1}{ \ln ^ p (x) x^s } dx = \int_{\ln a}^\infty \frac{e^{(1 -s)u }}{ u^p } du $$ If $1 -s >0$, the exponential blows up, so the integral diverges. If $1- s < 0$, the integral will converges, why? If $1-s =0$, we have $$ \int_{\ln a} ^\infty \frac{1}{u^p} du $$ you should know this converges when $p >1$ and diverges otherwise.