Improper integral problem.

Question

How to find divergence/convergence condition for $p$ on

$$\int\limits_{2}^{\infty} \frac{1}{{(\ln x)}^p} \, \mathrm d x$$

I tried comparison test , but failed.

Answer 1

Set $\ln(x)=t$. We then have $x=e^t$, i.e., $dx=e^tdt$. Hence, we have the integral to be $$\int_{\ln(2)}^{\infty} \dfrac{e^tdt}{t^p}$$ which clearly diverges for all $p$.

Answer 2

If $p\leqslant 0$, the integral is divergent because the integrand is greater than $1$ if $x\geqslant e$.

If $p\lt 0$, then $\int_t^{2t}\frac 1{(\log x)^p}\mathrm dx\geqslant \frac t{(\log(2t))^p} $; if $\int_2^{\infty}\frac 1{(\log x)^p}\mathrm dx$ was convergent, then we would have $\lim_{t\to\infty} \int_t^{2t}\frac 1{(\log x)^p}\mathrm dx=0$, which is not the case.