I see this question and the answer by joriki.
However I cannot understand joriki's argument that $$\int_2^\infty \left(\frac1{x\log^2x}\right)^p\mathrm dx$$ diverges for p>1.
So I try to show that by myslef, by finding a function $f(x)<\left(\frac1{x\log^2x}\right)^p$ for sufficiently large $x$ and $\int f(x) dx=\infty$, but it is difficult.
Please help me to understand that argument or give me some hints.
For $p>1$, the integral converges, since $$\dfrac1{x\log^2(x)} < \dfrac1x\quad\text{for }x>e$$ and $$\int_2^{\infty} \dfrac{dx}{x^p} = \dfrac1{(p-1)2^{p-1}}$$
Meanwhile I was working the problem came user141421's answer which shows that, for $p \gt 1$, $$I(p)=\int_2^\infty \left(\frac1{x\log^2x}\right)^p\mathrm dx$$ converges. Just for your curiosity, I give you a few values of this integral $$I(1)=\frac{1}{\log (2)}\simeq 1.4427$$ $$I(2)=\frac{\log (2) \left(2 \log ^2(2) \text{Ei}(-\log (2))-1+\log (2)\right)+2}{12 \log ^3(2)} \simeq 0.384129$$ $$I(3)=\frac{2 \log ^2(2) \left(16 \log ^3(2) \text{Ei}(-\log (4))+1+\log (2) (\log (4)-1)\right)+6-\log (8)}{120 \log ^5(2)} \simeq 0.235996$$ For higher values of $p$, a CAS gives long formulas and I shall just report numerical values for the next integrals $$I(4) \simeq 0.174447$$ $$I(5) \simeq 0.140805$$ $$I(6) \simeq 0.119689$$ $$I(7) \simeq 0.105277$$ $$I(8) \simeq 0.0948764$$ $$I(9) \simeq 0.0870681$$
