How can I study the convergence of the improper integral $\int_{2}^{+\infty} \frac{dx}{x\ln^2x}$?

Question

I've tried applying Comparison and Limit Criteria (reducing it to some form of the $p$-series) to no avail.

Rather than finding out this particular solution, I'd like to know how to even approach these kinds of exercises.

Answer 1

$u=ln(x)$, $xdu=dx$, $\int_{ln(2)}^{+\infty}{1\over {u^2e^u}}e^udu$ converges

Answer 2

Notice that every integral of the form $$ \int_{2}^\infty \dfrac{dx}{x \ln^p(x)} $$ can be explicitly computed via substitution $u = \ln(x)$.

Answer 3

Note that for $x\ge2$, $\frac{1}{x\ln^2(x)}>0$. One way you could try it is using the fact that: $$\int_2^\infty\frac{1}{x^a}dx=\frac{2^{1-a}}{a-1},\,\,\,a>1$$ so if you can show that: $$x\ln^2(x)>x$$ for $x\ge2$ then you can use this integral as a bound

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As others have mentioned, you can also show that: $$\int_2^\infty\frac{dx}{x\ln^2(x)}dx=\int_{\ln(2)}^\infty\frac{1}{u^2}du$$ which clearly has a finite value and can be calculated using the formula I gave above

Answer 4

Note that, if $p > 0$ then $\left(\dfrac1{\log^p(x)}\right)' = -p \dfrac{dx}{x\log^{p + 1}(x)} $ so that $\int \dfrac{dx}{x\log^{p + 1}(x)} =-\dfrac1{p\log^p(x)} $.

Therefore $\int_a^{\infty} \dfrac{dx}{x\log^{p + 1}(x)} =-\dfrac1{p\log^p(x)}|_a^{\infty} =\dfrac1{p\log^p(a)} $.

Your case is $p=1$.

Also note that $\left(\dfrac1{\log(\log(x))}\right)' = -\dfrac1{x \log(x) \log^2(\log(x))} $ so $\int \dfrac1{x \log(x) \log^2(\log(x))} =-\dfrac1{\log(\log(x))} $.

See if you can establish what $\left(\dfrac1{\log(\log(\log ... \log(x)))}\right)' $ is where there are $n$ nested $\log$s.