Why does $\int_{x=2}^{\infty} \frac{1}{\log(x)^2 x}$ converge? My math script has the integral $\int_2^\infty \frac{1}{x \log(x)^2}dx$ in it and then simply states that it converges to (at?) $\frac{1}{\log(2)}$ without any explanation.
I have found this math SE comment  which explains that the almost identical integral $\int_2^\infty \frac{1}{x \log(x)}dx$ diverges because of the integral convergent test and cauchy condensation test. I tried applying the same tests which resulted in $\sum_2^\infty \frac{1}{\log(2^n)^2}= \frac{1}{2\log(2)}  \sum_2^\infty \frac{1}{n}$ which diverges. So now I am wondering how they concluded that the integral converges at all.
Any help would be appreciated
 A: You approach is correct the Cauchy condensation test is a way for ensure the convergence. However your conclusion when you use the test is not correct because you wrote "$\sum_{2}^{\infty} \frac{1}{\log(2^{n})^{2}}=\frac{1}{2\log(2)}\sum_{2}^{\infty}\frac{1}{n}$" think about that lines and find the mistake.
For see the convergence, first notice that the mapping
\begin{align*}
f: [2,+\infty[&\longrightarrow \Bbb{R},\\ x&\longmapsto \frac{1}{x\log^{2}x}
\end{align*}
satisfies:

*

*The mapping $f$ is continuous over $[2,+\infty[$.

*For all $x\in [2,+\infty[$, we have $f(x)\geqslant 0$, i.e., $f$ is non-negative.

*For all $x\leqslant y$, we have $f(x)\geqslant f(y)$, i.e., $f$ is monotone decreasing.

Hence, $$\int_{2}^{+\infty}f(x)\, {\rm d}x<+\infty \iff \sum_{n=2}^{+\infty}f(n)<+\infty \iff \sum_{n=1}^{+\infty}2^{n}f(2^{n})<+\infty.$$
Then,
$$\sum_{n=1}^{+\infty} 2^{n}f(2^{n})=\sum_{n=1}^{+\infty} 2^{n}\left(\frac{1}{2^{n}\log^{2}2^{n}}\right)=\sum_{n=1}^{+\infty}\frac{1}{\log^{2}2^{n}}.$$
Now, by limit comparison test $$\lim_{n\to +\infty} \frac{\frac{1}{\log^{2}2^{n}}}{\frac{1}{n^{2}}}=\frac{1}{\log^{2}2}>0$$
implies $\sum_{n=1}^{+\infty}\frac{1}{\log^{2}2^n}$ converges because $\sum_{n=1}^{+\infty}\frac{1}{n^{2}}$ converges. Therefore by Cauchy condensation test $\int_{2}^{+\infty}\frac{1}{x\log^{2}x}\, {\rm d}x$ converges because $\sum_{n=1}^{+\infty}\frac{1}{\log^{2}2^{n}}$ converges.
For see $\int_{2}^{+\infty}\frac{1}{x\log^{2}x}\, {\rm d}x=\frac{1}{\log 2}$ just use the hint given by Robert z so via substitution $x\mapsto \log x$ over $[2,+\infty[$ we have for $N>0$,
$$\int_{2}^{+\infty}\frac{1}{x\log^{2}x}\, {\rm d}x=\lim_{N\to +\infty} \int_{\log 2}^{\log N}\frac{1}{x^{2}}\, {\rm d}x=\frac{1}{\log 2}.$$
