Solve integral with ln(x) for the integral criteria

Question

I want to check if $\sum\limits_{n=2}^{\infty}\frac{1}{n\ln^2(n)}$ is convergent by using the integral criteria.

Therefore I want to solve the integral $\int_{2}^{\infty}\frac{1}{x\ln^2(x)} dx$

First I tried to start only with the denominator and to use partial integration there, but I didn't succeeded, as the formula gets more complex with every duration and after using it for three times, I didn't had the feeling that it will end. (With every duration I got the term $x*ln(x)-x$ one time more multiplied inside the integral.)

Therefore I tried to split it first into $\int_{2}^{\infty}\frac{1}{x}\times \frac{1}{\ln^2(x)} dx$ and then to use partial integration on the whole term. But even with this I couldn't get to a result. I am also not sure about how to integrate $\ln^2(x)$. It doesn't seem to be correct to just integrate $\ln (x)$ and then take the square of the result. Also when I want to solve $\int \ln (x)\times\ln (x)dx$ first, by using partial integration, I encounter the same problem as before: I can't get to an end as the part under the integral grows with every iteration.

How can I solve the integral correctly to know if the sum converges? In this example I think that it should converge, but if I had a divergent series, how can I find out, if the integral exists or not? Just by trying to integrate it?

Answer 1

Note that $\frac1x=\ln'x$. Therefore$$\int\frac1{x\ln^2x}\,\mathrm dx=\int\frac{\ln'x}{\ln^2x}\,\mathrm dx.$$So, if you do $\ln x=t$ and $\frac1x\,\mathrm dx=\mathrm dt$, you get$$\int\frac1{t^2}\,\mathrm dt=-\frac1t=-\frac1{\ln x}.$$

Answer 2

Hint:

$$\int\frac{dx}{x\log^2x}=\int\frac{d(\log x)}{\log^2x}.$$

Answer 3

You can indeed use integration by parts. As $(\ln x)' = \frac{1}{x}$ and $(\frac{1}{(\ln x)^2})' = -2\frac{1}{x(\ln x)^3}$ we have for $R \gt 0$ $$ \int_2^R \frac{1}{x(\ln x)^2} \,\mathrm{d}x = \left[\frac{1}{\ln x}\right]_2^R + 2 \int_2^R \frac{1}{x(\ln x)^2} \,\mathrm{d}x, $$ hence $$ \int_2^R \frac{1}{x(\ln x)^2} \,\mathrm{d}x - 2 \int_2^R \frac{1}{x(\ln x)^2} \,\mathrm{d}x = \frac{1}{\ln R} - \frac{1}{\ln 2} $$ and therefore $$ \int_2^R \frac{1}{x(\ln x)^2} \,\mathrm{d}x = \frac{1}{\ln 2} - \frac{1}{\ln R} \to \frac{1}{\ln 2} \lt \infty $$ for $R \to \infty$.