Integral of 1/ln(x). Convergence test

[0kelvin]

Homework Statement

Integral of ln(x) from 1 to infinite diverges. But how do I know if the 1/ln(x) will diverge too?

Relevant Equations

1/ln(x)

Some functions have straight foward integrals, but they get complicated if you take the inverse of it. 1/f(x) for instance.

The primitive of 1/x is ln(x). In this case it's easy to check that the integral of 1/x or ln(x) from 1 to infinite diverges.

$\int_1^\infty (\ln(x))^n dx$

If n = 0, I have f(x) = 1. This cannot converge.

If n = 1, I have that the integral diverges.

If n < 0, then I have no idea except to let wolfram tell me.

If 1 < n < 0, the integral of ln(x) already diverges, taking the root of it just slows down a bit but still diverges.

 

[BvU]

For 1 < x you have ln(x) < x $\Rightarrow$ 1/ln(x) > 1/x.
If the integral of 1/x diverges, the integral of 1/ln(x) certainly diverges ...

 

[haruspex]

For 1 < x you have ln(x) < x $\Rightarrow$ 1/ln(x) > 1/x.
If the integral of 1/x diverges, the integral of 1/ln(x) certainly diverges ...

... and the next challenge is to show that asymptotically $\ln^n(x)<x$.

 

[archaic]

... and the next challenge is to show that asymptotically $\ln^n(x)<x$.

If I guessed correctly, would this be by finding this : $\frac{n!}{x^n}=0$ at $\infty$?