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I'm quite lost on the following problem:

For $p \ge 0$, for what values of $p$ does the integral converge (the answer given is for any value of $p$).

$$\int_{1}^{\infty} \frac{\text{d}x}{\ln^p(x)} $$

I can't figure out how to work out the given answer. My thought is to split this up into two integrals to consider the bounds, but then I get lost as to what comparisons I might make.

If there are multiple ways to approach this, I would appreciate seeing them, as I'm struggling with these types of problems.

User 123732
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Kevin
  • 31

3 Answers3

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Since (using for example the L'Hôpital's rule) $$\lim_{t\to\infty}\frac t{\ln^pt}=+\infty$$ then for $t$ sufficiently large (say for $t\ge t_0$) we have $$\frac{1}{\ln^p t}\ge \frac1t,\quad \forall t\ge t_0$$ hence the given integral is divergent.

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Under the substitution $x=e^y$ the integral becomes $$ \int_0^\infty e^y\cdot y^{-p}\,dy, $$ which does not converge because the integrand diverges to infinity.

user134824
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First take $p>0$. For large $x$ we have $$0<\ln x<x^{1/p}$$ and so $$\frac{1}{(\ln x)^p}>\frac{1}{x}\ .$$ Therefore $$\int_e^\infty \frac{dx}{(\ln x)^p}$$ diverges for all $p>0$, and so the bit near $x=1$ doesn't matter. If $p=0$ then the integrand is constant and so the integral diverges.

David
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