Divergence of an integral

Question

I am sure this is a familiar example to many, as it comes up as an example of a function which belongs to $L^p$ for $p=1$ but not $L^p$ for $p < 1$, but I am having a hard time seeing why it is "trivially" divergent.

By direct calculation $$ \int_2^\infty \frac{dx}{x \ln^2 x} < \infty$$ Now consider $$ I(p) = \int_2^\infty \left( \frac{1}{x \ln^2 x} \right)^{p} \ dx$$ Now $I(p)$ should be divergent for $p < 1$, but I can't quite see how. For instance by writing $p = 1 - \varepsilon$, I am able to show that the integral diverges for $p < \frac{1}{2}$, but this isn't enough. The square in the logarithm is throwing me off, and I can't seem to find a smart lower bound for the function $\ln^2 x$.

Any help would be nice, I would prefer a hint to a direct answer though.

I suggest to use the integral test for the convergence of $\sum_{n=2}^{\infty} \frac{1}{n^p \ln^{2p} n}$. This is divergent by the Cauchy condensation test. — Crostul, Mar 14 '16 at 14:30
Does this answer your question? $\int_2^\infty \left(\frac1{x\log^2x}\right)^p\mathrm dx$ diverges for p>1 — hbghlyj, Jul 12 '23 at 10:50

score 2 · Accepted Answer · answered Mar 14 '16 at 14:25

We have $\ln(x)^p \le x^{\delta}$ for any $p, \delta > 0$ and $x$ sufficiently large. Then $\frac{1}{\ln(x)^p} \ge \frac{1}{x^\delta}$. If $p = 1-\epsilon$ for some $\epsilon \in (0,1)$ then we have $$I(1-\epsilon) = \int^\infty_2\frac{1}{x^{1-\epsilon} \ln(x)^{2-2\epsilon}} dx \ge \int^{x^*}_2 \frac{1}{x^{1-\epsilon} \ln(x)^{2-2\epsilon}} dx + \int^\infty_{x^*} \frac{1}{x^{1-\epsilon + \delta}} dx$$ for any $\delta > 0$ where $x^*$ is the point where $x^\delta$ overtakes $\ln(x)^{2-2\epsilon}$.

Nice and simple, thank you! – Kayle of the Creeks Mar 14 '16 at 14:28 — Kayle of the Creeks, Mar 14 '16 at 14:28

Divergence of an integral

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