Example of an integrable function $p(x)$ such that $p(x)\log(p(x))$ is Not integrable

Question

For a mesurable function $f: \mathbb{R} \to \mathbb{R}$, we say that $f$ is integrable if $$ \int_{\mathbb{R}} |f(x)|\,dx < \infty . $$

Is there a integrable function $p:\mathbb{R} \to \mathbb{R}_{\geq 0}$ such that $ p(x)log(p(x))$ is NOT integrable?

If exists, I’d appreciate it if you could provide a concrete example.

I want to know whether entropy can be defined for any probability density function.

Adapt https://math.stackexchange.com/questions/279304/can-the-entropy-of-a-random-variable-with-countably-many-outcomes-be-infinite — Eric Towers, Dec 08 '24 at 03:21
https://math.stackexchange.com/questions/574503/infinite-series-sum-n-2-infty-frac1n-log-n — Eric Towers, Dec 08 '24 at 03:22
https://math.stackexchange.com/questions/878050/partition-with-infinite-entropy — Eric Towers, Dec 08 '24 at 03:24
This question is similar to: Can the entropy of a random variable with countably many outcomes be infinite?. If you believe it’s different, please [edit] the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. — Eric Towers, Dec 08 '24 at 03:33

score 2 · Accepted Answer · answered Dec 08 '24 at 04:18

2

To answer the question directly, it looks like $p(x) = \frac1{x\log(x)^2}$ (restricted over $[2, \infty)$ say) fits the bill. One obtains

$$p(x) \log p(x) = -\frac1{x\log(x)} - \frac{2\log \log x}{x\log(x)^2}$$

where the second term may be safely ignored but the first gives a divergent integral.

answered Dec 08 '24 at 04:18

user43208

+1. To make this a probability density on $[b, \infty)$ for $b>1$, it becomes $p(x)= \frac{\log_e(b)}{x\log_e(x)^2} = \frac{1}{x\log_b(x)^2\log_b(e)}$ so it has CDF $F(x)=1-\frac{\log_e(b)}{\log_e(x)} = 1-\frac{1}{\log_b(x)}$. The derivative of any CDF with a heavier tail will also work. – Henry Dec 08 '24 at 09:59

1 Answers1