Example of a random variable that is integrable, but p-th moments for any p > 1 do not exist

Question

I have seen examples of random variables that are integrable, but the second moment does not exist. Is there an example of a random variable that is integrable, but the p-th moment, for any p > 1, does not exist?

Edit: To be precise, what I require is a random variable (ideally symmetric) such that $\mathbb{E}|X| < +\infty$ and, for any $p \in (1,2]$ (or more generally, $p > 1$, not necessarily integer), we have $\mathbb{E}|X|^p = \infty$.

Answer 1

You could take the probability density function

$$\rho(x) = \frac{c}{(x^2+1) \log^2(x^2+2)}$$

with $c$ that makes the integral over $\mathbb{R}$ equal $1$. Let $\mu$ be the probability measure on $\mathbb{R}$ given by $\rho$. Now take on $(\mathbb{R}, \mathcal{B}, \mu)$ the random variable $X(x) = x$. Then

$$E(|X|) < \infty \\ E(|X|^p) = \infty \textrm{ for all } p> 1$$

Note: in general, the set of $p$'s such that $E(|X|^p)< \infty$ is convex (due to Hölder's inequality; so a segment containing $0$, infinite or not ), and every such segment appears in this way. In our example the set of $p$'s is $(-1,1]$.

Answer 2

The Pareto distribution with shape $\alpha =2$ (or any $1<\alpha\leq 2$) has a well-defined first moment, but all other moments diverge. By appropriate choices of $\alpha$ you can also find distributions with $\mathbb{E}[X^p]$ being finite only up to some $p=k$.