Let $f$ be a probability density. We know $f \in L^1$ since by definition, it integrates to 1.
But is $f \in L^p$ for any $p > 1$?
It seems logical to assume so? We know that densities approach zero for $|x| >> 0$. So if you square it, or take it to some other power, it will become smaller and smaller, hence making it more and more integrable?
So all densiites are in $L^p$ for all $p > 1$!?
Consider the pdf $$f(x)=\frac1{\sum_{j=2}^\infty \frac1{j\ln^2j}}\cdot\sum_{k=2}^\infty k\cdot 1_{\left[k,\,k+\frac1{k^2\ln^2 k}\right)}(x)$$
Then, $$\int_{-\infty}^\infty \lvert f(t)\rvert^p\,dt=\left(\sum_{j=2}^\infty\frac1{j\ln^2j}\right)^{-p}\sum_{k=2}^\infty\frac{1}{k^{2-p}\ln^2k}=\begin{cases}1&\text{if }p=1\\ \infty&\text{if }p>1\end{cases}$$