Recently I tried to give an argument for why, if $Z \sim N(0,1)$ denotes the standard normal, $\mathbb{E}|\frac{1}{Z}| = \infty$. However, in hindsight this argument does not seem to rely on many special properties of $Z$. As a test of my understanding, I have tried to make the following generalization:
Claim (Sufficient Condition for $\mathbb{E}|\frac{1}{X}| = \infty$): For any random variable $X$ with a density $p(x)$ strictly positive and continuous at every point of $[0,1]$, it is the case that $\mathbb{E}|\frac{1}{X}|=\infty$.
Question: Is this claim true?
Purported proof: By the law of the unconscious statistician, $$\mathbb{E}\left[\left|\frac{1}{X}\right|\right] = \int_{\mathbb{R}} \frac{1}{|x|}p(x) dx = \int_{-\infty}^0 \frac{1}{|x|}p(x) dx + \int_0^1 \frac{1}{x}p(x) dx + \int_1^{\infty} \frac{1}{x}p(x) dx $$
Because $p(x)$ is strictly positive and continuous at every point of $[0,1]$, by the extreme value theorem there exists an $x_0 \in [0,1]$ such that $p(x_0)>0$ and for all $x \in [0,1]$, $$p(x) \ge p(x_0)\,,$$ i.e. $\min\limits_{x \in [0,1]}p(x) =: p(x_0)$ can be well-defined.
Therefore, by monotonicity and linearity of integrals we have:
$$\mathbb{E}\left[\left|\frac{1}{X}\right|\right] \ge \int_0^1 \frac{1}{x} \left(\min\limits_{x \in [0,1]}p(x) \right) dx + \dots = \left( \min\limits_{x \in [0,1]} p(x) \right) \int_0^1 \frac{1}{x}dx + \dots = \infty $$ since the integral $\int_0^1 \frac{1}{x}dx$ is well known to diverge. "$\square$"
Can this actually be correct? It seems odd to me that the behavior of the random variable on the interval $[0,1]$ alone can determine whether its reciprocal has finite expectation or not, especially for distributions which are not symmetric about $0$. It also seems too generally/widely applicable to too many important examples of (continuous) distributions to go completely unmentioned, for example, in the Wikipedia page on inverse distributions.
