I'm trying to answer a question where the sample space is $\mathbb{R}$, $X(x) = \min\{2, x\}$, a random variable, $f_X(x) = 2(1 + x)^{-3}$ if $x \geq 0$ and $0$ everywhere else. Find $\mathbb{E}(\frac{1}{X})$.
$X(0) = \min\{2, 0\} = 0$, so $\frac{1}{x}f_X(x)$ does not look integrable over an interval containing $0$.
But $f_X(0) = 0$, so maybe use L'Hôpital's rule. The trouble is that the limits of $f_X$ tending to zero from below, $0$, and above, $2$ are unequal.
As the integral doesn't converge, is $\mathbb{E}(\frac{1}{X}) = \infty$?