Assume we have a random variable X with some arbitrary distribution F(x). How can we calculate $\mathbf{E}[X^{-n}]$ for $n\geq 0$?
For example, let $X \sim Rayleight(\sigma)$. If we use the definition of $\mu_k = \sigma^k 2^\frac{k}{2}\,\Gamma\left(1 + \frac{k}{2}\right)$ with $k=-n$, that would give $\sigma^{-n} 2^\frac{-n}{2}\,\Gamma\left(1 - \frac{n}{2}\right)$ which is not defined for many values of $n$, e.g., $n=2,4,6,8, ...$. Furthermore, it is negative some range of values of $n$ which makes no sense since $X$ is a positive random variable and so does $X^{-n}$!
What am I missing here?
The requested negative moments are not defined, for the simple reason that if $$f_X(x) = \frac{x}{\sigma^2} e^{-x^2/(2\sigma^2)}, \quad x > 0,$$ then the integral $$\operatorname{E}[X^{-2k}] = \int_{x=0}^\infty \frac{1}{\sigma^2 x^{2k-1}} e^{-x^2/(2\sigma^2)} \, dx$$ is not convergent for $k \in \mathbb Z^+$: the integrand is unbounded at $x = 0$ and behaves as $x^{-2k+1}$, and the integral is obviously divergent. In fact, the odd negative moments are also unbounded for $n \ge 3$ (the first negative moment does exist).