Let $X\sim N(0,\sigma^2)$ normally distributed rv. 1) What is $E\left[\frac{1}{X^2}\right]$? 2) What about $E\left [\frac{1}{X^4} \right]$?
Entering the first in Wolfram Alpha or Mathematica yields a seemingly nonsensical answer, $E\left[\frac{1}{X^2}\right] = - \frac{1}{\sigma^2}$ (negative number!). Trying a quick integration by parts also gives the same answer (see below).
The second one gives $E\left [\frac{1}{X^4} \right]=\frac{1}{3\sigma^4}$ which also seems off given $E\left [{X^4} \right]=3 \sigma^4$, which would imply $E\left [\frac{1}{X^4} \right]=\frac{1}{E\left [{X^4} \right]}$
Why would integration by parts lead to wrong answers? Here are the line by line steps for the first one with $\sigma=1$ to simplify: Using $\int v'u = [uv] - \int vu'$:
$E\left[\frac{1}{X^2}\right] = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} x^{-2} e^{-x^2/2} dx$,
with \begin{align} \int_{-\infty}^{\infty} x^{-2} e^{-x^2/2} dx &= [(-\frac{1}{x})e^{-x^2} ]_{-\infty}^{\infty} - \int_{-\infty}^{\infty} (\frac{1}{x}) (x) e^{-x^2} dx \\ &= [(-\frac{1}{x})e^{-x^2} ]_{-\infty}^{\infty} - \int_{-\infty}^{\infty} e^{-x^2/2} dx\\ &= 0 - \sqrt{2\pi}\\ &= - \sqrt{2 \pi}. \end{align} Implying $E\left[\frac{1}{X^2}\right]=-\frac{1}{\sqrt{2\pi}} \sqrt{2\pi}=-1$.
See any mistakes anywhere?
For convenience let $\sigma=1$ and observe that:
$$\int_0^1 x^{-\alpha}e^{-\frac12x^2}dx\geq e^{-\frac12}\int_0^1x^{-\alpha}dx=+\infty$$ for $\alpha>1$.
This indicates that $X^{-\alpha}$ is not integrable if $\alpha>1$.
Even stronger (see the comment of @Did on this question).