Vtyjkyuk

Let $Xsim N(0,sigma^2)$ normally distributed rv.
1) What is $Eleft[frac1X^2right]$?
2) What about $Eleft [frac1X^4 right]$?

Entering the first in Wolfram Alpha or Mathematica yields a seemingly nonsensical answer, $Eleft[frac1X^2right] = - frac1sigma^2$ (negative number!). Trying a quick integration by parts also gives the same answer (see below).

The second one gives $Eleft [frac1X^4 right]=frac13sigma^4$ which also seems off given $Eleft [X^4 right]=3 sigma^4$, which would imply $Eleft [frac1X^4 right]=frac1Eleft [X^4 right]$

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 $int v'u = [uv] - int vu'$:

$Eleft[frac1X^2right] = frac1sqrt2 pi int_-infty^infty x^-2 e^-x^2/2 dx$,

with
beginalign
int_-infty^infty x^-2 e^-x^2/2 dx &= [(-frac1x)e^-x^2 ]_-infty^infty - int_-infty^infty (frac1x) (x) e^-x^2 dx \
&= [(-frac1x)e^-x^2 ]_-infty^infty - int_-infty^infty e^-x^2/2 dx\
&= 0 - sqrt2pi\
&= - sqrt2 pi.
endalign
Implying $Eleft[frac1X^2right]=-frac1sqrt2pi sqrt2pi=-1$.

See any mistakes anywhere?

For convenience let $sigma=1$ and observe that:

$$int_0^1 x^-alphae^-frac12x^2dxgeq e^-frac12int_0^1x^-alphadx=+infty$$ for $alpha>1$.

This indicates that $X^-alpha$ is not integrable if $alpha>1$.

Even stronger (see the comment of @Did on this question).