On page 18 of Conditional Measures and Applications(2005) by M.M.Rao, there is an example that $\Bbb E(X^n\mid Y=y)$ always exists, but $\Bbb E(X^n)$ doesn't. It is mentioned that
$\Bbb E(X^n\mid Y=y)$ can't be a conditional expectation of $X$ given $Y$, since (9)(Law of iterated expection) does not hold.
Here's the screenshot of the example:
I'm not sure whether it means that $\Bbb E(\Bbb E(X^n\mid Y)) <\Bbb E(X^n)$ in this example.
So much for the background.
The question can be simplified as:
Can we decide that whether $\int_{(0, +\infty)} y^{-\frac{1}{2}} (\int_{[0, +\infty)} x^{2m} e^{-x^{2}y}dx)dy$ is finite for $m \in {\Bbb Z}^+$?
The change of variable $x\to z=x\sqrt{y}$ shows that the integral is $$\int_0^\infty y^{-1/2}\int_0^\infty z^{2m}y^{-m}\mathrm e^{-z^2}y^{-1/2}\mathrm dz\mathrm dy=\int_0^\infty y^{-1-m}\mathrm dy\cdot\int_0^\infty z^{2m}\mathrm e^{-z^2}\mathrm dz,$$ which is infinite for every $m\geqslant0$.