= Solution
{c}
Absolute <integrability> now requires
$$
\mathbb E|Ze^{\alpha Z^4}|
=\frac2{\sqrt{2\pi}}\int_0^\infty z e^{\alpha z^4-z^2/2}\,dz<\infty.
$$
For $\alpha\le0$, the <integrand> is bounded by $ze^{-z^2/2}$, whose <integral> is finite. For $\alpha>0$, once $z^2\ge1/\alpha$, the exponent is at least $\alpha z^4/2$, so the <integral> diverges. Hence
$$
\boxed{\mathbb E[Ze^{\alpha Z^4}]\text{ exists exactly for }\alpha\le0,
\quad\text{and then equals }0.}
$$
Again the zero follows from the <odd function> and <symmetry>; outside the <integrable> range both signed parts diverge, so the <expected value> is undefined.
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