 Integrability threshold for exponential powers of a Gaussian

Let $Z$ have standard normal distribution, let $m\ge 1$ be an integer and let $\alpha \in \mathbb{R}$. A finite expected value of $Z{e}^{\alpha Z^{2m}}$ requires absolute integrability, which reduces to finiteness ofFor $m=1$ it is finite exactly when $\alpha <1/2$. For every $m\ge 2$ it is finite exactly when $\alpha \le 0$: the higher power dominates the negative quadratic when $\alpha >0$, and for $\alpha \le 0$ the integrand is bounded by $z{e}^{-z^2/2}$. In the finite cases the signed expected value is zero by symmetry. Otherwise each signed part has infinite integral; a formal symmetric cancellation is only a Cauchy principal value.