The expectation values for integer powers of a Poisson random variable $X\sim Poiss(\lambda)$ are well known.
I'm interested in the expectation value of $X^\alpha$ for arbitrary rational $\alpha$. I'm not familiar enough with the literature to gauge whether this is a simple problem with an exact solution or not.
Wikipedia mentions the special case $\alpha=1/2$: it is known that the distribution of $X^{1/2}$ is approximately normal with mean $\lambda^{1/2}$ (and constant variance $1/4$).
This makes me think there is no simple answer to my question. Then again, in the references cited by Wikipedia, the authors were interested in more than the expectation value, so perhaps the answer to the question here is simpler.
I doubt that there is a simple solution. See eg here.
One approximate approach: letting $Y=X^a=g(X)$ and doing a Taylor expansion around the mean
$$\mu_Y \approx g(\mu_X) + \frac{1}{2!} g^{''}(\mu_X) m_2 + \frac{1}{3!} g^{[3]}(\mu_X) m_3 + \cdots$$ where $\mu_x=\lambda$ and $m_n$ is the centered $n$-th centered moment. Further $g^{[n]}(\mu_X)= (a)_n \lambda^{a-n}$. A first order approximation would be:
$$E(X^a) \approx \lambda^a + \frac{a(a-1)(a-2) }{2} \lambda^{a-1} $$ which for $a=1/2$ gives me $\sqrt{\lambda} + \frac{3}{16 \sqrt{\lambda}} $