On the derivation of the values of $E[X^s]$ with $s \in C$ from the moments. Take $s \in C$ and an absolutely continuous random variable $X$ for which all moments exist and are given by $E[X^n] = f(n)$ for some continuous function $f$.
Then can we always say that $E[X^s] = f(s)$ where $f(s)$ is well defined?
To give an example say we have a random variable $B\sim Beta(a,b)$ we know that $E[B^n] = \frac{\beta(a+n,b)}{\beta(a,b)} \ \forall{n}\in N $ is it true that $E[B^s] = \frac{\beta(a+s,b)}{\beta(a,b)}$ where $s$ is a complex number with $Real(s) > -a$?
 A: To give the process in my comment in slightly more detail,


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*Given $E[X^n]$ for all $n \in \{0,1,2,\dotsc\}$, the characteristic function of the distribution is uniquely determined by its series expansion,
$$ \varphi_X(t) = E[e^{itX}] = \sum_{n=0}^{\infty} \frac{(it)^n}{n!} E[X^n]. $$

*The characteristic function determines the distribution uniquely; since the random variable is absolutely continuous, the density can be recovered by what is essentially Fourier inversion,
$$ f(x) = \frac{1}{2\pi}\int_{\mathbb{R}} e^{-itx} \varphi_X(t) \, dt. $$

*The non-integral moments are then simply $E[X^s] = \int_{\mathbb{R}} x^s f(x) \, dx$ by the usual formula. Note that this only really makes sense if $X$ is nonnegative, since $x^s$ can have different definitions for negative $x$ and it's not necessarily obvious which you want.
This is admittedly a rather crude approach, but it suffices for your example.
If the random variable is nonnegative and you don't have all the moments, $s \mapsto X^s$ is log-convex, which you can use to find a simple bound on moments smaller than the largest one you have:
$$ E[X^s] \leq (E[X^a])^{\alpha/a} (E[X^b])^{\beta/b}, $$
where $\alpha+\beta=s$ and $\alpha/a+\beta/b=1$.
