Laplace transform and fractional moments Is there any "easy" way to calculate fractional moments from Laplace transform.
To be more specyfic let us consider the following example. Let $X$ be a positive random variable and
$L(\theta) := E[\exp (-\theta X)]$
be its Laplace transform. Of course it is easy to calculate $E[X^n]$ where $n$ is a natural number but what with e.g. $E[ X^{1/2}]$.
 A: If $X$ is positive the following works.   Let $F(\theta)=E(e^{-\theta X})$ be the Laplace transform. Given $s\in \mathbb{R}$, write $s=n-\alpha$ with $n$ an integer and $\alpha >0$.  Then
$$E(X^{s}) = (-1)^n\frac{1}{\Gamma(\alpha)} \int_0^\infty F^{(n)} (\theta) \theta^{\alpha-1} d \theta$$
with $\Gamma$ the usual Gamma function,
$$\Gamma(\alpha) = \int_0^\infty \theta^{\alpha -1} e^{-\theta} d \theta.$$
Indeed by Tonelli's theorem and then a change of variable,
$$\int_0^\infty F^{(n)}(\theta) \theta^{\alpha-1} d \theta = (-1)^n E \Bigl(\int_0^\infty X^n e^{-\theta X} \theta^{\alpha-1} d \theta \Bigr)= (-1)^n E(X^{n-\alpha}) \int_0^\infty \theta^{\alpha-1} e^{-\theta} d \theta,$$ and so long as $\alpha >0$ the integral on the right is convergent.
A: For your example of $X^{1/2}$ you can evaluate the fractional half derivative of the Laplace transform (see for example the Wikipedia article on fractional calculus) at $\theta = 0$.
A: A version of the Laplace transform is the moment generating function. A paper answer to this question is The moment generating function has its moments by Noel Cressie and Marinus Borkent.
An example is at https://stats.stackexchange.com/a/317475
