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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 $$E(X^{s}) = \frac{1}{\Gamma(\alpha)} (-1)^n\frac{1}{\Gamma(\alpha)} \int_0^\infty F^{(n)} (\theta) \theta^{\alpha} d \theta$$ with $\Gamma$ the usual Gamma function, $$\Gamma(\alpha) = \int_0^\infty \theta^{\alpha -1} e^{-\theta} d \theta.$$
Indeed by Fubini, $$\int_0^\infty F^{(n)}(\theta) \theta^{\alpha-1} d \theta = (-1)^n E (\int_0^\infty X^n e^{-\theta X} \theta^{\alpha-1} d \theta )= (-1)^n E(X^{n-\alpha}) \int_0^\infty \theta^{\alpha-1} e^{-\theta} d \theta,$$ and so long as $\alpha >0$ the integral on the right is convergent.
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 $$E(X^{s}) = \frac{1}{\Gamma(\alpha)} \int_0^\infty F^{(n)} (\theta) \theta^{\alpha} d \theta$$ with $\Gamma$ the usual Gamma function, $$\Gamma(\alpha) = \int_0^\infty \theta^{\alpha -1} e^{-\theta} d \theta.$$
Indeed by Fubini, $$\int_0^\infty F^{(n)}(\theta) \theta^{\alpha-1} d \theta = (-1)^n E (\int_0^\infty X^n e^{-\theta X} \theta^{\alpha-1} d \theta )= E(X^{n-\alpha}) \int_0^\infty \theta^{\alpha-1} e^{-\theta} d \theta,$$ and so long as $\alpha >0$ the integral on the right is convergent.