Expectation $\alpha-$stable random variable

Question

I'm trying to solve the following exercise, but I'm lost.

Let $\sigma(\alpha)$ an $\alpha-$stable random variable of positive values, for some $\alpha\in(0,1),$ i.e., its Laplace Transform is given by $$E(\exp(-\lambda\sigma(\alpha)))=\exp(-\lambda^{\alpha}),\quad\lambda\geq 0.$$

Computing $E((\sigma(\alpha))^{-\beta}),\quad\beta>0.$

I was trying to get an expression with Mittag-Leffler function but everything that I do is useless. Perhaps if we get the distribution function of random variable $\sigma(\alpha)$ it could be a little easy calculate the required expectation.

Any suggestion, help or idea is thanked in advance.

Answer 1

Since $x^{-\beta} = \frac{1}{\Gamma(\beta)}\int_0^\infty \lambda^{\beta-1} e^{-\lambda x} d\lambda$ for any $x>0$, we have $$ E [\sigma(\alpha)^{-\beta}] = \frac{1}{\Gamma(\beta)}\int_0^\infty \lambda^{\beta-1} E[e^{-\lambda \sigma(\alpha)}]d\lambda = \frac{1}{\Gamma(\beta)}\int_0^\infty \lambda^{\beta-1} e^{-\lambda^\alpha} d\lambda = \frac{\Gamma(\beta/\alpha)}{\alpha\Gamma(\beta)}. $$