# Laplace exponent of a standard $\alpha-$stable subordinator

I'm trying to calculate the Laplace exponent of a standar $\alpha-$stable subordinator.

An $\alpha-$stable subordinator has Lévy measure $\frac{c}{x^{1+\alpha}}dx,$ where $\alpha\in(0,1)$ and $c$ is a positive constant.

Now, if the subordinator is $\alpha-$stable then has drift zero. So Laplace exponent has the form $$\Phi(\lambda)=c\int_{0}^{\infty}(1-e^{-\lambda x})\frac{dx}{x^{1+\alpha}}.$$

I've read on a book that Laplace exponent of this kind of subordinators have the form $$\Phi(\lambda)=\frac{\alpha}{\Gamma(1-\alpha)}\int_{0}^{\infty}(1-e^{-\lambda x})\frac{dx}{x^{1+\alpha}}.$$

I've proved that such measure is Lévy measure; the hypotesis $\alpha\in(0,1)$ is indispensable to have finite integrals independent of the constant $c.$ So I'd like to prove that constant $c$ is equal to $\frac{\alpha}{\Gamma(1-\alpha)}$ but I don't get it.I was trying to use Gamma function but I don't find something useful.

Any kind of help is thanked in advanced.

• I don't really get what you are asking. The measure $c/x^{1+\alpha} \, 1_{(0,\infty)}(x) dx$ is the Lévy measure of a subordinator for any $c>0$. Why would you expect to be able to show that $c$ takes some specific value...? – saz May 14 '18 at 19:07
• I'm expecting that,,using gamma function, find the Laplace exponent of such subordinator. Even more, $\Phi(\lambda)=\Gamma(-\alpha)\lambda^{\alpha}$ @saz – Squird37 May 14 '18 at 22:23