Identity for stable Lévy subordinator

Question

I want a proof or a reference for the identity $$ \int_0^\infty \frac{s^{n-1}}{\Gamma(n)} p_\beta(s,x)\,ds =\frac{x^{n\beta-1}}{\Gamma(\beta n)},\quad x>0, \,n\in\mathbb N, $$ where $x\mapsto p_\beta(s,x)$, $x>0$ is the density of the $\beta$-stable Lévy subordinator at time $s>0$, $\beta\in(0,1)$.

A possible proof is to take the Laplace transforms of both sides above, which are the same, using $\int_0^\infty e^{-\lambda x}p_\beta(s,x)\,dx=e^{-\lambda^\beta s}$, which also proves that the left hand side is finite almost everywhere. If one can prove that the left hand side is also continuous, then uniqueness of Laplace transforms proves the identity.

Answer 1

It is of course possible to show that the left-hand side is continuous (e.g. using bounds for the derivative of $p_\beta(s, x)$), but there is a more direct way. Observe that $$p_\beta(s, x) = s^{-1/\beta} p_\beta(1, s^{-1/\beta} x).$$ Thus, $$ \int_0^\infty s^{n - 1} p_\beta(s, x) ds = \int_0^\infty s^{n - 1/\beta - 1} p_\beta(1, s^{-1/\beta} x) ds = \beta x^{n \beta - 1} \int_0^\infty t^{-n \beta} p_\beta(1, t) dt . $$ There are likely many ways to do the integral in the right-hand side; for example, one can write $$ t^{-n \beta} = \frac{1}{\Gamma(n \beta)} \int_0^\infty e^{-t u} u^{n \beta - 1} du $$ and use Fubini's theorem: $$\begin{aligned} \int_0^\infty t^{-n \beta} p_\beta(1, t) dt & = \frac{1}{\Gamma(n \beta)} \int_0^\infty \biggl(\int_0^\infty e^{-t u} p_\beta(1, t) dt\biggr) u^{n \beta - 1} du \\ & = \frac{1}{\Gamma(n \beta)} \int_0^\infty e^{-u^\beta} u^{n \beta - 1} du \\ & = \frac{1}{\beta \Gamma(n \beta)} \int_0^\infty e^{-v} v^{n - 1} dv = \frac{\Gamma(n)}{\beta \Gamma(n \beta)} \end{aligned}$$ (this is the usual way to write the Mellin transform in terms of the Laplace transform).