I ran into this problem for a while and did not figure out which book I shall refer to.
Suppose $E_t^\beta$ is an inverse $\beta$ stable subordinator, we know for 0<$\beta$<1, \begin{align} f_{E_t^{\beta }}(x)=&\frac{1}{x}H_{1,1}^{0,1}\left(\frac{x^{\beta }}{t}\mid \begin{array}{c} (1,1) \\ (1,\beta ) \\ \end{array} \right)\\ =&\frac{1}{x}\frac{1}{2\pi i}\int _{c-i \infty }^{c+i \infty }\frac{\Gamma[-s]}{\Gamma[-\beta s]}\left(\frac{x^{\beta }}{t}\right)^{-s}ds\\ =&\frac{1}{x}\sum _{n=0}^{\infty } \frac{(-1)^n}{n!}\frac{1}{\Gamma[-\beta n]}\left(\frac{x^{\beta }}{t}\right)^{-n}\\ =&\frac{1}{x}\sum _{n=1}^{\infty } \frac{\left(-\frac{x^{\beta }}{t}\right)^{-n}}{n!}\frac{1}{\Gamma[-\beta n]}. \end{align}
My questions are
- Is the residue calculation correct? It looks $s=0$ shall not be a singularity since we have $\frac{\Gamma[0]}{\Gamma[0]}$. Further, for large $n$, being a multiple of 10,100,1000,..., which makes $\beta n\in \mathbb{N}$, similar situation happens again. For example, suppose $\beta=0.1$, then when $n=10$, the integrand becomes $\frac{\Gamma[-10]}{\Gamma[-1]}$. How do we deal with these $\frac{\infty}{\infty}$ cases?
- In the last equality, although the summation starts from $n=0$ and $n=1$ do not make any difference, since $\frac{1}{\Gamma[0]}=0$. However, it does make difference when we are calcuating the Laplace transform. In more details, when $n$ starts from 0, \begin{align*} \mathcal{L}_x\left(f_{\beta }(x,t),u\right)=\sum _{n=0}^{\infty } \frac{\left(-\frac{1}{t}\right)^{-n} u^{n \beta }}{n!}\{\Re[n\beta ]<0\&\Re[u]>0\}=e^{-t u^{\beta }}. \end{align*} I believe the summation condition can be ignored by the analytical continuity. In that case, $n$ starts from 1 will lead to $e^{-t u^{\beta }}-1$. So, when shall we add up these zero items automatically? Usually, books and papers admit this Laplace transform and the series expansion in which $n$ starts from 1.
Many thanks in advance. Any reference materials are also appreciated.