I came across a paper writing about continuous-time random walk, which derived the number of distinct sites visited by a random walker. It says that given the waiting time distribution $\psi(t) \sim t^{-(1+\beta)}$ with $0 <\beta<1$, the Laplace transform of $\psi(t)$, denoted as $\tilde{\psi(u)}$ follows:

$$\tilde{\psi(u)} \sim 1 - (Au)^{\beta}\tag{1}$$

I did a simple Laplace transform of it and it seems to give $\tilde{\psi(u)} \sim u^{\beta}\Gamma(-\beta)$. Perhaps I was doing it wrong because the input of a Gamma function should be larger than $0$? So I also did an integral by part to get a $\Gamma(1-\beta)$ to make it seem legal. But however I did it I cannot get the above result $(1)$ given by the paper. I then followed a reference cited in the paper and noted there that $\psi(t)$ has a more specific form with a constant term:

$$\psi(t) \sim \frac{\beta\tau^{\beta}}{\Gamma(1-\beta)}t^{-(1+\beta)}\tag{2}$$

where $\tau$ is a constant. The constant has a Gamma function in its denominator, which makes me think the way I approached it to have a Gamma function in the Laplace transform may be correct. Because then the Gamma function term gets cancelled out in this more specific form of $\psi(t)$.

However, I can only get this result: $\tilde{\psi(u)} \sim -(\tau u)^{\beta}$ instead of the $\tilde{\psi(u)} \sim 1-(\tau u)^{\beta}, u\rightarrow 0$. What am I doing wrong?

Furthermore, I don’t know how to analyze the range of $u$ in the Laplace transformed domain. Any hints? Thanks!


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