Laplace transform of inverse power law: $t^{-(1+\beta)}$ for $t > 0$ and $0 < \beta < 1$

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!