I need to calculate the inverse Laplace transform of $F(s)=\frac{1}{s^\alpha-\lambda}$ where $\lambda$ is a complex number and $\alpha$ is a real numbers.
I know that it is going to envolves the gamma function but I have no idea what to do with that $\lambda$.
Can someone give me any hint?
I'll assume $\alpha > 0$, so that we have $\lim\limits_{\operatorname{Re} s \to +\infty} F(s) = 0$, as it should be for the Laplace transform of a function with at most exponential growth. We can expand $F(s)$ into a geometric series,
$$F(s) = \frac{1}{s^{\alpha} - \lambda} = s^{-\alpha}\frac{1}{1 - \lambda s^{-\alpha}} = \sum_{k = 0}^{\infty} \lambda^k s^{-(k+1)\alpha},$$
that converges uniformly on $\mathbb{C}\setminus D_R(0)$ for sufficiently large $R$ (depending on $\lambda$), and apply the inverse Laplace transformation termwise. Since
$$\int_0^{\infty} t^{\beta - 1} e^{-st}\,dt = s^{-\beta}\Gamma(\beta)$$
for $\beta > 0$, we get the series expansion
$$f(t) := \sum_{k = 0}^{\infty} \frac{\lambda^k}{\Gamma((k+1)\alpha)}t^{(k+1)\alpha-1}\tag{$\ast$}$$
for the (potential) inverse Laplace transform of $F$. Now one can verify that the series in $(\ast)$ converges for all $t > 0$ and all $\lambda \in \mathbb{C}$, and that $f(t)e^{-st} \in L^1((0,+\infty))$ for $\operatorname{Re} s$ large enough. The dominated convergence theorem then shows that
$$\mathcal{L}[f](s) = \int_0^{\infty} f(t)e^{-st}\,dt = \sum_{k = 0}^{\infty} \frac{\lambda^k}{\Gamma((k+1)\alpha)} \int_0^{\infty} t^{(k+1)\alpha-1}e^{-st}\,dt = F(s).$$
For $\alpha \in \mathbb{N}\setminus \{0\}$, one can give a closed-form expression of $f$ in terms of $\exp \bigl(\rho_m t\bigr)$, where $\rho_m$ ranges over the possible values of $\lambda^{1/\alpha}$, but for general $\alpha \in (0,+\infty)$ I'm not aware of a nicer expression than $(\ast)$.