I am looking for a function $f(t)$ whose Laplace transform is given by $F(s)$: \begin{equation} F(s)=\frac{(1+\rho s)^\alpha}{(1+\rho s)^\alpha-1} \end{equation}
Now, I know from this table the following inverse transform: $$ \frac{s^\alpha}{s^\alpha-1}\to \frac{1}{t}\varepsilon_{1,0}(-t^\alpha) $$
Where $\epsilon$ is the generalised Mittag-Leffler function and $\alpha,\rho>0$.
Therefore I could compute the inverse transform of my original expression by using the shift and scaling properties of the Laplace transform:
Let $s^\prime=1+\rho s$, then $F\left(\frac{s^\prime-1}{\rho}\right)$ should give $\rho f(\rho t) e^{-\rho t}$. But when I compare numerical results with my answer, it is completely wrong... What am I doing wrong?
Using: $$\sum _{j=0}^{\infty } (1+s \rho )^{-j \alpha }=\frac{(1+\rho s)^{\alpha }}{(1+\rho s)^{\alpha }-1}$$
so:
$$\mathcal{L}_s^{-1}\left[(1+s \rho )^{-j \alpha }\right](t)=\frac{e^{-\frac{t}{\rho }} \left(\frac{t}{\rho }\right)^{j \alpha }}{t \Gamma (j \alpha )}$$ and then: $$\mathcal{L}_s^{-1}\left[\frac{(1+\rho s)^{\alpha }}{(1+\rho s)^{\alpha }-1}\right](t)=\sum _{j=0}^{\infty } \frac{e^{-\frac{t}{\rho }} \left(\frac{t}{\rho }\right)^{j \alpha }}{t \Gamma (j \alpha )}=\frac{e^{-\frac{t}{\rho }} E_{\alpha ,0}\left(\left(\frac{t}{\rho }\right)^{\alpha }\right)}{t}$$
Mathematica code:
Sum[(E^(-(t/\[Rho])) (t/\[Rho])^(j \[Alpha]))/( t Gamma[j \[Alpha]]), {j, 0, Infinity}] == (E^(-(t/\[Rho]))) /t* MittagLefflerE[\[Alpha], 0, (t/\[Rho])^\[Alpha]]