I am analyzing creep functions for some rock mechanics problem, and I found that the function: \begin{equation} J(t)=1/E +Bt^p, \end{equation} is the best choice for my dataset, where $E$, $B$ and $p$ are positive constants, and additionally $p$ is between 0 and 1. Now, I'd like to invert it to get the relaxation modulus. Thus first, I take the usual viscoelastic inversion formula: \begin{equation} M(t) = \mathcal{L}^{-1}\bigg[\frac{1}{\widetilde{J}(s)s^2} \bigg]; \end{equation} in my case, the function to invert equals: \begin{equation} \widetilde{M}(s)=\frac{Es^{p-1}}{s^p+a}=E\bigg[\frac{1}{s}-\frac{1}{s+s^p/a}\bigg], \end{equation} where $a=EB\Gamma(p+1)$. The $1/s$ part is obvious, but I wonder if there is a way to invert $1/(s+s^p/a$). It looks like a mixture of a power law with exponential function, but I've got no idea how to proceed to get some result. The $p$ values I obtained experimentally were between 0.1 and 0.4 approximately. I found out that inversion is possible (of course) for $p=0$ and $p=1$, and also for $p=1/2$. Maybe, there are some other fixed $p$ values with existing inverse transforms.
I will appreciate any help or hints on this problem. Thanks!
This is a half answer and little more.
Not that:
$$\sum _{n=0}^{\infty } (-x)^n=\frac{1}{1+x}$$
so, $$\color{red}{\mathcal{L}_s^{-1}\left[\frac{1}{s+\frac{s^p}{a}}\right](t)}=\mathcal{L}_s^{-1}\left[\frac{1}{s \left(1+\frac{s^{-1+p}}{a}\right)}\right](t)=\mathcal{L}_s^{-1}\left[\frac{\sum _{n=0}^{\infty } \left(-\frac{s^{-1+p}}{a}\right)^n}{s}\right](t)=\sum _{n=0}^{\infty } \mathcal{L}_s^{-1}\left[(-1)^n a^{-n} s^{-1+n (-1+p)}\right](t)=\color{red}{\sum _{n=0}^{\infty } \frac{(-1)^n a^{-n} t^{-n (-1+p)}}{\Gamma (1-n (-1+p))}}$$
The answer is a $\color{red}{red}$ Sum,but I can't find closed form it.
Only for given parameter $\color{red}{p}$ CAS can find solution.
Mathematica code:
for p=1/10:
for p=2/10:
for p=3/10:
for p=4/10:
EDITED: