Series expansion using an exponential basis

Question

I am currently trying to Laplace invert an expression with the following pattern

$$ \frac{s \alpha \text{Cosh}[s(L-x)]+\beta \text{Sinh}[s(L-x)]}{s(\gamma \text{Cosh}[sL]+s \delta \text{Sinh}[sL])} $$

where $s$ is the complex variable and $x\in[0,L]$ a real variable; all other parameters are real constants.

One path I foresee in achieving the inversion is by expressing the denominator in the form of a power series of exponential terms, as these correspond to shifts in the time domain.

For instance, this approach can successfully be applied to

$$ \frac{1}{s(1+\mathrm{e}^{-s})}=\frac{1}{s}\sum_{n=0}^{\infty}(-1)^n\mathrm{e}^{-ns} $$

which upon inversion yields a train of Heaviside functions

$$ \mathcal{L}^{-1}\left(\frac{1}{s(1+\mathrm{e}^{-s})}\right)=\sum_{n=0}^{\infty}(-1)^n H(t-n) $$

Therefore, my question is, using Mathematica, how can I transform $(\gamma \text{Cosh}[sL]+s\delta\text{Sinh}[sL])^{-1}$ into a series of exponential functions; that is,

$$ \frac{1}{\gamma \text{Cosh}[sL]+s\delta\text{Sinh}[sL]} = \sum_{n=0}^\infty a_n(s) \mathrm{e}^{-n sL} $$

Any help will be appreciated.

Answer 1

You'd have to work out whether this is well behaved and convergent on your own so taking your own word on how sane the approach is, you can define your expression:

ClearAll[expr];
expr = (γ Cosh[s L] + s δ Sinh[s L])^(-1);

then replace -Log[q] for s and expand around 1 to some order and replace back q as E^(-s) (these are inverses of each other in the domain of interest). I also collect the exponentials for aesthetic reasons.

Normal@Series[expr /. s -> -Log[q], {q, 1, 5}] /. q -> E^(-s) // 
 Collect[#, E^(-s)] &

Finally, it is this over s that you are interested in:

InverseLaplaceTransform[%/s, s, t]