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.


1 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:

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]
  • $\begingroup$ The code seems to do the job, thanks. The tricky part now is to identify the recurrence relationship behind the coefficients of the exponentials as the infinite series (infinite order in the series expansion your propose) is required for full definition of the inverse transform. $\endgroup$
    – Alex
    Feb 25, 2014 at 14:42
  • $\begingroup$ @Alex Have you tried FindSequenceFunction? $\endgroup$
    – Michael E2
    Feb 25, 2014 at 16:26
  • $\begingroup$ Negative, I have gone another direction to solve the problem, bypassing the calculation of that inverse Laplace transform. I did not know about that function though. I'm sure it'll prove handy another time. Thanks! $\endgroup$
    – Alex
    Feb 27, 2014 at 19:14

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