It appears possible to compute the transform for ``zeta > 1`` by replacing
``zeta`` with ``1 + zeta``.
In [22]: G = K/(tau**2*s**2 + 2*tau*(1 + zeta)*s + 1)
In [24]: print(inverse_laplace_transform(G, s, t))
K*(exp(t*sqrt(zeta)*sqrt(zeta + 2)/tau) - 1)*(exp(t*sqrt(zeta)*sqrt(zeta +
2)/tau) + 1)*exp(-t*(sqrt(zeta)*sqrt(zeta + 2) + zeta +
1)/tau)*Heaviside(t)/(2*tau*sqrt(zeta)*sqrt(zeta + 2))
The transform of the original is obtained by replacing ``zeta`` with ``zeta
- 1``.
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