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| Here is an even simpler example: LaplaceTransform[Exp[-t] Cos[2 t], t, s] returns (1 + s)/(4 + (1 + s)^2) as expected. However, InverseLaplaceTransform[(1 + s)/(4 + (1 + s)^2), s, t] returns 1/2 E^((-1 - 2 I) t) (1... |