0
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I tried this:

F = InverseLaplaceTransform[
  1/((1 + s^2) (1 + (s + 1/2)^2)) + 
   Exp[-Pi*s]*1/((1 + s^2) (1 + (s + 1/2)^2)), s, t]

But it returns an answer with a lot of complex numbers. Is there a way that I can get a strictly real answer?

Using Mathematica 11.2.0

Thanks.

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  • 4
    $\begingroup$ \\ComplexExpand\\FullSimplify yields 4/17 ((4 Cos[t] - Sin[t]) (-1 + UnitStep[-\[Pi] + t]) - E^(-t/2) (4 Cos[t] + Sin[t]) (-1 + E^(\[Pi]/2) UnitStep[-\[Pi] + t])), which should be good enough. $\endgroup$ – AccidentalFourierTransform Apr 8 '18 at 1:11
  • 1
    $\begingroup$ @AccidentalFourierTransform Nice! Thanks for the help. $\endgroup$ – David Apr 8 '18 at 2:30

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