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I want to compute the the Laplace inverse transform of the following

$$\bar f(s)=\frac{1}{s^2+a s+b}$$

which clearly depends on $a^2-4b$. When I use Mathematica, it just gives the answer for the case $a^2-4b>0$. I couldn't resolve the issue by using Assuming or Assumptions. I mean how can I get the answer for the cases $a^2-4b=0$ and $a^2-4b<0$.

$Assumptions = {a^2 - 4 b == 0};
f = 1/(s^2 + a s + b);
InverseLaplaceTransform[f, s, t]
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f = 1/(s^2 + a s + b);
inv = InverseLaplaceTransform[f, s, t]

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FullSimplify[inv, a^2 - 4 b > 0]

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FullSimplify[inv, a^2 - 4 b < 0]

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assumming a and b are Reals:

FullSimplify[ComplexExpand@inv, a^2 - 4 b < 0]

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Limit[inv, b -> a^2/4]

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  • $\begingroup$ (+1) Thanks for the attention. The result for $a^2-4b<0$ is incorrect! :) Indeed, what you have obtained for $a^2-4b>0$ and $a^2-4b<0$ is the same $\endgroup$ – H. R. May 28 '18 at 13:39
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    $\begingroup$ @H.R. - The result for a^2 - 4b < 0 is correct, it merely involves a complex argument to Sinh. If you want an explicitly real form, use Assuming[a^2 - 4 b < 0, inv // ComplexExpand[#, TargetFunctions -> {Re, Im}] & // FullSimplify] which evaluates to (2*Sin[(1/2)*Sqrt[-a^2 + 4*b]*t])/ (E^((a*t)/2)*Sqrt[-a^2 + 4*b]) $\endgroup$ – Bob Hanlon May 28 '18 at 13:52
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    $\begingroup$ @H.R. The result is correct, o.k., for Realsdo this FullSimplify[ComplexExpand@inv, a^2 - 4 b < 0] $\endgroup$ – rmw May 28 '18 at 14:04
  • $\begingroup$ Please add your last comment to the answer so that I can accept it. :) $\endgroup$ – H. R. May 28 '18 at 15:05

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