In[1]:= InverseLaplaceTransform[LaplaceTransform[a^2/(a^2 + t^2), t, s], s, t]
Out[1]= 1/2 a InverseLaplaceTransform[(2 CosIntegral[a s] - Log[1/a^2] - 
  2 Log[a]) Sin[a s] + Cos[a s] (Sqrt[1/a^2] a \[Pi] - 2 SinIntegral[a s]), s, t]

Or should I expect InverseLaplaceTransform can always recover the original expression?

Version:; Platform: Microsoft Windows (64-bit)

  • 2
    $\begingroup$ Try this: Assuming[ a > 0, InverseLaplaceTransform[ LaplaceTransform[ a^2 / (a^2 + t^2), t, s], s, t] ] $\endgroup$ – LouisB Dec 12 '17 at 3:50
  • $\begingroup$ @LouisB Good. You can write it as an answer and I will accept it. $\endgroup$ – ziyuang Dec 12 '17 at 9:44

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