# Laplace transform giving incorrect result

In Mathematica:

LaplaceTransform[Exp[-Exp[-t]], t, s]
Out:= Gamma[s]-Gamma[s,1]


But performing InverseLaplaceTransform on the result does not give me any output. Even InverseLaplaceTransform[Gamma[s], s, t] does not give any output, but InverseLaplaceTransform[-Gamma[s,1], s, t] returns Exp[-Exp[-t]] which is identical to my input. Does that mean InverseLaplaceTransform[Gamma[s], s, t]=0 according to Mathematica?

Integration by summing residues shows that inverse Laplace transform of Gamma function is: $$\mathbf{L^{-1}}[\Gamma(s)]=\int_{\gamma-i\infty}^{\gamma+i\infty}ds~e^{st}\Gamma(s)\\ =\sum_{n=0}^\infty Res_{s=-n}[e^{st}\Gamma(s)]\\ =\sum_{n=0}^\infty e^{-nt}\frac{(-1)^n}{n!}=\sum_{n=0}^\infty \frac{(e^{-t})^n}{n!}=e^{-e^{-t}}$$

in which Gamma[s,1] has no role! I have verified by numerically inverting the Laplace transform (using Talbot's method) that presence of Gamma[s,1] indeed doesn't matter at all (i.e. whether I invert Gamma[s] or Gamma[s]-Gamma[s,1] I get the same number and this matches with Exp[-Exp[-t]]). I know this is a broad question, but what is going on with Mathematica here?

• I confirm that it did not work for me (Mma 11.3, Win7). – Alexei Boulbitch Nov 28 '18 at 9:26

sol = Sum[LaplaceTransform[(-Exp[-t])^j/j!, t, s], {j, 0, Infinity}]

• My main objection is why InverseLaplaceTransform[-Gamma[s,1], s, t]=Exp[-Exp[-t]] without Gamma[s] which is also a part (and an important part, as integration by residue shows) of its own output? – Deep Nov 28 '18 at 11:55