# InverseLaplaceTransform does not evaluate [closed]

I am very new to Mathematica and not a mathematician, so the question may be very trivial.

I am trying to find the function $$f(t)$$ such that

$$\qquad e^{-e^{s}} = \int_0^{+\infty} e^{-st}f(t)dt$$,

so I tried using the inverse Laplace transform function.

InverseLaplaceTransform[Exp[-Exp[s]], s, t ]


However, the output I obtain is simply the command itself. I don't understand if such function simply does not exist (not even in complex numbers) or if I need to introduce some kind of constraint.

• If it returns unevaluated, it means Mathematica cannot compute an explicit formula for the inverse transform. – Sjoerd Smit Oct 1 at 9:01
• $$\mathcal{L}_s^{-1}[\exp (-\exp (s))](t)=\sum _{j=0}^{\infty } \frac{(-1)^j \delta (j+t)}{j!}$$ where: $\delta (j+t)$ is DiracDelta[j + t] for t>0 sum is: 0 – Mariusz Iwaniuk Oct 1 at 15:46
• Thank you! How did you obtain it? I'd like to understand what I was doing wrong – caup Oct 2 at 9:07
• $$\mathcal{L}_s^{-1}[\exp (-\exp (s))](t)=\mathcal{L}_s^{-1}\left[\sum _{j=0}^{\infty } \frac{(-\exp (s))^j}{j!}\right](t)=\sum _{j=0}^{\infty } \mathcal{L}_s^{-1}\left[\frac{(-\exp (s))^j}{j!}\right](t)=\sum _{j=0}^{\infty } \frac{(-1)^j \delta (j+t)}{j!}=0$$ then execute: Assuming[t > 0, Sum[((-1)^j DiracDelta[j + t])/j!, {j, 0, Infinity}]] – Mariusz Iwaniuk Oct 2 at 14:53