# Inverse Laplace transform

For this expression: $(1 - Exp[-Sqrt[1+s] x])/(1+s)$

How to make the Inverse Laplace Transform analytically?

InverseLaplaceTransform[(1 - Exp[-Sqrt[1+s] x])/(1+s),s,t] does not directly give solution

• Unfortunately, InverseLaplaceTransform[(1 - Exp[-Sqrt[1 + s] x])/(1 + s), s, t, Assumptions -> x > 0] does not work. Jun 15, 2014 at 21:25
• Mathematica 13.3 gives answer. Jun 29, 2023 at 13:59

I'm not a mathematician so this may be "smoke and mirrors."

You are looking for the inverse Laplace transform of

g[s_] = (1 - Exp[-Sqrt[1 + s] x])/(1 + s);


Looking at the simpler problem

InverseLaplaceTransform[g[s - 1], s, t] // Simplify[#, x > 0] &


1 - Erfc[x/(2*Sqrt[t])]

With x > 0, let

f[t_] = E^-t (1 - Erfc[x/(2*Sqrt[t])]);

g[s] == LaplaceTransform[f[t], t, s] // Simplify[#, x > 0] &


True

So f[t] is the inverse Laplace transform of g[s]

• Your answer is correct. Verified it with Maple. Screen shot: !Mathematica graphics the command is: inttrans[invlaplace]( (1- exp(-sqrt(1+s)*x))/(1+s),s,t) assuming positive; note that Maple can do it directly, just needed assumption on x>0 Jun 16, 2014 at 8:00