# Convolution using the Laplace integral transform of certain functions

I am trying to convolve two functions:

$$f(t) = e^{- t}$$

$$g(t) = e^{-(e^{-t})^2}$$

$$(f*g)(t) = \int_{0}^{t} f(t-\tau)g(\tau) d\tau = \int_{0}^{t} e^{-(t-\tau)} e^{-(e^{-\tau})^2} d\tau$$

Using the convolution property, we can transform it into a product of folding functions, each of which is obtained using the direct Laplace transform of each of the original functions, i.e.:

$$(f*g)(t) \rightarrow L^{-1}(L(f(t)) \cdot L(g(t)))$$

But it is known that the original functions must meet certain requirements.

With these functions, the Laplace transform will look like:

ut = Exp[-Exp[-t]^2]
LaplaceTransform[ut, t, s] LaplaceTransform[Exp[-t], t, s]


And this is result:

$$\frac{\Gamma \left(\frac{s}{2}\right)-\Gamma \left(\frac{s}{2},1\right)}{2 s+2}$$

InverseLaplaceTransform[(Gamma[s/2] - Gamma[s/2, 1])/(
2 + 2 s), s, t]

Out[85]= InverseLaplaceTransform[(Gamma[s/2] - Gamma[s/2, 1])/(
2 + 2 s), s, t]


But the inverse Laplace transform of this function does not work. What could be the reason and are there any ways to find the inverse Laplace transform of this function?

I will be glad to any advice and help.

• "Unfortunately, neither Mathematica nor Maple mastered this integral. " Integrate gives the result within 2 seconds, how do you code it?: f[t_] = E^-t; g[t_] = Exp[-Exp[-t]^2]; AbsoluteTiming[Integrate[f[t - \[Tau]] g[\[Tau]], {\[Tau], 0, t}]] – xzczd Jul 4 at 7:41
• I fixed it. The main problem is with the inverse Laplace transform. – dtn Jul 4 at 7:44
• I want to note that the integral is not calculated if you select the function Exp[-2 t] – dtn Jul 4 at 7:47
• Not too surprsing. Somewhat related: mathematica.stackexchange.com/q/163949/1871 Sadly the method there doesn't work for this problem. – xzczd Jul 4 at 8:27
• A workaround: f2[t_] = InverseLaplaceTransform[-Gamma[s/2, 1], s, t]; g2[t_] = InverseLaplaceTransform[1/(2 s + 2), s, t]; Integrate[f2[t - \[Tau]]*g2[\[Tau]], {\[Tau], 0, t}] // FullSimplify – Mariusz Iwaniuk Jul 4 at 10:50