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.

  • 1
    $\begingroup$ "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}]] $\endgroup$ – xzczd Jul 4 at 7:41
  • 1
    $\begingroup$ I fixed it. The main problem is with the inverse Laplace transform. $\endgroup$ – dtn Jul 4 at 7:44
  • $\begingroup$ I want to note that the integral is not calculated if you select the function Exp[-2 t] $\endgroup$ – dtn Jul 4 at 7:47
  • 3
    $\begingroup$ Not too surprsing. Somewhat related: mathematica.stackexchange.com/q/163949/1871 Sadly the method there doesn't work for this problem. $\endgroup$ – xzczd Jul 4 at 8:27
  • 2
    $\begingroup$ 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 $\endgroup$ – Mariusz Iwaniuk Jul 4 at 10:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.