I can't even get an inverse Laplace for this expression numerically in mathematica, is there a way to inverse this equation below?

I have also tried to use fixt talbot package for a numerical calculation and it was not possible to get an answer. I also struggled with the options presented here, but I was not successful

How can I invert a Laplace transform numerically?

    1/s^(5/4) (0.0048093124845766414` + 0.02440766170480293` s^(1/4) + 
       0.012905621492995754` E^(-35.` s) s^(1/4) + 
       0.0032916706480934036` E^(0.7079967811634376` s) s^(1/4)
         Gamma[0.`, 35.70799678116344` s] - 
       0.0053059348427644286` Gamma[1.25`, 35.` s])


    InverseLaplaceTransform[  s^(5/4)/(0.0048093124845766414 +
    0.02440766170480293s^(1/4) + 0.012905621492995754 E^(-35. s) s^(1/4)
    +0.0032916706480934036 E^(0.7079967811634376 s) s^(1/4)Gamma[0, 35.70799678116344 s]
     - 0.0053059348427644286 Gamma[1.25, 35. s]), s, t]

Update: The first one works in Mathematica version 13.1. but the second one still not.

I think because the inverse of some functions is conditional in this case, we get InverseLaplaceTransform [of something that is defined in some cases and undefined in others] printed after we excite the code.

Getting an answer is only possible if the basis of the variables is defined. I think that is the point, but it is only possible in version 13.1, not in the previous version.

  • 1
    $\begingroup$ Your question boils down to: How can I calculate the inverse Laplace transform of Gamma[5/4,s]/s^(5/4). All other terms seem to be ok, see Map[InverseLaplaceTransform[#,s,t]&, {s^(-5/4),s^(-1),Gamma[0,s]/s}]. But I am a little confused by the fact that InverseLaplaceTransform[Exp[s]*Gamma[0,s]/s,s,t] gives 1-EulerGamma+1/t-Log[s] which still contains an s, in Version 12.3. $\endgroup$
    – user293787
    Aug 1 at 13:37
  • 4
    $\begingroup$ @user, that looks like an old bug that is now fixed; the correct answer Log[1 + t] is now returned. $\endgroup$ Aug 1 at 13:51
  • 2
    $\begingroup$ With 13.1, I obtain for the InverseLaplaceTransform of the expression in the question, 0.0244077 + 0.00530593 t^(1/4) + 0.0129056 HeavisideTheta[-35. + t] - 0.00530593 t^0.25 HeavisideTheta[-35. + t] + 0.00329167 InverseLaplaceTransform[(E^(0.707997 s) Gamma[0., 35.708 s])/s, s, t]. $\endgroup$
    – bbgodfrey
    Aug 1 at 14:04
  • 1
    $\begingroup$ Not sure. Let's see what others, like @J.M., has to say. $\endgroup$
    – bbgodfrey
    Aug 1 at 14:43
  • 1
    $\begingroup$ I'll only note that InverseLaplaceTransform[(E^(a s) Gamma[0, b s])/s, s, t] evaluates, so you can plug appropriate values of a and b into the result of that expression. $\endgroup$ Aug 1 at 15:11


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