1
$\begingroup$

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])

and


    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.

Improve this question
$\endgroup$
11
  • 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. $\endgroup$
    – J. M.'s persistent exhaustion ♦
    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. $\endgroup$
    – J. M.'s persistent exhaustion ♦
    Aug 1 at 15:11

0

Reset to default

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.