Hi, a year ago, within this thread, Mariusz Iwaniuk posted a code combining Laplace and Mellin transforms, that effectively obtained the inverse Laplace transform of a certain complicated function that I considered. I am hopelessly trying to apply the same code to one more similar function, namely:

F[s_]:=((Sqrt[s] - Sqrt[s+1])^2)/((th*Sqrt[s] + Sqrt[s+1])^3)/s

where th>0 is a parameter and s is a Laplace variable. The program produces some output, seemingly without error messages, but the result appears incorrect, so that I must be doing something wrong. Unfortunately, I am unable to make meaningful interactive tests, as I have only access to Mathematica 12 in batch mode. Mariusz: if you still read these messages, may I ask you to try to apply your program to the above function? I would be happy to learn what I do wrong. Leslaw

I have found only:

HoldForm[InverseLaplaceTransform[(1 + s + s th)/(
   2 s (Sqrt[1 + s] + Sqrt[s] th)^2), s, t] == 
  Integrate[(
   E^-x (-2 th Sqrt[x] + 
      E^(th^2 x)
        Sqrt[\[Pi]] (1 + 2 th^2 x) Erfc[
        th Sqrt[x]]) ((1 + th) Hypergeometric0F1[1, -t x] + 
      t Hypergeometric0F1[2, -t x]))/(
   2 Sqrt[\[Pi]]), {x, 0, Infinity}] == 
  Sum[1/2 (-1)^n (1 + n) th^
    n ((1 + th) Hypergeometric1F1[1 + n/2, 1, -t] + 
      t Hypergeometric1F1[1 + n/2, 2, -t]), {n, 0, Infinity}]]

for: Integral: -1 < th < Infinity

for: Sum: -1< th <1

Probably no-closed form solution.

Mariusz: I am trying to understand your analytical inversion code, but I found a certain problem. In the application of the Mellin transform you make an assumption that 0 < Re[q] < 1. Later on, in the integral formula for the hypergeometric function you assume that Re[b] > Re[a]. However, some of the hypergeometric terms have a = 2-q/2 and b=3/2, so that for q sufficiently close to zero the condition Re[b] > Re[a] would not be satisfied. Is this indeed a problem? Leslaw

Hi Leslaw,

-Do I understand correctly that it does a sort of a parsing of the input expression into addends and addends into products of constant coefficients and s-dependent functions, and then applies InverseLaplaceTransform to the s-dependent functions, with the exception of the particular function from the table of Laplace transforms, for which the inverse is directly provided?

Yes, that is exactly what ILT function is doing.

Basically it is just the InverseLaplaceTransform function augmented with the particular table look-up function that I have mentioned in my previous reply.

Hrach

I tried your code formulation, but it produces only the inverses of the first two expansion terms, so it is worse than my original code. I am not sure which version of MATHEMATICA I have access to; it is probably 10.2.0. Is there any other approach? Leslaw

f = (((1 + s)^2 + 3*s^(3/2)*(1 + s)^(1/2)*th - 4*s*(1 + s)*th + 
      3*s^(1/2)*(1 + s)^(3/2)*th + 
      s^2*th^2))/(8 (s^(3/2)*(1 + s)^(1/2)*((1 + s)^(1/2) + 
          s^(1/2)*th)^3));
ff = InverseLaplaceTransform[Series[f, {s, Infinity, 20}] // Normal, 
   s, t];(*With 20 terms*)
N[ff /. t -> 1 /. th -> 1, 20](*for th=1, t=1*)
(*0.065780620406027739102*)

We can now compute InverseLaplaceTransfrom numerically in Mathematica 12.2.0:

InverseLaplaceTransform[f /. th -> 1, s, 1.0, WorkingPrecision -> 20](*for th=1, t=1*)
(*0.065780620406027739079*)

We can see 18 correct digits with 20 terms.