See attached file.

Regards.

Attachments:

InverseLaplace A...nb

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.

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

Well, thank you very much, your result is absolutely amazing and turns upside down all my recent work! But I am afraid it is out of my reach to understand what you have done. Is this a 100% exact result, or a very accurate approximation (I see that some sort of expansions is involved)? Another question is why the InverseLaplaceTransform[] command is not able to derive such a result directly from F[s]? I am also not sure about the range of applicability. For example, I notice that there are expressions involving Sqrt[1-th^2]. What happens if th > 1? Is there also no restriction on t? Your result also makes me wonder about the possibilities of obtaining the analytical inverse of a simpler function that I considered several months ago:

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

Would that be possible to use the same approach in this case? Leslaw

Many thanks, the proposal of Hrach seems to work like a charm. If only I could understand what this ILT function actually does! 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?

I also have another problem related to my F[s] function, but I put it into a separate question. Leslaw

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

function ff is analytical expressions for the inverses of the successive series expansion terms ?

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.