0
$\begingroup$

How do I obtain the Inverse Laplace transform of the following expression? $$\frac{\tan ^{-1}\left(\sqrt{s}\right)}{\sqrt{s}}$$

Using Mariusz Iwaniuk recipes with Feynmann trick and "Mellin Transform trick" in earlier question I get the same result as inverse Laplace Transform of $\frac{\tan ^{-1}\left(\frac{1}{\sqrt{s}}\right)}{\sqrt{s}}$ which seems wrong since these two functions are distinct.

ClearAll["Global`*"];
ilaplace[expr_] := InverseLaplaceTransform[expr, s, t];

(*Custom Laplace transforms from "Hypergeometric" post*)

(*Mariusz solution from \
https://mathematica.stackexchange.com/a/285338/217*)
augmentInv[expr_, var_] := 
  Module[{a1, a2, a3, a4, expra}, 
   expra = expr /. {ArcTan[a1_] -> ArcTan[a1/var], 
      Log[a1_] -> Log[a1/var], 
      Hypergeometric2F1[a1_, a2_, a3_, a4_] -> 
       Hypergeometric2F1[a1, a2, a3, a4/var]};
   If[MemberQ[Reduce`FreeVariables[expra], var], expra, expra/var]];

ilaplaceMellin0[expr_] := 
  Block[{expra, mellin, ilap, imellin, s, t, a}, 
   expra = augmentInv[expr, a];
   mellin = FunctionExpand@MellinTransform[expra, a, q];
   ilap = InverseLaplaceTransform[mellin, s, t];
   imellin = InverseMellinTransform[ilap, q, a] /. a -> 1;
   FullSimplify@imellin];

SetAttributes[ilaplaceMellin0, Listable];
ilaplaceMellin[expr_] := 
  Block[{dummy}, 
   Distribute@dummy@Expand[expr] /. dummy -> ilaplaceMellin0];

(*https://zackyzz.github.io/feynman.html*)
augment[expr_, var_] := 
  Module[{a1, a2, a3, a4, expra}, 
   expra = expr /. {ArcTan[a1_] -> ArcTan[a1 var], 
      Log[a1_] -> Log[a1 var], 
      Hypergeometric2F1[a1_, a2_, a3_, a4_] -> 
       Hypergeometric2F1[a1, a2, a3, a4 var]};
   If[MemberQ[Reduce`FreeVariables[expra], var], expra, var*expra]];

ilaplaceFeynmann0[expr_] := 
  Block[{repl, ilap, a1, a2, a3, a4, a}, expra = augment[expr, a];
   ilap = InverseLaplaceTransform[D[expra, a] // Factor, s, t];
   Assuming[{t > 0}, Integrate[ilap, {a, 0, 1}]]];
SetAttributes[ilaplaceFeynmann0, Listable];
ilaplaceFeynmann[expr_] := 
  Block[{dummy}, 
   Distribute@dummy@Expand[expr] /. dummy -> ilaplaceFeynmann0];


expr1 = ArcTan[Sqrt[2]/Sqrt[s]]/(Sqrt[2] Sqrt[s]);
expr2 = Hypergeometric2F1[1, 1/3, 4/3, s];

expr3 = ArcTan[1/Sqrt[s]]/ Sqrt[s];
expr4 = ArcTan[Sqrt[s]]/ Sqrt[s];

exprs = {expr1, expr2, expr3, expr4};
methods = {ilaplace, ilaplaceFeynmann, ilaplaceMellin};
results = Outer[TimeConstrained[#1[#2], 10] &, methods, exprs];

TableForm[results, 
 TableHeadings -> {{"default", "Feynmann", "Mellin"}, exprs}]

enter image description here

Improve this question
$\endgroup$
3
  • $\begingroup$ Are you asking why you are not getting the result you want using these methods/code you are using? Since using the direct command InverseLaplaceTransform in Mathematica works as is as shown in the answer below? $\endgroup$
    – Nasser
    Commented Jun 23, 2023 at 23:40
  • $\begingroup$ @Nasser you are right, I'm getting incorrect result due to augmentation method, which was required in order to perform inverse Laplace of ArcTan[1/Sqrt[s]]/Sqrt[s]. Here it appears default method works $\endgroup$
    – Yaroslav Bulatov
    Commented Jun 24, 2023 at 0:19
  • $\begingroup$ Mathematica 13.3 expanded on LaplaceTransform and InverseLaplaceTransform capabilities significantly but cannot solve this one it seems. $\endgroup$
    – flinty
    Commented Jun 30, 2023 at 18:50

1 Answer 1

Reset to default
2
$\begingroup$

InverseLaplaceTransform[ArcTan[Sqrt[s]]/Sqrt[s], s, t] //Simplify gives

$$\frac{e^{-t}-1}{2 t}$$.

Improve this answer
$\endgroup$
7
  • $\begingroup$ Actually on V 13.2.1 only Simplify not FullSimplify is needed. I wonder in this case which version of Mathematica OP used. $\endgroup$
    – Nasser
    Commented Jun 23, 2023 at 23:37
  • $\begingroup$ Thanks it appears the default method works on this expression (the extra tricks in code were needed to invert ArcTan[1/Sqrt[s]]/Sqrt[s] but apparently need extra work) $\endgroup$
    – Yaroslav Bulatov
    Commented Jun 24, 2023 at 0:27
  • $\begingroup$ The result is incorrect, looks like a bug in Mathematica actually, workaround wolframcloud.com/obj/yaroslavvb/nn-linear/… $\endgroup$
    – Yaroslav Bulatov
    Commented Jun 25, 2023 at 17:29
  • $\begingroup$ Going the other way LaplaceTransform[(Exp[-t] - 1)/(2 t), t, s] // Simplify produces 1/2 Log[s/(1 + s)] in Mathematica 13.3 so is this correct? It could be, as there are Logs if we TrigToExp of the arctan expression but I can't tell. $\endgroup$
    – flinty
    Commented Jun 30, 2023 at 18:42
  • 1
    $\begingroup$ @flinty you can tell it's incorrect by plugging numeric values and comparing against numerical InverseLaplaceTransform $\endgroup$
    – Yaroslav Bulatov
    Commented Jul 1, 2023 at 6:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.