Any tips how to massage the following to get computed by Mathematica for $p>1$? I suspect the result should be expressible in terms of exponential integral

InverseLaplaceTransform[Hypergeometric2F1[1, 1/p, 1 + 1/p, s], s, t]

The following equality was obtained using "guess-and-check"

LaplaceTransform[ -ExpIntegralE[1/p, -x]/p, x, s] == 
 Hypergeometric2F1[1, 1/p, 1 + 1/p, s]  (* True *)
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    $\begingroup$ Given that the code won't work even for a fixed value such as $p=2$, I think this cannot work in the general case. Perhaps try forming the explicit transform integral. Might work. $\endgroup$ May 18 at 2:05
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    $\begingroup$ @DavidG.Stork In this particular way there's a convoluted indirect way of getting from domain-specific knowledge (LaplaceTransform[(d^(1 - p) ExpIntegralE[1/p, -a d^-p t])/p, t, s]), but was curious about a general technique for this $\endgroup$ May 18 at 2:09
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    $\begingroup$ Check for the quotient to be a constant. Equality can fail for many reasons. $\endgroup$
    – Roland F
    May 18 at 13:48

3 Answers 3



MellinTransform[Hypergeometric2F1[1, 1/p, 1 + 1/p, s/a], a, q] // 
FunctionExpand, s, x], q, a] /. a -> 1]

(*-(ExpIntegralE[1/p, -x]/p) *)

TL;DR: Convert to a MeijerG function using MeijerGReduce and apply the general formula found at : https://functions.wolfram.com/HypergeometricFunctions/MeijerG/22/03/0001/ using the identities found at https://functions.wolfram.com/HypergeometricFunctions/MeijerG/16/01/01/

Hypergeometric2F1[1, 1/p, 1 + 1/p, s] can be written as a MeijerG function using MeijerGReduce. Then, with some extra work, one can apply the formula for the Laplace transform of a MeijerG function found at https://functions.wolfram.com/HypergeometricFunctions/MeijerG/22/03/0001/. That said, the notation kp+l for the number of arguments on the right-hand side does not seem coherent with the number of arguments shown. I will consider that a mistake and that it should be kp instead. It is unclear to me what some of the extra variables that appear in the condition are or what the $c^*$ is. I will assume they can be ignored in the case at hand where, as we will see, $l=k=\alpha=1$ in that formula.

First, the MeijerG representation of Hypergeometric2F1[1, 1/p, 1 + 1/p, s] :

Hypergeometric2F1[1, 1/p, 1 + 1/p, s] // 
  MeijerGReduce[#, s] & // FullSimplify

Inactive[MeijerG][{{0, (-1 + p)/p}, {}}, {{0}, {-(1/p)}}, -s]/p

The MeijerG function should look like the right-hand side from the link to apply the inverse Laplace transform. Using the identities found at https://functions.wolfram.com/HypergeometricFunctions/MeijerG/16/01/01/ one obtains an equivalent form:

(-1/s)*Inactive[MeijerG][{{0}, {1/p}}, {{0, 1/p - 1}, {}}, -(1/s)]/p


Inactive[MeijerG][{{0, (-1 + p)/p}, {}}, {{0}, {-(1/p)}}, -s]/
   p == (-1/s)*
    Inactive[MeijerG][{{0}, {1/p}}, {{0, 1/p - 1}, {}}, -(1/s)]/p // 
  Activate // FullSimplify

(* True *)

The general formula for the Laplace transform restricted to the case $l=k=\alpha=1$ shows that to obtain the inverse Laplace transform from (-1/s)*Inactive[MeijerG][{{0}, {1/p}}, {{0, 1/p - 1}, {}}, -(1/s)]/p, it is sufficient to :

  • Multiply the overall expression by $s^\alpha=s$

  • Change $s$ to $1/s$ in the argument of the MeijerG function

  • Drop the $1-\alpha=0$

The inverse Laplace transform is then:

-Inactive[MeijerG][{{}, {1/p}}, {{0, -1 + 1/p}, {}}, -x]/p // 
  Activate // FullSimplify 

(* -(ExpIntegralE[1/p, -x]/p) *)

I put together recipes provided by Mariusz Iwaniuk into standalone functions ilaplaceFeynmann and ilaplaceMellin. The Mellin transform approach seems to be the fastest

ilaplace[expr_] := InverseLaplaceTransform[expr, s, t];

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

(*Mariusz solution from \
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;

SetAttributes[ilaplaceMellin0, Listable];
ilaplaceMellin[expr_] := 
   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_] := 
   Distribute@dummy@Expand[expr] /. dummy -> ilaplaceFeynmann0];

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

results = 
  Outer[TimeConstrained[#1[#2], 60] &, {ilaplace, ilaplaceFeynmann, 
    ilaplaceMellin}, {expr1, expr2}];

 TableHeadings -> {{"default", "Feynmann", "Mellin"}, {expr1, expr2}}]

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