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I am trying to write a function that generalizes the Laplace and Fourier transforms, but I'm having trouble getting Integrate to play along. The function is called FLTransform in the following snippet, where FTransform and LTransform are the equivalent Fourier and Laplace transforms, respectively:

FTransform[expr_, t_, \[Omega]_] := FourierTransform[expr, t, \[Omega], FourierParameters -> {1, -1}];
LTransform[expr_, t_, s_] := BilateralLaplaceTransform[expr, t, s];
FLTransform[expr_, t_, s_] := Integrate[expr Exp[-s t], {t, -\[Infinity], \[Infinity]}];

First, I tried Cos[t]. The Fourier transform is defined for this function, but the Laplace transform is not (well, it is only defined for Re[s] == 0):

f[t] := Cos[t];
FTransform[f[t], t, \[Omega]]
LTransform[f[t], t, s]
FLTransform[f[t], t, s]

FTransform and LTransform yield the following outputs:

\[Pi] DiracDelta[-1 + \[Omega]] + \[Pi] DiracDelta[1 + \[Omega]]
BilateralLaplaceTransform[Cos[t], t, s]

However, FLTransform fails with the message Integrate::idiv, as the integral does not technically converge (thus the DiracDelta). Is it possible to rectify this somehow?

QuaternionsRock
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1 Answers1

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This is not the realy answer, just the beginning...

ClearAll["`*"]; Remove["`*"]

FTransform[expr_, t_, s_] := FourierTransform[expr, t, s, FourierParameters -> {1, -1}];
LTransform[expr_, t_, s_] := (BilateralLaplaceTransform[expr, t, s] /. s -> Is);
LDTransform[expr_, t_, s_] := (LaplaceTransform[expr /. t -> -t, t, s] /. 
 s -> Is) + (LaplaceTransform[expr /. t -> t, t, s] /. s -> -Is);
FLTransform[expr_, t_, s_] := Integrate[exprExp[I s t], {t, -[Infinity], [Infinity]}];
MTransform[expr_, t_, s_] := (MellinTransform[expr /. t -> Log[t], t, s] /. s -> I*s);


f[t] := 1/(1 + t^2);(* For: Cos[t] some methods don't work*)


FTransform[f[t], t, s]
LTransform[f[t], t, s]
LDTransform[f[t], t, s]
FLTransform[f[t], t, s]
MTransform[f[t], t, s]
Mariusz Iwaniuk
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  • Sorry, perhaps I should clarify - I would like to define FLTransform in terms of Integral such that it does work for the input Cos[t]. – QuaternionsRock Apr 10 '23 at 20:29
  • @QuaternionsRock it is better to clarify this also in the OP by editing the question and adding this clarification to the question itself. Comments can be easily overlooked. – CA Trevillian Apr 11 '23 at 04:02