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my question has two parts (please check everything for any mistakes):

The Fourier transform of

FourierTransform[Exp[-a Abs[t]], t, ω, 
 FourierParameters -> {1, -1}]=(2 a)/(a^2 + ω^2)

which is the correct answer.

1.How to get its Inverse Fourier Transform correctly?

InverseFourierTransform[2 a/((a^2) + (ω^2)), ω, t, 
 Assumptions -> a > 0 , FourierParameters -> {1, -1}]=E^(a t) HeavisideTheta[-t] + E^(-a t) HeavisideTheta[t]

is it correct?

  1. How to display the Inverse Fourier Transform without any HeavisideTheta?(or anything else which makes the answer different from the initial function) so the answer looks like the initial theoretical notation: Exp[-a Abs[t]].

Thanks

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$Version

"12.1.0 for Mac OS X x86 (64-bit) (March 14, 2020)"

Clear["Global`*"]

expr = Exp[-a Abs[t]];

ft = FourierTransform[expr, t, ω, FourierParameters -> {1, -1}]

(* (2 a)/(a^2 + ω^2) *)

ift[t_] = InverseFourierTransform[ft, ω, t, Assumptions -> a > 0,
  FourierParameters -> {1, -1}]

(* E^(a t) HeavisideTheta[-t] + E^(-a t) HeavisideTheta[t] *)

Since HeavisideTheta is undefined at zero, the ift is equal to expr everywhere except for t == 0.

Simplify[expr == ift[t], #] & /@ {t < 0, t == 0, t > 0}

(* {True, 2 HeavisideTheta[0] == 1, True} *)

This is due to the fact that expr is not smooth for t == 0

Plot[expr /. a -> 1, {t, -3, 3}]

enter image description here

EDIT: For FourierParameters -> {1, -1}] the InverseFourierTransform is given by

ift[t_] = 1/(2 Pi) Integrate[
   ft E^(I ω t), {ω, -Infinity, Infinity},
   Assumptions -> {a > 0, Element[t, Reals]}]

(* E^(-a Abs[t]) *)

ift[t] == expr

(* True *)
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    $\begingroup$ what should I add to the InverseFourierTransform to not calculate for t=0? $\endgroup$ – Haley May 11 '20 at 15:59
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    $\begingroup$ InverseFourierTransform does not calculate the inverse for t == 0 since the inverse is not defined there. If you want to define ift at zero, use Limit, i.e., ift[0] = Limit[ift[t], t -> 0] $\endgroup$ – Bob Hanlon May 11 '20 at 17:02
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    $\begingroup$ Yeah, I got that part, but please just mention this how can I recreate the exact initial function Exp[-a Abs[t]] ? $\endgroup$ – Haley May 11 '20 at 17:09

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