I would like to calculate the FBI transform of a set of Fourier series functions.

The formula is:

enter image description here

and here, $a$ and $n$ are integers, where $n=2$ since I have a 2-dimensional function to consider and $a\ne0$. However there are three variables, $t,x,y$, and indeed the domain would determine if this integral converges.

I cannot find any data on the site on how to calculate this nor any good examples of that. Does anyone have an idea how that can be done on a function as:

  f[x] == (1 + I) E^(5 I)

Integration by Mathematica on $[a,b]=[0.1,5]$ should be:

Integrate[(1/(2 Pi)) (1 + I) (E^(5 I)*(E^(-a (Abs[x - y]^2/2)) E^(I x t))), {x, 0.1, 5}]

But how can one vary $a$ and evaluate the output as $a\longrightarrow \infty$? This would be a certain simulation of the change of the integral according to the change of a.


  • 1
    $\begingroup$ There is information missing from this post. (1) It seems x is supposed to be a vector. Is that the case? (2) your f(x) is not defined using x. (3)) Any inputs should be in copy-pastable format, and desired results should be included in the post. (4) Where it is stated that n=2 I assume what was meant is n is even, Please edit. $\endgroup$ Commented Mar 26, 2023 at 20:23
  • $\begingroup$ See update..... $\endgroup$ Commented Mar 27, 2023 at 8:39

1 Answer 1


Maybe this helps:

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

n = 2;
f[x_] := (1 + I)^(5 I)
F[a_, y_, t_] := (2 Pi)^(-n/2)*FourierTransform[f[x]*Exp[-a*Abs[x - y]^2/2], x, t, FourierParameters -> {1, -1}, 
Assumptions -> {a >= 0, y \[Element] Reals}]

F[0, y, t](*If a -> 0*)
(*(1 + I)^(5 I) DiracDelta[t]*) 

G = F[a, y, t](*General formula*)
(*((1 + I)^(5 I) E^(-((t (t + 2 I a y))/(2 a))))/(Sqrt[a] Sqrt[2 \[Pi]])*)

Limit[G, a -> Infinity, Assumptions -> {t \[Element] Reals, y \[Element] Reals}] (*If a -> Infinity*)
(* 0 *)

Table[Plot3D[G // Re, {t, -1, 1}, {y, -2, 2}], {a, 10^-6, 5, 1/2}] // Quiet (* a = [0, 5] ,Reals Part*)
Table[Plot3D[G // Im, {t, -1, 1}, {y, -2, 2}], {a, 10^-6, 5, 1/2}] // Quiet (* a = [0, 5] ,Imaginary Part*)
  • $\begingroup$ Thanks Mariusz, this looks perfect. I will test this on various types of functions. $\endgroup$ Commented Mar 27, 2023 at 8:42
  • $\begingroup$ I found an misinterpretation in the OP that gave an error in the code. $a$ should indeed be studied from $0, ...,\infty$, but that $a$ is not the same as $a$ in $[a,b]$. My fault, $a$ in the exponent should be $\alpha$. $\endgroup$ Commented Mar 27, 2023 at 18:15

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