Similar to this question on Math StackExchange, I'd like to show that the Fourier transform of a Gaussian is the same function. Specifically, I'd like to use Mathematica's FourierTransform to take the Fourier transform of $f(t) = \exp(-\frac{1}{2}t^2)$ is the same function.

In Mathematica, the Fourier transform computed by FourierTransform is $$\sqrt{\frac{|b|}{(2\pi)^{1-a}}} \int_{-\infty}^{\infty} f(t)\exp(ib\omega t).$$

One can use FourierParameters to specify values for $a$ and $b$. As the Mathematica documentation explains, common choices for $(a,b)$ are $(0,1)$ (default; modern physics), $(1,-1)$ (pure math; systems engineering), $(-1,1)$ (classical physics), and $(0,-2\pi)$ (signal processing).

Meanwhile, this question on Math StackExchange appears to use the following definition of the Fourier transform of a function $f(t)$: $$\hat{f}(\xi) = \int_{\infty}^{\infty}f(t)\exp(-2\pi it\xi) dt.$$

What values of $(a,b)$ are being used in the previous equation?

Clearly, the equation uses $\xi$ in place of $\omega$, and it looks like $b = -2\pi$ (as seen in the exponential). So, what value of $a$ is being used?

It appears that the square root factor in the Mathematica generalized form is 1 in the new equation, so it must be that $$\sqrt{\frac{|b|}{(2\pi)^{1-a}}} = 1.$$ Solving for $a$ (with $b = -2\pi$), I think we obtain $a = 0$.

Thus, $(a,b)=(0,-2\pi)$, and this is the common choice for signal processing. Now let's use FourierTransform with FourierParameters -> {0, -2*Pi} to compute the Fourier Transform of $f(t)$:

FourierTransform[Exp[-(1/2)*r^2], r, q, FourierParameters -> {0, -2*Pi}]

which gives the output:


This is equivalent to $$\sqrt{2\pi}\exp(-2\pi^{2}q^{2}) = \sqrt{2\pi}\exp(-\pi(\sqrt{2\pi}q)^2).$$ That last expression, however, does NOT appear to be equivalent to the Fourier transform of: $$\exp(-\pi(\frac{t}{\sqrt{2\pi}})^2)$$

which is inconsistent with the result stated in this answer on Math StackExchange.

Am I incorrect that $(a,b) = (0,-2\pi)$ is implied in the question on Math StackExchange? Where am I going wrong?

  • 1
    $\begingroup$ I am not sure I understand where the problem is. FourierTransform[Exp[-(1/2)*r^2], r, q, FourierParameters -> {0, -2*Pi}] and FourierTransform[Exp[-Pi (r/Sqrt[2 Pi])^2], r, q, FourierParameters -> {0, -2*Pi}] give the same result ... Perhaps you forgot to change t to r? $\endgroup$
    – Domen
    May 16, 2023 at 20:57
  • $\begingroup$ f[t] == FourierTransform[f[t], t, w] /. w -> t evaluates to True so you want the default parameters. Then Options[FourierTransform, FourierParameters] evaluates to {FourierParameters -> {0, 1}} $\endgroup$
    – Bob Hanlon
    May 16, 2023 at 21:02
  • $\begingroup$ @BobHanlon If they are using the default parameters in math.stackexchange.com/a/2799976/34691 , then why does FourierTransform[f[t], t, w] == Sqrt[2*Pi]*Exp[-2*(Pi^2)*w^2] not evaluate to True? $\endgroup$
    – Andrew
    May 16, 2023 at 21:18
  • $\begingroup$ The question asked for the “same function”. What I showed is the “same function” not just the same general form. $\endgroup$
    – Bob Hanlon
    May 16, 2023 at 21:47

1 Answer 1


This is true if you use the default parameters: {0,1}

FourierTransform[Exp[-t^2/2], t, \[Omega]]
(* Out[]= E^(-(\[Omega]^2/2)) *)

% /. \[Omega] -> t
(* Out[]= E^(-(t^2/2)) *)

The definition you posted will NOT give equality:

FourierTransform[Exp[-t^2/2], t, \[Omega], 
  FourierParameters -> {0, -2*Pi}]
(* Out[]= E^(-2 \[Pi]^2 \[Omega]^2) Sqrt[2 \[Pi]] *)

% /. \[Omega] -> t
(* Out[]= E^(-2 \[Pi]^2 t^2) Sqrt[2 \[Pi]] *)

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