I have been thinking about the $2\pi$ factor in the various conventions of the Fourier transform. For example, I was looking for a way to justify the following:

$(*)$ If we define $\hat f(\xi) = \int f(x) e^{-ix\xi} \, dx$, then the following equation is false: $f(x) = \int \hat f(\xi) e^{ix\xi} \, d\xi$

Obviously, one way to prove $(*)$ is to derive the correct form of the Fourier inversion formula, but I wanted a heuristic argument to convince myself quickly that $(*)$ is true. Here is what I came up with:

  1. Suppose $x$ has units of time, and $f(x)$ is unitless. In the expression $e^{i\theta}$, the variable $\theta$ should have units of radians. Hence $\xi$ must have units of radians/time.
  2. From $\hat f(\xi) = \int f(x) e^{-ix\xi} \, dx$, it follows that $\hat f(\xi)$ must have units of time (coming from the $dx$).
  3. Combining 1 and 2, we can conclude $\int \hat f(\xi) e^{ix\xi} \, d\xi$ has units of time*radians/time = radians, which does not match the units of $f(x)$. Thus $f(x) = \int \hat f(\xi) e^{ix\xi} \, d\xi$ must be false, since the units on both sides do not agree. This "proves" $(*)$.

(On the other hand, if we define $\hat f(\xi) = \int f(x) e^{-2\pi ix\xi} \, dx$, then we can think of the $2\pi$ as having units of radians, so $\xi$ now has units of 1/time. As a result, there are no unit agreement issues with $f(x) = \int \hat f(\xi) e^{2\pi ix\xi} \, d\xi$.)

I have two questions:

  1. Is there a way to make the above argument more precise? I would like to think of it as a dimensional analysis argument, but radians are dimensionless.
  2. Is there a way to give a heuristic argument to see that the following is true?

$(**)$ Let $L > 0$. Suppose we define $\hat f(\xi) = \int f(x) e^{-Lix\xi} \, dx$, and suppose that $f(x) = \int \hat f(\xi) e^{Lix\xi} \, d\xi$ holds. Then $L$ must be $2\pi$.

(I asked this question on math.SE, but the responses there did not address my question.)

  • 1
    $\begingroup$ Another way to ask it: we incorrectly think that $[0, 1]$ has measure $1$ on both the time side and the spectral side. One or the other must have measure $1/(2\pi)$, or both might have measure $1/\sqrt{2\pi}$. Heuristically, why? $\endgroup$
    – LSpice
    Aug 5, 2020 at 4:15
  • 1
    $\begingroup$ And, of course, any approach to this has to answer why $x \mapsto e^{-x^2/(2\pi)}$ is an eigenfunction (or maybe I've got my factors off; as you say, this is why one wants an argument like this, so as not to have to check every time!). To me that 'feels' more like the place for a heuristic argument than Fourier inversion. $\endgroup$
    – LSpice
    Aug 5, 2020 at 4:19
  • 4
    $\begingroup$ The definition of what counts as an eigenfunction of the Fourier transform is tied up with how you define it. Letting $\hat{f}(y) = \int_{\mathbf R} f(x)e^{-2\pi ixy}\,dx$, $e^{-\pi x^2}$ is its own Fourier transform, but this is not true when the Fourier transform uses $e^{-ixy}$ in the integral. $\endgroup$
    – KConrad
    Aug 5, 2020 at 5:03
  • 7
    $\begingroup$ Does this answer your question? The $2\pi$ in the definition of the Fourier transform $\endgroup$ Aug 5, 2020 at 5:33
  • 1
    $\begingroup$ Specifically, from @FrancoisZiegler's reference, @‍WhatsUp's answer and the associated comments talk about "why $2\pi$?". $\endgroup$
    – LSpice
    Aug 5, 2020 at 11:47


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Browse other questions tagged or ask your own question.