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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.)

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    $\begingroup$ 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 '20 at 4:15
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    $\begingroup$ 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 '20 at 4:19
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    $\begingroup$ 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 '20 at 5:03
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    $\begingroup$ Does this answer your question? The $2\pi$ in the definition of the Fourier transform $\endgroup$ – Francois Ziegler Aug 5 '20 at 5:33
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    $\begingroup$ Specifically, from @FrancoisZiegler's reference, @‍WhatsUp's answer and the associated comments talk about "why $2\pi$?". $\endgroup$ – LSpice Aug 5 '20 at 11:47

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