# The $2\pi$ factor in the Fourier transform and dimensional analysis

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.)

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