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:

- 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.
- 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 $dx$).
- 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:

- 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.
- 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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