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 more heuristic way to easily convince myself 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 $2\pi$.

 A: One of the most interesting and oldest observations concerning a self-adjoint operator $A : \mathcal{D}(A)\subseteq \mathcal{H}\rightarrow\mathcal{H}$ on a complex Hilbert space $\mathcal{H}$ is that $R(\lambda)=(\lambda I-A)^{-1}$ has a residue at $\infty$ equal to $I$ because
$$
               \lambda (\lambda I-A)^{-1}=I+(\lambda I-A)^{-1}A\rightarrow I
$$
as $\lambda\rightarrow \infty$ along any $\lambda =re^{i\theta}$ for fixed $0 < \theta < \pi$ or $\pi < \theta < 2\pi$ as $r\uparrow\infty$. Because of this, the following strong (i.e. vector) limit exists:
$$
                  I = \mbox{s-}\lim_{\epsilon\downarrow 0}\frac{1}{2\pi}\int_{-\infty}^{\infty}((u+i\epsilon)I-A)^{-1}-((u-i\epsilon)I-A)^{-1} du
$$
This equates the residue at $\infty$ to the integral around the residues on the real axis. The $2\pi$ comes from integrating over a circle. This is generally true for all self-adjoint operators on a Hilbert space.
For $A=\frac{1}{i}\frac{d}{dx}$, the above gives
$$
                      f =\frac{1}{2\pi}\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}f(y)e^{-iuy}dy\right)e^{iux}dx.
$$
The same type of argument applies to the Fourier series as well, except that the singularities of the resolvent on $[0,2\pi]$ are discrete poles of order $1$, which gives rise to a discrete outer sum instead of an integral
$$
                      f = \frac{1}{2\pi}\sum_{n=-\infty}^{\infty}\left(\int_{0}^{2\pi}f(y)e^{-iny}dy\right) e^{inx}
$$
The factor of $2\pi$ relates an integral over expanding circles to a difference of integrals over an interval of the real axis. This factor of $2\pi$ is present for all self-adjoint operators: it relates the residue at $\infty$ (which is always $I$) to an integral over the real axis.
This observation is at the core of earliest derivations of Spectral Theory. This question goes to the heart of Spectral Theory, and is an important part of how the subject evolved. It's the right question to be asking from an Historical perspective.
A: $$2\pi = \int_{-\infty}^\infty 2\frac{\sin(x)}{x}dx$$ it doesn't come from anywhere else.

$f \ast \frac1{2\pi}2\frac{\sin(ax)}{x}$ is the inverse Fourier transform of $\hat{f}1_{|t|<a}$ where $2\frac{\sin(ax)}{x}$ is the Fourier transform of $1_{|t|<a}$. The Fourier inversion theorem is $$f = \lim_{a\to\infty}f \ast \frac1{2\pi}2\frac{\sin(ax)}{x}$$ which is mostly easy (integration by parts) when $f,f'\in  L^1$.

