Are the coefficients on the Fourier transform arbitrary?

Question

I was just wondering the other day about a convention I'd always taken for granted. I've seen the Fourier transform written a lot of ways. The first way (which, for reference, I'll call Scheme 1) is: $$ \begin{align} f \left ( x \right ) & = \int_{-\infty}^{\infty} e^{i k x} \tilde{f} \left ( k \right ) \; dk \\ \tilde{f} \left ( k \right ) & = \frac{1}{2 \pi} \int_{-\infty}^{\infty} e^{-i k x} f \left ( x \right ) \; dx \end{align} $$ Where $ f \left ( x \right ) $ is our function and $ \tilde{f} \left ( x \right ) $ is the Fourier transform of our function. I've also seen what I'll call Scheme 2: $$ \begin{align} f \left ( x \right ) & = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} e^{i k x} \tilde{f} \left ( k \right ) \; dk \\ \tilde{f} \left ( k \right ) & = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} e^{-i k x} f \left ( x \right ) \; dx \end{align} $$ Thirdly, I've encountered Scheme 3: $$ \begin{align} f \left ( x \right ) & = \frac{1}{2 \pi} \int_{-\infty}^{\infty} e^{i k x} \tilde{f} \left ( k \right ) \; dk \\ \tilde{f} \left ( k \right ) & = \int_{-\infty}^{\infty} e^{-i k x} f \left ( x \right ) \; dx \end{align} $$ and Scheme 4: $$ \begin{align} f \left ( x \right ) & = \int_{-\infty}^{\infty} e^{2 \pi i k x} \tilde{f} \left ( k \right ) \; dk \\ \tilde{f} \left ( k \right ) & = \int_{-\infty}^{\infty} e^{-2 \pi i k x} f \left ( x \right ) \; dx \end{align} $$ However, no matter how you put it, I've never seen a form without a $ 2 \pi $ popping up somewhere. I was wondering why. As an attempt to satisfy my question, I set out to show that $ 2 \pi $s pop up naturally out of the property that the Fourier transform of the inverse Fourier transform of a function gives back the original function. I'm going to use either Scheme 1, 2, or 3, but assume that the exponent does not absorb the factor as in Scheme 4. That's just a u-substitution. Let's call the coefficient on the $ f \left ( x \right ) $ formula $ \alpha $ and the coefficient on the $ \tilde{f} \left ( k \right ) $ formula $ \beta $. I wish to show that $ \alpha \beta = \frac{1}{2 \pi} $. Here's my algebra: $$ \begin{align} f \left ( 0 \right ) & = \alpha \int_{-\infty}^{\infty} e^{i k 0} \tilde{f} \left ( k \right ) \; dk \\ & = \alpha \int_{-\infty}^{\infty} \tilde{f} \left ( k \right ) \; dk \\ & = \alpha \int_{-\infty}^{\infty} \beta \int_{-\infty}^{\infty} e^{-i k x} f \left ( x \right ) \; dx \; dk \\ & = \alpha \beta \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-i k x} f \left ( x \right ) \; dk \; dx \\ & = \alpha \beta \int_{-\infty}^{\infty} f \left ( x \right ) \int_{-\infty}^{\infty} e^{-i k x} \; dk \; dx \\ & = \alpha \beta \int_{-\infty}^{\infty} f \left ( x \right ) \gamma \delta \left ( x \right ) \; dx \\ & = \alpha \beta \gamma f \left ( 0 \right ) \end{align} $$ Where $ \int_{-\infty}^{\infty} e^{-i x y} \; dy = \gamma \delta \left ( x \right ) $ for a constant $ \gamma $ I'm about to address in a moment. If $ f \left ( 0 \right ) = 0 $ we get no condition, but if $ f \left ( 0 \right ) \neq 0 $ we get that $ \alpha \beta \gamma = 1 $, or $ \alpha \beta = \frac{1}{\gamma} $. Indeed, that integral evaluates to $ 2 \pi $. $ \gamma = 2 \pi $! Yay! That's what we wanted to show!
Well, when you look up a proof of this fact, you run into a brick wall. Proof by Fourier Theorem. So, that fact was using the $ \frac{1}{2 \pi} $ already in the formula. Is that number just arbitrary, carried over from discrete Fourier series for convenience? Or is there some more fundamental result that introduces the $ 2 \pi $s into the equations?
Just curious. Thanks!

Answer 1

The normalization comes from the value of $$ \int_{-\infty}^{\infty}\frac{\sin(Rx)}{x}dx = \pi,\;\;\; R > 0. $$ For example, suppose $f$ is differentiable at $x$ and integrable on $\mathbb{R}$. Then \begin{align} & \int_{-R}^{R}e^{isx}\int_{-\infty}^{\infty}f(y)e^{-isy}dyds \\ & = \int_{-\infty}^{\infty}\int_{-R}^{R}e^{is(x-y)}dsf(y)dy \\ & = \int_{-\infty}^{\infty}2\frac{\sin(R(x-y))}{x-y}f(y)dy \\ & = 2\int_{-\infty}^{\infty}\frac{\sin(Ry)}{y}f(y+x)dy \\ & = 2\int_{-\infty}^{\infty}\frac{\sin(Ry)}{y}\{f(y+x)-f(x)\}dy +2\pi f(x)\\ & = 2\int_{\infty}^{\infty}\sin(Ry)\left[\frac{f(y+x)-f(x)}{y}\right]dy+2\pi f(x) \\ & \rightarrow 2\pi f(x) \mbox{ as $R\rightarrow\infty$ }. \end{align} To Change variables, let $s = 2\pi v$. Then $ds=2\pi dv$ and \begin{align} f(x) & = \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{isx}\int_{-\infty}^{\infty}e^{-isy}f(y)dy ds\\ f(x) & = \int_{-\infty}^{\infty}e^{i2\pi vx}\int_{-\infty}^{\infty}e^{-i2\pi vy}f(y)dy dv \end{align}

Answer 2

It comes up in the math like you showed, but it could be interpreted as a scaling constant to acknowledge that taking the Fourier transform changes the domain of a function to a signals domain. Since signals are measured in sinusoidal waves with period $2\pi$.

Answer 3

The whole Fourier inversion theorem is an elaboration of the computation of the integral $\int_{-\infty}^\infty e^{-x/2}\,\mathrm dx$, which is $\sqrt{2\pi}$, and that is where the constant comes out from.

It is a fact of nature, and cannot be gotten rid of. Some people like to redefine Lebesgue's measure on the line to incorporate the factor $1/\sqrt{2\pi}$ appearing into your scheme 2 into the measure itself, and then you get constant-free formulas —this is, of course, just cheating. Rudin does this in his book on functional analysis, for example.