# Fourier Transforms via magic

A while ago I found a series of papers which do some weird stuff with derivative operators:

- New Dirac Delta function based methods with applications to perturbative expansions in quantum field theory by Kempf/Jackson/Morales, 2014
- How to (Path-) Integrate by Differentiating also by Kempf/Jackson/Morales, 2015
- Integration by differentiation: new proofs, methods and examples by Jia/Tang/Kempf, 2016

The general theme is: evaluating functions on derivative operators \(f(\p)\), and applying this to delta functions \(f(\p_x) \delta(x)\), is occasionally useful and can give weird alternate characterizations of the Fourier transform and can be used to efficiently solve integrals.

The authors are physicists, unsurprisingly, and I’m sure there are a bunch of reasons why these results are either not that surprising or surprising-yet-not-useful, but I found them remarkable. But the whole thing is confusing and hard to make sense of. Here’s a… totally different take, in which I rederive the main result by poking around.

tldr: the Fourier transform of \(f(x, \p_x)\) is \(f(i \p_k, -ik) 2 \pi \delta(k)\), whatever that means.

## 1.

First let’s fix a Fourier transform convention:

\[\hat{f}(k) = \F{f(x)} = \hat{f}(k) = \int f(x) e^{- ik x} dx\] \[f(x) = \mathcal{F}^{-1}[\hat{f}(k)] = \frac{1}{2 \pi} \int \hat{f}(k) e^{ ik x} dk\]I prefer not to use the ones that put \(2 \pi\) in the exponent because it’s distracting.

Here are a few common Fourier transform formulas in this convention, for reference:

\[\begin{aligned} \F{1} &= 2 \pi \delta(k)\\ \F{\delta(x)} &= 1 \\ \F{\p_x^n f(x)} &= (ik)^n \hat{f}(k) \\ \F{x^n f(x)} &= (i \p_k)^n \hat{f}(k) \end{aligned}\]## 2

It is common to take Fourier transforms of operators acting on functions, like \(\F{\p_x f(x)} = ik \hat{f}(k)\), in order to solve differential equations. This can be computed using integration by parts inside the transform:

\[\F{\p_x f(x)} = \int \p_x f(x) e^{-ikx} dx = - \int f(x) \p_x e^{-ikx} dx = (ik) \F{f(x)}\]It seems plausible to use the same argument to Fourier-transform a “freestanding” derivative operator, like \(\p_x\):

\[\F{\p_x} = \int \p_x e^{-ikx} dx = (-ik) \F{1} = (-ik) 2 \pi \delta(k)\]I find this compelling, because it seems to work. Note that the minus sign is related to integration by parts. We might rewrite \(\p_x f(x)\) as an operator \(- f(x) \p_x\). These are different in general, but under an integral where the boundary vanishes they are the same, which is an assumption we’ll make liberally.

We can also find \(\F{x}\) by rewriting it as a derivative in \(k\):

\[\F{x} = \int x e^{-ikx} dx = \int (i \p_k) e^{-ikx} dx = (i \p_k) \F{1} = 2 \pi i \delta'(k)\]Armed with these, we may transform any function in \(\p_x\) or \(x\) which has a Taylor series:

\[\begin{aligned} \F{f(\p_x)} &= \int f(\p_x) e^{-ikx} dx = \int f(-ik) e^{-ikx} dx = f(-ik) (2 \pi \delta(k)) \\ \F{f(x)} &= \int f(x) e^{-ikx} dx = \int f(i \p_k) e^{-ikx} dx = f(i \p_k) (2 \pi \delta(k)) \end{aligned}\]Or even both at once, as long as we are careful with what all of the operators act on:

\[\F{f(x, \p_x)} = f(i \p_k, -ik) 2 \pi \delta(k)\]In this expression, the \(\p_x\) and \(\p_k\)s are acting to the right, *not* on internal members of the expression. If they act on an internal member they pick up a minus sign, like we saw above:

All of this mostly seems to work if we allow negative powers of \(x\) and \(\p_x\) also, but there is some funny business around integration bounds.

\[\begin{aligned} \F{1/x} &= \int \frac{1}{x} e^{-ikx} dx \\ &= \int (i \p_k)^{-1} e^{-ikx} dx \\ &= (i \p_k)^{-1} \int e^{-ikx} dx \\ &= (i \p_k)^{-1} 2 \pi \delta(k) \\ &= -2 \pi i (\theta(k) + c) \end{aligned}\]What value of \(c\) should be used? To get the same value as Wikipedia’s table of Fourier transforms it should be \(c = \frac{1}{2}\). This makes \(\theta(k) + \frac{1}{2}\) an odd function, which means it is a choice that \(\frac{1}{x} \vert_{x=0} = 0\). This seems somewhat arbitrary, and I suspect that one could get away with just not choosing at all. If we do use \(c = \frac{1}{2}\), we get:

\[\F{1/x} = -2 \pi i (\theta(k) + \frac{1}{2}) = - i \pi \sgn(k)\]The other direction is simpler:

\[\F{\p_x^{-1}} = \int \p_x^{-1} e^{-ikx} dx = \frac{1}{ik} \F{1} = \frac{1}{ik}2 \pi \delta(k)\]In summary we have a rough hand-waving method for – well, maybe not for rigorously deriving Fourier transforms, but at least for guessing at them, for derivative operators can be written as Laurent series (Taylor series with negative powers). Just swap \(f(x, \p_x) \ra f(i \p_k, - ik)\).

In a sense this is a quarter rotation in the \((x, \p_x)\) plane, followed by multiplying by \(i\) and relabeling \(x \ra k\). That is:

\[\begin{pmatrix} k \\ \p_k \end{pmatrix} = i R \begin{pmatrix} x \\ \p_x \end{pmatrix}\]I don’t now what it means to rotate in the \((x, \p_x)\) plane, but it turns out that you can do other linear transformations in this plane as well – fractional rotations, or arbitrary matrices. Incidentally, the Laplace transform is \((t, \p_t) \ra (-\p_s, -s)\), although the two-sided transform is better behaved than the more common one-sided transform. The one-sided version produces a bunch of integration bounds \(f(0)\) and such in the process, which is useful because it’s used for signals that turn ‘on’ at \(t=0\), but not too helpful for understanding as a rotation.

## 3. An integration technique

These are the main results of the papers mentioned above, I guess because papers have to justify their existence.

Recall that the integral of a function over the real line is equal to its Fourier transform evaluated at \(0\):

\[\int_{-\infty}^{\infty} g(x) dx = \hat{g}(0)\]Using our form of \(\hat{g}\) this is:

\[\hat{g}(0) = 2 \pi g(- i \p_k) \delta(k) \vert_{k = 0}\]This is readily computable:

\[\begin{aligned} \int_{-\infty}^{\infty} \frac{\sin x}{x} dx &= 2 \pi \frac{e^{-\p_x} - e^{\p_x}}{2 i} \frac{1}{i \p_x} \delta(x) \vert_{x =0} \\ &= \pi [e^{\p_x} - e^{- \p_x}] \theta(x) \vert_{x = 0} \\ &= \pi [ \theta(x + 1) + c - \theta(x-1) - c]_{x = 0} \\ &= \pi [ \theta(x + 1) - \theta(x-1) ]_{x = 0} \\ &= \pi \end{aligned}\]This is so much easier and more *algebraic* than doing a limit and taking the principal value or whatever you normally have to do. As a bonus you get to see the Fourier transform as an intermediate step (it’s \(\pi [ \theta(x + 1) - \theta(x-1)]\))

There are also versions for integrating over finite intervals, doing Laplace transforms, and a bunch of other stuff. I’ll probably write more about them later. There are more tricks – solving bounded integrals, for instance, amounts to evaluating \(\int (\theta(b) - \theta(a)) f(x) dx\), and using the fact that we know the Fourier transform of \(\theta(x)\). Although it is messy: \(\sgn(x)\) has a clean transform, \(\F{\sgn(x)} = \frac{1}{ik}\). Then you solve for \(\theta(x) = \frac{1}{2}(\sgn(x) - 1)\).

Anyway, I wanted to write this up so I don’t forget about it or how to understand it. Hope it’s useful to somebody else.