I wasn't sure whether to post this under physics or math (and landed on physics due to fear of being crucified for lack of rigor on math.stackexchange).

In field theory, when we encounter divergences that arise when we consider infinitely small spacings between momenta, we often use a regularization $$ \delta^{(N)}(\mathbf{0}) \propto \text{Volume of an $N$-dimensional Box} $$ where $\delta^{(N)}(\mathbf{x}) = \int \frac{d^N \mathbf{x}}{(2\pi)^N} e^{i \mathbf{p} \cdot \mathbf{x}}$ is the $N$-dimensional Dirac delta function. The idea in using this regularization is to give the walls of the box a length $L$ and write $$ \delta^{(N)}(\mathbf{0}) = \int_{[-L/2,L/2]^{N}} \frac{d^N \mathbf{x}}{(2\pi)^N} = \frac{L^N}{(2\pi)^N} \ . $$ Note that this is of course divergent when one takes $L \to \infty$.

Now I'm doing a calculation in which I encounter the object $\delta'(0)$ (in one-dimension). In the context of the calculation it seems to make sense to interpret $\delta'(0) \to 0$, and I was wondering if this makes sense in any way? We know that $$ \delta'(0) = \int_{-\infty}^{\infty} \; \frac{dp}{2\pi} i p \ , $$ which is implied by differentiating $\delta(x) = \int_{-\infty}^{\infty} \frac{dp}{2\pi} {e^{i p x}}$. The above makes it seem like $\delta'(0) = 0$ if you use the earlier regularization and write the integral as $\lim\limits_{L \to \infty} \int_{-L/2}^{L/2} \frac{dp}{2\pi} i p = 0$ (since $p$ is odd when reflected about the origin).

Another way in which this seems to make sense is noting that the Dirac delta function can be written as $\delta(x) = \int \frac{dp}{2\pi} \cos(p x)$ which implies $\delta'(x) = - \int \frac{dp}{2\pi} p \sin(p x)$, and so at $x=0$ it again seems like $\delta'(0) = 0$.

Does this make sense? Definitely not formally in the sense of distributions...

  • $\begingroup$ Related: physics.stackexchange.com/q/161275/2451 , physics.stackexchange.com/q/419668/2451 and links therein. $\endgroup$
    – Qmechanic
    Jan 13, 2021 at 17:30
  • 1
    $\begingroup$ Note that your argument only works with a symmetric interval $[-L/,L/2]$. $\endgroup$
    – fqq
    Jan 13, 2021 at 17:40
  • $\begingroup$ Recall this is the midpoint of the odd operator, $-\delta(x)\partial_x$, so your 0 is multiplied by $\infty$! It certainly depends on the symmetry properties of its argument, as confirmed by @Charles Francis. $\endgroup$ Jan 19, 2021 at 1:32

2 Answers 2


No. Integrate by parts, with $a<0<b$

$$ \int_a^b f(x)\delta'(x) dx = [f(x)\delta(x)]_a^b - \int_a^b f'(x)\delta(x) dx = -f'(0)$$

(this result can be done more rigorously).

  • $\begingroup$ Nitpick: there's a minus sign missing in the last equaity. $\endgroup$
    – jacob1729
    Jan 13, 2021 at 17:31
  • $\begingroup$ Oops.. I even remembered that I had forgotten the minus sign before posting, and still forgot to insert it when I posted! $\endgroup$ Jan 13, 2021 at 17:34

Fun question. Let me add a legal disclaimer first by saying that of course there is a rigorous way to deal with the derivative of the delta function and I mention in it this answer. But I agree that it's fun to see how much you can get away with not being rigorous. And that, in my reading, is the spirit of your question, so that's how I'll answer.

I'll start by saying how I usually think of the derivative of the delta function non-rigorously in the $x$ basis. There the delta function is zero everywhere but "infinity" at the origin. Imagine taking the derivative. A hair to the left of the origin, the function increases "infinitely fast." A hair to the right, again "infinitely fast" but negative. So when I integrate this against a function, it picks up a point a hair to the left of the origin with positive weight, and a hair to the right with negative weight. In other words, this is like the negative of the derivative of the function. And if you treat things rigorously, you indeed find that the derivative of the delta function is a distribution that gives you minus the derivative of the function it acts on, as someone will tell you in a less fun answer.

Now let's think about it your way. You wrote $$\delta'(x)=\int \frac{dp}{2\pi}\,ipe^{ipx}.$$ Sure, at $x=0$ this is zero. My picture in the above paragraph is also zero at the origin, because it's symmetric, and it jumps from "$+\infty$" just left of the origin to "$-\infty$" just right. Now let's look at your expression just left of the origin $x=-\epsilon$. $$\int \frac{dp}{2\pi}\,ipe^{-ip\epsilon}= \int \frac{dp}{2\pi}\,ip(1-ip\epsilon)$$ The first term is odd and integrates to zero. The second term is the integral of a quadratic over the real line so it's "infinity." Do the same thing for $x=+\epsilon$ and you get "minus infinity." So it all works out the same way as I mentioned above.


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