The well known scaling property for the Dirac $\delta$-function (distribution) states (see the Wikpedia entry)

$$\int_{-\infty}^\infty dx\:\delta(a x) =\frac{1}{|a|}$$

Now we can generally consider $\delta(x)$ as the limiting case of a probability density function

$$p(x,\epsilon)=\frac{1}{\epsilon \sqrt{\pi}} e^{-x^2/\epsilon^2}$$


$$\ \int_{-\infty}^\infty dx\:p(x,\epsilon) = 1$$


$$\delta(x)=\lim_{\epsilon\to 0}\:p(x,\epsilon)$$

Assume now we spread the events over a different $x$-range by doing the linear scale transformation $x\rightarrow ax$, so $p(x,\epsilon)$ transforms to

$$p(ax,\epsilon)=\frac{1}{\epsilon \sqrt{\pi}} e^{-a^2x^2/\epsilon^2}$$

However, this function is not normalized anymore, as

$$\ \int_{-\infty}^\infty dx\:p(a x,\epsilon) = \frac{1}{|a|}$$

Imagine $p(x)$ describes the way you spend a given amount of money over time, then here, by increasing the length of time ($|a|<1$), one would have miraculously gained money! Obviously, this is not possible. In order to qualify as a probability density function $p(a x)$ must be properly normalized (this is expressed in the 'Location-Scale Family' theorem of probability theory, see also this link) i.e.

$$p(ax,\epsilon)=\frac{|a|}{\epsilon \sqrt{\pi}} e^{-a^2x^2/\epsilon^2}$$

With this normalized function, the increased period of time the money is spent is compensated by the smaller amounts you have to spend, so you correctly end up with the same total spent, i.e.

$$\ \int_{-\infty}^\infty dx\:p(a x,\epsilon) = 1$$

However, if one generates the $\delta$-function from this, one finds

$$\int_{-\infty}^\infty dx\: \delta(a x)=\lim_{\epsilon\to 0}\: \int_{-\infty}^\infty dx\: p(a x,\epsilon) =1 $$

rather than $\int_{-\infty}^\infty dx\: \delta(a x)= \frac{1}{|a|}$ as stated throughout in the literature.

Can anybody see anything wrong with my argument?

EDIT: My argument above used the (by definition) normal property of the generating Gaussian distribution function if one generally interprets the Dirac $\delta$-function as the limit of a probability density function. In this sense it was an additional constraint that enables the use of a continuous mathematical function (in this case the Gaussian function) to approximate the actually discrete probability distribution. However, this can actually also be derived without resorting to probability theory by considering $\delta$ as a distribution that acts as a 'unit impulse' function when applied to a test function $f(x)$ i.e.

$$\int_{-\infty}^\infty dx\: \delta(x)f(x) = f(0) $$

Also with this interpretation, $\delta(x)$ can be interpreted as the limit of $\delta$-sequences, for instance

$$\delta_n(x)=\frac{n}{\sqrt{\pi}} e^{-n^2x^2}$$

with the notation as defined in the link. If we set $n=\frac{a}{\epsilon}$ this becomes instead

$$\delta_{a,\epsilon}(x)=\frac{a}{\epsilon\sqrt{\pi}} e^{-a^2x^2/\epsilon^2}$$

which means

$$\ \int_{-\infty}^\infty dx\:\delta_{a,\epsilon}(x) = 1 \neq \frac{1}{|a|}$$

  • $\begingroup$ densities are not functions, they are forms. $\rho(x)$ is meaningless, only $\rho(x)\mathrm dx$ has intrinsic meaning. $\endgroup$ Commented Apr 4, 2021 at 16:45
  • $\begingroup$ They are called probability density functions, in every book I have ever read about them. They have inputs and outputs. I'm not a philosopher, and probability is particularly slippery in that regard. But if you're going to go against standard terminology, you need to at least give some reason why you are doing so. $\endgroup$ Commented Apr 4, 2021 at 16:47
  • $\begingroup$ The point is that mathematical functions (like the Gaussian) are continuous, but probability has a discrete nature. Particle density functions in physics for instance only approximate the average discrete number of particles at a given point. In order for this to work, the continuous function (the probability density function) needs to be normalized. $\endgroup$
    – Thomas
    Commented Apr 4, 2021 at 16:57

2 Answers 2


You define $\delta(ax)$ to be a sequence of normalized functions. But that is not the definition of $\delta(ax)$.

The definition of $\delta$ is that (with some abuse of notation) $$\int_{-\infty}^{\infty}\delta(x)\,f(x)\,dx = f(0)$$ for all functions $f$ continuous at $x=0.$ (The real definition actually limit $f$ to a smaller set of functions, but this works for now.)

The definition of scaling the argument, i.e. $\delta(ax)$ (with $a>0$) is that a variable change should work: $$ \int_{-\infty}^{\infty}\delta(ax)\,f(x)\,dx = \{ \text{ set } y = ax \} = \int_{-\infty}^{\infty}\delta(y)\,f(y/a)\,dy/a = f(0/a)/a = f(0)/a \\ = \int_{-\infty}^{\infty}\frac{1}{a}\delta(x)\,f(x)\,dx. $$ Since this should be true for all $f$ (continuous at $x=0$) we have $$\delta(ax) = \frac{1}{a}\delta(x).$$ For $a<0$ we get $$\delta(ax) = -\frac{1}{a}\delta(x)$$ so in general, for $a\neq0$ we have $$\delta(ax) = \frac{1}{|a|}\delta(x).$$

Your formula $\int_{-\infty}^\infty dx\,p(a x,\epsilon) = \frac{1}{|a|}$ is correct and gives the correct formula for $\delta(ax)$: $$ \int_{-\infty}^\infty dx \, \delta(a x) \, f(x) = \lim_{\epsilon\to 0} \int_{-\infty}^\infty dx \, p(a x, \epsilon) \, f(x) = \{ y := ax \} = \lim_{\epsilon\to 0} \int_{-\infty}^\infty \frac{dy}{|a|} \, p(y, \epsilon) \, f(y/a) = \int_{-\infty}^\infty \frac{dy}{|a|} \, \delta(y) \, f(y/a) = \frac{1}{|a|} f(0/a) = \frac{1}{|a|} f(0) = \int_{-\infty}^\infty \frac{dy}{|a|} \, \delta(y) \, f(y). $$

  • 1
    $\begingroup$ @Thomas. No, that last equality is not correct. Here is the correct one: $$ \lim_{\beta\to 0}\:\frac{|a|}{\beta \sqrt{\pi}}\int_{-\infty}^\infty dx\: e^{-(ax)^2/\beta^2} = |a| \int_{-\infty}^\infty dx\: \delta(ax). $$ You have $$ \delta(x) = \lim_{\beta\to 0} \frac{1}{\beta\sqrt{\pi}} e^{-x^2/\beta^2} $$ so, replacing $x$ with $ax$ we get $$ \delta(ax) = \lim_{\beta\to 0} \frac{1}{\beta\sqrt{\pi}} e^{-(ax)^2/\beta^2}. $$ Please follow the rules of mathematics! $\endgroup$
    – md2perpe
    Commented Apr 5, 2021 at 10:49
  • 1
    $\begingroup$ If $f$ is an ordinary function such that $\int_{-\infty}^{\infty} f(x) \, dx = 1$ then $\int_{-\infty}^{\infty} f(ax) \, dx = 1/|a|.$ If you care about consistency then you would accept this to be true also for a distribution like $\delta.$ Are you confusing the generalized function concept 'distribution' with 'probability distributions'? $\endgroup$
    – md2perpe
    Commented Apr 5, 2021 at 12:08
  • 1
    $\begingroup$ @Thomas. If you require distributions to be normalized then you are defining something else than distributions. Then, for every distribution $u$ you would have $2u=u$ and normal mathematical laws would not apply. Obviously we have to differ between "ordinary distributions" and "Thomas distributions". For $\delta$ as an ordinary distribution, $\delta(ax)=\delta(x)/|a|.$ For $\delta$ as a Thomas distribution strange laws apply. $\endgroup$
    – md2perpe
    Commented Apr 6, 2021 at 18:08
  • 1
    $\begingroup$ @Thomas. The $\delta$-distribution is not defined as a limit of normalized functions. It is defined as a linear functional that maps a "test function" $\varphi$ to $\varphi(0).$ The "Thomas distribution" $\delta$ stays normalized under scaling of its argument because you define it that way and introduce an extra factor to make it work. Please differ between "ordinary distributions" and "Thomas distributions"! $\endgroup$
    – md2perpe
    Commented Apr 6, 2021 at 21:28
  • 1
    $\begingroup$ @Thomas. You are right, $\delta(ax)$ doesn't map $\varphi$ to $\varphi(0).$ That's because $\delta(ax) \neq \delta(x).$ Instead it maps $\varphi$ to $\varphi(0)/|a|.$ That's because $\delta(ax) = \delta(x)/|a|.$ $\endgroup$
    – md2perpe
    Commented Apr 7, 2021 at 22:06

Just about everything you wrote is correct. However, you assumed that to produce $\delta(ax)$, you needed to use probability densities. You don't. $\delta(x)$ is the limit of probability densities, but $\delta(ax)$ is not. So I think you are just having an interpretation issue. $\delta(ax)$ is not normalized, and isn't supposed to be: as you said, its integral should be $\frac{1}{|a|}$.

  • $\begingroup$ On what grounds are you saying that $\delta(ax)$ can not be seen as the limit of probability densities? Just define $y=ax$, and by definition $\delta(y)$ can be described in the same way as the limit of e.g.a normal Gaussian function as $\delta(x)$. If you scale the overall size of the distribution without scaling the density in the inverse sense, you end up with a total probability larger or smaller than 1 though. $\endgroup$
    – Thomas
    Commented Apr 2, 2021 at 14:15
  • $\begingroup$ But you would have to integrate over $dy$, not $dx$. I was using the condition that a probability density, specifically, must integrate to $1$, otherwise it's not a probability density but rather something proportional to one. $\endgroup$ Commented Apr 2, 2021 at 14:22
  • $\begingroup$ If you scale both the distribution and the units by the same amount, that wouldn't be a scale transformation anymore. It would be the same distribution. $\endgroup$
    – Thomas
    Commented Apr 2, 2021 at 14:46

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