I am trying to proof the following identity: \begin{align} \delta(ax) = \frac{1}{|a|}\delta(x). \end{align} where $\delta$ represents the dirac delta function. I found that \begin{align} \int_{-\infty}^\infty f(x) \delta(ax) \,\,dx = \frac{1}{|a|} \int_{-\infty}^\infty f(x)\delta(x)\,\,dx. \end{align} Consequently we get \begin{align} \delta(ax) = \frac{1}{|a|}\delta(x). \end{align} by "removing" the integrals. Is this step actually valid?

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    $\begingroup$ These are "distributions". So you need the definition of equality for distributions. I think trying to work with equations like $\delta(ax) = \frac{1}{|a|}\delta(x)$ without knowledge of distributions cannot be done rigorously. Of course physicists do this all the time, but they are often not concerned about mathematical rigor. $\endgroup$
    – GEdgar
    Nov 16, 2020 at 22:11
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    $\begingroup$ Even if this is valid what purpose does it have? the dirac is still infinite at $x=0$ and $0$ everywhere else $\endgroup$
    – Henry Lee
    Nov 16, 2020 at 22:13
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    $\begingroup$ @HenryLee $\delta$ is not defined this way. It is the limit in the sense of distributions of the function sequence $h_n(x)=2n\ 1_{[-1/n,1/n]}$. This limit is something that we can integrate against any continuous function: $\lim_n \int_{-\infty}^\infty h_n(x)\phi(x)dx=\phi(0)$. $\endgroup$
    – reuns
    Nov 17, 2020 at 1:25
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    $\begingroup$ Does this answer your question? How to make a change of variable inside the Dirac delta? $\endgroup$
    – LL 3.14
    Nov 18, 2020 at 0:57

2 Answers 2


As pointed out by @GEdgar in the comments, equality here is in the sense of distribution, and integration of a Dirac delta against a function is not really a Riemann or Lebesgue integral. It's an abuse of notation. But what does all of that mean?

The space of distributions, often denoted $\mathcal{D}^{'}(\mathbb{R})$, is the space of all maps $f: \mathcal{C}_c ^\infty (\mathbb{R}) \to \mathbb{R}$ that are linear and continuous (whatever that means for now) where $\mathcal{C}_c ^\infty (\mathbb{R})$ is the space of smooth functions with compact support. For example, the dirac delta takes any function $\phi \in \mathcal{C}_c ^\infty (\mathbb{R})$ and maps it to its value at zero: $\phi(0) \in \mathbb{R}$. The action of dirac delta on $\phi$ is denoted: $\langle \delta, \phi \rangle$. It can be shown that such a map is linear and continuous, and hence, the dirac delta is a distribution. If a map $f$ is smooth enough, for example, a continuous function, then you can write: $\langle f, \phi \rangle = \int_{-\infty} ^{\infty} f \phi \; dx$ for any $\phi \in \mathcal{C^\infty _c(\mathbb{R})}$. However, this is not necessarily the case. With that in mind, let's move on to clarify your question.

Two distributions $f,g$ are said to be equal if $\forall \phi \in \mathcal{C^\infty _c(\mathbb{R})}$, we have: $$ \langle f, \phi \rangle = \langle g, \phi \rangle $$ In other words, both distributions induce the same effect on $\phi$. So, the equality: $\delta(ax) = \frac{1}{|a|} \delta(x)$ actually reads: the action of the distribution $\delta(ax)$ on an arbitrary fixed $\phi$ is equal to the action of the distribution $\frac{1}{|a|} \delta(x)$ on $\phi$. Formally; $\langle \delta(ax), \phi(x) \rangle = \langle \frac{1}{|a|}\delta(x), \phi(x) \rangle $. In order to formally prove this result we need to know what is the composition of a distribution.

The left hand-side of your equality is $\delta(ax)=\delta \circ g$ where $g(x)=ax$. Now that we know what a distribution is, we can move on and ask ourselves what the composition of the Dirac delta distribution with a function $g$ is. Is the composition of a distribution with a function $g$ a distribution? It turns out that the answer is yes if $g$ and its inverse are smooth. Which is true for our case (assuming $a \neq 0)$. This composition is defined as: $$\begin{aligned} \langle \delta \circ g, \phi \rangle &= \langle \delta, \phi \circ g^{-1} |\text{det } dg^{-1}| \rangle \\ & = \langle \delta, \phi (\frac{y}{a}) \frac{1}{|a|} \rangle \\ & = \frac{1}{|a|} \langle \delta, \phi (\frac{y}{a}) \rangle \\ & = \frac{1}{|a|} \phi(0) \\ & = \langle \frac{1}{|a|}\delta, \phi \rangle \\ \end{aligned} $$

If we allow for an abuse of notation, the definition of a composition of a distribution with a smooth function is equivalent to changing the variabe of integration! $$ \int_{-\infty}^{\infty} \delta(g(x)) \phi (x) \; dx = \int_{-\infty}^{\infty} \delta(y) \phi (\frac{y}{a}) \frac{1}{|a|} \; dy $$


Shortly: yes, but talking about "removing the integrals" is an oversimplification.

A bit more complete answer: $\delta(x)$ is defined by $$\int_{-\infty}^\infty f(x)\,\delta(x)\,dx=f(0)$$ (it must hold for any sufficiently regular $f(x)$). You found that, for any sufficiently regular $f(x)$, $$\int_{-\infty}^\infty f(x)\,\delta(ax)\,dx = \frac{1}{|a|}\int_{-\infty}^\infty f(x)\,\delta(x)\,dx = \frac{1}{|a|}f(0)\;,$$ so, after multiplying both sides by $|a|$, you obtain $$\int_{-\infty}^\infty f(x)\,|a|\delta(ax)\,dx=f(0)\;.$$ Since it holds for any $f(x)$, it means (by definition) that $$|a|\delta(ax)=\delta(x)\;.$$


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