# Dirac Delta Function "removing integral"

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?

• 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. Nov 16, 2020 at 22:11
• Even if this is valid what purpose does it have? the dirac is still infinite at $x=0$ and $0$ everywhere else Nov 16, 2020 at 22:13
• @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: $\lim_n \int_{-\infty}^\infty h_n(x)\phi(x)dx=\phi(0)$. Nov 17, 2020 at 1:25
• Does this answer your question? How to make a change of variable inside the Dirac delta? Nov 18, 2020 at 0:57

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 $$|a|\delta(ax)=\delta(x)\;.$$