# How is the Dirac delta $\delta (x^3)$ different from $\delta(x)$?

How is Dirac delta $$\delta (x^3 )$$ different from $$\delta (x)$$? It is my understanding that $$\delta [\psi]:=\langle\delta,\psi\rangle=\int_{\Omega}\delta(x)\psi(x)\,dx= \begin{cases} \psi(0) & \text{if 0\in \Omega}\\ 0 & \text{otherwise} \end{cases}$$ But $$x^3=0$$ iff $$x=0$$ so does this not imply, $$\delta(x^3) = \delta(x) = 0$$ Apologies if I am completely misunderstanding the topic, I am very new to distributions.

• Good question. The delta distribution with nonlinear arguments is a tricky concept. This answer of robjohn has been extremely useful to me in order to understand that. Jan 22 '21 at 19:55
• Compare $\int_{\Bbb R}f(x)\delta(x^3)\,dx=\int_{\Bbb R}u^{-2/3}f(u^{1/3})\delta(u)/3\,du$ with $\int_{\Bbb R}f(u)\delta(u)\,du$. Jan 22 '21 at 20:35
• It appears to me that you are using conflicting notation. The "$\delta(x)$" appearing inside the integral is not the same as the "$\delta$" that is applied to $\psi$ on the left hand side. Jan 22 '21 at 20:37
• Maybe too naive but øan't you substitute $y = x^3$ in $\int \delta(x^3) f(x) dx$ and see what happens? Jan 22 '21 at 20:54
• @pregunton I've updated the question to fit exactly what is written in my lecture handout Jan 22 '21 at 21:39

## 1 Answer

Let $$\psi$$ be a test function, i.e. $$\psi \in \mathcal{C}^{\infty} _{c}(\mathbb{R})$$, then: $$\int \delta(x^3)\psi (x) dx = \int \delta(y) \psi (y^{\frac{1}{3}}) \frac{1}{3y^{\frac{2}{3}}} dy$$ Note that our new test function $$\psi (y^{\frac{1}{3}}) \frac{1}{3y^{2/3}}$$ is not smooth, hence we cannot define $$\delta(x^3)$$ as a distribution. In general, if $$u$$ is a distribution, then $$u \circ g$$ is a distribution as well if both $$g$$ and its inverse are smooth. In such a case: $$\langle u \circ g, \psi \rangle := \langle u, \psi \circ g^{-1} |\text{det } dg^{-1}| \rangle$$. Please note that the integration above is an abuse of notation.