This is related to my previous question here. Here I am evaluating (in the sense of distributions) for a vector $\mathbf{r} = (x,y,z)$ $$ \int \frac{d^3 \mathbf{p}}{(2\pi)^3} e^{i \mathbf{r} \cdot \mathbf{p}} = \delta(x) \delta(y) \delta(z) \ . $$ At the same time, you can evaluate the above integral as $$ \int \frac{d^3 \mathbf{p}}{(2\pi)^3} e^{i \mathbf{r} \cdot \mathbf{p}} = \frac{1}{2 \pi^2 |\mathbf{r}|} \int_0^\infty dP \; P \sin(|\mathbf{r}| P) = - \frac{\delta'(|\mathbf{r}|)}{2\pi|\mathbf{r}|} $$ where the last line uses the representation $\int_0^\infty \frac{dP}{\pi} P \sin( \alpha P) = - \delta'(\alpha)$ (which follows from $\int_0^\infty \frac{dP}{\pi} \cos( \alpha P) = \delta(\alpha)$). There is additionally the identity $\alpha \delta'(\alpha) = - \delta(\alpha)$ would imply that $$ \cdots = \frac{\delta(|\mathbf{r}|)}{2\pi|\mathbf{r}|^2} = 2 \delta(x)\delta(y)\delta(z) $$ where in the last line I have used the identity $\delta(x) \delta(y) \delta(z) = \frac{\delta(|\mathbf{r}|)}{4\pi|\mathbf{r}|^2}$ which is equation (3.1.30) from Kanwal's book on distributions.

The above seems to say that $\delta(x)\delta(y) \delta(z) = 2 \delta(x)\delta(y) \delta(z)$, which is obviously wrong. Where am I making a mistake in my second way of computing?

EDIT: Note that the $3D$ Fourier transform of a rotationally invariant function $f(|\mathbf{p}|)$ is also rotationally invariant $F(|\mathbf{x}|)$, $$ F(|\mathbf{x}|) = \int \frac{d^3 \mathbf{p}}{(2\pi)^3} f(|\mathbf{p}|) e^{i \mathbf{r} \cdot \mathbf{p}} = \int \frac{d^3 \mathbf{p}}{(2\pi)^3} f(|\mathbf{p}|) e^{i |\mathbf{r}| |\mathbf{p}| \cos\theta } $$ where I have taken $z \to |\mathbf{r}|$ (free to do so by rotationally symmetry) and used spherical coordinates. From here the $\phi$ integral evaluates to $2\pi$, then doing the $\theta$ integral gives the $\propto \sin(|\mathbf{r}|P)/|\mathbf{r}|$ function in my formula above. I then use $f(|\mathbf{p}|) =1$ in the above.

  • $\begingroup$ Could you please show how you do this step? $$\int \frac{d^3 \mathbf{p}}{(2\pi)^3} e^{i \mathbf{r} \cdot \mathbf{p}} = \frac{1}{2 \pi^2 |\mathbf{r}|} \int_0^\infty dP \; P \sin(|\mathbf{r}| P)$$ $\endgroup$
    – md2perpe
    Dec 9, 2020 at 17:27
  • $\begingroup$ @md2perpe Sure thing, I added an edit in the above. This is a standard result for rotationally invariant 3D Fourier transforms $\endgroup$ Dec 9, 2020 at 18:27

1 Answer 1


Your computation of the integral using spherical coordinates $$\int \frac{d^3 \mathbf{p}}{(2\pi)^3} e^{i \mathbf{r} \cdot \mathbf{p}}=\frac1{(2\pi)^2}\int_{\Bbb R^+}\int_{[0,\pi]} r^2\sin\theta\cdot e^{i|{\bf r}|P\cos\theta}\,d\theta\,dP=\frac{\delta(|{\bf r}|)}{2\pi|{\bf r}|^2}$$ is correct as are the identities you have used up to this point. However, it is important to note the lack of consensus in the proportionality constant of $|{\bf r}|^{-2}\delta(|{\bf r}|)$. For example,

  • Kanwal (2004) defines $\delta(x) \delta(y) \delta(z) = \dfrac{\delta(|\mathbf{r}|)}{4\pi|\mathbf{r}|^2}$ as you have stated, whereas

  • Bracewell (1999) defines $^3\delta(x,y,z)=\dfrac{\delta(\rho)}{2\pi\rho^2}$ where $\rho=|{\bf r}|$ (see equation $(51)$ of this link).

This primarily stems from the ill-defined nature of $x^{-\kappa}\delta(x)$ when $\kappa\ge1$ as discussed in Delta function at the origin in polar coordinates and Delta function of the Euclidean norm $\delta(|\mathbf x|)$ / in polar coordinates at origin $\delta(r)$ for polar coordinates representations.


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