Delta function of the Euclidean norm $\delta(|\mathbf x|)$ / in polar coordinates at origin $\delta(r)$

Solution 1:

I will write $\omega_d := s_{d-1} = \frac{2\,\pi^{d/2}}{\Gamma(d/2)}$.

If $f(x) = \tilde{f}(|x|)$ with $\tilde{f}\in L^1_{\mathrm{loc}}$, then for any $\varphi\in C^\infty(\mathbb{R})$, $$ \langle \tilde{f},\varphi\rangle = \int_{\mathbb{R}_+} \tilde{f}(r)\,\varphi(r)\,\mathrm{d}r = \frac{1}{\omega_d}\int_{\mathbb{R}^d} f(x)\frac{\varphi(|x|)}{|x|^{d-1}}\,\mathrm{d}x = \frac{1}{\omega_d}\left\langle f, \frac{\varphi(|x|)}{|x|^{d-1}}\right\rangle $$ so if we want to define a generalization of the notion of radial change of variable, we might want to define $\tilde{f}(r)$ by setting $$ \langle \tilde{f},\varphi\rangle := \frac{1}{\omega_d}\left\langle f, \frac{\varphi(|x|)}{|x|^{d-1}}\right\rangle $$ in the general case. In particular, we will have the relation $\omega_d\,|x|^{d-1} \tilde{f}(|x|) = f(x)$. Remark however that the above definition does not make sense if we replace $f$ by $\delta_0$. $$ \frac{\delta_0(x)}{|x|^{d-1}} \text{ is not a priori a well defined distribution} $$ Defining $g(r) = \tilde{f}(|r|)$ for $r\in\mathbb{R}$, we can however try to solve the equation $\omega_d\,r^{d-1} g(r) = \delta_0$. Taking the Fourier transform ($\hat{g}(y) = \int_{\mathbb{R}} g(x)\,e^{-2i\pi x\cdot y}\,\mathrm{d}x$) we get $$ \frac{\omega_d}{(-2i\pi)^{d-1}} \hat{g}^{(d-1)} = 1 $$ so by integrating $d-1$ times, we get $$ \hat{g}(y) = \frac{(-2i\pi)^{d-1}}{\omega_d} \left(\frac{y^{d-1}}{(d-1)!} + \sum_{k=1}^{d-1} a_k\,x^{k-1}\right) $$ so that (if I did not messed up with the power on the $-1$) $$ g(r) = \frac{(-1)^{d-1}}{\omega_d} \left(\frac{\delta_0^{(d-1)}}{(d-1)!} + \sum_{k=1}^{d-1} C_k\,\delta_0^{(k-1)}\right) $$ (where $\delta_0^{(n)}$ is the $n$-th derivative of the Dirac delta centered in $0$) and we can see that there is no uniqueness of $\delta_0(r)/r^{d-1}$ (which is natural since $x^n \delta_0^{(n-1)} = 0$). However, the Dirac is homogeneous, and thus so is its radial representation. Therefore $$ g = \frac{(-1)^{d-1}}{\omega_d\,(d-1)!} \,\delta_0^{(d-1)} $$ which can be written in a more informal way with $r=|x|$ $$\boxed{ \delta_0(x) = \frac{(-1)^{d-1}}{\omega_d\,(d-1)!} \,\delta_0^{(d-1)}(r) }$$

Remark: Of course, we can now choose to define $$ \frac{\delta_0(r)}{r^{d-1}} := \frac{(-1)^{d-1}}{\,(d-1)!} \,\delta_0^{(d-1)}(r) $$ as the homogeneous solution of the equation $r^{d-1} g(r) = \delta_0$, and so we find as you was writing $\omega_d\,\delta_0(x) = \frac{\delta_0(r)}{r^{d-1}}$, but now we know the true meaning of this notation.