Strong definition of the radial Dirac delta function and its derivative

Question

The G. Barton textbook “Elements of Green’s functions and Propagation,”, Oxford University Press, 1991 has a very nice introduction on the Dirac delta function. When the 3-D delta function is expressed in spherical coordinates, the author uses what it calls the “strong definition” of the radial delta function: $$\int_{0}^{\eta}\delta \left ( r \right )=1$$ which leads to $$\delta \left ( \mathbf{r} \right )=\frac{\delta \left ( r \right )}{4\pi r^{2}}$$ Other authors use the definition $$\int_{0}^{\eta}\delta \left ( r \right )=\frac{1}{2}$$ which leads to $$\delta \left ( \mathbf{r} \right )=\frac{\delta \left ( r \right )}{2\pi r^{2}}$$ as can be seen, for example, in the Mathematica web site. On the other hand, the Barton textbook shows that the derivative of the radial delta function is $$ {\delta}'\left ( r \right )=-\frac{\delta \left ( r \right )}{r}$$ My problem is that I find that this expression for the delta function derivative is incompatible with the strong definition of the radial delta function.
Here is my proof. I start from the standard Fourier representation of the 3D delta function: $$ \delta \left ( \mathbf{r} \right )=\frac{1}{\left ( 2\pi \right )^3}\int d\mathbf{k}\exp \left ( i\mathbf{k\cdot \mathbf{r}} \right )$$ Carrying out the integral in spherical coordinates and choosing the $\mathbf{r}$ vector along the $z$ axis, this becomes $$ \delta \left ( \mathbf{r} \right )=\frac{1}{\left ( 2\pi \right )^2}\int_{0}^{\infty }dkk^{2}\int_{0}^{\pi }d\theta \sin \theta \exp \left ( ikr\cos \theta \right )$$ which is trivially integrated to give $$\delta \left ( \mathbf{r} \right )=\frac{1}{\left ( 2\pi \right )^2ir}\int_{0}^{\infty }dkk\left [ \exp \left ( ikr \right ) -\exp \left ( -ikr \right ) \right ]$$ This can be re-written as $$ \delta \left ( \mathbf{r} \right )=\frac{1}{\left ( 2\pi \right )^2ir}\left [ \frac{1}{i}\frac{\partial }{\partial r}\int_{0}^{\infty }dk\exp \left ( ikr \right )-\left ( -\frac{1}{i}\frac{\partial }{\partial r} \right )\int_{0}^{\infty }dk\exp \left ( -ikr \right ) \right ]$$ which can be simplified to $$\delta \left ( \mathbf{r} \right )=-\frac{1}{\left ( 2\pi \right )^2r}\left [ \frac{\partial }{\partial r}\int_{-\infty }^{\infty } dk\exp \left ( ikr \right )\right ]$$ If I now invoke the Fourier representation of the delta function, I can write this as $$\delta \left ( \mathbf{r} \right )=-\frac{1}{\left ( 2\pi \right )^2r}\left [ \frac{\partial }{\partial r}\left ( 2\pi \delta \left ( r \right ) \right )\right ]$$ And using the above expression for the derivative of the radial delta function we finally get: $$ \delta \left ( \mathbf{r} \right )=\frac{1}{2\pi r^{2}}\delta \left ( r \right )$$ Therefore, unless I am making a mistake in my derivation which smart readers may discover, my result implies that the expression $ {\delta}'\left ( r \right )=-\frac{\delta \left ( r \right )}{r}$ is incompatible with the strong definition of the radial delta function $\delta \left ( \mathbf{r} \right )=\frac{\delta \left ( r \right )}{4\pi r^{2}}$.

Answer 1

I don't agree with what you wrote, because you don't define any rigorous setting.

If you want a continuous functional $T$ such that for $\phi$ smooth on $\Bbb{R}^3$ $$\int_{-\infty}^\infty \int_0^{2\pi} \int_0^\pi T(r,\theta,\phi)\phi(r \sin\theta \cos \phi, r \sin\theta \sin \phi, r \cos \theta) r^2 \sin(\theta) dr d \theta d\phi = \phi(0,0,0)$$ Then you can take $T(r,\theta,\phi) =\delta(r) \frac{ 1}{\pi^2 r^2}$, and think to the division by $r^2$ as the linear map sending smooth functions $f : \Bbb{R} \times \Bbb{R/2\pi Z} \times \Bbb{R/2\pi Z}\to \Bbb{C}$, $f(0,\phi,\theta)=\partial_r f(0,\phi,\theta) = 0$ to smooth functions $\Bbb{R} \times \Bbb{R/2\pi Z} \times \Bbb{R/2\pi Z} \to \Bbb{C}$.

Answer 2

Note that while $$\int_{-\infty}^\infty\left( \frac{1}{2\pi} \int_{-\infty}^\infty e^{i k x} dk \right)dx = 1,$$ that $$\int_{0}^\infty\left( \frac{1}{2\pi} \int_{-\infty}^\infty e^{i k r} dk \right)dr = \frac{1}{2}.$$ Thus, for $r\in[0,\infty)$, $$\delta(r) = \frac{1}{\pi} \int_{-\infty}^\infty e^{i k r} dk.$$