# Strong definition of the radial Dirac delta function and its derivative

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 $$\delta \left ( \mathbf{r} \right )=\frac{\delta \left ( r \right )}{4\pi r^{2}}$$.

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 $$\Bbb{R} \times \Bbb{R/2\pi Z} \times \Bbb{R/2\pi Z} \to \Bbb{C}$$.
• I apologize for my lack of rigor. ( I am a physicist.) However, I believe that the issue here is more mundane, and it probably has to do with the integration-by-parts approach customarily used to derive expressions for the derivative of the delta "function". This becomes problematic for the radial component because you have to evaluate the delta function at $r=0$. – Jose Menendez Aug 4 at 17:02