How can you justify the delta function equation $\int_0^\infty~\delta(r)~dr = 1$? How can you have, when working with spherical polar coordinates ($r,\theta,\phi$), that
\begin{equation*}
\int_0^\infty~\delta(r)~dr = 1
\end{equation*}
When $\delta(x)$ is defined as
\begin{equation*}
\delta(x)=0~~~~~~if ~~x\neq0;~~~~~~\int_{- \infty}^{ \infty}~\delta (x)~dx=1
\end{equation*}
Hence,
\begin{equation*}
\delta(r)=0~~~~~~if ~~r\neq0;~~~~~~\int_{- \infty}^{ \infty}~\delta (r)~dr=1
\end{equation*}
Also please note that the above does not really make any sense, as a “radial coordinate” r, cannot take on negative values.
 A: You can’t justify the delta function equation $\int_0^\infty~\delta(r)~dr = 1$, working with the usual definition of the delta function.
You have to redefine what you mean by your symbol $\delta(x)$, hence $\delta(r)$.
To eventually work with a radial coordinate, r, you could take the following, as your basic definition of $\delta(x)$,
\begin{equation*}
\delta(x)=0~~~~~~if ~~x\neq0;~~~~~~\int_0^{ \infty}~\delta (x)~dx=1
\end{equation*}
Hence, for $\delta(r)$, we have the definition
\begin{equation*}
\delta(r)=0~~~~~~if ~~r\neq0;~~~~~~\int_0^{ \infty}~\delta (r)~dr=1~~~~~~~~~~~~~~(1)
\end{equation*}
where $r\in [0,\infty)$.
Barton, Reference 1, pg 33 , would call the above definition a ‘one-sided’ definition.
This definition contains the delta function equation that was to be justified.
Reference:

*

*G.Barton, Elements of Greens Functions and Propagation. Potentials,
Diffusion, and Waves, Clarendon Press Oxford, 1989.

NB: Pgs7-40 , contain material on the ‘Dirac Delta Function’
Other Information
You would then also have the rule,
\begin{equation*}
\int_0^{ \infty}~f(r)~\delta (r)~dr=f(0)~~~~~~~~~~~~~~~~~~~~~(2)
\end{equation*}
For those with access to Barton, (1) is almost the same as (1.4.12) on pg 33, and (2) is very similar to an equation on pg 9.
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