How can you justify the delta function equation $\int_0^\infty~\delta(r)~dr = 1$?

Question

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.

Answer 1

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:

1. 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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