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Dirac delta in spherical coordinate is $\delta(r^\to-r_0^\to)= \frac{1}{r^2 \sin(\theta)}\delta(r-r_0)\delta(\theta-\theta_0)\delta(\phi-\phi_0) $, and the element of volume is $dV = r^2\sin(\theta)\,dr\,d\theta \,d\phi $. We have:

$$\int dV \, \delta(r^\to-r_0^\to) = 1 \Rightarrow \int_0^\infty dr \, \delta(r-r_0)=1 $$

I want to know about the behavior of Delta Dirac function under integral in this situation: $$\int_{x_0}^\infty dx\, \delta(x-x_0)$$

What is the result of the above integral?

What would happen if in the spherical coordinate we integrate over the origin and hence we would have $\int_0^\infty dr \, \delta(r) $. Is this integral one? Clearly when we integrate over space around the origin, the origin is included in the integration volume and $\int_0^\infty dr \, \delta(r) $ must be one. However, I am not sure what happens in this case $\int_{x_0}^{\infty} dx \, \delta(x-x_0)$ when we are in Cartesian coordinate.

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    $\begingroup$ If $x_0\in(0,\infty)$ then yes. The case of $x_0$ also being the lower bound for integration is marginal: $x_0-\epsilon$ would give zero and $x+\epsilon$ would give 1 (for $\epsilon>0$.) $\endgroup$ – Semiclassical Jul 29 '14 at 15:31
  • $\begingroup$ @Semiclassical. I want to know the result of the integral in the marginal case. If in the marginal case the result is not one, why does the integral of the Dirac delta in the spherical coordinate give the result one? $\endgroup$ – MOON Jul 29 '14 at 15:36
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $$ \mbox{In spherical coordinates,}\quad \delta\pars{\vec{r}}={\delta\pars{r}\delta\pars{\cos\pars{\theta}}\delta\pars{\phi} \over r^{2}}\quad \mbox{such that} $$

\begin{align} \color{#66f}{\large\int_{{\mathbb R}^{3}}\delta\pars{\vec{r}}\,\dd^{3}\vec{r}} &=\int_{0^{-}}^{\infty}\dd r\,r^{2}\int_{0}^{\pi}\dd\theta\,\sin\pars{\theta} \int_{0}^{2\pi}\dd\phi\,{\delta\pars{r}\delta\pars{\cos\pars{\theta}}\delta\pars{\phi} \over r^{2}} \\[3mm]&=\underbrace{\bracks{\int_{0^{-}}^{\infty}\delta\pars{r}\,\dd r}} _{\ds{=\ 1}}\ \underbrace{\bracks{% \int_{0}^{\pi}\delta\pars{\cos\pars{\theta}}\sin\pars{\theta}\,\dd\theta}} _{\ds{=\ 1}}\ \underbrace{\bracks{\int_{0}^{2\pi}\delta\pars{\phi}\,\dd\phi}}_{\ds{=\ 1}}\ \\[3mm]&=\ \color{#66f}{\Large 1} \end{align}

$$\mbox{Note that}\quad \int_{{\mathbb R}^{3}}\delta\pars{\vec{r} - \vec{r}_{0}}\,\dd^{3}\vec{r} =\int_{{\mathbb R}^{3}}\delta\pars{\vec{r}}\,\dd^{3}\vec{r} $$

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  • $\begingroup$ Why did you let the lower limit of the radial integration get the negative value by this sign 0-? The variable r cannot be negative. $\endgroup$ – MOON Jul 29 '14 at 18:24
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    $\begingroup$ @yashar It's a limiting case. Note that $\displaystyle{\large\int_{0^{-}}^{\infty}\delta\left(r\right)\,{\rm d}r = 1}$ and $\displaystyle{\large\int_{0^{+}}^{\infty}\delta\left(r\right)\,{\rm d}r = 0}$. The cartesians coordinates 'include' $\displaystyle{\large\vec{r} = \vec{0}}$. When we switch to sphericals we have to remember that point must be 'included'. The $\displaystyle{\large 0^{-}}$ 'artifact' takes care of that. $\endgroup$ – Felix Marin Jul 29 '14 at 18:48

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