[Math] Dirac delta function under integral

dirac delta

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.

Best Answer

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $$ \mbox{In spherical coordinates,}\quad \delta\pars{\vec{r}}={\delta\pars{r}\delta\pars{\cos\pars{\theta}}\delta\pars{\phi} \over r^{2}}\quad \mbox{such that} $$

\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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