Dirac delta function under integral 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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$$
\mbox{In spherical coordinates,}\quad
\delta\pars{\vec{r}}={\delta\pars{r}\delta\pars{\cos\pars{\theta}}\delta\pars{\phi} \over r^{2}}\quad
\mbox{such that}
$$

\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}

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