How to calculate this integral in 3 dimensions involving the Dirac delta function? How would I go about calculating the integral $ \int d^3 \mathbf r {1\over 1+ \mathbf r \cdot \mathbf r} \delta(\mathbf r - \mathbf r_0) $ where $\mathbf r_0 = (2,-1,3)$
My attempt so far:
I have been trying to do this in spherical polar coordinates.
${\rm d}^3 \mathbf r = r^{2}\sin\left(\theta\right)\,{\rm d}r\,{\rm d}\theta\,{\rm d}\phi $
$\mathbf r \cdot\mathbf r = r^2  $ 
I think $\delta(\mathbf r - \mathbf r_0)={1\over r^2\sin\left(\theta\right)}\, \delta(r-r_0)\delta(\theta-\theta_0)\delta(\phi-\phi_0) $ ?
Therefore the integral I am attempting is:
$ \int {1\over r^2 }\delta(r-r_0)\delta(\theta-\theta_0)\delta(\phi-\phi_0) dr d\theta d\phi $
Presumably the $\phi$ integral will just give $2\pi$ , and the $ \theta $ integral will just give $\pi$ ? so then we have
$2\pi^2 \int {1\over r^2 }\delta(r-r_0)\ dr  $
So where do I go from here?
Thanks
 A: Observe that 
$$
\int \operatorname{d}^3\pmb r\; f(\pmb r)δ(\pmb r−\pmb r_0)=f(\pmb r_0)
$$
so that
$$
\int \operatorname{d}^3\pmb r\frac{1}{1+\pmb r\cdot\pmb r}δ(\pmb r−\pmb r_0)=\frac{1}{1+|\pmb r_0|^2}
$$
In spherical coordinates $(r,\theta,\phi)$, where $r \in [0,\infty)$, $\theta\in [0,\pi]$, and $\phi\in [0,2\pi)$, 
$$ \delta(\pmb r -\pmb r_0)=\frac{\delta(r-r_0)}{r^2}\frac{\delta(\theta-\theta_0)}{\sin^2\theta}\delta(\phi-\phi_0)\quad ~\textrm{Eq. 1 (cf. [1])}$$
and $\operatorname{d}^3\pmb r=r^2\sin^2\theta \operatorname{d}r\operatorname{d}\theta\operatorname{d}\phi$ so that
$$
\int \operatorname{d}^3\pmb r\frac{1}{1+\pmb r\cdot\pmb r}δ(\pmb r−\pmb r_0)=
\int_0^{2\pi}\operatorname{d}\phi\,\delta(\phi-\phi_0)
\int_0^{\pi}\operatorname{d}\theta \delta(\theta-\theta_0)\int_0^{\infty}\operatorname{d}r \frac{1}{1+r^2}\delta(r-r_0)
$$
then
$$
\int \operatorname{d}^3\pmb r\frac{1}{1+\pmb r\cdot\pmb r}δ(\pmb r−\pmb r_0)=
\int_0^{\infty}\operatorname{d}r \frac{1}{1+r^2}\delta(r-r_0)=\frac{1}{1+r_0^2}
$$
Alternatively, observe that if the problem involves spherical coordinates, but with no dependence on either $\theta$ or $\phi$, one has
$$
\delta(\pmb r -\pmb r_0)=\frac{\delta(r-r_0)}{4\pi r^2}
$$
and
$$
\int \operatorname{d}^3\pmb r\frac{1}{1+\pmb r\cdot\pmb r}δ(\pmb r−\pmb r_0)=
\int_0^{2\pi}\operatorname{d}\phi
\int_0^{\pi}\operatorname{d}\theta \sin^2\theta
\int_0^{\infty}r^2\operatorname{d}r \frac{1}{1+r^2}\frac{\delta(r-r_0)}{4\pi r^2}=\frac{1}{1+r_0^2}
$$
Bibliography
[1] fen.bilkent.edu.tr/~ercelebi/mp03.pdf
