# 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

• $\large{\bf r}\cdot{\bf r} = r^{2}$. Also, $\large\int{\rm f}\left({\bf r}\right)\delta\left({\bf r} - {\bf r_{0}}\right)\,{\rm d}^{3}{\bf r} = {\rm f}\left({\bf r_{0}}\right)$ Jan 12, 2014 at 17:38
• Ok thanks. I've edited the question to include this. Obviously that is correct - I think I confused myself. Jan 12, 2014 at 17:38
• Am I correct in my thoughts on the phi and theta integrals? Jan 12, 2014 at 17:46

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 $$\phi\in [0,2\pi)$$, $$\delta(\pmb r -\pmb r_0)=\frac{\delta(r-r_0)}{r^2}\frac{\delta(\theta-\theta_0)}{\sin^2\theta}\delta(\phi-\phi_0)\quad ~\textrm{Eq. 1 (cf. )}$$ 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 $$\int \operatorname{d}^3\pmb r\frac{1}{1+\pmb r\cdot\pmb r}δ(\pmb r−\pmb r_0)= \int_0^{2\pi}\operatorname{d}\phi \int_0^{\pi}\operatorname{d}\theta \sin^2\theta \int_0^{\infty}r^2\operatorname{d}r \frac{1}{1+r^2}\frac{\delta(r-r_0)}{4\pi r^2}=\frac{1}{1+r_0^2}$$
• @user1887919 yes! $r_0^2=2^2+(-1)^2+3^2=14$ so the integral is $f(\pmb r_0)=\frac{1}{1+14}=\frac{1}{15}$. Jan 13, 2014 at 20:51