Integrals with Dirac delta functions in 3-D

Question

I've been trying to make sense of these two integrals, somehow the result seems intuitive, yet very hard to compute. We define

$$ f(x)=\frac{1}{4\pi}\delta(|x|-R)$$ and then note that $$ -\frac{1}{2}\int\int\frac{f(x)f(y)}{|x-y|}=-\frac{1}{2R}$$ and

$$\int \frac{f(x)}{|x-y|}dx=\frac{1}{|y|}\quad \text{if }|y|\geq R$$

the integration is over $R^3$, and $\delta$ is the Dirac delta function. I'd appreciate any help with this.

Answer 1

WARNING: I am aware this is not the actual solution to the problem, but it may give you ideas.

I am going to call $|{\bf x}| = x$, it is relatively straightforward to show that

$$ |{\bf x} - {\bf y}| = \sqrt{x^2 + y^2 - 2 xy\cos(\theta - \theta')} $$

where $\theta$ is the polar angle of ${\bf x}$, and $\theta'$ is the polar angle of ${\bf y}$. With this in mind

\begin{eqnarray} \int {\rm d}^3{\bf x}~{\rm d}^3{\bf y} \frac{f({\bf x}) f({\bf y})}{|{\bf x} - {\bf y}|} &=& \frac{1}{(4\pi)^2}\int {\rm d}\cos\theta ~{\rm d}\phi ~{\rm d}x~ x^2 \int {\rm d}\cos\theta' ~{\rm d}\phi' ~{\rm d}y~ y^2 \frac{\delta(x - R)\delta(y - R)}{\sqrt{x^2 + y^2 - 2 xy\cos(\theta - \theta')}} \\ &=& \frac{(2\pi)(4\pi)R^4}{(4\pi)^2}\int {\rm d}\cos(\theta - \theta') \frac{1}{\sqrt{R^2 + R^2 - 2R^2 \cos(\theta - \theta')}} \\ &=& \frac{R^3}{2\sqrt{2}} \int_{-1}^{+1}{\rm d}z \frac{1}{\sqrt{1 - z}} = \frac{R^3}{8} \end{eqnarray}

Answer 2

Since $R^2 \sin \theta \, d\theta d\phi$ can be viewed as an area element, integrating over $\boldsymbol x$ gives $$g(\boldsymbol y) = \frac 1 {4 \pi} \int_{\mathbb R^3} \frac {\delta(x - R)} {|\boldsymbol x - \boldsymbol y|} \, d\boldsymbol x = \frac 1 {4 \pi} \int_{x = R} \frac {dS(\boldsymbol x)} {|\boldsymbol x - \boldsymbol y|} = \frac 1 {4 \pi} \int_{x = R} \frac {dS(\boldsymbol x)} {|\boldsymbol x - (0, 0, y)|} = \\ R \, [y \leq R] + \frac {R^2} y \, [y > R].$$ As the result is continuous in $y$, we can also write $$\frac 1 {4 \pi} \int_{\mathbb R^3} \delta(y - R) \, g(\boldsymbol y) \, d\boldsymbol y = \frac 1 {4 \pi} \int_{y = R} g(\boldsymbol y) \, dS(\boldsymbol y) = R^3.$$