# Integrating product of Dirac deltas and step functions

I have the following integral

$$\int d^4\boldsymbol{x}' \,\delta\big[(\boldsymbol{x}-\boldsymbol{x'})^2+\alpha^2\big]\,\Theta(-x_0+x'_0)\,\delta\big[(\boldsymbol{x'})^2+\alpha^2\big]\,\Theta(-x_0'),\tag{1}\label{1}$$

where $$\boldsymbol{x}=\{x_0,x_1,x_2,x_3\}$$, the square is $$(x')^2=x_\mu'\cdot x'^{\mu}$$, and the metric convention used is $$\{-1,1,1,1\}.$$

First I wanted to do it in the limit $$\alpha\to0$$, where then I used the second dirac delta and theta function to impose $$x_0'=-\sqrt{(x_1')^2+(x_2')^2+(x_3')^2}.$$

Then performing the first integral over $$dx_0'$$ we would end up with

$$\int d^3\boldsymbol{x}'\,\frac{\delta\big[\boldsymbol{x}^2-2x_0\sqrt{(x_1')^2+(x_2')^2+(x_3')^2}-2x^1 x_1'-2x^2 x_2'-2x^3 x_3'\big]}{2\sqrt{2}\sqrt{(x_1')^2+(x_2')^2+(x_3')^2}}\,\times\Theta\bigg(-x_0-\sqrt{(x_1')^2+(x_2')^2+(x_3')^2}\bigg),\tag{2}\label{2}$$

where the denominator came from the expansion of the second dirac delta via $$\int_{\mathbb{R}^n}f(x)\,\delta\big(g(x)\big)dx=\int_{g^{-1}(0)} \frac{f(x)}{|\nabla g|}d\sigma(x).\tag{3}\label{3}$$

This is not a very nice expression to work with so then I tried to go to spherical coordinates, however this ran into some issues.

I tried

$$\int dr'\int d\phi_1 d\phi_2\,\times\frac{(r')^2\,\text{sin}(\phi_1)}{2\sqrt{2}\,r'}\delta\big[\boldsymbol{x}^2-2r'\big(x_0+x_1 \text{sin}(\phi_1) \text{cos}(\phi_2)+x_2 \text{sin}(\phi_1) \text{sin}(\phi_2)+x_3 \text{cos}(\phi_1)\big)\big]\,\times\Theta\big(-x_0-r'\big),\tag{4}\label{4}$$

I don't think this factor of $$r'$$ should be there, as the final result should be proportional to (updated 28/09) $$\Theta\big(-x_0-\sqrt{x_1^2+x_2^2+x_3^2}\big)$$. Furthermore, since we normally have

$$\delta(\boldsymbol{x}-\boldsymbol{x}_0)=\frac{1}{r^2\text{sin}(\phi_1)}\delta(r-r_0)\delta(\phi-\phi_0)\delta(\theta-\theta_0),\tag{6}\label{6}$$

I am not sure I am even expressing the Dirac delta correctly in \eqref{4}.

I am also a bit lost in the difference between the usual coordinate change involving the Jacobian, and the relation \eqref{3}, obviously one is a coordinate change and one is not, but they seem to have very similar effects.

So I my main questions are; what is the correct way to transform the expression \eqref{2} into polar coordinates, and am I using \eqref{3} correctly in tandem with the coordinate change?

• Hint: Use 3D rotational symmetry to simplify the 4-vector $x^{\mu}=(ct,0,0,r)$. Sep 24 at 15:22
• I'm sorry but I just cannot see how that helps exactly. I still have this factor of $r'$ outside that I don't understand, and if I do this won't I just get something like $r'=\frac{-x_0^2+r^2}{2(x_0+r\text{cos}(\phi_1))}$? which still looks odd to me. Sep 27 at 16:49
• The final result should actually be proportional to $\Theta(-x_0-\sqrt{x_1^2+x_2^2+x_3^2})$, (see my update), which seems to make more sense, but still I am not quite there. Sep 28 at 17:39

If we use 3D rotational symmetry and define

$$x^{\mu}~=~(T,0,0,\underbrace{R}_{\geq 0})\quad\text{and}\quad x^{\prime \mu}~=~(t,r\sin\theta\cos\phi,r \sin\theta \sin\phi,r\cos\theta),$$

and

$$\rho~=~\sqrt{r^2+\alpha^2}\quad\text{and}\quad \Delta~=~R^2-T^2,$$

then OP's integral (1) becomes

\begin{align} I(R,T,\alpha)~=~& \int_{\mathbb{R}} \!\mathrm{d}t \int_{\mathbb{R}_+} \!r^2 \mathrm{d}r \int_0^{\pi} \!\sin\theta \mathrm{d}\theta\int_0^{2\pi} \!\mathrm{d}\phi~\cr &\delta(\rho^2+R^2-2Rr\cos\theta-(T-t)^2)~\Theta(T-t) ~\delta(\rho^2-t^2)~\Theta(t)\cr ~\stackrel{R\neq 0}{=}& 2\pi \Theta(T) \int_0^T \!\mathrm{d}t \int_{\mathbb{R}_+} \!r^2 \mathrm{d}r \int_{-1}^1 \! \mathrm{d}\cos\theta~\cr &\frac{1}{2Rr}\delta\left(\frac{\rho^2+R^2-(T-t)^2}{2Rr}-\cos\theta\right) ~\frac{\delta(\rho-t)}{\rho+t}\cr ~=~& 2\pi \Theta(T) \int_{\mathbb{R}_+} \!\frac{r^2 \mathrm{d}r}{2Rr~2\rho} ~\Theta\left(\left|\frac{\rho^2+R^2-(T-\rho)^2}{2Rr}\right|<1\right) ~\Theta(0<\rho

where

$$A~=~4\Delta, \quad B~=~-4T\Delta \quad\text{and}\quad C~=~-\Delta^2 -4R^2\alpha^2.$$

The discriminant is

$$D~=~B^2-4AC~=~16R^2\Delta(\Delta +\alpha^2).$$

The roots are

$$\rho_{\pm}~=~\frac{-B\pm\sqrt{D}}{2A}~=~\frac{T\pm R\sqrt{1+\alpha^2/\Delta}}{2}.$$

• Case $$-\alpha^2\leq \Delta\leq 0$$: Then $$D\leq 0$$ and the integral $$I(R,T,\alpha)=0$$ vanishes.

• Case $$\Delta> 0$$: Then $$D>0$$ and the roots $$\rho_{\pm}$$ are outside the integration region $$[\alpha,T]$$, so the integral $$I(R,T,\alpha)=0$$ vanishes.

• Case $$\Delta \leq -\alpha^2$$: Then $$R\leq T$$, and therefore $$R\sqrt{1+\alpha^2/\Delta}\leq T$$, so the root $$\rho_+< T$$. $$I(R,T,\alpha)~=~\frac{\pi}{2R} \Theta(T-\alpha)~\Theta(-\Delta-\alpha^2)~\underbrace{m([\rho_-,\rho_+]\cap [\alpha,T])}_{= \text{ length of the interval}} ,$$ where $$m$$ denotes the Lebesgue measure.

Let us for the remainder of this answer assume that $$\alpha=0$$. Then the discriminant $$D\geq 0$$. The roots are $$\rho_{\pm}~=~\frac{-B\pm\sqrt{D}}{2A}~=~\frac{T\pm R}{2}.$$ It follow that the integral is supported in the future light-cone: $$I(R,T,\alpha=0)~=~\frac{\pi}{2R} \Theta(T)~R\Theta(-\Delta)~=~\frac{\pi}{2}\Theta(T)\Theta(-\Delta).$$ (The case $$R=0$$ follows e.g. from continuity of the limit $$R\to 0^+$$.)

• Thank you very much. I have one question though; when going from the 2nd to the 3rd '=' in the integral, I assume you use a generalisation of $\delta(x)=\frac{d}{dx}H(x)$? Do you know where I could find this? I have been unable to so far. Sep 30 at 15:12
• Another question; in going from the 1st to 2nd '=' sign, I can see that $$\int_{\mathbb{R}}dt \,\Theta(T-t)\Theta(t)=\int_{0}^{\infty}dt\,\Theta(T-t),$$ after which I would have thought became $$\int_{0}^{\infty}dt\,\Theta(T-t)=\int_{0}^{R}dt,$$ but you still have a $\Theta(T)$ remaining? Is it because we still might have $T<0$ and need to account for that possibility? Sep 30 at 15:40
• 1. No, the Theta function comes from the integration limits. 2. Yes. Sep 30 at 17:57
• Ah I see now, thank you :) Sep 30 at 19:20