I have a function of two two-component vectors with this shape $$F(\vec{p},\vec{q})=G(\vec{p},\vec{q})\times\delta\Big(a+b(|\vec{p}|+|\vec{q}|)-\vec{v}\cdot(\vec{p}+\vec{q})\Big),\tag{1}$$ with $a>0$ and $0<b<|\vec{v}|<1$. Besides, $G(\vec{p},\vec{q})$ is symmetric in both arguments. I want to compute the following integral $$I=\int d^2\vec{p}\int d^2\vec{q}~F(\vec{p},\vec{q})\tag{2}.$$ The delta function constrains the integration region, but I can't see how it does completely. My attempt is the following:

First, I have the freedom to choose $\vec{v}=(v,0)$, so the argument of the delta function is $$\chi(\vec{p},\vec{q})=a+b\left(\sqrt{p_x^2+p_y^2}+\sqrt{q_x^2+q_y^2}\right)-v(p_x+q_x)=0.\tag{3}$$ Now I can use $$\delta\Big(g(x)\Big)=\sum_i\frac{\delta(x-x_i)}{|g'(x_i)|}~~~~,~~~~\text{with }g(x_i)=0\tag{4}$$ to get simple deltas. For example, if I want to fix the value of $p_y$, I get $$p_y^{\pm}=\pm\sqrt{\left[\frac{v}{b}(p_x+q_x)-\frac{a}{b}-|\vec{q}|\right]^2-p_x^2}\tag{5}.$$ However, this expression takes imaginary values in the integration region. How can I correctly compute $(2)$?

  • $\begingroup$ for integration to be defined, the function must be continuous. Diract delta fails in that in single point. $\endgroup$
    – tp1
    Feb 4, 2023 at 13:16
  • $\begingroup$ @tp1 No, it is NOT a necessary condition that a function be continutous in order for that function to be integrable. Moreover, the Dirac Delta is NOT a function, it is a distribution, and the object writtin $\int_{-\infty}^\infty $ is NOT an integral, it is a linear functional form. The notation is an abuse of notation. $\endgroup$
    – Mark Viola
    Feb 4, 2023 at 19:40

1 Answer 1


I think I worked it out:

With $(5)$, if I want that $(3)$ is satisfied, I need $$\left[\frac{v}{b}(p_x+q_x)-\frac{a}{b}-|\vec{q}|\right]\geq0.$$ With that, if I want $p_y$ to be real, I also need $$\frac{v}{b}(p_x+q_x)-\frac{a}{b}-|\vec{q}|\geq|p_x|.$$ Now I define $s(\vec{q})=a+b|\vec{q}|-vq_x$, so the condition is $$\text{If}~~p_x>0~:~~p_x\geq\frac{s(\vec{q})}{v-b}\tag{C1}$$ $$\text{If}~~p_x<0~:~~p_x\geq\frac{s(\vec{q})}{v+b}\tag{C2}$$

A priori the integral in $p_x$ is from $-\infty$ to $\infty$, but now this condition constrains the integration region: If $s(\vec{q})>0$, we see from (C$2$) that there won't be contribution of negative $p_x$ to the integral. Therefore, if $s(\vec{q})>0$, the region of integration for $p_x$ will be from $s(\vec{q})/(v-b)$ to $\infty$. On the other hand, if $s(\vec{q})<0$, the integration region for $p_x<0$ is from $s(\vec{q})/(v+b)$ to 0, and for positive $p_x$ from $0$ to $\infty$, so $$\int\limits_{-\infty}^\infty dp_x\rightarrow\Theta\left[-s(\vec{q})\right]\hspace{-4mm}\int\limits_{\displaystyle{\frac{s(\vec{q})}{v+b}}}^\infty\hspace{-3mm}dp_x+\Theta\left[s(\vec{q})\right]\hspace{-4mm}\int\limits_{\displaystyle{\frac{s(\vec{q})}{v-b}}}^\infty\hspace{-3mm}dp_x$$


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