Proving a useful identity with the Dirac Delta function

Question

I am studying QFT and renormalization of QED, and in a passage is exploited the following identity: $$\frac{1}{ABC}=2∫_{0}^{1} dxdydzδ(x+y+z-1)\frac{1}{(Ax+By+Cz)^{3}}$$ for every $A,B,C\in\mathbb{C}$.Does someone know how to prove it? I know it's just to compute the integral but I'm struggling with the fact that I have a finite domain. Should I use the step function?

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ Hereafter, $\bracks{\cdots}$ is an $Iverson\ Bracket$. \begin{align} & \color{#44f}{\left. 2\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}{\delta\pars{x + y + z - 1} \over \pars{Ax + By + Cz}^{\,3}}\dd x\,\dd y\,\dd z\,\right\vert_{\ A, B, C\ \in\ \mathbb{C}\setminus\braces{0}}} \\[5mm] = & \ 2\int_{0}^{1}\int_{0}^{1}{\bracks{0 < 1 - y - z < 1} \over \braces{A\pars{1 - y - z} + By + Cz}^{\,3}}\,\dd y\,\dd z \\[5mm] = & \ 2\int_{0}^{1}\int_{0}^{1}{\bracks{y < 1 - z} \over \braces{A + \pars{B - A}y + \pars{C - A}z}^{\,3}}\,\dd y\,\dd z \\[5mm] = & \ {1 \over A - B}\int_{0}^{1}\left.{1 \over \braces{A + \pars{B - A}y + \pars{C - A}z}^{\,2}}\right\vert_{y\ =\ 0}^{y\ =\ 1-z}\,\,\,\,\dd z \\[5mm] = & \ {1 \over A - B}\int_{0}^{1}\braces{% {1 \over \braces{B + \pars{C - B}z}^{2}} - {1 \over \braces{A + \pars{C - A}z}^{2}}}\dd z \\[5mm] = & \ {1 \over A - B}\braces{% -\,{1 \over C - B}{1 \over B + \pars{C - B}z} + {1 \over C - A}{1 \over A + \pars{C - A}z}}_{0}^{1} \\[5mm] = & \ {1 \over A - B}\ \times \\ & \braces{\!\!% -\,{1 \over \pars{C - B}C} + {1 \over \pars{C - A}C} + {1 \over \pars{C - B}B} - {1 \over \pars{C - A}A}\!\!} \\[5mm] = & \ \bbx{\color{#44f}{1 \over ABC}} \\ & \end{align}