Where do the step function integral boundaries come from?

Question

EDIT: I have a confusion about Heavyside step function. Suppose I have integral like $$ \int_{0}^{\infty}dE_1\int_{0}^{\infty}dE_2\int_{0}^{\infty}dE_3 \delta(2- \gamma-E_1-E_2-E_3) $$ my first attempt, to solve this integral, would be $$ \int_{0}^{\infty}dE_1\int_{0}^{\infty}dE_2\theta(2-\gamma-E_1-E_2) $$ but I also should change the boundaries and here what I get confused. There are some constraints on the variables: $$ 0 \le E_1 \le 1- \gamma $$ $$ 0 \le E_2 \le 1-\gamma $$ $$ 0\le E_3 \le 1 $$ and the interval should be such like below $$ \int_{0}^{1-\gamma}dE_1\int_{1-E_1-\gamma}^{1-\frac{\gamma}{1-E_1}}dE_2 $$ This is not what I expect actually. Can anyone explain the last integral boundaries?

Answer 1

Note that the constant $a:=1-\gamma\geq 0$ is assumed to be non-negative. OP's integral then becomes

$$I~:=~\iiint_{[0,a]\times[0,a]\times[0,1]}\!\! dx~dy~dz ~\delta(1+a-x-y-z)$$ $$~=~\int_0^a\!dx\int_0^a\!dy\int_0^1\!dz ~\delta(1+a-x-y-z)$$ $$~\stackrel{\begin{matrix}x^{\prime}=a-x\\z^{\prime}=1-z\end{matrix}}{=}~ \int_0^a\!dx^{\prime} \int_0^1\!dz^{\prime} \int_0^a\!dy~\delta(x^{\prime}+z^{\prime}-y)$$ $$~=~\int_0^a\!dx^{\prime} \int_0^1\!dz^{\prime} \int_0^{\infty}\!dy~\theta(a-y)~\delta(x^{\prime}+z^{\prime}-y)$$ $$~=~\int_0^a\!dx^{\prime} \int_0^1\!dz^{\prime} ~\theta(a-x^{\prime}-z^{\prime})$$ $$ ~=~\frac{a^2}{2}-\frac{(a-1)^2}{2}\theta(a-1) .$$ In the last equality we used the interpretation of the double integral as an area of a polygon in the $(x^{\prime},z^{\prime})$ plane.