Suppose we have a definite integral like $$ I=\int_0^\infty dx \int_0^\infty dy \, Θ(α-x-y) $$ where $a \in R_+^*$ and $Θ$ is the Heaviside step function. Of course this is easy in that we can find the answer without working with integrals, since it's simply the area of a triangle with vertices $O(0,0)$ , $A(a,0)$, $B(0,a)$, which is $\frac{a^2}{2}$.
However trying to work the integral out doesn't seem so trivial (to me at least). One could, naively, do for example $I=\int_0^\infty dx \int_0^{a-x} dy = -\infty$ (since $Θ=0$ for $y \ge a-x$) which is obviously wrong- somehow $Θ$ should affect both integral limits, but I can't see how this can happen. Probably I'm misunderstanding the definition of $Θ(α-x-y)$.
EDIT : As @John Polcari pointed out, in the above exaple, we should limit $x$ such that $a-x \ge 0$, so we should have $I= \int_0^\infty dx \int_0^{a-x}Θ(a-x) \, dy $ which indeed gives the right result. However this seems to me like adding the $Θ(α-x)$ by hand, which is no different from the first argument with the triangle's area. Is there a prettier-more strict- way of doing it?
The formally correct result is $$\int\limits_0^\infty {dz\,\Theta \left( {a - z} \right)} = a\,\Theta \left( a \right)$$
In case it is unclear to see how to apply this:
By implication from above $$\int\limits_0^\infty {dz\,z} \,\Theta \left( {a - z} \right) = \frac{{{a^2}}}{2}\Theta \left( a \right)$$ Then $$\begin{array}{l} \int\limits_0^\infty {dx\int\limits_0^\infty {dy\,\Theta \left( {a - x - y} \right)} } = \int\limits_0^\infty {dx\left( {a - x} \right)\Theta \left( {a - x} \right)} \\ = a\int\limits_0^\infty {dx\,\Theta \left( {a - x} \right)} - \int\limits_0^\infty {dx\,x\;\Theta \left( {a - x} \right)} \\ = {a^2}\Theta \left( a \right) - \frac{{{a^2}}}{2}\Theta \left( a \right) = \frac{{{a^2}}}{2}\Theta \left( a \right) \end{array}$$ The final step function is appropriate because the if $a<0$, the answer really is zero. $\hspace{0pt}$