I'm struggling to understand the bounds on integration region after performing integral over Delta function.
Correct result from book:
$$
\int_0^1dz \int_0^1 dy \int_0^1 dx \delta(x+y+z-1) = \int_0^1dz \int_0^{1-z}dy
$$
I do not understand why second integral has bound $0<y<1-z$ ?
My attempt (incorrect somewhere):
Since $$\int_a^b \delta(x-x_0)=
\begin{cases}
1,& \text{if } a<x_0<b\\
0,& \text{otherwise}
\end{cases}$$
Doing first x integral gives
$$\int_0^1 dx \delta(x+y+z-1) =
\begin{cases}
1,& \text{if } 0<1-z-y<1\\
0,& \text{otherwise}
\end{cases}
$$
Which would imply the bounds on y to be: $-z<y<1-z$
Best Answer
You have shown
$$\int_0^1dz \int_0^1 dy \int_0^1 dx \delta(x+y+z-1) = \int_0^1dz \int_0^1dy\chi_z(y)$$
Where $$\chi_z(y) = \begin{cases}1,&-z < y < 1-z\\0, &y\le -z \text{ or } 1-z \le y\end{cases}$$
But what you've overlooked is that you are only integrating $y$ from $0$ to $1$. It doesn't matter that $\chi_z(y) = 1$ for values of $y < 0$, because the integral does not extend to that region. When restricted to $[0,1]$, the function is $$\chi_z(y)\mid_{[0,1]} = \begin{cases}1,& 0 \le y < 1-z\\0, &1-z \le y\le 0\end{cases}$$