Region of integration problem in higher dimensions Let $A_{(4)} = \{ (x_1, x_2, x_3,x_4) \in (\mathbb{R}_{<0})^4 : a \leq x_1 + x_2 + x_3 + x_4 \leq 0 \}$. It has been some time since I've done these integral computations where you need variable bounds, and I'm wondering whether there's an easy way to see what the variable bounds need to be for $$\int_A f.$$
For example, if we consider $A_{(2)} = \{ (x_1, x_2) \in (\mathbb{R}_{<0})^2 : a \leq x_1 +x_2 \leq 0 \}$, then $$\int_A f dx_2 dx_1 = \int_a^0 \int_{a-x_1}^0 f dx_2 dx_1.$$ In the next dimension up, it is not clear (at least to me) what the bounds should be.
 A: We have that the range of $x_1$ is from $a$ to $0$. When $x_1$ is fixed, $x_2$ has the range from $a-x_1$ to $0$. When $x_1$ and $x_2$ are fixed, $x_3$ has the range from $a-x_1-x_2$ to $0$, and so forth.
A: At first we take a somewhat closer look at the case $n=2$.
Case n=2:
We have
\begin{align*}A_{(2)}=\{(x_1,x_2)\in\left(\mathbb{R}_{<0}\right)^2:a\leq x_1+x_2<0\}
\end{align*}
where we use $<$ for the upper bound of $x_1+x_2$, since both $x_1$ and $x_2$ are less than zero. The region of interest can be equivalently transformed to:
\begin{align*}
&\qquad(I)&&\qquad(II)&&\qquad(III)\\
\\
&&&a\leq x_1<0&&a\leq x_1<0\\
&a\leq x_1+x_2<0&\quad\Leftrightarrow\qquad&a-x_1\leq x_2<-x_1&\qquad\Leftrightarrow\qquad&a-x_1\leq x_2<0\\
&x_1,x_2<0&&x_1,x_2< 0&\\
\end{align*}

*

*(I) to (II): We see from the conditions in (I) that $a\leq x_1<0$. We obtain the second inequality chain in (II) by subtracting $x_2$ from $a\leq x_1+x_2<0$.


*(II) to (III): We  simplify the inequality chains  since $x_1,x_2<0$  follows from $a\leq x_1<0$ and $x_2<\min\{-x_1,0\}=0$.

We obtain
\begin{align*}
\int_{A_{(2)}}f(x)\,dx_2dx_1=\int_{a}^{0}\int_{a-x_1}^{0}f(x)\,dx_2dx_1
\end{align*}

Case n=3:
We start similarly as before: We have $A_{(3)}=\{(x_1,x_2,x_3)\in\left(\mathbb{R}_{<0}\right)^3:a\leq x_1+x_2+x_3<0\}$ . The region of interest can be equivalently transformed to:
\begin{align*}
&\quad(I')&&\quad(II')&&\quad(III')\\
\\
&&&a\leq x_1<0&&a\leq x_1<0\\
&a\leq x_1+x_2+x_3<0&\Leftrightarrow\ &a-x_1\leq x_2+x_3<-x_1&\quad\Leftrightarrow\ &\color{blue}{a-x_1\leq x_2+x_3<0}\\
&x_1,x_2,x_3<0&&x_1,x_2,x_3< 0&&\color{blue}{x_2,x_3< 0}\\
\end{align*}

*

*(I') to (III'): Here we do the same steps as we did in the case $n=2$. We see the blue marked inequality chains indicate the same situation as in the case $n=2$ with
\begin{align*}
&a\to a-x_1\\
&x_1\to x_2\\
&x_2\to x_3
\end{align*}
So, when considering the blue marked inequality chains we have iteratively reduced the case $n=3$ to the case $n=2$ and   we can proceed as before:
\begin{align*}
&\quad(III')&&(IV')&&(V')\\
\\
&a\leq x_1<0&&a\leq x_1<0&&a\leq x_1<0\\
&\color{blue}{a-x_1\leq x_2+x_3<0}&\Leftrightarrow\ &\color{blue}{\left(a-x_1\right)\leq x_2<0}&\Leftrightarrow\ &\color{blue}{a-x_1\leq x_2<0}\\
&\color{blue}{x_2,x_3< 0}&&\color{blue}{\left(a-x_1\right)-x_2\leq x_3< -x_2}&&\color{blue}{a-x_1-x_2\leq x_3<0}\\
&&&\color{blue}{x_2,x_3< 0}\\
\end{align*}

We finally obtain:
\begin{align*}
\int_{A_{(3)}}f(x)\,dx_3dx_2dx_1=\int_{a}^{0}\int_{a-x_1}^{0}\int_{a-x_1-x_2}^{0}f(x)\,dx_3dx_2dx_1
\end{align*}

