Region des Integrationsproblems in höheren Dimensionen

Question

Lassen $A_{(4)} = \{ (x_1, x_2, x_3,x_4) \in (\mathbb{R}_{<0})^4 : a \leq x_1 + x_2 + x_3 + x_4 \leq 0 \}$. Es ist einige Zeit her, dass ich diese integralen Berechnungen durchgeführt habe, bei denen Sie variable Grenzen benötigen, und ich frage mich, ob es eine einfache Möglichkeit gibt, zu sehen, wofür die variablen Grenzen sein müssen$$\int_A f.$$

Zum Beispiel, wenn wir überlegen $A_{(2)} = \{ (x_1, x_2) \in (\mathbb{R}_{<0})^2 : a \leq x_1 +x_2 \leq 0 \}$, dann $$\int_A f dx_2 dx_1 = \int_a^0 \int_{a-x_1}^0 f dx_2 dx_1.$$ In der nächsten Dimension ist (zumindest für mich) nicht klar, wo die Grenzen liegen sollten.

Answer 1

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*}

Answer 2

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.