# Is it true in general that $\int \dots \int_{0 \le x_1 \le \dots \le x_n,\ 0 \le x_n\le1}dx_1\dots dx_n=\left(\frac{1}{2}\right)^n?$

I was looking at an example with the following integral: $$\iiiint_{0 \le x \le y \le z \le t,\ 0 \le t \le \frac{1}{2}} 1 \,dx\,dy\,dz\,dt = \frac{1}{16}$$ Is it true in general that $$\int \dots \int_{0 \le x_1 \le \dots \le x_n,\ 0 \le x_n\le1}dx_1\dots dx_n=\left(\frac{1}{2}\right)^n?$$

(edit: here $0 \le x_n \le 1$ but in the example $0 \le t \le \frac{1}{2}$ so this can't be true...). Intuitively it would seem so since each dimension is "cut in half" by each inequality. Are there some other results for multiple integrals with this domain that have other integrand that is not a constant? Is it true that $$\int \dots \int_{0 \le x_1 \le \dots \le x_n,\ a \le x_n\le b}f(\boldsymbol{x})\,dx_1\dots dx_n=\int_a^b \int_0^{x_{n-1}} \int_0^{x_{n-2}} \dots \int_0^{x_2}f(\boldsymbol{x})\,dx_1\dots dx_n$$

• Did you mean $0 \leq x \leq y \leq z \leq t \leq \frac12$, because as you have it written the integral is $+\infty$. – nullUser Jun 25 '13 at 16:13
• Yes, sorry about that. – Lotus3000 Jun 25 '13 at 16:20

To find the value of: $$\int \dots \int_{0 \le x_1 \le \dots \le x_n,0 \le x_n\le1}dx_1\dots dx_n=?$$ First note that the first integral gives $x_1 |^{x_2}_0 = x_2$. The next one gives $x_3^2/2$ and so on. This leads to: $$\int \dots \int_{0 \le x_1 \le \dots \le x_n,0 \le x_n\le1}dx_1\dots dx_n=\frac{L^{n-1}}{(n-1)!}$$ Where $L$ is the upper limit, in your case $1$, and in the original case, $1/2$.

Note that the integral may be written as $$\int_0^{1/2}\int_0^{x_n}\int_0^{x_{n-1}}\cdots\int_0^{x_2}dx_1\cdots dx_{n-2}dx_{n-1}dx_{n}$$ $$= \int_0^{1/2}\int_0^{x_n}\int_0^{x_{n-1}}\cdots\int_0^{x_3}x_2dx_2\cdots dx_{n-2}dx_{n-1}dx_{n}$$ $$= \int_0^{1/2}\int_0^{x_n}\int_0^{x_{n-1}}\cdots\int_0^{x_4}\frac{x_3^2}{2}dx_3\cdots dx_{n-2}dx_{n-1}dx_{n}$$ $$= \cdots$$ $$=\int_0^{1/2}\frac{x_n^{n-1}}{(n-1)!}dx_n$$ $$= \frac{x_n^n}{n!}\biggr|_0^{1/2}$$ $$= \frac{1}{2^n n!}$$

The formulas in the question, as well as the accepted answer, are wrong.

Given an $L>0$, let $C_L:=[0,L]^n$ be the the $n$-cube with side length $L$. The integral $$\int_{C_L} 1\>{\rm d}(x_1,\ldots,x_n)$$ is of course equal to the volume of $C_L$, and therefore $=L^n$. If we take only the part $S\subset C_L$ on which the coordinates satisfy $x_1\leq x_2\leq\ldots\leq x_n$ we obtain $$\int_S 1\>{\rm d}(x_1,\ldots,x_n)={\rm vol}(S)={{\rm vol}(C_L)\over n!}={L^n\over n!}\ ,$$ by symmetry.