Multidimensional integral induction

Question

I am trying to integrate the region under $\sum_{i=1}^{n}{x_i}=1$. What I've done is write out the multiple integral: $$\int_{0}^{1}\int_{0}^{1-x_1}...\int_{0}^{1-\sum_{i=1}^{n-1}{x_i}}1 \,dx_n\,dx_{n-1}...\,dx_1$$ Then by evaluating the first few terms I got a recursive relationship for the $k$-th term, such that it produces a constant equal to $\sum_{i=0}^{k}{(-1)^i\frac{C^k_i}{i+1}}$, which I evaluated to be equal to $\frac{1}{k+1}$, thus I got that the final answer is $\frac{1}{n!}$. I just want to check that this is indeed correct, since I deduced it just from the values I got. A formal derivation would also be most welcome, or any general suggestions as to how I could go about formalizing this.

Answer 1

You are looking at an $n$-dimensional simplex, which is (as long as $n>0$) an $n$-dimensional hyperpyramid of height $1$, and with a base that is an $n-1$ simplex with recursively the same properties: for this it suffices to take as apex any vertex other than the origin. The formula for the $n$-volume of such a hyperpyramid is $\frac1n$ times the $n-1$-volume of the base times the height. You can easily work out this recursion to give $\frac1{n!}$ as answer.

A geometric way to understand this result is to cut up the unit cube by the $\binom n2$ hyperplanes of the form $x_i=x_j$ for certain $i<j$, into $n!$ pieces, each given by one specific relative ordering of the $n$ coordinates $x_i$. Each of these pieces is an $n$-dimensional simplex with the same properties as mentioned above; although they are not isometric to the simplex you are considering, though do have equal volume to it, which clearly must be $\frac1{n!}$.

A fun observation is that you can deduce the volume of a regular simplex (in which all edges have the same length $l$) from this. If you choose the origin as apex of you simplex, the base $B$ is a regular simplex with edges all of length $\sqrt2$, and the height (the distance of the origin to the hyperplane containing $B$) is easily seen to be $\frac{\sqrt n}n=\frac1{\sqrt n}$. Using the known $n$-volume and the given formula one finds that the $n-1$-volume of $B$ is $\frac{\sqrt n}{(n-1)!}$. Now turning to a regular $n$-simplex with edges of length $l$, the formula for its $n$-volume must involve a factor $l^n$, and is easily deduced to be $$\frac{\sqrt{n+1}}{2^{n/2}n!}\,l^n$$(check that for $n=2$ this gives $\frac{\sqrt3}4l^2$).