Are there standard formulas for the integral over a simplex of a monomial in the barycentric coordinates? Can someone supply a reference? I think I have seen such formulas, but I am unable to find such material by searching on the web. This doesn't seem to be a difficult problem, but maybe there is a neat and memorable way of presenting the answer.



Integrating a barycentric monomial over a simplex
Not for the first time in my life, and almost certainly not the last, I have to eat humble pie. I thought that Igor had missed the point in his comment. In fact, it was I who missed the point. My answer has the advantage that it uses only very elementary methods. It needed a strong nudge and explanations from Igor before I realized the relevance of his different approach. What is great about Igor's answer is that it links a rather mundane question of mine to some very beautiful mathematics that was previously unfamiliar to me and probably to others. I have reversed my previous downvote (which would in any case have been uncalled for, even if his answer had been wrong) into an upvote. My own answer follows. See Igor's answer in his comments.


Thanks to Tim Goodman whose remarks eased my task of finding (a proof of) the formula.
The following material is presumably all standard, but I have not been
able to find a reference. Let the simplex $\sigma$ have dimension $n$,
and let $x\_0,\dots,x\_n$ be barycentric coordinates. The expansion of
$(x\_0 + \dots x\_n)^d=1^d=1$ is the sum of terms of the form
$$\frac{d!}{i\_0!\dots i\_n!} x\_0^{i\_0}\dots x\_n^{i\_n}.$$
These terms form the *Bernstein basis* for the vector space of
homogeneous polynomials of degree $d$ in the barycentric coordinates.
The sum of the basis elements is 1, and there are $n+d \choose d$ of
them (tip: Google "stars and bars"). It will turn out that the integral
over $\sigma$ is independent of the decomposition $d=i\_0+\dots+i\_n$.
It will follow that the integral of any Bernstein basis vector is
$${\text{vol}(\sigma)}\bigg/{{n+d}\choose d}.$$


To prove the above assertion, we only need to prove it for a single $n$-simplex. The formula for other simplices will then follow by the change of variable formula for integration.
First note that
$$\frac{1}{i!j!}\int\_{t=s}^{t=1} (t-s)^i (1-t)^j\,dt
=\frac{(1-s)^{i+j+1}}{(i+j+1)!},$$
which is obvious when $i=0$, and, for $i>0$ follows by induction on
$i$, using integration by parts to reduce $i$ and increase $j$.
Taking $s=0$, we get the required result when $n=1$.


We have coordinates $0\le t\_1\le\dots\le
t\_n\le 1$ for a certain standard $n$-simplex, with respect to which $x\_i = t\_{i+1}-t\_i$ for
$1\le i < n$, $x\_0=t\_1$ and $x\_n=1-t\_n$. We want to evaluate
$$\frac{d!}{i\_0!\dots i\_n!}\int\_{0\le t\_1 \le \dots\le t\_n\le 1}
t\_1^{i\_0}(t\_2-t\_1)^{i\_1}\dots (t\_n-t\_{n-1})^{i\_{n-1}}(1-t\_n)^{i\_n}
\,dt\_1\dots dt\_n .$$


From the integral with
respect to $t$ above, we see that we can eliminate $t\_n$ by
integrating over the interval $[t\_{n-1},1]$. We obtain the integral
of another Bernstein basis element of degree $d+1$ over the standard simplex of dimension $n-1$. Explicitly, we
obtain
$$\frac{d!}{i\_0!\dots i\_{n-2}!(i\_{n-1}+i\_n+1)!}\int\_{0\le t\_1 \le \dots
\le t\_{n-1}\le 1}
t\_1^{i\_0}(t\_2-t\_1)^{i\_1}\dots (1-t\_{n-1})^{i\_{n-1}+i\_n+1}
\,dt\_1\dots dt\_{n-1} ,$$
which is $1/(d+1)$ times the integral of a Bernstein basis vector in
dimension $n-1$ of degree $d+1$. This proves by induction on $n$ that the result is
independent of the decomposition $d=i\_0+\dots i\_n$ and the proof is
complete.


