Is the integral of $(x_1,\ldots,x_n)\mapsto \prod_{i=1}^n x_i$ over the unit disk in $\mathbb R^n$ is zero?

Question

Let $n$ be a positive integer, $f:\mathbb R^n\to\mathbb R$ be the map $(x_1,\ldots,x_n)\mapsto \prod_{i=1}^n x_i$, and $$\overline D_1(0) = \left\{x\in\mathbb R^n: \sum_{i=1}^n x_i^2\leqslant 1 \right\}$$be the closed unit disk centered at $0$. Then $$\int_{\overline D_1(0)}f\ \mathsf d\lambda=0 $$ where $\lambda$ is Lebesgue measure.

My argument: The magnitude of the integral of $f$ over each orthant of $\mathbb R^n$ is the same, and $f$ is positive on half of them and negative on the rest, and hence summing the integrals of $f$ over each orthant yields $0$.

I verified the result for $n=1,2,3$ but I'm not entirely convinced my proof is rigorous.

Answer 1

Your analysis is just fine.

If you call the product $f$, then it's clear that $$ f(x_1, \ldots, x_n) = - f(x_1, \ldots, -x_n) $$ so that integrating with respect to $x_n$, from $-1$ to $1$, gives you zero. Since your whole integral can be written as an iterated integral, you're done.

(essentially this is your "even/odd" argument, but done along one direction only.)

Answer 2

Each orthant corresponds to a $\sigma\in\{-1,1\}^n$. Let $|\sigma|$ denote the product of all coordinates of $\sigma$.

Call $A_\sigma$ the orthant corresponding to $\sigma$. Then, the integral on the unit ball is $$\sum_{\sigma\in \{-1,1\}^n}\int_{A_\sigma}x_1x_2\ldots x_nd\lambda$$

There is an obvious transformation from $A_\sigma$ to $A_1$ by $x_i\to|x_i|$.

So, $$\int_{A_\sigma}x_1x_2\ldots x_nd\lambda=|\sigma|\int_{A_1}x_1x_2\ldots x_nd\lambda$$ Define $\phi:\{-1,1\}^n\to\{-1,1\}^n$ by taking $(a_1,a_2,\ldots,a_n)\to(-a_1,a_2,\ldots,a_n)$. Then, $|\sigma|+|\phi(\sigma)|=0$. Moreover, $\phi$ is an involution, thus, we can cancel the integral from $A_\sigma$ by $A_{\phi(sigma)}$.