Exercise 4.1.29* (Evaluate $\int_0^1 \int_0^1 \int_0^1 \lfloor x + y + z \rfloor dx\, dy\, dz$)

Proof.

(a)

Let $f:{[0,1]}^3\to ℝ$ denote the function defined by $$f(x,y,z)=\lfloor x+y+z\rfloor .$$

Then $f$ takes only the values $0,1,2,3$.

Consider the cube $(A,B,C,D,E,F,G,H)$, where

$$\begin{array}{llll}\hfill & A=(0,0,0),B=(1,0,0),C=(1,1,0),D=(0,1,0),& \hfill & \\ \hfill & E=(0,0,1),F=(1,0,1),G=(1,1,1),H=(0,1,1).& \hfill & \end{array}$$

Put

$$A_k=\{(x,y,z)\in {[0,1]}^3\mid \lfloor x+y+z\rfloor =k\},k=0,1,2,3.$$

The $A_k$ are measurable, and $f$ takes only a finite set of values, so $f$ is a simple function (Rudin, Real and complex analysis, p. 15).

We write $\mu (A_k)$ the volume of $A_k$. Note that $\mu (A_3)=\mu (\{(1,1,1)\})=0$, so

$$\begin{array}{llll}\hfill J& ={\int \; <!--\; nolimits\; -->}_0^1{\int \; <!--\; nolimits\; -->}_0^1{\int \; <!--\; nolimits\; -->}_0^1\lfloor x+y+z\rfloor dxdydz=0\cdot \mu (A_0)+1\cdot \mu (A_1)+2\cdot \mu (A_2).& \hfill & \\ \hfill & & \hfill \end{array}$$

We note that $A_0$ is the tetrahedron $\mathit{AEBD}$, and $A_2$ the tetrahedron $\mathit{GCHF}$, which are symmetric relatively to the center of the cube, so $\mu (A_0)=\mu (A_2)$. This gives

$$J=\mu (A_0)+\mu (A_1)+\mu (A_2)=\mu ({[0,1]}^3)=1.$$

(b)

Bizarrely, this trick can be generalized to $n$-dimensions.

We write $f:{[0,1]}^n\to ℝ$ the function defined by

$$f(x_1,x_2,\dots ,x_n)=\lfloor x_1+x_2+\cdots +x_n\rfloor ,$$

and

$$\begin{array}{llll}\hfill A_k& =\{(x_1,x_2,\dots ,x_n)\in {[0,1]}^n\mid \lfloor x_1+x_2+\cdots +x_n\rfloor =k\},k=0,1,2,3.& \hfill & \\ \hfill & =\{(x_1,x_2,\dots ,x_n)\in {[0,1]}^n\mid k\le x_1+x_2+\cdots +x_n<k+1\}.& \hfill & \end{array}$$

Then $f$ is a simple function, and

$$\begin{array}{llllll}\hfill J_n& ={\int \; <!--\; nolimits\; -->}_0^1{\int \; <!--\; nolimits\; -->}_0^1\cdots {\int \; <!--\; nolimits\; -->}_0^1\lfloor x_1+x_2+\cdots +x_n\rfloor d{x}_{1}d{x}_2\cdots d{x}_n& \hfill & & \hfill & \\ \hfill & ={\int \; <!--\; nolimits\; -->}_{{[0,1]}^n}fd\mu & \hfill & & \hfill & \\ \hfill & =\underset{k=0}{\overset{n-1}{\sum }}\mathit{k\mu }(A_k)& \hfill (\mu (A_n)=0).& & \hfill & & \hfill \end{array}$$

We show that $\mu (A_k)=\mu (A_{n-1-k})$ for $k=0,1,\dots ,n-1$. We note first that $\mu (A_k)=\mu (\stackrel{\circ }{A_k})$, where

$$\stackrel{\circ }{A_k}=\{(x_1,x_2,\dots ,x_n)\in {[0,1]}^n\mid k<x_1+x_2+\cdots +x_n<k+1\},$$

because the set of points $(x_1,x_2,\dots ,x_n)$ such that $k=x_1+x_2+\cdots +x_n$ is a zero measure set.

Consider the symmetry $\phi$ relative to the center of the hypercube ${[0,1]}^n$:

$$\phi \{\begin{array}{ccc}\hfill {[0,1]}^n\hfill & \hfill \to \hfill & {[0,1]}^n\hfill \\ \hfill (x_1,x_2,\dots ,x_n)\hfill & \hfill \mapsto \hfill & (1-x_1,1-x_2,\dots ,1-x_n)\hfill \end{array}$$

Then, for all $(x_1,x_2,\dots ,x_n)\in {[0,1]}^n$,

$$\begin{array}{llll}\hfill (x_1,x_2,\dots ,x_n)\in \stackrel{\circ }{A_k}& ⟺k<x_1+x_2+\cdots +x_n<k+1& \hfill & \\ \hfill & ⟺n-k-1<(1-x_1)+(1-x_2)+\cdots +(1-x_n)<n-k& \hfill & \\ \hfill & ⟺\phi (x_1,x_2,\dots ,x_n)\in {\stackrel{\circ }{A}}_{n-1-k}.& \hfill & \end{array}$$

Since $\phi$ is an isometry, $\mu (\stackrel{\circ }{A_k})=\mu ({\stackrel{\circ }{A}}_{n-1-k})$, so

$$\mu (A_k)=\mu (A_{n-1-k}),k=0,1,\dots ,n-1.$$

With the trick of the child K.F. Gauss to sum the integers between $1$ and $100$, we write

$$\begin{array}{llll}\hfill J_n& =0\cdot \mu (A_0)+1\cdot \mu (A_1)+\cdots +\mathit{k\mu }(A_k)+\cdots +(n-1)\mu (A_{n-1})& \hfill & \\ \hfill J_n& =(n-1)\mu (A_{n-1})+(n-2)\mu (A_{n-2})+\cdots +(n-1-k)\mu (A_{n-1-k})+\cdots +0\cdot \mu (A_0)& \hfill & \\ \hfill & =(n-1)\mu (A_0)+(n-2)\mu (A_1)+\cdots +(n-1-k)\mu (A_k)+\cdots +0\cdot \mu (A_{n-1}).& \hfill & \end{array}$$

Therefore

$$\begin{array}{llll}\hfill 2J_n& =(n-1)\left(\mu (A_0)+\mu (A_1)+\cdots +\mu (A_{n-1})\right)& \hfill & \\ \hfill & =(n-1)\mu ({[0,1]}^n)& \hfill & \\ \hfill & =(n-1).& \hfill & \end{array}$$

So

$$\begin{array}{llll}\hfill J_n& ={\int \; <!--\; nolimits\; -->}_0^1{\int \; <!--\; nolimits\; -->}_0^1\cdots {\int \; <!--\; nolimits\; -->}_0^1\lfloor x_1+x_2+\cdots +x_n\rfloor d{x}_{1}d{x}_2\cdots d{x}_n=\frac{n-1}{2}.& \hfill & \\ \hfill & & \hfill \end{array}$$