An $n$ dimensional integral

[Tolaso J Kos]

Let $\lfloor \cdot \rfloor$ denote the floor function and let $n \in \mathbb{N}$ such that $n \geq 3$. Evaluate the integral:

$$\mathcal{J}=\int \limits_{[0, 1]^n} \left \lfloor x_1+x_2+\cdots+x_n \right \rfloor\, {\rm d}\left ( x_1, x_2, \dots, x_n \right )$$

Answer

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$\mathcal{J}=\frac{n-1}{2}$.

[Demetres]

Changing variables we have
\[ I = \int_{[0,1]^n} \lfloor n-(y_1+\cdots+y_n)\rfloor \, dy_1 \, dy_2 \, \cdots \, dy_n\]

The set of points $(y_1,\ldots,y_n) \in [0,1]^n$ for which $y_1+\cdots + y_n$ is an integer has measure zero. For all other points we have
\[ \lfloor n-(y_1+\cdots+y_n)\rfloor = n-1 - \lfloor (y_1+\cdots+y_n)\rfloor\]
Thus
\[ I = \int_{[0,1]^n} \left(n-1 - \lfloor (y_1+\cdots+y_n)\rfloor \right)\, dy_1 \, dy_2 \, \cdots \, dy_n = n-1-I\]
and so $I = (n-1)/2$.

[Tolaso J Kos]

Yet another one.

We are making use of a lemma stating that:

Lemma: For any real constant $c$ it holds that:

$$\int_{0}^{1}\left \lfloor x+c \right \rfloor\, {\rm d}x = c$$

Proof:
Split $c$ into its integer part $n$ and fractional part $c'$. For $0\leq x<1-c'$, we have $\lfloor x+c\rfloor=n$. For $1-c'\leq x<1$, we have $\lfloor x+c\rfloor=n+1$. Thus $$ \int_0^1 \lfloor x+c\rfloor\,{\rm d}x=\int_0^{1-c'}n\,{\rm d}x+\int_{1-c'}^1 (n+1)\,{\rm d}x = n(1-c')+(n+1)c'=n+c'=c$$

Now making use of the lemma we have for the $n$ dimensional integral that:

\begin{align*}
\int \limits_{[0, 1]^n}\left \lfloor x_1+x_2+\cdots+x_n \right \rfloor \, {\rm d}x &=\underbrace{\int_{0}^{1}\int_{0}^{1}\cdots \int_{0}^{1}}_{n-1}\left ( x_2+x_3+\cdots+x_n \right ) \, {\rm d}\left ( x_2, x_3, \dots, x_n \right ) \\
&= \frac{n-1}{2}
\end{align*}