An $n$ dimensional integral

Let $n \geq 3$ be a natural number. Evaluate the integral:

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


Solution

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:

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