Compute $\int_0^1\int_0^1...\int_0^1\lfloor{x_1+x_2+...+x_n}\rfloor dx_1dx_2...dx_n$ Compute $\int_0^1\int_0^1...\int_0^1\lfloor{x_1+x_2+...+x_n}\rfloor dx_1dx_2...dx_n$ where the integrand consists of the floor (or greatest integer less than or equal) function.
The case $n=1,2,3$ all can be solved geometrically. Actually the case $n=3$ is pretty fun where you get to dissect the unit cube. I guess the cases $n\ge4$ don't have useful geometric interpretations... so how to approach them (well, you still get to divide the sum into different cases: sum between 0 and 1, between 1 and 2, ..., between $n-1$ and $n$, and multiply the floor of the sum with the measure of the corresponding region) ?
 A: Let $f$ be the integrand; let $\Omega_k$ be the region of the unit hypercube over which the integrand has value $k$: that is, the region over which the sum of the coordinates lies between $k$ and $k+1$.
Note that the unit hypercube is invariant under the transformation which sends each coordinate $x_i$ to $1-x_i$. It follows that $\Omega_k$ (the region over which $\sum x_i$ lies between $k$ and $k+1$) and $\Omega_{n-1-k}$ (the region over which it lies between $n-k-1$ and $n-k$) have equal measure. 
Thus $$\int_{\Omega_k+\Omega_{n-1-k}}f = k \mu(\Omega_k)+(n-1-k)\mu(\Omega_{n-k-1})=(n-1)\mu(\Omega_k) \, ,$$
and so
$$\int_{[0,1]^n} f = \sum_{k=0}^{n-1} \int_{\Omega_k} f = \frac{1}{2}\sum_{k=0}^{n-1} \int_{\Omega_k+\Omega_{n-k-1}} f = \frac{n-1}{2} \sum_k \mu(\Omega_k) \, .$$
As the unit cube has measure $1$ and the $\Omega_k$ form a decomposition of it, it follows that the integral has value $\dfrac{n-1}{2}$.

Note: Let $\left<a\right>$ denote the fractional part of $a$. Then, by using the identity $\left<a\right>+\lfloor a \rfloor=a$, you can see that the above computation is equivalent to the following fact:

If $X_1,\dots, X_n$ are independent uniform random variables on $[0,1]$, the expected fractional part of their sum is $\frac{1}{2}$.

You could prove this directly by a similar symmetry argument (passing from the random variables $X_1,\dots,X_n$ to the random variables $1-X_1,\dots,1-X_n$), but I also think it's a more intuitively clear statement than the original one even before proof...
A: This can be done entirely through regular algebra and calculus.
Consider that
$$
\int_0^1 \lfloor x+a\rfloor dx = \int_a^{\lceil a\rceil}\lfloor a\rfloor dx+\int_{\lceil a\rceil}^{a+1} \lceil a\rceil dx \\
= \lfloor a\rfloor(\lceil a\rceil-a)+\lceil a\rceil(a+1-\lceil a\rceil)\\
= (\lceil a\rceil -\lfloor a\rfloor)a + \lceil a\rceil(\lfloor a\rfloor+1-\lceil a\rceil)\\
=a
$$
(if $a\in\mathbb{Z}$, then $\lceil a\rceil - \lfloor a\rfloor = 0$ and $\lfloor a\rfloor+1-\lceil a\rceil = 1$; otherwise, $\lceil a\rceil - \lfloor a\rfloor = 1$ and $\lfloor a\rfloor+1-\lceil a\rceil = 0$ - in either case, the result is $a$)
As such, only the inner integral involves the floor function, and we get
$$
\int_0^1\!\!\! \cdots\! \int_0^1\!\!\! \int_0^1 \lfloor x_1\!\!+\!x_2\!\!+\!\cdots\!+\!x_n\rfloor dx_1 dx_2 \cdots dx_n= \int_0^1\!\!\! \cdots\! \int_0^1 (x_2\!\!+\!\cdots\!+\!x_n)dx_2\cdots dx_n
$$
Which can be easily evaluated as
$$
1^{n-2}\int_0^1 x_2 dx_2+1^{n-2}\int_0^1 x_3 dx_3 + \cdots = \frac{n-1}2
$$
