Back some time ago, a friend challenged me to find this integral $$I = \int_0^1\int_0^1\cdots\int_0^1\min(x_1+x_2+\cdots+x_n, 1)\text{ d}x_1\text{ d}x_2\cdots\text{ d}x_n$$
I started off by trying to find a pattern by solving smaller integrals, which led me to find that each integral is basically the multidimensional "volume" of a $n$-dimensional "cube" with side length $1$, except with a small region sliced off, and this sliced region tends to $0$ as $n$ goes to infinity. So, I can say when $n\to\infty$, the integral is $1$, but this isn't helpful at all lol.
I have not made any further progress this way, so I just tried numerically bashing everything. After iterating up to $x_{15}$, I found what seems to be a pattern.
I've made the following conjecture $$I \stackrel{?}{=} \frac{(n+1)!-1}{(n+1)!}$$
However, I am unable to prove this. Anyone able to help?
We can rewrite the integral as
$$\int\limits_{[0,1]^n\cap\sum_i x_i \leq 1}\sum_i x_i \ d^n \mathbf{x} + \int\limits_{[0,1]^n\cap\sum_i x_i \geq 1}d^n \mathbf{x} = 1 - \int\limits_{[0,1]^n\cap\sum_i x_i \leq 1}1-\sum_i x_i \ d^n \mathbf{x}$$
$$ = 1 - \int\limits_{[0,1]^{n+1}\cap\sum_i x_i \leq 1} d^{n+1} \mathbf{x} \equiv 1 - \operatorname{vol}(n+1)$$
in other words, $1$ minus the volume of the sliced corner in $n+1$ dimensions. You can show that
$$\operatorname{vol}(n+1) = \int\limits_0^1\cdots \int\limits_0^{1-x_1-\cdots-x_{n-1}}(1-x_1-\cdots-x_n) \:dx_n\cdots dx_1$$ $$ = \int\limits_0^1\cdots\int\limits_0^{1-x_1-\cdots-x_{n-2}}\frac{(1-x_1-\cdots-x_{n-1})^2}{2}\:dx_{n-1}\cdots dx_1$$
$$= \cdots = \int_0^1 \frac{(1-x_1)^n}{n!}\:dx_1 = \frac{1}{(n+1)!}$$
Therefore the integral in question is $\boxed{1-\frac{1}{(n+1)!}}$