TITLE: Multiple integral of min
QUESTION [6 upvotes]: How can this be proven?
$$
\int_0^1 \int_0^1...\int_0^1 min(x_1,x_2,...,x_n) dx_1dx_2...dx_n = \frac1{n+1} $$
I tried to split the last integral in $$\int_0^{min(x_1,x_2,...,x_{n-1})} x_ndx_n + \int_{min(x_1,x_2,...,x_{n-1})}^1 min(x_1,x_2,...,x_{n-1})dx_n$$
And then continue in this manner. 
So the function that remains to be integrated satisfies this recurrence relation
$$ f_n = (1-x)f_{n-1} + \int f_{n-1}  $$
I then thought to use the following notation 
$$I_n = \frac{ (-1)^{n+1}x^n }{n} +  \frac{ (-1)^{n}x^{n-1} }{n-1} $$
If we compute the last integral, the n-1 integrals apply to $I_2$
If we apply another integral we get $I_3+I_2$ and so on, which will look like Pascal's triangle. But I can't go any further.

REPLY [8 votes]: Here is another opproach:
First consider the set $A=\{(t_1,\dots,t_n):t_1\le\dots\le t_n\}$. In this set, your integral becomes
$$
\int_0^1 \int_0^{t_n}...\int_0^{t_2} \int_0^{t_1} x_1 dx_1dx_2...dx_n = \frac1{(n+1)!}.
$$
Since there are $n!$ sets like $A$ which are the union is $[0,1]^n$ then 
$$
\int_0^1 \int_0^1...\int_0^1 min(x_1,x_2,...,x_n) dx_1dx_2...dx_n = n!\times  \frac1{(n+1)!}=\frac1{n+1}.
$$