$ \int_{[0,1]^n} \min(x_1,\ldots,x_n) \, dx_1\cdots dx_n $

Question

I need to compute:

$$ \int_{[0,1]^n} \min(x_1,\ldots,x_n) \, dx_1\cdots dx_n $$

I have shown that:

$$\int _{0}^{1} \min( x_{k} ,\dots,x_{n})^{m} dx_{k} = \min( x_{k+1} ,\dots,x_{n})^{m} -\frac{m}{m+1} \min( x_{k+1} ,\dots,x_{n})^{m+1}$$

and tried to do it recursively:

$ \begin{array}{l} \int _{0}^{1} ...\int _{0}^{1} \min( x_{1} ,...,x_{n}) dx_{1} ...dx_{n} =\int _{0}^{1} ...\int _{0}^{1} \min( x_{2} ,...,x_{n}) -\frac{1}{2} \min( x_{2} ,...,x_{n})^{2} dx_{2} ...dx_{n} =\\ =\int _{0}^{1} ...\int _{0}^{1} \min( x_{3} ,...,x_{n}) -\frac{2}{2} \min( x_{3} ,...,x_{n})^{2} +\frac{1}{3} \min( x_{3} ,...,x_{n})^{3} dx_{3} ...dx_{n} = \end{array}$

but I don't see how I can get something from this. Can anyone help?

Answer 1

Let $f(\mathbf{x})=\min(x_1,\ldots,x_n)$, where $\mathbf{x}=[x_1, \ldots,x_n]^\intercal\in\mathbb{R}^n$. There are $n$ possible cases to consider: $$\begin{align} E_j&=\{\mathbf{x}\in[0,1]^n: x_j=f(\mathbf{x})\}\\ &=\{\mathbf{x}\in[0,1]^n: x_j\leq x_1,\ldots, x_j\leq x_n\} \end{align} $$ The sets $E_j$ are mutually exclusive almost surely (they intersect in hyperplanes, and so the intersections have zero Lebesgue). By Fubini's theorem, $$ \begin{align} \int_{E_j}f(\mathbf{x})\,d\mathbf{x}&=\int^1_0\Big(\int^1_{x_1}\cdots\int^1_{x_1} x_1 \,dx_2\ldots dx_n\Big)\,dx_1\\ &=\int^1_0x_1(1-x_1)^{n-1}\,dx_1=B(2,n) \end{align} $$ where $B(m,n)$ is the beta function. Putting things together, $$\int_{[0,1]^n}f(\mathbf{x})\,d\mathbf{x}=nB(2,n)=n\frac{\Gamma(2)\Gamma(n)}{\Gamma(n+2)}=\frac{n!}{(n+1)!}=\frac{1}{n+1}$$