Evaluating $\int_{I^n} \left( \min_{1\le i \le n}x_i \right)^{\alpha}\,\, dx$

Let $\alpha \in \mathbb R$ and let's call $I:=[0,1]$. Evaluate
$$ \int_{I^n} \left( \min_{1\le i \le n}x_i \right)^{\alpha}\,\, dx. $$

Well, the case $n=1$ is easy and the integral equals $\frac{1}{\alpha+1}$, for every $\mathbb R \ni \alpha \ne - 1$.

I've done also the case $n=2$ and, if I'm not wrong, it's $\displaystyle \frac{2}{(\alpha+1)(\alpha+2)}$.

My big problem is that I cannot understand how to deal with the general case. Any ideas?

Thanks in advance.

Solution 1:

Denote $a_n^{\alpha}$ the integral. We have, since $\alpha\neq -1$, that \begin{align} a_n^{\alpha}&=n!\int_{\{0<x_1<\dots<x_n<1\}}(\min_{1\leq j\leq n}x_j)^{\alpha}dx_1\dots dx_n\\ &=n!\int_{\{0<x_1<\dots<x_n<1\}}x_1^{\alpha}dx_1\dots dx_n\\ &=n!\int_{\{0<x_2<\dots<x_n<1\}}\frac{x_2^{\alpha+1}}{\alpha+1}dx_2\dots dx_n\\ &=\frac 1{\alpha+1}na_{n-1}^{\alpha+1}. \end{align} By induction, $$a_n^{\alpha}=n!a_1^{\alpha+n-1}\prod_{j=1}^{n-1}\frac 1{\alpha+j}=n!\prod_{j=1}^n\frac 1{\alpha+j}.$$