Find multiple integrals $I_{\max}(k,n)$ and $I_{\min}(k,n)$ in various ways

Question

1. $I_{\max}(k,n)=\underbrace{\int\limits_0^1\int\limits_0^1\dots\int\limits_0^1}_k\left(\max\limits_{1\le i\le k}x_i\right)^n\,dx_1dx_2\dots dx_k$
2. $I_{\min}(k,n)=\underbrace{\int\limits_0^1\int\limits_0^1\dots\int\limits_0^1}_k\left(\min\limits_{1\le i\le k}x_i\right)^n\,dx_1dx_2\dots dx_k$

Various solutions are welcome.

P.S. I added my solution as an answer.

Answer 1

1) Evaluation of $I_{\max}(k,n)$

It is clear that $\max x = x$ so $I_{\max}(1,n)=\int\limits_0^1 x_1^n\,dx_1=\frac{1}{n+1}$.

Also it is obvious that $\max\limits_{1\le i\le k}x_i=\max\left(x_k,\max\limits_{1\le i\le k-1}x_i\right)$. Then

$$\begin{align}I_{\max}(k,n)&=\underbrace{\int\limits_0^1\int\limits_0^1\dots\int\limits_0^1}_k\left(\max\left(x_k,\max\limits_{1\le i\le k-1}x_i\right)\right)^n\,dx_1dx_2\dots dx_k=\\ &=\underbrace{\int\limits_0^1\int\limits_0^1\dots\int\limits_0^1}_{k-1}\left(\int\limits_0^{\large\max\limits_{1\le i\le k-1}x_i}\left(\max\limits_{1\le i\le k-1}x_i\right)^n \,dx_k+\int\limits_{\large\max\limits_{1\le i\le k-1}x_i}^1 x_k^n \,dx_k\right)\,dx_1dx_2\dots dx_{k-1}=\\ &=\underbrace{\int\limits_0^1\int\limits_0^1\dots\int\limits_0^1}_{k-1}\left(\left.\left(\max\limits_{1\le i\le k-1}x_i\right)^n x_k\right|_0^{\large\max\limits_{1\le i\le k-1}x_i}+ \left.\frac{x_k^{n+1}}{n+1}\right|_{\large\max\limits_{1\le i\le k-1}x_i}^1\right)\,dx_1dx_2\dots dx_{k-1}=\\ &=\underbrace{\int\limits_0^1\int\limits_0^1\dots\int\limits_0^1}_{k-1}\left(\left(\max\limits_{1\le i\le k-1}x_i\right)^{n+1} +\frac{1-\left(\max\limits_{1\le i\le k-1}x_i\right)^{n+1}}{n+1}\right) \,dx_1dx_2\dots dx_{k-1}=\\ &=\frac{1}{n+1}\cdot\left(n\underbrace{\int\limits_0^1\int\limits_0^1\dots\int\limits_0^1}_{k-1}\left(\max\limits_{1\le i\le k-1}x_i\right)^{n+1}\, dx_1dx_2\dots dx_{k-1}+1\right)=\\ &=\frac{1}{n+1}\cdot\left(nI_{\max}(k-1,n+1)+1\right)\end{align}$$ By induction it can be shown that $I_{\max}(k,n)=\frac{k}{k+n}$:

• $\left.I_{\max}(k,n)\right|_{k=1}=\frac{1}{1+n}=\frac{1}{n+1}$
• $I_{\max}(k+1,n)=\frac{1}{n+1}\cdot\left(nI_{\max}(k,n+1)+1\right)=\frac{1}{n+1}\cdot\left(n\cdot\frac{k}{k+(n+1)}+1\right)=\frac{nk+k+n+1}{(n+1)(k+n+1)}=\\=\frac{(n+1)(k+1)}{(n+1)(k+n+1)}=\frac{k+1}{(k+1)+n}$

2) Evaluation of $I_{\min}(k,n)$

It is clear that $I_{\min}(1,n)=\int\limits_0^1 x_1^n\,dx_1=\frac{1}{n+1}$. We will find $I_{\min}(k,n)$ in the same manner as we did it for $I_{\max}(k,n)$:

$$\begin{align}I_{\min}(k,n)&=\underbrace{\int\limits_0^1\int\limits_0^1\dots\int\limits_0^1}_k\left(\min\left(x_k,\min\limits_{1\le i\le k-1}x_i\right)\right)^n\,dx_1dx_2\dots dx_k=\\ &=\underbrace{\int\limits_0^1\int\limits_0^1\dots\int\limits_0^1}_{k-1}\left(\int\limits_0^{\large\min\limits_{1\le i\le k-1}x_i} x_k^n \,dx_k+\int\limits_{\large\min\limits_{1\le i\le k-1}x_i}^1 \left(\min\limits_{1\le i\le k-1}x_i\right)^n \,dx_k\right)\,dx_1dx_2\dots dx_{k-1}=\\ &=\underbrace{\int\limits_0^1\int\limits_0^1\dots\int\limits_0^1}_{k-1}\left(\left.\frac{x_k^{n+1}}{n+1}\right|_0^{\large\min\limits_{1\le i\le k-1}x_i}+ \left.\left(\min\limits_{1\le i\le k-1}x_i\right)^n x_k\right|_{\large\min\limits_{1\le i\le k-1}x_i}^1\right)\,dx_1dx_2\dots dx_{k-1}=\\ &=\underbrace{\int\limits_0^1\int\limits_0^1\dots\int\limits_0^1}_{k-1}\left(\left(\min\limits_{1\le i\le k-1}x_i\right)^n-\frac{n}{n+1}\cdot\left(\min\limits_{1\le i\le k-1}x_i\right)^{n+1}\right) \,dx_1dx_2\dots dx_{k-1}=\\ &=I_{\min}(k-1,n)-\frac{n}{n+1}\cdot I_{\min}(k-1,n+1)\end{align}$$

By induction it can be shown that $I_{\min}(k,n)=\frac{k!n!}{(k+n)!}=\binom{k+n}{k}^{-1}$:

• $\left.I_{\min}(k,n)\right|_{k=1}=\frac{1!n!}{(1+n)!}=\frac{1}{n+1}$
• $I_{\min}(k+1,n)=I_{\min}(k,n)-\frac{n}{n+1}\cdot I_{\min}(k,n+1)=\frac{k!n!}{(k+n)!}-\frac{n}{n+1}\cdot\frac{k!(n+1)!}{(k+(n+1))!}=\\=\frac{k!n!}{(k+n+1)!}\cdot((k+n+1)-n)=\frac{(k+1)!n!}{((k+1)+n)!}$