$$ I = \int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \text{Max}(|(x_1-x_2)(y_1-y_2)|,|(x_2-x_3)(y_2-y_3)|,|(x_3-x_1)(y_3-y_1)|) \ dx_1 \ dy_1 \ dx_2 \ dy_2 \ dx_3 \ dy_1$$

I am trying to find the exact value of this integration. I started with a bit simpler ones.

$$ I_1=\int_{0}^{1}\int_{0}^{1} |x_1-y_1|\ dx_1 \ dy_1=\frac{1}{3}$$ $$ I_2=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \text{Max}(|(x_1-y_1)|,|(x_2-y_2)|)\ dx_1 \ dy_1 \ dx_2 \ dy_2=\frac{7}{15}$$

Using Mathematica I got $$ I_3=\int_0^1....\int_0^1 \text{Max}(|(x_1-y_1)|,|(x_2-y_2)|,|(x_3-y_3)|)\ dx_1 \ dy_1 \ dx_2 \ dy_2 \ dx_3 \ dy_3=\frac{19}{35}$$

$I_1$ and $I_2$ can be calculated by carefully setting the limits of the variables. But this method is too complicated for solving $I$. Is there any approch to evaluate it? Any help is much appreciated.

EDIT: $$I_n=\int_0^1....\int_0^1 \text{Max}(|(x_1-y_1)|,...,|(x_i-y_i)|)\ dx_1 \ dy_1 ... \ dx_n \ dy_n=2^{2n+1}n\left(\text{B}\left(\frac{1}{2};n+1,n\right)-2\text{B}\left(\frac{1}{2};n+2,n\right)\right)$$ where $\text{B}(x;a,b)$ is the incomplete beta function.

In the original question, $I$ is integration over maximum of the pairwise product of difference of two variables.

  • $\begingroup$ Your statement is confusing. For $I$, you have $Max(x_i-x_j)(y_i-y_j)|$ while for $I_2$ you have $Max(x_i-y_i)(x_j-y_j)|$. Which is it? $\endgroup$ Commented Jan 15, 2020 at 4:48
  • $\begingroup$ Does not really matter. You can interchange the variables wlog. $\endgroup$
    – piepie
    Commented Jan 15, 2020 at 4:54
  • 2
    $\begingroup$ Try changes of variables $u_i=x_i-y_i$ and $v_i=x_i+y_i$. $\endgroup$ Commented Jan 15, 2020 at 5:04
  • $\begingroup$ @herbsteinberg, can you please elaborate a little? $\endgroup$
    – piepie
    Commented Jan 15, 2020 at 11:42
  • $\begingroup$ The change of variables simplifies the integration by cutting the number of arguments for integration in half. For each pair of $(x_i,y_i)$ the integration domain gets rotated by $45^o$ so it consists of two isosceles right triangles, and symmetry allows you to drop one and multiply by two. $\endgroup$ Commented Jan 15, 2020 at 19:18

1 Answer 1


Change of variables: Let $u=x-y$ and $v=x+y$. $\int_0^1\int_0^1f(|x-y|)dxdy=\frac{1}{2}\int_{-1}^1 f(|u|)\int_{|u|}^{2-|u|}dvdu=\int_0^1f(u)(2-2u)du$.

Apply to your question: Let $M=max(u_1,...,u_n)$.

$I_n=\int_0^1....\int_0^1 M\prod_{k=0}^n (2-2u_k)du_k$

The domain of integration can be divided into $n$ parts, where in each part $M=u_k$. Let $J_k$ be the integral over the $k^{th}$ part. Then $J_k=\int_0^1u_k(2-2u_k)(\prod_{j\ne k} \int_0^{u_k}(2-2u_j)du_j )du_k=\int_0^1 u_k(2-2u_k)(2u_k-u_k^2)^{n-1}du_k$.

Since all $J_k$ are equal, $I_n=n\int_0^1 u(2-2u)(2u-u^2)^{n-1}du$

  • 1
    $\begingroup$ Note: I checked for n=1 and n=2. Answers agree. $\endgroup$ Commented Jan 16, 2020 at 22:35
  • $\begingroup$ Thank you so much. My original question was different ($I$ in the question is different from $I_n$). But I guess this approach can be used to solve that. Will give it a try. $\endgroup$
    – piepie
    Commented Jan 17, 2020 at 3:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .