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I was wondering whether there is a closed-form solution for this (nested) integral:

$$ \int_{-1}^{1}\int_{t_{0}}^{1}\int_{t_{1}}^{1}...\int_{t_{a-2}}^{1}\prod_{\begin{array}{c} i<j\\ j=\{0,..,a-1\}\\ i=\{0,..,a-1\} \end{array}}\left(t_{i}-t_{j}\right)^{4}dt_{a-1}dt_{a-2}...dt_{0} $$

These are the results I get for $a=2$ and $a=3$:

$a=2$: $$ \int _{-1}^1\int _{t_0}^1(t_0-t_1)^4 dt_1 dt_0 = \frac{32}{15} $$

$a=3$: $$ \int _{-1}^1\int _{t_0}^1\int _{t_1}^1(t_0-t_1)^4 (t_0-t_2)^4 (t_1-t_2)^4dt_2dt_1dt_0 = \frac{8192}{33075} $$

Is there a known closed-form solution $\forall a$?

Edit: As noted by @Steven Stadnicki in the comments, the function inside the integrals can also be written as $$ \prod_{\begin{array}{c} i\ne j\\ j=\{0,..,a-1\}\\ i=\{0,..,a-1\} \end{array}}(t_{i}-t_{j})^{2} $$

Thanks!

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    $\begingroup$ One easy note: by symmetrization, the product can be written as $\prod_{i\neq j}(t_i-t_j)^2$... $\endgroup$ Jul 8 '20 at 4:56
  • $\begingroup$ Yeah, thanks, good point, I've just included it in the question $\endgroup$
    – Lab
    Jul 10 '20 at 17:51
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    $\begingroup$ Another note related to symmetry: the integral in question is equal to $\frac{1}{a!} \int_{-1}^{1} \cdots \int_{-1}^{1} \prod_{i < j} (t_i - t_j)^{4} \, dt_{a-1} \cdots dt_{0}$. $\endgroup$
    – Jason
    Jul 12 '20 at 4:49
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I'm using (verbatim) the easy-to-read paper [1] to show that $$J_n:=\idotsint\limits_{-1<t_1<\ldots<t_n<1}\prod_{1\leqslant j<i\leqslant n}(t_i-t_j)^4\,dt_1\ldots dt_n\color{blue}{=2^{n(2n-1)}\prod_{0<j<i<2n}\frac{i-j}{i+j}}.\tag{*}\label{result}$$ [1] N.G. de Bruijn, On some multiple integrals involving determinants, 1955

Observe that $J_n=\frac{1}{n!}\idotsint_{[-1,1]^n}=\frac{2^{n(2n-1)}}{n!}\idotsint_{[0,1]^n}$ (the first equality follows from the symmetry of the integrand; the second one is obtained after substituting $t_i=2x_i-1$ and renaming $x_i$ back to $t_i$).

Now the relevant result from the paper (see section $7$) is as follows. For $1\leqslant i,j\leqslant2n$, define $$F_{i,j}(t_1,\ldots,t_n)=\begin{cases}\varphi_i(t_k),&j=2k-1\\\psi_i(t_k),&j=2k\end{cases},\quad G_{i,j}=\int_\Omega\varphi_i(t)\psi_j(t)\,dt,$$ where $\varphi_k,\psi_k : \Omega\to\mathbb{R}$ are good enough (for all the integrals to exist), and let $F(t_1,\ldots,t_n)$ and $G$ be the corresponding $2n\times2n$ matrices. Then $$\idotsint_{\Omega^n}\det F(t_1,\ldots,t_n)\,dt_1\ldots dt_n=2^n{n!}\operatorname{Pf}\widehat{G},$$ where $\operatorname{Pf}\widehat{G}$ is the Pfaffian of $\widehat{G}=\frac12(G-G^\mathsf{T})$.

In our case, we choose $\Omega=[0,1]$, $\varphi_k(t)=t^{k-1}$ and $\psi_k(t)=\varphi_k'(t)$; from the answer to a recent question of yours (see also this one by myself), we know that $\det F(t_1,\ldots,t_n)$ is precisely our integrand in $J_n$. Thus, $J_n=2^{2n^2}\operatorname{Pf}\widehat{G}$ where $G_{1,1}=\widehat{G}_{1,1}=0$ and, otherwise, $$G_{i,j}=(j-1)\int_0^1 t^{i+j-3}\,dt=\frac{j-1}{i+j-2}\implies 2\widehat{G}_{i,j}=\frac{j-i}{i+j-2}.$$

So, it remains to compute $\operatorname{Pf}\widehat{G}$ or its square $\det\widehat{G}$. Now the relevant result from the paper is $$\det_{1\leqslant i,j\leqslant 2n}\frac{x_i-x_j}{x_i+x_j}=\prod_{1\leqslant i<j\leqslant 2n}\left(\frac{x_i-x_j}{x_i+x_j}\right)^2$$ (stated near the beginning of section $9$; see this question). This leads to \eqref{result} directly.

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  • $\begingroup$ If we use $b_k$ to denote $\prod_{\ell=1}^{k} \ell^{k-\ell+1} = \prod_{\ell=1}^{k} \ell!$, then we can write $\prod_{0 < j < i < 2n} \frac{i-j}{i+j} = 2^{n-1} \frac{b_{2n-2} b_{2n-1}}{b_{4n-3}^{1/2}} \left( \frac{(2n-2)!}{(4n-3)!} \right)^{1/2}$. $\endgroup$
    – Jason
    Jul 13 '20 at 7:24
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    $\begingroup$ @Jason: I tried to play it around, but all the alternate forms look weird, so I leave it as is. $\endgroup$
    – metamorphy
    Jul 13 '20 at 8:24
  • $\begingroup$ @metamorphy: Nice answer and interesting reference. (+1) $\endgroup$
    – epi163sqrt
    Jul 13 '20 at 16:20
  • $\begingroup$ @MarkusScheuer: Thank you. In fact I took it from this question. And it feels like one is missing a tiny step to make a complete answer there. $\endgroup$
    – metamorphy
    Jul 18 '20 at 13:50
  • $\begingroup$ @metamorphy: You're welcome. Interesting reference, thanks for the hint. $\endgroup$
    – epi163sqrt
    Jul 19 '20 at 18:38

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