Closed-Form solution for nested integrals of this polynomial?

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

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!

• One easy note: by symmetrization, the product can be written as $\prod_{i\neq j}(t_i-t_j)^2$... Jul 8 '20 at 4:56
• Yeah, thanks, good point, I've just included it in the question
– Lab
Jul 10 '20 at 17:51
• Another note related to symmetry: the integral in question is equal to $\frac{1}{a!} \int_{-1}^{1} \cdots \int_{-1}^{1} \prod_{i < j} (t_i - t_j)^{4} \, dt_{a-1} \cdots dt_{0}$. Jul 12 '20 at 4:49

I'm using (verbatim) the easy-to-read paper [1] to show that $$J_n:=\idotsint\limits_{-1 [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 $$\det_{1\leqslant i,j\leqslant 2n}\frac{x_i-x_j}{x_i+x_j}=\prod_{1\leqslant i (stated near the beginning of section $$9$$; see this question). This leads to \eqref{result} directly.

• If we use $b_k$ to denote $\prod_{\ell=1}^{k} \ell^{k-\ell+1} = \prod_{\ell=1}^{k} \ell!$, then we can write $\prod_{0 < j < i < 2n} \frac{i-j}{i+j} = 2^{n-1} \frac{b_{2n-2} b_{2n-1}}{b_{4n-3}^{1/2}} \left( \frac{(2n-2)!}{(4n-3)!} \right)^{1/2}$. Jul 13 '20 at 7:24
• @Jason: I tried to play it around, but all the alternate forms look weird, so I leave it as is. Jul 13 '20 at 8:24
• @metamorphy: Nice answer and interesting reference. (+1) Jul 13 '20 at 16:20
• @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. Jul 18 '20 at 13:50
• @metamorphy: You're welcome. Interesting reference, thanks for the hint. Jul 19 '20 at 18:38