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Let $w>0$, $\lbrace k_{ij} \rbrace_{i,j\in [1,n]}$ be numbers taking either $0$ or $1$ as value and $i$ the usual complex number. I am trying to prove the following identity.

$$\int_{]-\infty,0]^n}dx_1\dots dx_n\, \prod_{j<l}\left[\mathrm{sgn}(x_l-x_j)\right]^{k_{jl}}\prod_{l=1}^n e^{(ia_{l}+w)x_l}=\prod_{l=1}^n\frac{1}{ia_l +w}\prod_{j<l}\left(\frac{i\, a_l-i\, a_j}{ia_l+ia_j+2w}\right)^{k_{jl}}$$

In the case where all $k_{ij}$ are equal to $0$, the integral is trivial because we can integrate each variable independently. But whenever a $k_{ij}$ is non zero, the variables are coupled through the sign function and it seems way harder.

Alexandre Krajenbrink
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  • Is $\mathbb{R}_{-}^{n} = (-\infty, 0]^n$? – Sangchul Lee Jun 02 '17 at 16:21
  • Yes, sorry for the bad notation. – Alexandre Krajenbrink Jun 02 '17 at 16:21
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    No worries, the notation made sense but I was not completely sure as it is also used to denote half-plane in some literature. – Sangchul Lee Jun 02 '17 at 16:22
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    the right hand side seems to be zero if $a_1=a_2=...=a$ can you prove that from the lhs (if at least one of the $k_{lj} \neq 0$)? – tired Jun 02 '17 at 17:16
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    for only one $k_ij\neq 1$ this is not too hard to prove.. i can see what i get for the general case – tired Jun 02 '17 at 17:45
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    i don't have the time to really do the calculations but the identity $\text{sgn}(x)=2 \theta(x)-1$ seems to reduce the problem to an equivalent integral over the $n$-simplex which should be doable ($\theta(x)$ is the Heaviside function). – tired Jun 02 '17 at 17:58

1 Answers1

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Write $\alpha_l = w + ia_l$ and apply the substitution $x_l \mapsto -x_l$. Then the identity is equivalent to

$$ \int_{[0,\infty)^n} \prod_{j < l} \left( 1 - 2 k_{jl} \mathbf{1}_{\{x_l > x_j\}} \right) \prod_{l=1}^{n} \alpha_l e^{-\alpha_l x_l} \, dx_l = \prod_{j<l} \left(1 - 2 k_{jl} \frac{\alpha_j}{\alpha_j + \alpha_l} \right). \tag{*}$$

Now consider the case where $n = 3$ and $k_{12} = k_{23} = 1$ but $k_{13} = 0$. Then the LHS of $\text{(*)}$ is

$$ 1 - \frac{2\alpha_1}{\alpha_1 + \alpha_2} - \frac{2\alpha_2}{\alpha_2 + \alpha_3} + \frac{2\alpha_1}{\alpha_1 + \alpha_2 + \alpha_3} \cdot \frac{2\alpha_2}{\alpha_2 + \alpha_3} $$

while the RHS is

$$ 1 - \frac{2\alpha_1}{\alpha_1 + \alpha_2} - \frac{2\alpha_2}{\alpha_2 + \alpha_3} + \frac{2\alpha_1}{\alpha_1 + \alpha_2} \cdot \frac{2\alpha_2}{\alpha_2 + \alpha_3} $$

They cannot equal and hence the identity does not hold.

I suspect that the identity is true exactly when $(k_{jl})$ satisfies sort of transitivity: for any $j < l$,

$$ k_{jl} = 1 \quad \Leftrightarrow \quad k_{jm} = 1 = k_{ml} \quad \text{for all} \quad j < m < l. $$

I am currently trying to prove that this condition implies the identity $\text{(*)}$. This essentially boils down to proving the identity when $k \equiv 1$, but even this does not seem an easy task.

Sangchul Lee
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  • Actually for $k\equiv 1$, the integral is easily doable. Indeed, there exists a Pfaffian structure such that $\prod_{i<j} \mathrm{sgn}(x_j-x_i)=\mathrm{Pf}(\mathrm{sgn}(x_j-x_i))$. Then $\int_{]-\infty,0]^n}dx_1\dots dx_n, \mathrm{Pf}(\mathrm{sgn}(x_j-x_i))\prod_{l=1}^n e^{(ia_{l}+w)x_l}$ is by theorem equal to $\mathrm{Pf}(\int_{]-\infty,0]^2}dx dy; \mathrm{sgn}(y-x)e^{(ia_{l}+w)x+(ia_{k}+w)y})$ which resumes to a Schur Pfaffian and gives the expected result for $k \equiv 1$. The problem when $k$ is not equal to 1 is that the Pfaffian structure is not complete... – Alexandre Krajenbrink Jun 30 '17 at 06:10
  • The reference for the theorem I mention is the paper of Bruijn entitled "On some multiple integrals involving determinants" https://pure.tue.nl/ws/files/1920642/597510.pdf. – Alexandre Krajenbrink Jun 30 '17 at 06:17