Evaluation of an integral involving sign function 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.
 A: 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.
