Let $d\ge 1$ and $m\ge 0$ be integers and let $\left\{ A_j \right\}_{j=1}^d\in ({\mathbb R}^d - \vec{0})$ be real numbers. By solving Spectral densities of finite dimensional sample covariance matrices we discovered the following identity. We have: \begin{align} &(-1)^{m d+\binom{d}{2}}\int\limits_{0< \nu_1 < \cdots < \nu_d < \infty} \prod\limits_{1\le i < j \le d}(\nu_j-\nu_i) \cdot \left[\prod\limits_{j=1}^d \nu_j^m\right] \cdot e^{-\sum\limits_{j=1}^d A_j \nu_j}\cdot \prod\limits_{j=1}^d d\nu_j \\[10pt] = {} & \left| \begin{array}{ccc} \partial_{A_1}^{m+0} & \cdots & \partial_{A_1}^{m+d-1}\\ \partial_{A_2}^{m+0} & \cdots & \partial_{A_2}^{m+d-1}\\ \vdots& \ddots & \vdots\\ \partial_{A_d}^{m+0} & \cdots & \partial_{A_d}^{m+d-1} \end{array} \right| \frac{1}{\prod\limits_{j=1}^d (\sum\limits_{\xi=j}^d A_\xi)} \\[10pt] %= {} & {\mathfrak S}^{(d)}_m \cdot\frac{\prod\limits_{j=1}^{d}(m+j-1)!}{\prod\limits_{j=1}^d \left(\sum\limits_{\xi=j}^d A_\xi\right)^{m+d-1+j}}\\[10pt] = {} &\frac{P_{m,d}^{(\binom{d}{2} m + (d-2)^2)}(\vec{A})}{\prod\limits_{j=1}^d (\sum\limits_{\xi=j}^d A_\xi)^{(d-j+1)(m+j-1)+1}} \end{align} Here $$ P_{m,d}^{(\binom{d}{2} m + (d-2)^2)}(\vec{A}) $$ is a multinomial of order $(\binom{d}{2} m + (d-2)^2)$ in $\vec{A}:=\left(A_\xi\right)_{\xi=1}^d$.
Note: The old version that I posted yesterday was obtained by multiplying the partial derivatives in the expansion of the determinant being applied to the right hand side rather than by applying the product of the partial derivatives to the term on the right hand side as it should be. As such the old version was wrong. Unfortunately the result does not simplify to such an extent as I had it before. This version is right. I include a Mathematica code snippet that verifies it.
d = 4; m = RandomInteger[{0, 10}]; Clear[AA];
sigmas = Permutations[Table[i, {i, 1, d}]];
A = RandomReal[{1, 5}, d, WorkingPrecision -> 50];
NIntegrate[
Product[(nu[j] - nu[i]), {i, 1, d}, {j, i + 1, d}] Product[
nu[j]^m, {j, 1, d}] Exp[-Sum[A[[j]] nu[j], {j, 1, d}]],
Evaluate[Sequence @@
Table[{nu[xi], If[xi == 1, 0, nu[xi - 1]], Infinity}, {xi, 1, d,
1}]]]
(-1)^(m d +
Binomial[d,
2]) Expand[(Sum[
Signature[sigmas[[i]]] D[#,
Evaluate[
Sequence @@
Table[{AA[xi], m + sigmas[[i, xi]] - 1}, {xi, 1, d,
1}]]], {i, 1, Length[sigmas]}]) & /@ {1/
Product[Sum[AA[xi], {xi, j, d}], {j, 1, d}]}] /.
AA[n_] :> A[[n]]
I stumbled on this identity by simply generating the left hand side in Mathematica for particular values of $d$ and $m$ and then simplifying the result.
Now, my question is twofold. Firstly how would you go about pfinding the multinomial in the numerator. Secondly how would you generalize this identity in the case when the exponent $m$ in the integrand in the top equation is not constant but instead depends on the index $j$.
