Let $N \ge 2$ be an integer and let $\vec{A}:= \left( A_j \right)_{j=1}^N \in {\mathbb R}^N_+$. Consider a following integral:
\begin{equation} {\mathfrak I}_N(\vec{A}) := \int\limits_{{\mathbb R}_{\ge}^N} \prod\limits_{1 \le i < j \le N} (\lambda_i - \lambda_j) \cdot e^{-\vec{A} \cdot \vec{\lambda}} \cdot d^N \vec{\lambda} \tag{1} \end{equation}
Here ${\mathbb R}_{\ge}^N := \left\{\left. (\lambda_j)_{j=1}^N \right| \infty > \lambda_1 \ge \lambda_2 \ge \cdots \ge \lambda_N \ge 0\right\}$ is the Weyl chamber.
This integral is of interest in Random Matrix Theory, in particular in computing moments and/or correlation functions of the Wishart distribution.
Even by taking a glimpse of the formula above one can see that computing this integral isn't that hard conceptually. There are at least two different ways one might tackle it, firstly by doing elementary integrations over consecutive $\lambda$'s from the one with the smallest index to the one with the biggest index, or secondly by expanding the Vandermonde determinant in the integrand in a series and integrating term by term. I chose the first approach and , using Mathematica, I guessed the following result:
\begin{equation} {\mathfrak I}_N(\vec{A}) := \sum\limits_{\begin{array}{lll} l_1+l_2+ \cdots + l_N = \binom{N}{3} \\ l_1 \ge 0, \cdots l_N \ge 0 \end{array}} {\mathfrak A}^{(N)}_{l_1,l_2,\cdots,l_N} \frac{\prod\limits_{\xi=1}^N ({\bar A}_\xi)^{l_\xi} }{ \prod\limits_{\xi=1}^N ({\bar A}_\xi)^{\xi \cdot N -\xi^2+1}} \tag{2} \end{equation} where ${\bar A}_j := A_1+\cdots+A_j$ for $j=1,\cdots, N$ and ${\mathfrak A}^{(N)}_{l_1,l_2,\cdots,l_N} $ are some positive integers that do not depend on $\vec{A}$. We have worked out those coefficients for $N=2,3,4,5$ as you can see below:
Clear[lmb]; Clear[A]; NN = 6;
FF = Product[
lmb[i] - lmb[j], {i, 1, NN}, {j, i + 1,
NN}] Exp[-Sum[A[j] lmb[j], {j, 1, NN}]];
Do[
FF = Integrate[ FF, {lmb[j], If[j == NN, 0, lmb[j + 1]], Infinity},
Assumptions -> And @@ Table[A[j] > 0, {j, 1, NN}]];
PrintTemporary["j=", j, " done"];
, {j, 1, NN}];
FF
(Denominator[ress] /. A[j_] :> Subscript[A, j]) // MatrixForm
(Table[Expand[(Numerator[ress[[nn]]] /.
A[j_] :> A[j] - If[j == 1, 0, A[j - 1]])], {nn, 1,
Length[ress]}] /. A[j_] :> Subscript[A, j]) // MatrixForm
It was impossible to go beyond $N=5$ due to the sheer number of calculations that overwhelmed Mathematica.
In view of this my question is the following. Would it be possible to find some smarter way of computing this integral. In particular could we etsablish some recurrence relations in $N$ to do the computations?
Owing to the excellent advice of the user MathWonk I can now write the full answer to my question.
Using the change of variables given in the comments above one finds a closed form expression as an action of a differential operator on the product of inverses of the quantities $({\bar A}_j)_{j=1}^N$. From that the recurrence in question reads:
\begin{equation} {\mathfrak I}_N(\vec{A}) = (-1)^{N-1} \left[ \prod\limits_{i=1}^{N-1} \left( \partial_{{\bar A}_i} + \cdots + \partial_{{\bar A}_{N-1}} \right) \cdot {\mathfrak I}_{N-1} \left( A_1, \cdots, A_{N-1} \right) \right] \cdot \frac{1}{{\bar A}_N} \tag{3} \end{equation} subject to ${\mathfrak I}_2(\vec{A}) := 1/({\bar A}_1^2 {\bar A}_2) $.
Now we are going to show that the solution to the recurrence $(3)$ is equation $(2)$. Before doing this let us note that the differential operator in $(3)$ can be always written as ${\hat O}_{N-1} := \prod\limits_{i=1}^{N-1} \left( \partial_{{\bar A}_i} + \cdots + \partial_{{\bar A}_{N-1}} \right) = \sum\limits_{d_1=0}^1 \sum\limits_{d_2=0}^2 \cdots \sum\limits_{d_{N-2}=0}^{N-2} \sum\limits_{d_{N-1}=1}^{N-1} {\mathfrak N}^{(N-1)}_{d_1,\cdots,d_{N-1}} \partial^{(d_1)}_{{\bar A}_1} \partial^{(d_2)}_{{\bar A}_2}\cdots \partial^{(d_{N-1})}_{{\bar A}_{N-1}} $ where $d_1+\cdots+d_{N-1}=N-1$. The coefficients ${\mathfrak N}^{(N-1)}_{\cdots}$ can be easily generated from the following observation ${\hat O}_N = (\partial_1 + \cdots \partial_N) \left. {\hat O}_{N-1} \right|_{i \rightarrow i+1} $ where the last term denotes the operator at $N \rightarrow N-1$ with all the subscripts being increased by one.
Therefore we have:
\begin{eqnarray} {\mathfrak I}_N(\vec{A}) &=& (-1)^{N-1} \left( \sum\limits_{d_1=0}^1 \sum\limits_{d_2=0}^2 \cdots \sum\limits_{d_{N-2}=0}^{N-2} \sum\limits_{d_{N-1}=1}^{N-1} {\mathfrak N}^{(N-1)}_{d_1,\cdots,d_{N-1}} \delta_{d_1+d_2+\cdots+d_{N-1} , N-1} \partial^{(d_1)}_{{\bar A}_1} \partial^{(d_2)}_{{\bar A}_2}\cdots \partial^{(d_{N-1})}_{{\bar A}_{N-1}} \right) \cdot \left( \sum\limits_{l_1+\cdots+l_{N-1} = \binom{N-1}{3}} {\mathfrak A}^{(N-1)}_{l_1,\cdots,l_{N-1}} \prod\limits_{p=1}^{N-1} \frac{1}{({\bar A}_p)^{p(N-1)-p^2+1-l_p}} \right) \cdot \frac{1}{{\bar A}_N} \\ &=& \left( \sum\limits_{d_1=0}^1 \sum\limits_{d_2=0}^2 \cdots \sum\limits_{d_{N-2}=0}^{N-2} \sum\limits_{d_{N-1}=1}^{N-1} {\mathfrak N}^{(N-1)}_{d_1,\cdots,d_{N-1}} \delta_{d_1+d_2+\cdots+d_{N-1} , N-1} \right) \cdot \left( \sum\limits_{l_1+\cdots+l_{N-1} = \binom{N-1}{3}} {\mathfrak A}^{(N-1)}_{l_1,\cdots,l_{N-1}} \prod\limits_{p=1}^{N-1} \frac{(-1)^{d_p} (p(N-1)-p^2+1-l_p)^{(d_p)}}{({\bar A}_p)^{p N-p^2+1-l_p-p+d_p}} \right) \cdot \frac{1}{{\bar A}_N} \tag{4} \end{eqnarray} Now, we relabel the $l$-indices in the last equation in $(4)$ in an obvious way, i.e. ${\bar l}_p := l_p +p - d_p$ for $p=1,\cdots,N-1$ and ${\bar l}_N=0$. Then we have $\sum\limits_{p=1}^N {\bar l}_p = \sum\limits_{p=1}^{N-1} l_p + \binom{N}{2} - \sum\limits_{p=1}^{N-1} d_p = \sum\limits_{p=1}^{N-1} l_p + \binom{N}{2} - (N-1) = \sum\limits_{p=1}^{N-1} l_p + \binom{N-1}{2} = \binom{N-1}{3} + \binom{N-1}{2} = \binom{N}{3}$. We conclude that the equation in question has the same functional form as $(2)$ except that $N \rightarrow N+1$. Proof completed.
Below we show a code snippet that generates the expressions in questions for $N=3,\cdots,7$ and verifies two things, firstly that the results always have the same functional form as in $(2)$ and secondly that the results match those being obtained using the old primitive method. Here we go:
Update:
This is just a note to emphasize that the results derived in here are very useful. The quantity ${\mathfrak I}_N(\vec{A})$ is nothing else but a generating function of various spectral moments and/or spectral correlation coefficients of a sample correlation matrix (Wishart matrix). The aforementioned quantities can be generated by differentiating our generating function appropriately. For example, if the underlying correlation matrix (the oracle matrix) is an identity matrix then the spectral moments read:
\begin{eqnarray} m_p &=& \frac{{\mathfrak P}_{N,T}}{N \cdot T^p} \int\limits_{{\mathbb R}_{\ge}^N} \left( \sum\limits_{j=1}^N \lambda_j^p \right) \cdot \prod\limits_{j=1}^N \lambda_j^{\frac{T-N-1}{2}} \cdot \prod\limits_{1 \le i < j \le N} \cdot e^{-\frac{1}{2} \sum\limits_{j=1}^N \lambda_j} d^N\lambda \\ &=& \frac{{\mathfrak P}_{N,T}}{N \cdot T^p} \sum\limits_{j=1}^N \left(\prod\limits_{\xi=1}^N \partial^{ \frac{T-N-1}{2} + \delta_{\xi,j} \cdot p}_{A_\xi} \right) \left. {\mathfrak I}_N(\vec{A}) \right|_{\vec{A}=1/2(1,1,\cdots,1)} \tag{5} \end{eqnarray}
where $p=0,1,2,\cdots$.
To illustrate that this really works we Took $N=2,3,4$ and $T=2 \zeta+1+N$ for $\zeta=1,\cdots,5$ and for each value of $(N,T)$ we compute the first three spectral moments, i.e. $p=0,1,2$, using the last equation in $(5)$. The results are given below where the rows & the columns are labelled by $N$ and by $T$ respectively. Here we go:
The reader can check with
Repetowicz, Przemysław; Richmond, Peter, Statistical inference of multivariate distribution parameters for non-Gaussian distributed time series, Acta Phys. Pol. B 36, No. 9, 2785-2796 (2005). ZBL1372.91126. that the results are correct. Indeed $m_1 = 1$ and $m_2 = N/T + 1 + 1/T$ as expected.
Update 1:
This is to illustrate how to calculate the spectral moments if the oracle matrix $\Sigma$ is not an identity matrix. Here we only give the result for $N=3$. Without loss of generality we take $\Sigma = diag(s_1,s_2,s_3)$, we define (see my answer to this question for details) three auxiliary quantities as follows:
\begin{eqnarray} {\mathfrak A}_1(\vec{\alpha}) &=& \frac{1}{4} \left(\frac{1}{s_1}-\frac{1}{s_2}\right) \cos \left(2 \alpha _1\right)+\frac{1}{8} \left(\frac{1}{s_1}-\frac{1}{s_2}\right) \cos \left(2 \alpha _1-2 \alpha _2\right)+\frac{1}{8} \left(\frac{1}{s_1}-\frac{1}{s_2}\right) \cos \left(2 \alpha _1+2 \alpha _2\right)+\frac{1}{4} \left(\frac{1}{s_2}-\frac{2}{s_3}+\frac{1}{s_1}\right) \cos \left(2 \alpha _2\right)+\frac{1}{4} \left(\frac{1}{s_2}+\frac{2}{s_3}+\frac{1}{s_1}\right)\\ {\mathfrak A}_2(\vec{\alpha}) &=& \frac{1}{8} \left(\frac{1}{s_1}-\frac{1}{s_2}\right) \sin \left(2 \alpha _1+\alpha _2-2 \alpha _3\right)+\frac{1}{8} \left(\frac{1}{s_1}-\frac{1}{s_2}\right) \sin \left(2 \alpha _1-\alpha _2+2 \alpha _3\right)+\frac{1}{8} \left(\frac{1}{s_2}-\frac{1}{s_1}\right) \sin \left(2 \alpha _1-\alpha _2-2 \alpha _3\right)+\frac{1}{8} \left(\frac{1}{s_2}-\frac{1}{s_1}\right) \sin \left(2 \alpha _1+\alpha _2+2 \alpha _3\right)+\frac{1}{32} \left(\frac{1}{s_1}-\frac{1}{s_2}\right) \cos \left(2 \alpha _1-2 \alpha _2-2 \alpha _3\right)+\frac{1}{32} \left(\frac{1}{s_1}-\frac{1}{s_2}\right) \cos \left(2 \alpha _1+2 \alpha _2-2 \alpha _3\right)+\frac{1}{32} \left(\frac{1}{s_1}-\frac{1}{s_2}\right) \cos \left(2 \alpha _1-2 \alpha _2+2 \alpha _3\right)+\frac{1}{32} \left(\frac{1}{s_1}-\frac{1}{s_2}\right) \cos \left(2 \alpha _1+2 \alpha _2+2 \alpha _3\right)+\frac{1}{8} \left(\frac{1}{s_2}-\frac{1}{s_1}\right) \cos \left(2 \alpha _1\right)+\frac{1}{16} \left(\frac{1}{s_2}-\frac{1}{s_1}\right) \cos \left(2 \alpha _1-2 \alpha _2\right)+\frac{1}{16} \left(\frac{1}{s_2}-\frac{1}{s_1}\right) \cos \left(2 \alpha _1+2 \alpha _2\right)+\frac{3}{16} \left(\frac{1}{s_2}-\frac{1}{s_1}\right) \cos \left(2 \alpha _1-2 \alpha _3\right)+\frac{3}{16} \left(\frac{1}{s_2}-\frac{1}{s_1}\right) \cos \left(2 \alpha _1+2 \alpha _3\right)+\frac{1}{16} \left(\frac{1}{s_2}-\frac{2}{s_3}+\frac{1}{s_1}\right) \cos \left(2 \alpha _2-2 \alpha _3\right)+\frac{1}{8} \left(\frac{1}{s_2}-\frac{2}{s_3}+\frac{1}{s_1}\right) \cos \left(2 \alpha _3\right)+\frac{1}{16} \left(\frac{1}{s_2}-\frac{2}{s_3}+\frac{1}{s_1}\right) \cos \left(2 \alpha _2+2 \alpha _3\right)+\frac{1}{8} \left(-\frac{1}{s_2}+\frac{2}{s_3}-\frac{1}{s_1}\right) \cos \left(2 \alpha _2\right)+\frac{1}{8} \left(\frac{3}{s_2}+\frac{2}{s_3}+\frac{3}{s_1}\right)\\ {\mathfrak A}_3(\vec{\alpha}) &=& \frac{1}{8} \left(\frac{1}{s_1}-\frac{1}{s_2}\right) \sin \left(2 \alpha _1-\alpha _2-2 \alpha _3\right)+\frac{1}{8} \left(\frac{1}{s_1}-\frac{1}{s_2}\right) \sin \left(2 \alpha _1+\alpha _2+2 \alpha _3\right)+\frac{1}{8} \left(\frac{1}{s_2}-\frac{1}{s_1}\right) \sin \left(2 \alpha _1+\alpha _2-2 \alpha _3\right)+\frac{1}{8} \left(\frac{1}{s_2}-\frac{1}{s_1}\right) \sin \left(2 \alpha _1-\alpha _2+2 \alpha _3\right)+\frac{3}{16} \left(\frac{1}{s_1}-\frac{1}{s_2}\right) \cos \left(2 \alpha _1-2 \alpha _3\right)+\frac{3}{16} \left(\frac{1}{s_1}-\frac{1}{s_2}\right) \cos \left(2 \alpha _1+2 \alpha _3\right)+\frac{1}{8} \left(\frac{1}{s_2}-\frac{1}{s_1}\right) \cos \left(2 \alpha _1\right)+\frac{1}{16} \left(\frac{1}{s_2}-\frac{1}{s_1}\right) \cos \left(2 \alpha _1-2 \alpha _2\right)+\frac{1}{16} \left(\frac{1}{s_2}-\frac{1}{s_1}\right) \cos \left(2 \alpha _1+2 \alpha _2\right)+\frac{1}{32} \left(\frac{1}{s_2}-\frac{1}{s_1}\right) \cos \left(2 \alpha _1-2 \alpha _2-2 \alpha _3\right)+\frac{1}{32} \left(\frac{1}{s_2}-\frac{1}{s_1}\right) \cos \left(2 \alpha _1+2 \alpha _2-2 \alpha _3\right)+\frac{1}{32} \left(\frac{1}{s_2}-\frac{1}{s_1}\right) \cos \left(2 \alpha _1-2 \alpha _2+2 \alpha _3\right)+\frac{1}{32} \left(\frac{1}{s_2}-\frac{1}{s_1}\right) \cos \left(2 \alpha _1+2 \alpha _2+2 \alpha _3\right)+\frac{1}{8} \left(-\frac{1}{s_2}+\frac{2}{s_3}-\frac{1}{s_1}\right) \cos \left(2 \alpha _2\right)+\frac{1}{16} \left(-\frac{1}{s_2}+\frac{2}{s_3}-\frac{1}{s_1}\right) \cos \left(2 \alpha _2-2 \alpha _3\right)+\frac{1}{8} \left(-\frac{1}{s_2}+\frac{2}{s_3}-\frac{1}{s_1}\right) \cos \left(2 \alpha _3\right)+\frac{1}{16} \left(-\frac{1}{s_2}+\frac{2}{s_3}-\frac{1}{s_1}\right) \cos \left(2 \alpha _2+2 \alpha _3\right)+\frac{1}{8} \left(\frac{3}{s_2}+\frac{2}{s_3}+\frac{3}{s_1}\right) \end{eqnarray}
We also define an average over a cube as follows:
\begin{equation} \left< \cdots \right> := \int\limits_{[0,2\pi]^3} \cdots |\cos(\alpha_2)| d\alpha_1 d\alpha_2 d\alpha_3 \end{equation}
We also define an auxiliary constant:
\begin{eqnarray} {\mathfrak C}^{({\mathfrak a},p)}_{l_1,l_2,l_3} := (3 {\mathfrak a} + p - l_2-l_3)! (l_2-l_1+l_3+1)! (l_1+1) \cdot \left( \frac{({\mathfrak a}+p)!}{(l_2-l_1)! ({\mathfrak a}+p-l_2)!} \frac{{\mathfrak a}!}{l_3! ({\mathfrak a}-l_3)!} 1_{0 \le l_3 \le {\mathfrak a} } + \frac{({\mathfrak a})!}{(l_2-l_1)! ({\mathfrak a}-l_2)!} \frac{({\mathfrak a}+p)!}{l_3! ({\mathfrak a}+p-l_3)!} 1_{0 \le l_1 \le l_2 \le {\mathfrak a} } + \frac{({\mathfrak a})!}{(l_2-l_1)! ({\mathfrak a}-l_2)!} \frac{({\mathfrak a})!}{l_3! ({\mathfrak a}-l_3)!} 1_{0 \le l_3 \le {\mathfrak a} } 1_{0 \le l_1 \le l_2 \le {\mathfrak a} } \right) \end{eqnarray}
where ${\mathfrak a} := (T-N-1)/2$.
Then the spectral moments of the sample correlation matrix read:
\begin{eqnarray} m_p &=& \frac{1}{4 (2\pi)^2 \cdot \det(\Sigma)^{\frac{T}{2}} } \frac{{\mathfrak P}_{N,T}}{N \cdot T^p} \sum\limits_{j=1}^N \left< \left(\prod\limits_{\xi=1}^N \partial^{ \frac{T-N-1}{2} + \delta_{\xi,j} \cdot p}_{A_\xi} \right) \left. {\mathfrak I}_N(\vec{A}) \right|_{\vec{A}=1/2({\mathfrak A}_1(\vec{\alpha}),{\mathfrak A}_2(\vec{\alpha}),{\mathfrak A}_3(\vec{\alpha}))} \right> \\ m_p &=& \frac{1}{4 (2\pi)^2 \cdot \det(\Sigma)^{\frac{T}{2}} } \frac{{\mathfrak P}_{N,T}}{N \cdot T^p} 2^{6 + 3 {\mathfrak a}+p} \cdot \\ && \sum\limits_{0 \le l_1 \le l_1 \le {\mathfrak a}+p} \sum\limits_{l_3=0}^{{\mathfrak a}+p} {\mathfrak C}^{({\mathfrak a},p)}_{l_1,l_2,l_3} \cdot %(3 {\mathfrak a} + p - l_2-l_3)! (l_2-l_1+l_3+1)! (l_1+1) % % % \left< \left. \frac{\left(\left(A_1+A_2\right) \left(l_1+2\right)+A_1 \left(-l_1+l_2+l_3+2\right)\right) }{A_1^{l_1+3} \left(A_1+A_2\right)^{-l_1+l_2+l_3+3} \left(A_1+A_2+A_3\right)^{3 {\mathfrak a}+p-l_2-l_3+1}} \right|_{A_j\rightarrow {\mathfrak A}_j(\vec{\alpha})} \right> % \tag{5} \end{eqnarray}
We confirm, as always, the formula $(5)$ by comparing it with the exact result being obtained using the Isserlis' theorem. We took $N,T=3,8$ and $p=0,1,2,3$ and we get the following: