The question I have is quite specific. So in the hope that this post might help others in the future, my problem boils down to solving the following integral:


$$I\_\tau=\int \prod\_{i, j=1}^{N} d J\_{i j} \exp \left\{-\frac{N}{2(1-\tau^2)}\left( \sum\_{i, j, k} J\_{k i} A\_{i j} J\_{k j}-\tau\sum\_{ij}J\_{ij}J\_{ji}\right)\right\}$$
Assuming we can compute the determinant of $A$, is there a general identity to compute $I\_\tau$?




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Simplifying their argument, in the following paper [1], the authors compute the following Gaussian integral over all $N\times N$ real asymmetric matrices $J\_{ij}$: 


$$I=\int \prod\_{i, j=1}^{N}d J\_{i j} \exp \left\{-\frac{N}{2} \sum\_{i, j, k} J\_{k i} A\_{i j} J\_{k j}\right\}$$


Where $A\_{i j}=\frac{z\_{i}^{\*} z\_{j}}{N}+\frac{z\_{j}^{\*} z\_{i}}{N}+\delta\_{i j}$ and $z\_i\in \mathbb{C}$.


If I define the vector $\mathbf{x}$ such that:
$$\mathbf{x}:=\left(\begin{array}{c}
J\_{11} \\
J\_{12} \\
\vdots \\
J\_{1 n} \\
J\_{21} \\
J\_{22} \\
\vdots \\
J\_{n n}
\end{array}\right) \in \mathbb{R}^{n^{2}}$$
$$\implies x\_{N(i-1))+j}=J\_{ij}$$
Then we can represent $I$ as the following integral:
$$I=\int \left(\prod\_{i, j=1}^{N} \mathrm{d}x\_{N(i-1))+j} \right)\exp \left(-\frac{N}{2} \mathbf{x}^T\Sigma \mathbf{x}\right)$$
Where $\Sigma$ simply is:
$$\Sigma=\left(\begin{array}{cccc}
\mathbf{A} & 0 & \cdots & 0 \\
0 & \mathbf{A} & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & \mathbf{A}
\end{array}\right)$$


Thus $I\propto \frac{1}{\sqrt{|\Sigma|}}=|A|^\frac{-N}{2}$.


Defining $r\equiv\frac{1}{N}\sum\_i |z\_i|^2$, the authors observe that to order $1/\sqrt{N}$, the matrix $A$ has all but two of its eigenvalues equal to one. The two exceptions are the eigenvectors $z\_i$ and $z^\*\_i$ , both with eigenvalue $(1+r)$. This observation allows them to compute the determinant of $\Sigma$ and thus the Gaussian integral $I$.




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Now, considering the integral $I\_\tau$ using the same method:
$$I\_\tau=\int \left(\prod\_{i j} \mathrm{d} J\_{i j}\right)\exp \left\{-\frac{N}{2(1-\tau^2)}\left( \sum\_{i, j, k} J\_{k i} A\_{i j} J\_{k j}-\tau\sum\_{ij}J\_{ij}J\_{ji}\right)\right\}$$
$$I\_\tau=\int\left(\prod\_{i j} \mathrm{d} J\_{i j}\right) \exp \left\{-\frac{N}{2(1-\tau^2)} \sum\_{i j, k, l} J\_{i j}\left(A\_{j l} \delta\_{i k}-\tau\delta\_{i l} \delta\_{j k}\right) J\_{k l}\right\}$$
$$\implies I\_\tau=\int\left(\prod\_{i, j=1}^{N} \mathrm{d} x\_{N(i-1))+j}\right) \exp \left(-\frac{N}{2} \mathbf{x}^{T} \Sigma \mathbf{x}\right)$$
Where this time:
$$\Sigma\_{n(i-1)+j,n(k-1)+l}=A\_{j l} \delta\_{i k}-\tau\delta\_{i l} \delta\_{j k} \quad \forall i, j, k, l \in \mathbb{N} \cap[1, N]$$


For large $N$, how can we compute the determinant of $\Sigma$ this time? It seems to be a non-trivial question.


Any remark or comment is always appreciated, thank you.




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[1]: Rajan, K., & Abbott, L. F. (2006). Eigenvalue spectra of random matrices for neural networks. Physical review letters, 97(18), 188104.


Link: <https://www.researchgate.net/profile/Kanaka_Rajan/publication/6643146_Eigenvalue_Spectra_of_Random_Matrices_for_Neural_Networks/links/0912f50f7347f4863b000000.pdf>



A general formula for Gaussian integrals over matrix elements
Only the symmetric part of $A$ contributes to the integrand, so we may assume $A$ is symmetric and diagonalize it as $A=O\Lambda O^T$ with $O$ orthogonal and $\Lambda={\rm diag}\,(\lambda\_1,\lambda\_2,\ldots\lambda\_N)$. The orthogonal matrix may be removed by a change of variables, so without loss of generality we can take a diagonal $A$. So we seek the integral
$$I=\int \prod\_{i, j=1}^{N} d J\_{i j} \exp \left\{-\frac{N}{2(1-\tau^2)}\left( \sum\_{i, j} \lambda\_j J\_{ij}^2-\tau\sum\_{ij}J\_{ij}J\_{ji}\right)\right\}.$$
I now proceed similarly to [here](https://mathoverflow.net/a/350071/11260). Decompose the sum over $i,j$ as
$$\sum\_{ij}\lambda\_j J\_{ij}^2-\tau\sum\_{ij}J\_{ij}J\_{ji}=$$
$$\qquad\qquad=\sum\_{i}(\lambda\_i-\tau)J\_{ii}^2+\sum\_{i<j}\left(\lambda\_j J\_{ij}^2+\lambda\_i J\_{ji}^2-2\tau J\_{ij}J\_{ji}\right)$$
$$\qquad\qquad\equiv\sum\_{i} A\_i+\sum\_{i<j} B\_{ij}.$$
Then perform the Gaussian integrals separately for each term in the sum, 
$$I=\left(\prod\_{i=1}^N\int e^{-\beta A\_i}dJ\_{ii}\right)\left(\prod\_{i<j=1}^{N}\int\int e^{-\beta B\_{ij}}dJ\_{ij}dJ\_{ji}\right)$$
$$\qquad\qquad=(\pi/\beta)^{N^2/2}\left(\prod\_{i=1}^N(\lambda\_i-\tau)^{-1/2}\right)\left(\prod\_{i<j=1}^N(\lambda\_i\lambda\_j-\tau^2)^{-1/2}\right),$$
where I have defined $\beta=\frac{1}{2}N(1-\tau^2)^{-1}$, and assumed that $\beta>0$, $\lambda\_i>\tau$ for all $i$. 


So knowledge of only the determinant of $A$ is not sufficient to evaluate the integral, you need to know the individual eigenvalues -- not just their product. In particular, if $\tau$ happens to be close to one particular eigenvalue of $A$, then that eigenvalue will dominate the integral.


