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