Background:
In their paper, Sommers Crisanti Sompolinsky and Stein derive the spectral distribution of large random matrices $\mathbf{J}$ by studying the following integral: \begin{equation} I=\left[\int\left(\Pi \frac{d^{2} z_{i}}{\pi}\right) \exp \left\{-\epsilon \sum_{t}\left|z_{i}\right|^{2}-\sum_{i, J, k} z_{i}^{*}\left(\omega^{*} \delta_{i k}-J \bar{k}\right)\left(\omega \delta_{k j}-J_{k j}\right) z_{j}\right\}\right]_J \end{equation}
Where the brackets $[\dots]_J$ denote ensemble average with the following measure: $$ P(\mathbf{J})\propto \exp\left\{-\frac{N}{2\left(1-\tau^{2}\right)} \sum_{i j} J_{i j}^{2}+\frac{\tau N}{2\left(1-\tau^{2}\right)} \sum_{i j} J_{i j} J_{j i}\right\} $$
The parameter $-1<\tau<1$ represents the correlation between the off diagonal elements. When $\tau=1$ the matrix is symmetric.
By carrying out the average over the distribution of $J_{ij}$ and neglecting $\mathcal{O}(1/N)$ terms they find: $$ I=\int\left(\Pi \frac{d^{2} z_{i}}{\pi}\right) \exp \left\{-N\left(\epsilon r+\ln (1+r)+\frac{r x^{2}}{1+r(1+\tau)}+\frac{r y^{2}}{1+r(1-\tau)}\right)\right\}. $$ where $Nr\equiv\sum_i|z_i|^2$
My issue:
When I carry out the average over the $\mathbf{J}$ distribution I also find the $(1+r)$ term, but what escapes my understanding is how do they obtain $(1+r(1+\tau))$ and $(1+r(1-\tau))$ in the denominator and not $(1+r)$?
When I set $\tau=0$ I recover their results but when $\tau\neq0$ I cannot seem to understand how the value in the denominator is different from the value in the $\ln$. If someone understands or can guess how they did, I would appreciate it a lot if you could explain it to me.
Of course, any remark or advice is always appreciated. Thank you.
(Outline of my work): If this helps, here are the main steps of my derivation.
I expand the terms in the integral:
$$\begin{equation} I= \int\left(\prod_{i} \frac{d^{2} z_{i}}{\pi}\right) \exp \left\{-\epsilon \sum_{i}\left|z_{i}\right|^{2}-\sum_i |\omega|^2|z_i|^2+\sum_{ij}z_i^*\omega^*J_{ij}z_j+\sum_{ij}\omega z_i^* J_{ji}z_j-\sum_{i,j,k}z_i^*J_{ki}J_{kj}z_j -\frac{N}{2(1-\tau^2)}\sum_{ij}J_{ij}^2+\frac{\tau N}{2(1-\tau^2)}\sum_{ij}J_{ij}J_{ji}\right \} \end{equation} $$
Then, following their advice I decouple the terms quadratic in $J_{ij}$ with a complex Gaussian transformation, (based on the Hubbard Stratonovich method):
$$ \begin{equation} \exp \left(-\sum_{i,j,k}z_i^*J_{ki}J_{kj}z_j\right)=\exp \left(-\sum_{k}\left | \sum_j z_jJ_{kj}\right|^2\right)= \int \left(\prod_{k} d^2m_k\right) \exp \left(- \sum_k m_k m_k^* \pm i \sum_{kj}z_j^*J_{kj} m_k \pm i \sum_{kj}z_jJ_{kj} m_k^*\right ) \end{equation} $$
Then, I can perform the average over $\mathbf{J}$: \begin{equation} I\propto \int\int\left(\prod_{i} \frac{d^{2} z_{i}}{\pi}\right)\left(\prod_{i} d^{2} m_{i}\right) \exp \left\{-\sum_i|z_i|^2(\epsilon+|\omega|^2)-\sum_i|m_i|^2+\frac{1}{2N}\sum_{i j}b_{ij}^2 + \frac{\tau}{2N}\sum_{ij}b_{ij}b_{ji} \right \} \end{equation} where: \begin{equation} b_{ij}=(\omega^*z_i^*z_j+\omega z_iz_j^*+\mathrm{i}(m_iz_j^*+m_i^*z_j)) \end{equation}
Ignoring the $\mathcal{O}(1/N)$ terms, I integrate over the $\mathrm{d}^2m_i$ and finally obtain:
\begin{equation} = \int \exp \left\{-N\left ( \epsilon r+\log(1+r) +\frac{rx^2(1-r\tau + \tau r^2 (\tau +1))}{1+r}+\frac{ry^2(-1+r\tau + \tau r^2 (\tau -1))}{1+r}\right) \right \} \end{equation}
Which is correct for $\tau=0$ but clearly wrong when $\tau\neq0$. I went carefully over my computations so my mistake is more of understanding and doing something wrong rather than out of negligence.
Again, any remark or criticism that would help me improve, is appreciated. Thank you for your time and your help!
