Complex Gaussian integral I want to calculate this integral $$I=\int \prod\limits_{i=1}^N{d \bar z_{i}dz_{i}\exp[\sum_{i,j=1}^N \bar z_{i}M_{ij}z_{j}+\sum_{i=1}^N(\bar z_{i}f_{i}+\bar g_{i}z_{i}})],$$ where $z_{i}$ are complex variables, $g_{i}$ and $f_{i}$ are complex numbers. 
One can calculate this integral in case $f=g$, by simple transformation $z′=z−M−1gz′=z−M−1g$. 
But I can't find convenient transformation in case of $I$. 
 A: The answer to your question can be found in Condensed Matter Field Theory by Altland and Simons (3rd chapter, eq 3.18) which you can also find here for free. However in that book the derivation is just outlined and since my edition of the book also contains a typo in the outline, I will go through it.
You can recast your integral to look like this:
$I=\int d(v^{\dagger},v) exp^{-v^{\dagger}Av + w^{\dagger}v +v^{\dagger} w'}$
where $v$ is the complex vector conataining your variables $z_i$, A is the matrix containing your $M_{ij}$ and the integration measure is a shorthand for "integrating the Real and Imaginary parts of each variable seperatly over all $\mathbb{C}$ ".It can be easily seen that this integration measure is the same as teh one you had by writting yours as $d(x+iy)d(x+iy)^{*}$ and treating $x$ and $y$ as independent variables.(I tried to keep the same notation and names of quantities as my reference in case you want to give it a look after all)
From here you can employ the change of variables (this is where the typo occurs in the book):
$v=u+A^{-1}w'$
$v^{\dagger}=u^{\dagger}+w^{\dagger}{A^{-1}}^{\dagger}=u^{\dagger}+w^{\dagger}A^{-1}$
A couple of comments for that change of variables:


*

*Since you are integrating over all different values of $Re\{z_i\}$ and $Im\{z_i\}$ you can treat them as independent variables. Equivalently you can treat 
$v$ and $v^{\dagger}$ as independent variables (and change them thusly).

*I am assuming  $A$ is a Hermitian matrix and so ${A^{-1}}^{\dagger}=A^{-1}$. I have yet to find the proof for the general case (According to Altland and Simons, it exists somewhere in the literture)


Making the change of variables you get:
$-v^{\dagger}Av + w^{\dagger}v + v^{\dagger}w'= -u^{\dagger}Au +w^{\dagger}A^{-1}w'$
which turns the integral into 
$I=\int d(u^{\dagger},u) e^{-u^{\dagger}Au + w^{\dagger}A^{-1} w'}=(\int d(u^{\dagger},u) e^{-u^{\dagger}Au} )e^{w^{\dagger}A^{-1} w'} = \pi ^N detA^{-1}e^{w^{\dagger}A^{-1} w'}$
The last step can be also found in the book I referenced and essentially it utilizes the fact that there is a unitary matrix $U$ such that $A=U^{\dagger}D_A U$, where $D_A$ is the Diagonal matrix containing the eigenvalues of $A$.Making that substitution into your integral you will end up with a product of $2N$ real $1-D$ Gaussian integrals that will give you the $(\sqrt{\pi})^{2N} detA^{-1}=\pi ^N detA^{-1}$ factor.  
