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I am trying to solve the following integral, $$\int dz_1d\bar{z}_1\cdots dz_nd\bar{z}_n\:\exp(-\sum_{i,j}\bar{z}_i A_{ij}z_j),$$ where $A_{ij}$ is an $n\times n$ hermitian matrix. I know how to do this for the usual Gaussian integral with real coordinates, namely, let $A = O^{-1}\cdot D\cdot O$ where $D$ is a diagonal matrix; then let $y_i = O_{ij}x_j$, and simplify until you find $D_{ii}y^2_i$ and use a previous Gaussian integral which gives an answer of $\pi^{n/2}(\det{A})^{-1/2}$. I would like to do something similar for the complex integral, where I would perform the same change of variables, but then check for poles and use Cauchy's Residue theorem.

This was an identity given to me to use for solving path integrals, but I would like to know where it comes from (the above explantaion originally came from Zee's textbook, but for this more 'complex' integral (literally), I would like some guidance).

I would attempt what I said above, but I am not confident in using Residue Theorem and performing contour integrals (hence I am here). Any help/hints are appreciated, thanks!

Qmechanic
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MathZilla
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    You should be able to reduce this integral to a real integral by substituting $z = x + \mathrm{i} y$. – Janik Mar 25 '22 at 17:55
  • @Janik that didn't look like it would work initially, but at a closer look, it looks like I may get a product of two gaussian integrals (one with respect to $x$ and another with respect to $y$, and then use the original identity twice (posting this may have been premature). Thanks! – MathZilla Mar 25 '22 at 17:57
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    Possible duplicate: https://physics.stackexchange.com/q/451623/2451 – Qmechanic Mar 25 '22 at 18:44
  • @Qmechanic thanks for the comment, its not an Exact duplicate, but setting their two extra terms ($f^\dagger$ and $g$) to zero, I can follow it and complete the problem with the above suggestive comment as well. Thanks for point this out! (So I won't re-edit this question). – MathZilla Mar 25 '22 at 18:52
  • The generalization is treated in the answer. – Qmechanic Mar 25 '22 at 18:58

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