# Multi-dimensional gaussian integral with non-symmetric & non-hermitian coefficient matrix

There is a commonly used formula in quantum field theory, $$$$\int\prod_{i} \frac{dz_{i}^{\dagger} dz_{i}}{2\pi} e^{-z^{\dagger}Az}=\frac{1}{\det(A)}, (1)$$$$ where $$z=x+\text{i}y$$ is a complex vector, and $$A$$ is a matrix with positive definite hermitian part (We know a matrix can always be written as sumation of hermitian part and anti-hermitian part, $$A=\frac{A+A^{\dagger}}{2}+\frac{A-A^{\dagger}}{2}$$). The difficulty is coefficient matrix $$A$$ is neither hermitian nor complex symmetric.

For hermitian matrix $$A$$, Eq (1) is easy to verify, because a hermitian matrix can be diagonalized by a unitary matrix.

For complex symmetric $$A$$, the problem is a little bit more tougher although there are no less than two ways to manage.

1. The first way is analytical continuation from a real symmetric matrix. For real symmetric matrix A,

$$$$z^{\dagger}A z=[x^{T},y^{T}] \begin{bmatrix} A & \\ & A \\ \end{bmatrix} \begin{bmatrix} x \\ y \\ \end{bmatrix},$$$$

the integral is $$\sqrt{\det(A)}^{-1}\times\sqrt{\det(A)}^{-1}=\det(A)^{-1}$$.

If the imaginary part emerges, $$A\rightarrow A+iB$$, where $$B$$ is also a real symmetric matrix. The positive definiteness of the original real symmetric $$A$$ guarantees that there exists a finite range on the real axis, i.e., $$\epsilon\in[-\alpha,\alpha]$$, in which the matrix $$A+\epsilon B$$ is still positive definite so that the integral remains $$\det(A)^{-1}$$.

When the $$\epsilon$$ becomes complex number the real part of which is in the range, $$\text{Re}({\epsilon})\in[-\alpha,\alpha]$$, while the imaginary part is arbitary, the integral always converges and the determinant is not going to touch zero in the range, so the quantities on both sides of Eq (1) exist. According to the mathematical theorem (equal on piece, equal all over) for analytical functions, when $$\epsilon = 0+\text{i}$$, the equation still holds, so for complex symmetric $$A$$ with positive definite real part, Eq (1) is verified. Here we noticed that the original statement is A is a matix with positive definite hermitian part. For complex symmetric matrix, the hermitian part, $$\frac{A+A^{\dagger}}{2}$$ is the real part and the anti-hermitian part, $$\frac{A-A^{\dagger}}{2}$$ is the imaginary part.

2. The second way is based on the fact that if the real part of a complex symmetric matrix is positive definite, it can be diagonalized simultaneously with the imaginary part. There is a good description in section 3 and 4 from David K. zhang's note, enter link description here.

I can not figure out the proof for a general complex matrix with positive definite hermitian part.