If $A\in\operatorname{Mat}_{n\times n}(\mathbb{C})$
We define $A_+$ as the complex symmetric part(not to be confused with Hermitian part): $$A_+=\frac{A+A^T}{2}$$ and the antisymmetric part as $$A_-=\frac{A-A^T}{2}$$ If the matrix $\operatorname{Re}A_+$ is positive definite, then $$ \int_{\mathbb{R}^n} \! d^n x ~e^{-\frac{1}{2} x^T A x} ~=~ \sqrt{\frac{(2\pi)^n}{\det A_+}}\tag{B}$$ Here, I am thinking if, $$x^TA x=x^T\left(A_++A_-\right)x=x^TA_+x$$ where $$x^TA_-x=0$$ Is my line of reasoning correct ? Then can I claim $$ \int_{\mathbb{R}^n} \! d^n x ~e^{-\frac{1}{2} x^T A x+Bx} ~=~ \sqrt{\frac{(2\pi)^n}{\det A_+}}\tag{B}e^{\left(B^TA_+^{-1}B\right)}$$ where $B\in\operatorname{Mat}_{n\times 1}(\mathbb{C})$
It is sufficient to prove that $x^TA_-x=0$. I will use the Einstien's summation convention where \begin{align} x^TA_-x&=\sum_{i=1}^{n}\sum_{j=1}^{n}\left(A_-\right)_{ij}x_i x_j\\ &=\left(A_-\right)_{ij}x^ix^j \end{align} As $i$and $j$ are dummy variables, swapping $i$ and $j$ keeps keeps $x^TA_-x$ invariant, i.e, \begin{align} \left(A_-\right)_{ij}x^ix^j=\left(A_-\right)_{ji}x^jx^i \end{align}
Since, $A_-$ is antisymmetric, $\left(A_-\right)_{ij}=-\left(A_-\right)_{ji}$ and that $x^ix^j=x^jx^i$ \begin{align} x^TA_-x&=\frac{1}{2}\left[\left(A_-\right)_{ij}x^i x^j+\left(A_-\right)_{ji}x^jx^i\right]\\ &=\frac{1}{2}\left[\left\{\left(A_-\right)_{ij}-\left(A_-\right)_{ij}\right\}x^ix^j\right]\\ &=0 \end{align}