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Fix the orientation of the Berezin integral by defining its measure as the linear functional that extracts the coefficient of ${\prod }_{i=1}^N({\theta }_i{\bar{\theta }}_i)$; all monomials of lower degree integrate to zero. In particular $\int \mathcal{D}(\bar{\theta },\theta ){\prod }_i({\theta }_i{\bar{\theta }}_i)=1$. This specifies the sign convention in the notation $d^N\bar{\theta }{d}^N\theta$.

The Grassmann change-of-variables formula givesFor a unitary matrix, $\det U\det U^{\ast }=1$, so the combined measure is invariant. Diagonalize the Hermitian matrix $B$ by a unitary change of variables. If its eigenvalues are $b_i$, the integrand factors as ${\prod }_i(1-b_i{\bar{\theta }}_i{\theta }_i)={\prod }_i(1+b_i{\theta }_i{\bar{\theta }}_i)$. Extracting its top coefficient proves the Grassmann Gaussian integralThe determinant identity also holds without Hermiticity by direct expansion; Hermiticity makes the diagonalization proof immediate.

Let $C(x)={\sum }_k{x}_k{C}^{(k)}$. Integrating out the Grassmann variables first gives the exact polynomial expressionThe requested factorization requires $B$ to be invertible, an assumption omitted from the question. Under it, use the formal perturbative matrix logarithm near $g=0$:ConsequentlyExpanding $C(x)$ gives exactly the displayed coefficients $V_{k_1\dots k_m}^{(m)}=(m-1)!\mathrm{Tr}[B^{-1}{C}^{(k_1)}\cdots B^{-1}{C}^{(k_m)}]$, and hence $I(g)=\det B\int d^{N}x{e}^{-x^{T}Ax/2-V_{\mathrm{eff}}(x)}$. The logarithmic representation is a formal series, not a globally convergent logarithm for every $x$: the determinant can have zeros. The original determinant integral remains well-defined for singular $B$, but $I(0)=0$ and the requested normalized expansion then cannot be formed.

There is a second convention to specify. The $x_i$ are real, so a complex Hermitian $A$ gives $x^{T}Ax=x^T{A}_{R}x$, where $A_R=\mathrm{Re}A=(A+A^T)/2$ is real symmetric positive definite. Its Gaussian covariance matrix is $A_R^{-1}$. It equals $A^{-1}$ only if $A$ is real symmetric. ThusThe effective Feynman rules are: a bosonic contraction joins indices $k,\ell$ with $(A_R^{-1}{)}_{k\ell }$; an $m$-leg vertex carries $-g^m\mathrm{Sym}{V}_{k_1\dots k_m}^{(m)}$; sum all closed vacuum diagrams with their symmetry factors, including disconnected diagrams. The symmetrization is necessary because the $x_k$ commute: the trace tensor is cyclically invariant but need not be fully symmetric. The minus sign comes from expanding $e^{-V_{\mathrm{eff}}}$. Taking the logarithm of the normalized integral instead keeps only connected vacuum diagrams.

Equivalently, retain the fermions: an oriented contraction is $(B^{-1}{)}_{ij}=⟨{\theta }_i{\bar{\theta }}_j⟩$, the mixed vertex has weight $+g{C}_{ij}^{(k)}$, and every closed fermion loop gives a minus sign. The same bosonic covariance joins its $x$ legs. Expanding these contractions recovers the trace vertices above. This is bosonic trace vertices from a finite fermionic determinant.