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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.

Choose the orientation of the Berezin integral so thatHere the product has increasing $i$; this explicitly fixes the otherwise convention-dependent overall sign in the compact measure notation.

Since $A$ is a diagonalizable matrix, write $A=S\Lambda S^{-1}$ with $\Lambda =\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_N)$. Make the independent changes $\eta =S\xi$ and $\bar{\eta }=\bar{\xi }{S}^{-1}$. The Grassmann change-of-variables formula gives the two factors $(\det S{)}^{-1}$ and $\det S$, so the complete measure is unchanged. The exponent becomes ${\sum }_i{\lambda }_i{\bar{\xi }}_i{\xi }_i$. Each summand is even and squares to zero, and the different even summands commute. Consequently,Only the term containing every generator survives the Berezin integral. Hence the Grassmann Gaussian integral isZero eigenvalues give zero on both sides, so invertibility of $A$ is unnecessary. In fact the identity extends to all ordinary matrices: the top-degree coefficient of the exponential is the alternating determinant expansion. The assumption that $A$ is a diagonalizable matrix makes the proof especially transparent.

A Grassmann variable is an odd generator of a Grassmann algebra: ${\eta }_i{\eta }_j=-{\eta }_j{\eta }_i$, so ${\eta }_i^2=0$. For one generator, any function is $f(\eta )=a+b\eta$. The Berezin integral is the linear operationThus integration extracts a coefficient, rather than assigning a length or volume. For many generators it extracts the coefficient of the highest-degree monomial with the sign fixed by the order of the measure. Odd coefficients and Grassmann derivatives must retain their order; exchanging two odd objects changes the sign.

This operation is invariant under odd translations, because a translation only changes terms of lower degree. For an invertible ordinary matrix $S$ and $\eta =S\xi$, the Grassmann change-of-variables formula isThe inverse Jacobian determinant, rather than the ordinary commuting-variable Jacobian, compensates for the factor $\det S$ multiplying the top monomial. Integration agrees with the appropriate ordered Grassmann derivatives, but the orientation must be specified when combining barred and unbarred variables.

For an even-dimensional antisymmetric matrix $M$, put $\theta =(u+iv)/\sqrt{2}$ and $\bar{\theta }=(u-iv)/\sqrt{2}$, with independent Grassmann variables $u,v$. Anticommutation cancels the mixed terms, giving ${\bar{\theta }}^{T}M\theta =(u^{T}Mu+v^{T}Mv)/2$. The constant Grassmann change-of-variables formula changes only the normalization. The two real Pfaffians multiply to $\mathrm{Pf}(M{)}^2=\det M$, agreeing with the complex Grassmann Gaussian integral. In odd dimension $\det M=0$ and both top-degree integrals vanish.