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A Berezin integral over independent Grassmann variables selects the coefficient containing every variable. Fix the real orientation by $\int d{\omega }_1\cdots d{\omega }_{2m}{\omega }_{2m}\cdots {\omega }_1=1$. For an antisymmetric matrix $A$, only the order-$m$ term in the exponential survives:Anticommutation collects its top coefficient into the signed sum over pairings that defines the Pfaffian. Reversing the order of $2m$ variables gives $(-1{)}^{m(2m-1)}=(-1{)}^m$, cancelling the prefactor sign. The real Grassmann Gaussian integral is thereforewith this orientation, and a fixed sign times this answer with other orientations. In dimension two, write $A_{12}=a$; the exponential is $1-a{\omega }_1{\omega }_2=1+a{\omega }_2{\omega }_1$, and integration gives $a$. Although $\mathrm{Pf}(A{)}^2=\det A$, writing a positive square root $\sqrt{\det A}$ would give $|a|$ in this test and would lose the correct sign.

For the complex Grassmann Gaussian integral, treat ${\theta }_a$ and ${\bar{\theta }}_a$ as independent integration variables. Every surviving top term contains each $\theta$ and each $\bar{\theta }$ once. Expanding the exponential assigns one matrix entry from every row and column, and anticommutation supplies the permutation sign. Hencewhere $C_n$ is a nonzero constant fixed by the order and normalization of the differentials, independent of $M$. This argument works for singular matrices as well and does not assume that $M$ is Hermitian.

For the explicit $n=2$ check, put $Q={\bar{\theta }}_a{M}_{ab}{\theta }_b$. Moving all the $\theta$ variables before the barred ones givesThe linear and constant terms integrate to zero. Thus the integral is a fixed orientation sign times $M_{11}{M}_{22}-M_{12}{M}_{21}$, exactly the determinant. Choosing the top-monomial integral to be $-1$ gives $C_2=1$.

For the required real-complex compatibility of Grassmann Gaussian integrals, let $u_a={\omega }_a$ and $v_a={\omega }_{n+a}$. Substitution givesIndeed $v^{T}Mu=-u^T{M}^{T}v=u^{T}Mv$ because the variables anticommute and $M^T=-M$. The real quadratic matrix is consequently $A=\mathrm{diag}(M,M)$, of size $2n$. The linear change of Grassmann variables has a constant nonzero Berezin integration Jacobian, so it changes only the allowed normalization factor. For even $n$, the two blocks giveFor odd $n$, $\det M=\det (-M)=(-1{)}^n\det M$ implies $\det M=0$. Each of the separate $u$ and $v$ quadratic exponentials has only even degree and cannot saturate an odd number of variables, so the real integral vanishes too. This covers the odd case without attempting to define a Pfaffian of an odd-dimensional matrix.