I found this identity on Wikipedia: $$\int\exp\left[\theta^T A\eta+\theta^T J+K^T\eta\right]d\theta d\eta =\det A\exp\left[-K^TA^{-1}J\right]$$
where the integration variables are Grassmann variable.
Unfortunately the Wikipedia page gives little context, and I'm not clear if $K$ and $J$ are also supposed to be Grassmann variables.
Does anybody know how to prove this identity? Thanks in advance.
For any integer $n > 0$ and Grassmann variables $\eta_1, \eta_2, \cdots, \eta_n$ over $\mathbb{C}$.
Let $V$ be a vector space over $\mathbb{C}$ spanned by $\eta_1, \eta_2, \cdots, \eta_n$.
A function over $\eta_k$ can be viewed an element of the exterior algebra over $V$.
(with the operator $\wedge$ suppressed when one write down its formula).
$$f : \text{ function over } \eta_k \quad\longleftrightarrow\quad f \in \Lambda(V) = \bigoplus_{\ell=0}^n \Lambda^\ell (V) $$ Expand $f$ into of sums of exterior products of $\eta_k$,
$$f(\eta) = a_0 + \sum_{1\le i_1 \le n} a_{i_1} \eta_{i_1} + \sum_{1\le i_1 \le i_2 \le n } a_{i_1i_2} \eta_{i_1} \wedge\eta_{i_2} + \cdots + a_{12\ldots n} \eta_1 \wedge \cdots \wedge \eta_n $$ Up to a sign, the Grassmann integral over $f$ is the coefficient $a_{123\dots n}$ in front of $\eta_1 \wedge \cdots \wedge\eta_n$.
$$\int f(\eta) d\eta \stackrel{def}{=} a_{12\ldots n}$$
Let $X = \bigoplus_{\ell=0}^{n-1}\Lambda^\ell(V)$ be the collection of $f(\eta)$ whose $a_{12\cdots n} = 0$.
When we replace $\eta_k$ by $\eta_k + b_k$ for some $b_1,\ldots, b_n \in \mathbb{C}$, element $f(\eta)$ in $X$ get mapped to the element $f(\eta + b)$ which also belongs to $X$. Under same transformation, $\eta_1\wedge \cdots \wedge \eta_n$ get mapped to $$(\eta_1 + b_1)\wedge \cdots \wedge (\eta_n + b_n) = \eta_1 \wedge \cdots \eta_n + ( \cdots )$$ for some mess $(\cdots) \in \bigoplus_{\ell=0}^{n-1}\Lambda^\ell(V)$. In both cases, the coefficient $a_{12\cdots n}$ remains unchange. This leads to
$$\int f(\eta + b) d\eta = \int f(\eta) d\eta$$
Similarly, if we pick a $n\times n$ complex matrix $A = (A_{ij})$ and introduce a new set of variables $\zeta_i = \sum_{j=1}^n A_{ij} \eta_j$ fo, we find $f(\eta) \in X \implies f(\zeta) \in X$. Furthermore, $\eta_1\wedge \cdots \wedge \eta_n$ get mapped to
$$\sum_{1 \le i_1,\ldots,i_n \le n} A_{1i_1} A_{2i_2} \cdots A_{ni_n} \eta_{i_1}\wedge \cdots \wedge \eta_{i_n} = \sum_{\pi \in S_n} (-1)^\pi \prod_{i=1}^n A_{i\pi(i)} \eta_1 \wedge \cdots \wedge \eta_n = \det(A) \eta_1 \wedge \cdots \wedge \eta_n $$ where the middle term is a sum over all permutations $\pi$ of the $n$ indices $1,\ldots n$.
Notice in both cases, the coefficient of $a_{12\cdots n}$ get multiplied by the same factor $\det(A)$. This leads to
$$\int f(A\eta) d\eta = \det(A) \int f(\eta) d\eta$$
Combine these, we have following "change of variable" formula
$$\bbox[1em,border:1px solid blue]{ \int f(A\eta+b) d\eta = \det(A)\int f(\eta) d\eta }\tag{*1}$$
For the identity at hand, we have
$$\begin{align} & \int\exp\left[\theta^T A\eta+\theta^T J+K^T\eta\right]d\theta d\eta \stackrel{?}{=}\det A\exp\left[-K^T A^{-1} J\right]\\ \iff & \int\exp\left[\theta^T A\eta+\theta^T J+K^T\eta + K^T A^{-1} J\right]d\theta d\eta \stackrel{?}{=}\det A\\ \iff & \int \exp\left[(\theta + A^{-T}K)^T A(\eta + A^{-1}J)\right] d\theta d\eta \stackrel{?}{=} \det A \end{align} $$ Apply $(*1)$ for the transform $\theta \mapsto \theta + A^{-T} K$, $\eta \mapsto A\eta + J$, the last equality is equivalent to
$$\int \exp[ \theta^T \eta ] d\theta d\eta \stackrel{?}{=} 1\tag{*2}$$ Notice $$\begin{align} \exp[\theta^T\eta] &= \prod_{\ell=1}^n \exp[\theta_\ell \eta_\ell] = \prod_{\ell=1}^n (1 + \theta_\ell \eta_\ell)\\ &= (1 + \theta_1 \eta_1 ) \cdots (1 + \theta_n \eta_n ) = (\theta_1 \eta_1) \cdots (\theta_n \eta_n ) + (\cdots) \end{align} $$ where $(\cdots)$ is a mess contains terms with fewer than $2n$ products. Up to convention how to fix the sign in Grassmann integral, we find $(*2)$ is true. As a result, so does the identity at hand.
About the question whether $J, K$ are Grassmann variables or not. They can be but need not be.
In fact, $(*1)$ and $(*2)$ are derived under the assumption $A \in M^{n\times n}(\mathbb{C})$ and $b, J, K \in \mathbb{C}^n$.
To generalize $(*1)$, we replace $\mathbb{C}$ by any commutative ring $R$ and $V$ by $V_\eta \oplus V_\chi$, a direct sum of two $R$-module where $V_\eta$ is spanned by $\eta_1,\ldots,\eta_n$ and $V_\chi$ is spanned by $\chi_1,\cdots,\chi_m$. Instead of $\Lambda(V)$, functions over $\eta_k$ are now viewed as element of $\Lambda(V_\eta \oplus V_\chi) \simeq \Lambda(V_\eta) \otimes \Lambda(V_\chi)$. When one expand function $f(\eta)$ over exterior products of $\eta_k$, the coefficients $a_{i_1i_2\ldots i_k}$ are now taking values in $\Lambda(V_\chi)$. The Grassmann integral is still defined as the coefficient $a_{12\ldots n}$.
If one repeat the argument, one find $(*1)$ remains valid for $A \in M^{n\times n}(R)$ and $b \in \Lambda(V_\chi)^n$.
To generalize $(*2)$, we need one extra constraint $$\theta^T A\eta+\theta^T J+K^T\eta\quad\text{ commutes with }\quad K^T A^{-1} J$$ If this is satisfied we have $$\exp[\theta^T A\eta+\theta^T J+K^T\eta + K^T A^{-1} J] = \exp[\theta^T A\eta+\theta^T J+K^T\eta]\exp[K^T A^{-1} J]$$ Repeating the argument, one find $(*2)$ remains valid when $A \in M^{n\times n}(R)$ and components of $J, K$ coming from $\Lambda^{odd}(V_\chi) \stackrel{def}{=} \bigoplus\limits_{k=1,\text{ odd}}^{n}\Lambda^k(V_\chi)$.
What this means is aside from complex numbers, the components of $J, K$ can be Grassmann variables (or more generally, elements of $\Lambda^{odd}(\cdot)$) which anti-commute with $\theta, \eta$ and among themselves.