# How to understand following expression for gaussian integral of grassmannian function?

Let's have grassmannian numbers $\theta_{i}$: $$\theta_{i}\theta_{j} = -\theta_{j}\theta_{i}, \quad \theta_{i} x = x\theta_{i},$$ where x is just ordinary number.

How to understand (or how to show if it isn't the definition) that for two sets of grassmannian numbers $\theta_{i}, \eta_{j}$ $$\int e^{\eta_{i}A_{ij}\eta_{j}} d\eta = \sqrt{det A_{ij}} ? \qquad (1)$$ I know that for grassmannian numbers the integration is equal to the differentiation, but I don't understand how to apply this statement for my question.

Clearly it may be proved in the following way. First we "diagonalize" $A_{ij}$ to the form $$A_{ij} = \begin{pmatrix} 0 & \lambda_{1} & 0 & 0 & ... & 0 & 0 \\ -\lambda_{1} & 0 & 0 & 0 & ... & 0 & 0 \\ 0 & 0 & 0 & \lambda_{2} & \dots & 0 & 0 \\ 0 & 0 & - \lambda_{2} & 0 & \dots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \dots & \vdots & \vdots \\ 0 & 0& 0& 0 & \dots & 0 & \lambda_{n} \\ 0 & 0& 0& 0 & \dots & -\lambda_{n} & 0\end {pmatrix} .$$ Then I can expand the exponentials as $$e^{\eta_{i}A_{i, j}\eta_{j}} = 1 + \eta_{i}A_{i, i + 1}\eta_{j} + ... + \eta_{1}A_{12}\eta_{2}...\eta_{n - 1}A_{n - 1, n}\eta_{n}.$$ Finally, I use properties $\int 1 d \eta = 0, \quad \int \eta d \eta = 1$, so I will get $$\int e^{\eta_{i}A_{ij}\eta_{j}}d\eta = \prod_{i = 1}^{\frac{n}{2}}\lambda_{i} = \sqrt{det A_{ij}}.$$ Can you check it?
If you are integrating over an even dimensional Grassmann algebra than your reasoning is correct (to the best of my understanding you can only bring even dimensional skew-symmetric matrices to the block diagonal form you give). In this case your exponent can always be written in the form you give (lhs eq.1) where $$A_{ij}$$ is an $$N\times N$$ skew-symmetric matrix ($$N=2m$$ is even). This is true because one could write a general matrix as $$M_{ij}=(M_{ij})_{sym}+(M_{ij})_{anti-sym}$$ and the symmetric part would always vanish due to the Grassmann nature of the numbers ($$\theta_i\theta_i=0$$ and $$\theta_i\theta_j=-\theta_j\theta_i$$). If you want a more in dept derivation of the result you are asking about you can check my answer on a similar question here.
There's a problem with your last step. Even if $n$ is even so that the product makes sense, it's still unlikely that you'll get $\sqrt{\det A}$.