Integrals over Grassmann numbers I want to prove an identity from Peskin&Schroeder, namely that 
$$\left(\prod\limits_i^{} \int d \theta^*_i d\theta_i\right) \theta_m \theta_l^* \exp(\theta_j^* B_{jk} \theta_k)=\det(B) B^{-1}_{ml}\tag{9.70}$$
$B$ is a hermitean $N\times N$ matrix and $N$ should be even. $\theta_j$ and $\theta_j^*$ are $N$ complex Grassmann numbers.
I would like to do it similar to the case of complex and not Grassmann Gaussian integrals, where you artificially introduce a term in the exponential and then differentiate. I could prove, that
$$\left(\prod\limits_i^{} \int d \theta^*_i d\theta_i\right)\exp(\theta_j^* B_{jk} \theta_k+\eta_j^* \theta_j+\theta_j^*\eta_j)=\det(B)\exp(-\eta_j^*B_{jk}^{-1}\eta_k)$$
where $\eta$ and $\theta$ are (complex) Grassmann numbers.If they were just complex variables, the rest would be clear. I could get the first equation differentiating the second with respect to $\eta_m$ and $\eta_l^*$ and then let $\eta=\eta^*=0$. The problem is, that for a Grassmann number the exponential is $e^\theta=1+\theta$ since all higher order terms cancel. For a sum like we have here, there should also be terms of second order, but nevertheless the exponential is not reproduced by differentiation.
Can I nevertheless obtain the wished result in that way? It looks quite promising, I just don't see, how to come from the second equation (left side) to the first equation (left side). It should work with differentiating, but I don't see why.
 A: EDIT
The method you want to use is ok, and gives a quick result. Here it is:
$$I=\int\prod d\theta^{*}d\theta\,\theta_{k}^{*}\theta_{l}exp(\theta^{*}B\theta+\eta^{*}\theta+\theta^{*}\eta)\\=\left(\frac{\partial}{\partial\eta_{k}^{*}}\right)\left(\frac{\partial}{\partial\eta_{l}}\right)\int\prod d\theta^{*}d\theta\, exp(\theta^{*}B\theta+\eta^{*}\theta+\theta^{*}\eta)$$
from where it follows that $I$ is equal to 
$$\left.I=\mathrm{det}B\left(\frac{\partial}{\partial\eta_{k}^{*}}\right)\left(\frac{\partial}{\partial\eta_{l}}\right)\mathrm{exp}(\eta_{j}^{*}(B^{-1})_{ij}\eta_{i})\right\vert_{\eta^{*}=\eta=0}$$
$$I=\mathrm{det}B(B^{-1})_{kl}$$
Note: I use the shift $\theta_{i}\rightarrow\theta_{i} +(B^{-1})_{ij}\eta_{j}$ and $\theta_{i}^{*}\rightarrow\theta_{i}^{*} +\eta_{j}^{*}(B^{-1})_{ji}$. Here I used the fact that the second moments can be calculated as derivatives in the following way.
$$\left.\langle\eta_{l}\eta_{k}^{*}\rangle=\left(\frac{\partial}{\partial\eta_{k}^{*}}\right)\left(\frac{\partial}{\partial\eta_{l}}\right)\mathrm{exp}(\eta_{j}^{*}(B^{-1})_{ij}\eta_{i})\right\vert_{\eta^{*}=\eta=0}=(B^{-1})_{ij}$$
More or less, you can take this as a definition. But if you still want to prove this, do the following:
$$J=\left(\frac{\partial}{\partial\eta_{k}^{*}}\right)\left(\frac{\partial}{\partial\eta_{l}}\right)\mathrm{exp}(\eta_{j}^{*}(B^{-1})_{ij}\eta_{i})=\left(\frac{\partial}{\partial\eta_{k}^{*}}\right)\left(\frac{\partial}{\partial\eta_{l}}\right)\prod_{ij}(1-\eta_{j}^{*}(B^{-1})_{ij}\eta_{i})$$
After straight forward differentiation you get
$$J=\prod_{ij}\delta_{kj}\delta_{li}(B^{-1})_{ij}=(B^{-1})_{kl}$$
You can ignore whats below the line, that was my first answer. But I'll leave it because it may be instructive for others.


Since I cannot comment yet, i will sketch a proof for a simpler case
$$I=\int\prod_{i}d\theta_{i}^{*}d\theta_{i}exp(\theta_{i}^{*}B_{ij}\theta_{j})=det(B)$$
and hope that this will help you compute your integrals.  After expanding the all the exponential we arrive at
$$I=\frac{1}{N!}\int d\theta_{1}^{*}d\theta_{1}\dots d\theta_{N}^{*}d\theta_{N}(\theta_{i_1}^{*}B_{i_{1}j_{1}}\theta_{j_{1}})(\theta_{i_2}^{*}B_{i_{2}j_{2}}\theta_{j_{2}})\dots (\theta_{i_N}^{*}B_{i_{N}j_{N}}\theta_{j_{N}})$$
At this point some explanations are in order. The factor $1/N!$ appears from the expansion of the exponentials. To see how the integrand was obtained, let's look at the case when $N=2$. We will have something like this
$$\int d\theta_{1}^{*}d\theta_{1}d\theta_{2}^{*}d\theta_{2}\left(1+\theta_{i}^{*}B_{ij}\theta_{j}+\frac{(\theta_{i}^{*}B_{ij}\theta_{j})^2}{2!}+\cdots\right)$$
It is obvious that only the quadratic term will contribute to the above integral, because only this term can saturate the number of grassmann variables in the integral measure. 
$$\int d\theta_{1}^{*}d\theta_{1}d\theta_{2}^{*}d\theta_{2}\left(\frac{(\theta_{i}^{*}B_{ij}\theta_{j})^2}{2!}\right)=\frac{1}{2}\int d\theta_{1}^{*}d\theta_{1}d\theta_{2}^{*}d\theta_{2}\sum_{i_{1},i_{2},j_{1},j_{2}}^{2}(\theta_{i_1}^{*}B_{i_{1}j_{1}}\theta_{j_{1}})(\theta_{i_2}^{*}B_{i_{2}j_{2}}\theta_{j_{2}})$$
Expanding the sum and performing all four integrals, we get 
$$\int d\theta_{1}^{*}d\theta_{1}d\theta_{2}^{*}d\theta_{2}\left(\frac{(\theta_{i}^{*}B_{ij}\theta_{j})^2}{2!}\right)=\frac{1}{2!}[2(B_{11}B_{22}-B_{12}B_{21})]=detB$$
Now, let us return to our original integral $I$. The next step before performing the integrals is to reorder the integrals and the grassmann numbers. 
$$I=\frac{1}{N!}\int d\theta_{1}^{*}\dots d\theta_{N}^{*}\theta_{i_1}^{*}\dots\theta_{i_1}^{*}\int\theta_{1}\dots d\theta_{N}\theta_{j_1}\dots\theta_{j_1}B_{i_{1}j_1}\dots B_{i_{N}j_N}$$ 
From where we finally arrive at
$$I=\frac{1}{N!}\epsilon_{i_{1}\dots i_{N}}\epsilon_{j_{1}\dots j_{N}}B_{i_{1}j_1}\dots B_{i_{N}j_N}=\mathrm{det}B$$ 
(note: the ordering result $a_{1}b_{1}\dots a_{N}b_{N}=a_{1}\dots a_{N}b_{1}\dots b_{N}(-1)^{N(N-1)/2}$ has to be used twice for the integrals).
I found this method to be the most simple one of all when dealing with these sort of integrals. I hope this helps you prove those relations. The same method can be applied very easy in your case (just a strait forward extension). And with the method you described, it seams you are over complicating yourself. However, I'll work on it a bit and see what comes up.  
