Integral over Grassmann variable in the holomorphic representation

[Stalafin]

Homework Statement

Show that $$\int d\theta e^{\theta(\xi-\eta)}=\delta(\xi-\eta)$$,
where all of the above variable are Grassmann numbers. All this is in the holomorphic representation, where for some generic function:
$$f(\theta)=f_0 + f_1\theta$$

Homework Equations

How do I arrive at that bloody delta? As explained below, I see the analogy where theta all by itself acts like a delta function. But from there on...?

The Attempt at a Solution

Obviously, by expanding to first order according to the Grassmann algebra (all orders above the first vanish):
$$=\int d\theta 1+\theta(\xi-\eta) = \xi-\eta$$

What I do get is that the theta all by itself acts like a delta-function around 0 (taking the generic function above):
$$\int d\theta \theta f(\theta) = \int d\theta (f_1 + \theta f_2) = f_1 = f(0)$$

This we could have also written as:
$$\int d\theta f(\theta)\delta(\theta - 0) = f(0)$$

Where is the analogy?

 

[Johannes86]

Does not

$$\int d\xi\ (\xi-\eta)f(\xi)=\int d\xi\ (\xi-\eta)(f_1+\xi f_2)=\int d\xi\ (\xi f_1-\eta f_1+\xi\eta f_2)=f_1+\eta f_2=f(\eta)$$

imply

$$(\xi-\eta)=\delta(\xi-\eta)?$$

 

[Stalafin]

Hey! Thanks for your reply. Yes, I discussed this with a friend today, and we came to the same conclusion.

Thanks!