dimensional analysis of Grassmann integration/differentiation

There is another paradox that I need to resolve:

The Berezin integration rules for Grassmann odd variables give the same result as differentiation:

If $f=x+\theta\psi$ is a superfunction, the integral

$$\int d\theta(x+\theta\psi)=\psi$$

gives the same result as differentiation

$$\frac{d}{d\theta}(x+\theta\psi)=\psi.$$

How is this supposed to work in supersymmetry, where the Grassmann coordinates carry mass dimension -1/2? If I integrate, I expect a result to drop mass dimension by 1/2, whereas differentiation would lead to a gain. In the above example, I end up with the same object. What is its mass dimension?

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Grassman $d\theta$ has opposite mass dimension to $\theta$, which is why the notation is not 100% optimal, it confuses on this issue. But if you know how to evaluate the integral, that it goes like the derivative, then you know how change of scale works, and it's the opposite of normal change of scale:

$$\int d(k\theta) f(k\theta) = {1\over k} \int d\theta f(\theta)$$

and this is why the volume determinant for the integration ends up being the reciprocal of the Bose case.

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Oh! Makes perfect sense. So if you had to invent a new notation for the Grassmann-odd differential, what would it be? –  QuantumDot Oct 20 '12 at 1:23
@QuantumDot: I use $|_\theta$ instead of the integral at times, when I feel like it, but the current notation is good enough, you just have to be familiar with this oddity. –  Ron Maimon Oct 20 '12 at 3:09
Wow, then is there really any difference difference between Grassmann integration and Grassmann differentiation? –  QuantumDot Oct 20 '12 at 3:27
@QuantumDot: Nope. –  Ron Maimon Oct 20 '12 at 4:14
What about $\int d/\theta \cdot \theta =1$? –  Luboš Motl Oct 20 '12 at 8:44
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