Let's say f admits a taylor series f(ˉθθ)=A+Bˉθθ+Cˉθθˉθθ+…. Now, ˉθθˉθθ=−ˉθ2θ2=0, etc., so our function terminates at the linear term. Furthermore, the integral of f over dˉθdθ, by the rules of Berezin integration, is precisely equal to B.
Btw, the reason that integration is defined in this way is so that
∫dθ∂∂θf(θ)=0, which is something you might expect for "well-behaved" functions.
Getting back to your question, in physics, the way you would "integrate" over ˉθθ in a path integral is to define ˉψψ=eiϕ and then integrate over ϕ. This is called Bosonization, because it expresses the dynamics of fermions in terms of bosons. It isn't always possible, afaik.
This post imported from StackExchange Physics at 2014-03-05 14:51 (UCT), posted by SE-user lionelbrits