Example. Two generators fix the integration signs [fdip-008G]

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Take two generators and the alternating matrix \[A=\begin {pmatrix}0&a\\-a&0\end {pmatrix}.\] Anticommutation gives \[\tfrac 12\xi ^TA\xi =a\xi _1\xi _2.\] Since \((\xi _1\xi _2)^2=0\), \[e^{-\frac 12\xi ^TA\xi } =1-a\xi _1\xi _2=1+a\xi _2\xi _1.\] The reversed coefficient therefore gives \(I_\xi (e^{-\xi ^TA\xi/2})=a\). Selecting the increasing word \(\xi _1\xi _2\) instead would give \(-a\).

For one independent complex pair and any \(k\in \mathbb C\), \[e^{-k\bar \psi \psi }=1-k\bar \psi \psi =1+k\psi \bar \psi ,\qquad J_1(e^{-k\bar \psi \psi })=k.\] Both calculations use only the finite expansion and the ordered functionals in the definition, including when \(a=0\) or \(k=0\).