# Vanishing partition function [duplicate]

I am currently stuck with the following partition function

Let the action be $$S(X, \psi^1, \psi^2) = \frac{1}{2} (\partial h)^2 - \partial^2h\psi^1 \psi^2 ,\tag{9.29}$$ where $$h$$ is a real function of the bosonic variable $$X$$, $$\partial h = h'$$ and $$\psi^i$$ are Grassmann odd variables. The action is invariant under $$\delta X = \epsilon^1 \psi_1 + \epsilon^2 \psi_2$$ $$\delta \psi_1 = \epsilon^2 \partial h,\tag{9.30}$$ $$\delta \psi_2 = -\epsilon^1 \partial h,$$ with $$\epsilon^i$$ Grassmann odd variables.

Now Suppose $$\partial h$$ is nowhere $$0$$, i have to show that the corresponding partition function $$Z = \int dX d\psi^1 d\psi^2 e^{-S}= 0.\tag{9.31}$$

After a change of variables $$\hat{X} = X - \frac{\psi^1 \psi^2}{\partial h}$$ $$\hat{\psi}^1= \alpha(X) \psi^1\tag{9.32}$$ $$\hat{\psi}^2= \psi^1 + \psi^2,$$ such that $$S(X,\psi^1,\psi^2) = S(\hat{X}, 0,\hat{\psi}^2),\tag{9.33}$$ with $$\alpha$$ an arbitrary function of $$X$$, and some Grassmann integration, i get that

$$Z = -\int e^{-S(\hat{X}, 0, \hat{\psi}^2) } \frac{\partial^2h}{(\partial h)^2} \hat{\psi}^1 \hat{\psi}^2 d \hat{X} d \hat{\psi}^1 d \hat{\psi}^2\tag{9.35}$$

which should be zero. But i can not see why? I got the hint that the integrand is a total derivative in $$\hat{X}$$ but that didn't help me.

• I do not understand what you mean by $\partial^2{h}{(\partial h)^2}$, this should be zero. Commented May 17, 2019 at 15:15
• Possible duplicate: physics.stackexchange.com/q/376564/2451 , physics.stackexchange.com/q/324024/2451 Commented May 17, 2019 at 23:44
• It is not quite a duplicate since i can not see why this last integral is a total derivative. Commented May 18, 2019 at 7:44