Consider a time dependent Grassmann field i.e. $\theta(t)$. Now, consider the following Berezin integral $$\int [\mathcal{D}\theta] ~\prod_{t}\dot{\theta}$$ where $\dot{\theta}$ time derivative of $\theta$ and $[\mathcal{D}\theta]$ is a functional measure over the Grassmann field, something like a path integral measure and a product integral of $\dot{\theta}$. Now, what does the above integral evaluate to? One way I can think of evaluating the integral is that by recognizing that $\dot{\theta} = \delta(\dot{\theta})$ as $\dot{\theta}^2 = 0$. Therefore, we have $$\int [\mathcal{D}\theta] ~\prod_{t}~\dot{\theta} = \int [\mathcal{D}\theta]~\prod_{t} ~\delta(\dot{\theta}) = \det{\partial_t}$$ Is my evaluation of the integral correct?

Consider a time dependent Grassmann field i.e. $\theta(t)$. Now, consider the following Berezin integral $$\int [\mathcal{D}\theta] ~\prod_{t}\dot{\theta}\tag{1}$$ where $\dot{\theta}$ time derivative of $\theta$ and $[\mathcal{D}\theta]$ is a functional measure over the Grassmann field, something like a path integral measure and a product integral of $\dot{\theta}$. Now, what does the above integral evaluate to? One way I can think of evaluating the integral is that by recognizing that $\dot{\theta} = \delta(\dot{\theta})$ as $\dot{\theta}^2 = 0$. Therefore, we have $$\int [\mathcal{D}\theta] ~\prod_{t}~\dot{\theta} = \int [\mathcal{D}\theta]~\prod_{t} ~\delta(\dot{\theta}) = \det({\partial_t}).\tag{2}$$ Is my evaluation of the integral correct?

Consider a time dependent Grassmann field i.e. $\theta(t)$. Now, consider the following Berezin integral $$\int [\mathcal{D}\theta] ~\dot{\theta}$$ where $\dot{\theta}$ time derivative of $\theta$ and $[\mathcal{D}\theta]$ is a functional measure over the Grassmann field, something like a path integral measure. Now, what does the above integral evaluate to? One way I can think of evaluating the integral is that by recognizing that $\dot{\theta} = \delta(\dot{\theta})$ as $\dot{\theta}^2 = 0$. Therefore, we have $$\int [\mathcal{D}\theta] ~\dot{\theta} = \int [\mathcal{D}\theta] ~\delta(\dot{\theta}) = \det{\partial_t}$$ Is my evaluation of the integral correct?

Consider a time dependent Grassmann field i.e. $\theta(t)$. Now, consider the following Berezin integral $$\int [\mathcal{D}\theta] ~\prod_{t}\dot{\theta}$$ where $\dot{\theta}$ time derivative of $\theta$ and $[\mathcal{D}\theta]$ is a functional measure over the Grassmann field, something like a path integral measure and a product integral of $\dot{\theta}$. Now, what does the above integral evaluate to? One way I can think of evaluating the integral is that by recognizing that $\dot{\theta} = \delta(\dot{\theta})$ as $\dot{\theta}^2 = 0$. Therefore, we have $$\int [\mathcal{D}\theta] ~\prod_{t}~\dot{\theta} = \int [\mathcal{D}\theta]~\prod_{t} ~\delta(\dot{\theta}) = \det{\partial_t}$$ Is my evaluation of the integral correct?