In a statistics book I found the following definition for the Fisher information: $$I_{\theta} := \text{var} _{\theta} \left ( \frac{\partial }{\partial \theta} \log p_{\theta} (X)\right ), $$ where $X$ is a random variable that is distributed according to the density $p_{\theta}$. However, I also know the definition $$I_{\theta} := E _{\theta} \left ( \left ( \frac{\partial}{\partial \theta} \log p_{\theta} (X)\right ) ^2 \right ), $$ so I suspect that $$E _{\theta} \left ( \frac{\partial}{\partial \theta} \log p_{\theta} (X) \right ) = 0,$$ but I am not fully confident about it and I also do not know how to proceed if I want to prove this.
I have been thinking about the following. Write out $$E _{\theta} \left ( \frac{\partial}{\partial \theta} \log p_{\theta} (X) \right ) = \int p_{\theta} (x) \cdot \frac{\partial}{\partial \theta} \log p_{\theta} (x) \ dx.$$ Then use one of the following integration techniques: do a change of variables (maybe $y=p_{\theta} (x)$?) or use the method of integration by parts.
Does someone know more about this? A reference to an article, book, or website is also fine.
I already found the answer in another statistics book! We have: \begin{align*} E _{\theta} \left ( \frac{\partial}{\partial \theta } \log p_{\theta} (X) \right ) & = E _{\theta} \left ( \frac{1}{p_{\theta} (X) } \frac{\partial }{\partial \theta} p_{\theta} (X) \right ) \\ & = \int p_{\theta} (x) \frac{1}{p_{\theta} (x) } \frac{\partial }{\partial \theta} p_{\theta} (x) dx \\ & = \int \frac{\partial }{\partial \theta} p_{\theta} (x) dx \\ & = \frac{d}{d\theta} \int p_{\theta} (x) dx \\ & = \frac{d}{d\theta} 1 \\ & = 0. \end{align*} If someone wants to delete this question, then that is ok.