I am looking for help proving this proposition:
"If $Z$ is a standard normal random variable and $g(z)$ is differentiable, then $\mathbb{E}[g'(Z)]=\mathbb{E}[Zg(Z)]$"
I have tried evaluating $\mathbb{E}[g'(Z)]$ using integration by parts as follows. $$\mathbb{E}[g'(z)]=\int_{-\infty}^\infty{g'(z)}\frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}dz=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty{g'(z)}e^{-\frac{z^2}{2}}dz$$ Let $u=e^{-\frac{z^2}{2}}$ and $dv=g'(z)dz$. Now, $$\frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}g(z)\Big|_{-\infty}^{\infty}+\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}zg(z)e^{-\frac{z^2}{2}}dz$$ Interestingly the second half of this equation is equal to $\mathbb{E}[Zg(Z)]$. I am not sure how to show that the first half of the above equation equals zero and is thus equal to $\mathbb{E}[Zg(Z)]$. I am also not sure how to evaluate $\mathbb{E}[Zg(Z)]$ as to yield the same equation, so it seems I have hit a dead end. I would like some help finishing this proof. Preferably, I want to know the simplest method.