I've been asked to prove that if $X$ is a Gaussian random variable with zero mean and $F$ a function then:
$$\langle XF(X)\rangle = \langle X^2 \rangle \langle F'(X) \rangle $$
Where $\langle \cdot \rangle$ denotes the mean.
I see clearly that it must be some sort of consequence of integrating by parts but I haven't been able to prove it yet. I also don't get why I get the product of two integrals in the RHS instead of a single integral.
$$\langle XF(X)\rangle = \int_{-\infty}^\infty x F(x) \dfrac{1}{\sigma\sqrt{2\pi}}e^{-\tfrac{x^2}{2\sigma^2}}\,dx=-\sigma^2\int_{-\infty}^\infty \dfrac{1}{\sigma\sqrt{2\pi}}\, F(x)\,\, de^{-\tfrac{x^2}{2\sigma^2}}=$$ $$=-\sigma^2 \dfrac{1}{\sigma\sqrt{2\pi}} F(x)\,e^{-\tfrac{x^2}{2\sigma^2}}\biggm|_{-\infty}^\infty + \,\sigma^2\int_{-\infty}^\infty \dfrac{1}{\sigma\sqrt{2\pi}} \,e^{-\tfrac{x^2}{2\sigma^2}} dF(x)=$$ $$=\sigma^2\int_{-\infty}^\infty \dfrac{1}{\sigma\sqrt{2\pi}} \,e^{-\tfrac{x^2}{2\sigma^2}}F'(x) \,dx = \langle X^2 \rangle \langle F'(X)\rangle. $$
One need here that $\lim_{x\to\pm\infty}F(x)e^{-\tfrac{x^2}{2\sigma^2}}=0$.