Name of a proposition(?) related to Gaussian distribution

Question

While I was deriving some equation, I have found out that the following equation holds. The below integrations are calculated through wolfram alpha and their values are same, 1.51828. \begin{equation} \int_{\mathbb{R}}z\tanh(z)\exp\left(-\frac{z^2}{2}\right)\frac{dz}{\sqrt{2\pi}} = \int_{\mathbb{R}}\operatorname{sech}^2(z)\exp\left(-\frac{z^2}{2}\right)\frac{dz}{\sqrt{2\pi}} \end{equation} Since $\operatorname{sech}^{2}(z)$ is simply a derivative of $\tanh(z)$, is it generally true for the following form of equation? \begin{equation} \int_{\mathbb{R}}zf(z)\exp\left(-\frac{z^2}{2}\right)\frac{dz}{\sqrt{2\pi}} = \int_{\mathbb{R}}f'(z)\exp\left(-\frac{z^2}{2}\right)\frac{dz}{\sqrt{2\pi}} \end{equation} If it is true, what is the name of such proposition? In which book can it be found?

Answer 1

I doubt it has a name, but it follows trivially from integration by parts. Note that$$\left[f(z)\exp\left(-\frac{z^2}{2}\right)\right]_{-\infty}^\infty=\int_{-\infty}^\infty\left(f^\prime-zf\right)\exp\left(-\frac{z^2}{2}\right)dz,$$so if your integrals exist they're also equal, provided$$\lim_{z\to\infty}f(z)\exp\left(-\frac{z^2}{2}\right)=\lim_{z\to-\infty}f(z)\exp\left(-\frac{z^2}{2}\right)=0.$$