i am trying to prove the identity in the title. I strongly think I need to use the error function $$\text{erf}(x) = \frac{2}{\sqrt{\pi}} \int_0^x \exp\{-t^2\}\ dt$$ in some way. Best I have so far is replacing $F(x) = \frac{1+\text{erf}\left(\frac{x}{\sqrt{2}}\right)}{2}$ to end up with
$$2\int_{-\infty}^{\infty} x \, f(x) \, F(x) \ dx = \int_{-\infty}^{\infty} \text{erf}\left(\frac{x}{\sqrt{2}}\right) \, x\, \frac{1}{\sqrt{2\pi}} \exp\left\{-\frac{1}{2}x^2\right\}\ dx.$$
My attempts on partial integration have failed, any other ideas or does anyone succeed?
I have a document stating that the equality holds without any further remarks or calculations and I have confirmed it via integration by quadrature and am looking for an analytic proof.
Very thankful for any help.
Answer
We have
$$ \int x f(x) \, dx = \frac{1}{\sqrt{2\pi}}\int x e^{-x^2/2} \, dx = -\frac{1}{\sqrt{2\pi}}e^{-x^2/2}+C = -f(x)+C, $$
so integrating by parts gives
$$ 2\int_{-\infty}^{\infty} xf(x)F(x) \, dx = [-2f(x)F(x)]_{-\infty}^{\infty} + \int_{-\infty}^{\infty} 2f(x)^2 \, dx $$
since $F'=f$. But $f(x)^2 = e^{-x^2}$, which is only a scaling away from the standard normal, by putting $x=y/\sqrt{2}$, so $dx=dy/\sqrt{2}$, and hence
$$ 2\int_{-\infty}^{\infty} xf(x)F(x) \, dx = \frac{2}{\sqrt{2}\sqrt{2\pi}} = \frac{1}{\sqrt{\pi}}. $$
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