Let $g$~ $N(0,I_n)$ and $u,v\in\mathbb{R}^n$ with $\vert\vert v\vert\vert_2=1=\vert\vert u\vert\vert_2$. I want tho show that \begin{equation} \mathbb{E}[\langle g,u\rangle sign(\langle g,v\rangle)]=\sqrt(2/\pi)\langle u,v\rangle. \end{equation}
I am struggeling to find a angle to begin the proof. Mabye one approach is it to work with the definition of the dot product: $\langle g,u\rangle = \vert\vert g\vert\vert_2 cos(\theta)$ with $\theta$ being the angle between g and u.
Let $R\in\mathrm{O}(n)$ be an isometry which sends $v\mapsto e_1$ and $u\mapsto\cos(\theta)e_1+\sin(\theta)e_2$. (In particular, this means that $\langle u,v\rangle=\cos\theta$.) Then
$$\langle g,u\rangle\mathrm{sgn}\langle g,v\rangle=\langle Rg,Ru\rangle\mathrm{sgn}\langle Rg,Rv\rangle=\langle h,\cos(\theta)e_1+\sin(\theta)e_2\rangle\mathrm{sgn}\langle h,e_1\rangle$$
where $h=Rg$ has the same multivariate distribution as $g$ itself (since this distribution is invariant). Write out $h=(h_1,\cdots,h_n)$ with the $h_i$s being IID standard normals, so we are calculating
$$ \mathbb{E}\big[(\cos(\theta)h_1+\sin(\theta)h_2)\mathrm{sgn}(h_1)\big]=\cos(\theta)\mathbb{E}[h_1\mathrm{sgn}(h_1)]+\sin(\theta)\mathbb{E}[h_2\mathrm{sgn}(h_1)]. $$
Since $h_1,h_2$ are independent, the second term with $\sin(\theta)$ vanishes. The first term becomes $\cos(\theta)\mathbb{E}\big[|H|\big]$ with $H$ a standard normal, which we can compute with simple calculus.