Let $u = (u_1, \ldots, u_n)\in\mathbb{R}^n$ be a unit vector in $\mathbb{R}^n$, $Y_i$ be i.i.d standard normal
Is there any easy way to calculate
$$\mathbb{E} \left[ 1_{\displaystyle \left\{ \sum_{i=1}^n u_i \cdot Y_i > 0 \right\}} \sum_{i=1}^n u_i \cdot Y_i \right] $$
Progress: for $n=1$, it is an easy calculation. For $n \geq 2$, it seems to be messy. In fact, I think this is equal to $1/\sqrt{2\pi}$. I need this result to prove that the random line segment $L = (0, \sqrt{2\pi} Y)$ has the expectation = the unit ball in $\mathbb{R}^n$ where $Y$ is a standard gaussian in $\mathbb{R}^n$.
User Did is spot on. I'll just show you the details:
We can reformulate the expression within the square brackets as follows:
\begin{align*} 1_{\displaystyle \left\{ \sum_{i=1}^n u_i \cdot Y_i > 0 \right\}} \sum_{i=1}^n u_i \cdot Y_i = \max\left(\sum_{i=1}^n u_i \cdot Y_i, 0 \right) \end{align*}
We then have $\sum_{i=1}^n u_i \cdot Y_i \sim N(0, 1)$ by iid and the fact that $u$ is a unit vector. So, we are basically finding the expected value of the folded-normal distribution with $\mu = 0, \sigma^2 = 1$, and so the expected value is $\frac{1}{\sqrt{2\pi}}$.