Expected Value of Magnitude of Sum of Unit Vectors 
Given $n$ randomly selected unit vectors $v_1, v_2, \cdots, v_n$ in $\mathbb{R}^n$, how can we compute $\mathbb{E}[|v_1+v_2+\cdots+v_n|]$?

This is quite a simple question, but I believe it is actually quite difficult. I've tried elementary approaches, the most basic of which consists of just splitting the vectors into components, but ultimately we're taking the expected value of something that is not a linear function of the components, so we can't simplify it nicely with something like linearity of expectation. Intuitively, I suspect that the answer is just $\sqrt{n}$, assuming "on average", that the sum of the vectors is the sum of the basis vectors in $\mathbb{R^n}$. I thought of the problem after encountering the problem:

Let $v_1, \cdots, v_n\in \mathbb{R}^n$ be such that $|v_i|=1$ for each $i$. Prove that we can select $x_1, \cdots x_n$, each either $1$ or $-1$, such that $|\sum x_iv_i|\le \sqrt{n}$

The solution to this is quite a bit simpler, but I suppose it could possibly provide insight on how to solve my modification. The idea is to square the inequality and compute the expected value of $|\sum x_iv_i|^2$ by summing that over all choices of $x_i$. This is quite easy to do if you just split the vectors into components, as many of the terms vanish when summing.
Also, it's notable that there are many similar questions to this on MSE. I am posting this because I am unable to transfer any of the ideas over.
 A: This might give some idea:
For $n=1$ the answer is just $1$.
for $n=2$, the answer is $\frac{4}{\pi}$. The calculations are as follows:
Since $|v_1|=|v_2|=1$, we can write,
$\mathbb{E}[|v_1+ v_2|] = \mathbb{E}[\sqrt{(v_1+ v_2).(v_1+v_2)}] =\mathbb{E}[\sqrt{2+2\,v_1.v_2}]= \mathbb{E}[\sqrt{2+2\,\cos(\theta)}].$
So the problem reduces to finding the above expectation where $\theta$ is the angle between two vectors. Given that the vectors are uniformly distributed on a circle and they are chosen independently, simple calculations show that $\theta$ is uniformly distributed over $[0, \pi]$.
Having the distribution of $\theta$, we can easily calculate the expectation:
$$\mathbb{E}[\sqrt{2+2\,\cos(\theta)}] = \int_{0}^{\pi} \sqrt{2+2\,\cos(\theta)}\, \frac{1}{\pi}\; d\theta = \frac{4}{\pi} \approx1.273.$$
Which is a bit smaller than $\sqrt{2} \approx 1.414$.
For a general case, we can write:
$$\mathbb{E}[|v_1 + ...+v_n|]= \int ...\int\sqrt{n+2\,\cos(\theta_1)+...+2\,\cos(\theta_l)}\, f(\theta_1,...\theta_l) \;d\theta_1...d\theta_l. $$
Where $l$= $n\choose 2$, $f(\theta_1, ..., \theta_l)$ is the joint distribution and the integral is taken over the possible domain for $\theta=(\theta_1, ...,\theta_l)$ which should be calculated. For $n \geq 3$, not all angles are uniformly distributed and they are not independent.
