Expected length of sum of vectors 
Suppose we have $n$ arbitrary unit vectors $v_1, v_2, v_3, \dots, v_n$. (Here, a "random" unit vector is defined as $\langle \cos(x), \sin(x) \rangle$ for a random $x$ such that $0 \leq x < 2 \pi$). Evaluate the expected value of $$\left|\sum_{i=1}^{n}v_i\right|$$ where $|v|$ denotes the magnitude of $v$.


For $1$ vector, the answer is trivially $1$.
For $2$ vectors, we could fix $v_1$ to the $x$ axis, and the answer would be
$$\frac{1}{2\pi} \int_{0}^{2\pi}\left(\sqrt{\left(1+\cos\left(x\right)\right)^{2}+\sin\left(x\right)^{2}}\right)dx$$
First, let us evaluate the numerator.
$$\begin{align}&\int_{0}^{2\pi}\left(\sqrt{\left(1+\cos\left(x\right)\right)^{2}+\sin\left(x\right)^{2}}\right)dx \\
&=\int_{0}^{2\pi}\left(\sqrt{2+2\cos\left(x\right)}\right)dx \\
&=2\cdot\int_{0}^{2\pi}\left(\sqrt{\frac{1+\cos\left(x\right)}{2}}\right)dx \\
&=2\cdot\int_{0}^{2\pi}\left(\left|\cos\left(\frac{x}{2}\right)\right|\right)dx \\
&=2\cdot\int_{-\pi}^{\pi}\left(\cos\left(\frac{x}{2}\right)\right)dx \\
&=4\sin\left(\frac{\pi}{2}\right)-4\sin\left(-\frac{\pi}{2}\right) \\
&=8 \end{align}$$
Then, our answer for $2$ vectors would be $\frac{8}{2\pi} = \frac{4}{\pi}$.
For $3$ vectors, we have
$$\frac{1}{4\pi^{2}} \int_{0}^{2\pi}\int_{0}^{2\pi}\left(\sqrt{\left(1+\cos\left(x\right)+\cos\left(y\right)\right)^{2}+\left(\sin\left(x\right)+\sin\left(y\right)\right)^{2}}\right)dx\ dy$$
This seems hard to understand, does it have an elementary answer or a way of approaching?
 A: This is such a wonderful and thought provoking question. I have an idea on how we may be able to solve it in an elementary way.
I have an idea on how we can estimate the average magnitude for any number of vectors using a recursive method. Now, we know that for the case of one vector ($n=1$) the average magnitude  will be equal to $1$. For $n=2,$ it is the following case;

In this case you found the average as $\frac{4}{\pi}$ by letting
$$v_1=\binom{1}{0},$$
and integrating the magnitude of the vector $v_1+v_2$ with respect to the angle between the two vectors, and finally dividing it by $2\pi$ to get the average length of $v_1+v_2$ as the angle between them varies in $[0,2\pi).$
Now, for the $3$ vector case, we have $3$ vectors $v_1,v_2,$ and $v_3$ and we need to find the average magnitude of $v_1+v_2+v_3.$
WLOG, we let $v_1=\binom{1}{0}$ and we will vary the angles of the other three vectors (since they are unit vectors we know their magnitude). Here's a picture;

Lets only focus on the angle between the vectors $BC$ and $CD$ for a while. We know that the angle between them ranges from $0$ to $2\pi,$ so the magnitude of the vector $v_2+v_3=BD$ is also some number. But the average magnitude of $v_2+v_3=BD$ has to be $\frac{4}{\pi}$ as you earlier calculated in the $3$-vector case.
So, since we are considering an 'average' scenario, we can say that the magnitude of $BD=v_2+v_3$ is $\frac{4}{\pi}.$ But we need to be very careful. We are considering a superposition of all the possible angles between $v_2$ and $v_3.$

So we have just two vectors now! One is $v_1=\frac{1}{0}$ and the other is $v_2+v_3,$ where the second vector has a magnitude of $\frac{4}{\pi}.$ Now, we need to consider the angle between the two vectors $v_1$ and $v_2+v_3$ to range from $0$ to $2\pi$ and take the integral of the total vector magnitude for a given angle $\theta.$ Just like in the first case with two vectors.
Similarly, we can continue in a similar way to get a sequence of average vector magnitudes from for any number of vectors. Hope this works for you.
A: This problem seems to be posed by Pearson in 1905 to challenge the crowd (in the journal of Nature)
https://www.nature.com/articles/072294b0
The distribution of length is solved subsequently by J.C. Kluyver in 1905. (that document cannot be found in google...though many cited it  "a local probability problem").
A paper that discussed it more thoroughly is On the Problem of Random Flights in 1985. From what I read, the exact solution of the distrubtion of length is an integral of product of Bessel function, which could be numerically evaluated but may not be simplified..

A: Expanding on Binxu's comment on his post, although this isn't really an answer—just seeing how the result applies. For $n$ unit vectors, we have
$$F(n;r)=\Pr (|S_n|\leq r)=r\int_0^\infty J_1(rt)[J_0(t)]^n\, dt.$$
You can think of this probability function as the cumulative distribution function of the result. The probability density function (PDF) $f$ can be recovered by differentiation, and then the expected value found by its definition as the first moment around zero. To wit,
\begin{align*}
f(n;r) &= \partial_r F \\
&= \frac{1}{r} F(n;r)+r\partial_r\int_0^\infty J_1(rt)[J_0(t)]^n\, dt. \\
E_n &= \int_0^n f(n; r)\cdot r\, dr \\
&= \int_0^n F(n; r)\,dr+\int_0^nr^2\partial_r\int_0^\infty J_1(rt)[J_0(t)]^n\, dt\, dr.
\end{align*}
The second integral can be dealt with using integration by parts:
\begin{align*}
\int_0^nr^2\partial_r\int_0^\infty J_1(rt)[J_0(t)]^n\, dt\, dr &= r^2\left.\int_0^\infty J_1(rt)[J_0(t)]^n\, dt\right]^{r=n}_{r=0}-\int_0^n2F(n; r)\, dr.\\
\end{align*}
Thus we get
\begin{align*}
E_n=n^2\int_0^\infty J_1(nt)[J_0(t)]^n\, dt-\int_0^n F(n; r).
\end{align*}
For $n=2$ WolframAlpha gives $2$ for the first term and $2-\frac{4}{\pi}$ for the second one, so $E_2=\frac{4}{\pi}$. As to how the latter is derived... no clue. I am ignorant of the many identities and integrals involving Bessel functions. For $n=3$ WA fails to find the second integral. Personally I doubt there is a nice closed form. But at least now it's a double integral regardless of $n$....
Hopefully someone will have an actual answer.
