What is the average distance of point in hypercube to its center? How do I compute the average distance of point inside an hypercube to the center of the hypercube as a function of the dimensionality of the space?
Here I consider the hypercube defined as $C_n=\{x\in\mathbb{R}^n: -\frac{1}{2}\leq x_i\leq\frac{1}{2}, \forall_{i\leq n}\}$ with center $(0, ..., 0)\in\mathbb{R}^n $
Since for a random point in $C_n$ we have that each component $X_i$ is uniformly distributed between $-\frac{1}{2}$ and $\frac{1}{2}$. And since all such components are independent, does it follow that:
$$E\left[\sqrt{\sum_{i=1}^n X_i^2}\right] = \sqrt{E\left[\sum_{i=1}^n X_i^2\right]}=\sqrt{\sum_{i=1}^n E\left[X_i^2\right]}=\sqrt{\sum_{i=1}^n \frac{1}{12}}=\sqrt{\frac{n}{12}}$$ 
?
Edit:
Of course my calculation is wrong! the square root of the expected value is not the expected value of the square root!
But the question remains: What is the correct expression?
If we don't have closed form, could we even try to get the value recursively?
as 
$\begin{align}
A(n) &= \int_0^{\frac{1}{2}}...\int_0^{\frac{1}{2}}\sqrt{x_1^2 + ... + x_n^2}dx_1...dx_n\\
&= \int_0^{\frac{1}{2}}...\int_0^{\frac{1}{2}}x_n\sqrt{\frac{x_1^2 + ... + x_{n-1}^2}{x_n^2}+1}dx_1...dx_n \\
&= \int_0^{\frac{1}{2}}x_n\left(\int_0^{\frac{1}{2}}...\int_0^{\frac{1}{2}}\sqrt{\frac{x_1^2 + ... + x_{n-1}^2}{x_n^2}+1}dx_1...dx_{n-1}\right)dx_n \\
&=? \int_0^{\frac{1}{2}}x_ng(A(n-1), x_n)dx_n
\end{align}$
 A: For $n=2$, we have a square given by $$C_2 = [-1/2, 1/2] \times [-1/2, 1/2]$$ 
Let the portion of $C_2$ in the first quadrant be $$C_2' = [0,1/2]\times[0,1/2]$$
By symmetry, the average distance from a point $P$ (chosen uniformly randomly from the interior of $C_2$) to the origin, will be the same as the average distance of a point $P'$ (chosen uniformly randomly from the interior of $C_2'$) to the origin. 
We can find this average distance by integrating the distance formula over the quarter square, then dividing by the area of the region:
$$\displaystyle\frac{1}{\left(1/2\right)^2}\int_0^\frac{1}{2}\int_{0}^\frac{1}{2} \sqrt{x^2+y^2}\,dx\,dy = 4\int_0^\frac{1}{2}\int_{0}^\frac{1}{2} \sqrt{x^2+y^2}\,dx\,dy$$

In general, in $\mathbb{R}^n$, we can index points as $(x_1, x_2,x_3, ... , x_n)$. We have $$C_n = \left[\,-\frac{1}{2},\frac{1}{2}\,\right]^n$$
$$C_n' = \left[\,0,\frac{1}{2}\,\right]^n$$
The volume of the reduced portion of the $n$-dimensional hypercube, $C_n'$, is $1/2^n$. Therefore, we can compute the average distance as
$$\boxed{\mathbb{E}\,\left[\sqrt{\displaystyle\sum_{k=1}^n x_k^2}\,\right]=2^n\underbrace{\int_{0}^\frac{1}{2}\int_{0}^\frac{1}{2}\int_{0}^\frac{1}{2}\cdots \int_{0}^\frac{1}{2}}_{n\text{ integrals}}\left(\sqrt{\displaystyle\sum_{k=1}^n x_k^2}\,\right)dx_1\,dx_2\,dx_3\,\cdots\,dx_n\,\,}$$

In computation of the above integral, the following fact is useful: 

If $A \in \mathbb{R}$ is a non-negative constant, then for any constant $B \in \mathbb{R}$, we have the antiderivative $$\int\sqrt{x^2+A} \,\,dx = \frac{A\ln\left|x+\sqrt{x^2+A}\right| + x\sqrt{x^2+A}}{2} + B$$

This fact makes the computation possible, albeit extremely messy.
