Average distance from a square's perimeter to its center What is the average distance from any point on a unit square's perimeter to its center?
The distance from a square's corner to its center is $\dfrac{\sqrt{2}}{2}$ and from a point in the middle of a square's side length is $\dfrac{1}{2}$. A visual explanation of what I'm trying to explain
So, what would the average distance be, accounting for all the points along a square's perimeter?
Also if possible, a general formula for finding the average distance from center to edge of any $n$-sided regular polygon would be super awesome. 
 A: As mentioned in the comments, this answer requires calculus, I'll provide a general solution that should be applicable to all regular polygons.
Consider only $\frac 1{2n}$th of a n-sided polygon, from the centre of one side to the nearest vertex. Finding the average distance of this segment to the centre is sufficient, as the average will be the same for all other segments. 
If this segment subtends a total angle of $\theta$ at the centre, then the distance from centre of a point on this segment is given by
$$d = r\cos\theta \sec\alpha$$
Where $\alpha$ is the angle of this point. The average distance can be found by integrating through the angle $\theta$
$$\overline d = \frac{\int_0^\theta r\cos\theta \sec\alpha d\alpha}{\theta}\\
\overline d = \frac r \theta \cos\theta \ln|\sec\theta + \tan\theta|$$
For a n-sided polygon, $\theta = \frac \pi n$ and $r$ is the distance of a vertex from the centre. Substituting $n=4$ and $r=\frac {1}{ \sqrt{2}}$ should give you the distance for a square of unit side.
A: 
Consider a square ABCD of side $a$ centered at the origin $O$ (as shown in above figure)
By symmetry, the square ABCD is divided into $8$ congruent right triangles.
Consider any arbitrary point $P$ on square in first quadrant at an angle $x$ with +ve X-axis. The distance of point P from the center O is $\dfrac{a}{2}\sec x$. Taking average of distances of all the points on the perimeter of square (using symmetry of quadrants) as follows
$$D_\text{avg}=\frac{8\int_0^{\pi/4}\frac{a}{2}\sec x\ dx}{2\pi}$$
$$=\frac{2a}{\pi}\int_0^{\pi/4}\sec x\ dx$$
$$=\frac{2a}{\pi}\left[\ln\left|\tan\left(\frac{x}{2}+\frac{\pi}{4}\right)\right|\right]_0^{\pi/4}$$
$$=\frac{2a}{\pi}\ln(\sqrt2+1)$$ 
Therefore, the average distance from the center of all the points on the perimeter of a unit square ($a=1$) will be $$\frac{2}{\pi}\ln(\sqrt2+1)\approx 0.561099852 \ \mathrm{unit}$$
