The expected distance from a point to its neighbors in homogeneous spatial poisson process Let $n$ point are distributed as per a homogeneous spatial Poisson process of rate $λ$ in a square of side $2a$, and $R$ be the distance from a point to its  neighbors within a distance $r$. What is the expected value$E(R)$? 
Is the bellow proof correct?
Let $B(x,R')$ denote a disk centered at the position of a random point $x$ (consider this point as an origin) with radius $R'$. Let $m$ be the number of points inside the disk $B$. Given $m\geq 1$, let $\{r_1,r_2,…,r_m\}$ be the set of radial distances of points inside the disk respect to the origin.
Since a circle's circumference is proportional to its radius, the common density $f_R$ of $r_i$ for all $i$ is also proportional to its radius. That is, $f_R(r) = \alpha r$, for some constant $\alpha$. We find $\alpha$ as follows:
\begin{align}
1 &= \int_{0}^{R'} \alpha r\;dr \\
&= \left[ \dfrac{\alpha r^2}{2}\right]_{0}^{R'} \\
&= \dfrac{\alpha R'^2}{2}. \\
\therefore\quad \alpha &= \dfrac{2}{R'^2}
\end{align}
So $f_R(r) = \dfrac{2r}{R'^2}.$
If this is correct, what is the $E(R)$?

From this post on a similar topic:
The probability there is no neighbor within radius r is 
$$Pr(N=0)=e^{-λπr^2}$$
The probability the nearest neighbor is within radius r is 
$$F(r)=P(R≤r)=Pr(N≥1)=1−e^{-λπr^2}$$
So the probability density for the distance to the nearest neighbor is 
$$f(r)=\frac{d}{dx}{F(r)}=2λπre^{−λcπr^2}$$
You can integrate this PMF to get the expected value as follows
$$E[R]=\int rf(r) \, dr = \frac{1}{2\sqrt{λ}}$$
