Square distance of a 2D random walk with 180º angle choice Consider a random walk on the plane generated as follows:

*

*$(x_0, y_0)\in\mathbb{R}^2$ is the initial position.

*$\alpha_0\in(-\pi,\pi)$ is the initial direction that the walker is facing.

*The next position is generated by taking a length 1 step towards a random direction in "front" of the walker, that is $(x_{k+1}, y_{k+1}) = (x_k + \cos(\alpha_k), y_k + \sin(\alpha_k))$ and $\alpha_{k+1} = \alpha_k + \beta_k$, where $\beta_k\sim U(-\frac{\pi}{2},\frac{\pi}{2})$.


If $R_N$ is such a random walk with $N$ steps, I would like to know
what is the expected value of the square of the final distance from the initial point as a function of N (I'll denote this value by $D_N$).

It is obvious that $E(D_1)=1$ because we take a single step. Also, for $N=2$, since we can assume without loss of generality that $(x_0,y_0)=(0,0)$ and $\alpha_0=0$, one gets that $$E(D_2)=\frac{1}{\pi}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\|(1+\cos(t), \sin(t))\|_2^2\ dt=\frac{2}{\pi}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}1+\cos(t)\ dt = \frac{2}{\pi}(\pi + 2).$$
If I'm not wrong, in general we get
$$E(D_{N+1})=\frac{1}{\pi^N}\int_{[-\frac{\pi}{2},\frac{\pi}{2}]^N}\left(1+\sum_{k=1}^N\cos(t_1+\cdots+t_k)\right)^2+\left(\sum_{k=1}^N\sin(t_1+\cdots+t_k)\right)^2\ d(t_1,\ldots,t_N).$$
I know how to simplify the formula a bit more, or how to even run simulations in a computer to get an approximation of the final result, but would you be able to provide a closed formula for the value of such integral? Thanks!
 A: Let's use complex numbers to represent the position in 2D space. Assuming with out loss of generality that the starting angle is $\alpha_0=0$, then the position after $n$ steps is
$$
z_N = \sum_{n=1}^N \exp\left(i\sum_{k=1}^n \alpha_k\right),
$$
with $\alpha_k$ being uniformly random on $[-\pi/2,\pi/2]$. We then compute the mean square distance
$$
\begin{align}
\langle z_N\bar z_N\rangle &= \sum_{n=1}^N\sum_{n'=1}^N \exp\left(i\sum_{k=1}^n \alpha_k-i\sum_{k=1}^{n'} \alpha_k\right)\\
& = N+2(N-1)\langle e^{i\alpha}\rangle+2(N-2)\langle e^{i\alpha}\rangle^2+\ldots+2\langle e^{i\alpha}\rangle^{N-1}\\
&=N+\sum_{k=1}^{N-1}2(N-k)\langle e^{i\alpha}\rangle^k.
\end{align}
$$
The combinatorics are not too bad  and depends on how many $\alpha$s cancel in the square. Also we used that the $\alpha$s are uncorrelated to write $\langle e^{i\alpha_1+i\alpha_2}\rangle=\langle e^{i\alpha_1}\rangle\langle e^{i\alpha_2}\rangle$ etc.
Finally, we can compute
$$
 \langle e^{\pm i\alpha}\rangle= \frac{1}{\pi}\int_{-\pi/2}^{\pi/2}e^{\pm i\alpha}\, d\alpha= \frac{2}{\pi}
$$
and evaluate the sum
$$
\langle z_N\bar z_N\rangle = \frac{2^{2+N}+\pi^N(N(\pi^2-4)-4\pi)}{\pi^N(\pi-2)^2}.
$$
