Bounds on expected distance from start to end of $n$ line segments connected at random angles

Question

This is similar to this question, but I am interested in bounds or an asymptotic growth rate for $E(n)$. $E(n)$ is equal to $$\frac{1}{(2\pi)^n}\int_0^{2\pi}\int_0^{2\pi} ...\int_0^{2\pi}\sqrt{\left(\sum_{k=1}^{n} \cos \theta_k\right)^2+\left(\sum_{k=1}^{n} \sin \theta_k\right)^2} d\theta_1 d\theta_2...d\theta_n$$ which is the expected value of the distance from the origin to the end of the last line segment. Each line segment is of length $1$ and is connected end to end at an angle uniformly chosen from the range $0 < \theta < 2\pi$.

To start off, it is clear that $E(1) = 1$. $E(2) = \frac{4}{\pi} \approx 1.27$. $$E(3) = \frac{3\cdot2^{1/3}}{16 \pi^4} \Gamma(1/3)^6 + \frac{27 \cdot 2^{2/3}}{4 \pi^4}\Gamma[2/3]^6 \approx 1.57$$

I found the formula for $E(3)$ in OEIS entry A240946.

If anyone could provide an asymptotic growth rate or tight bounds for $E(n)$, that would be much appreciated.

Edit: On the OEIS page I linked earlier, this research paper was there. From that, I got that $p_n$, the PDF of $n$ line segments connected randomly together end to end, is $$p_n \approx \frac{2x}{n}e^{-\frac{x^{2}}{n}} $$. This then means that $$E(n) = \int_{0}^{\infty} xp_n \approx \frac{\sqrt{\pi}}{2} \sqrt{n}$$

From the same research paper, $$p_n(x) = \int_{0}^{\infty} xtJ_0(xt)J_0^n(t) dt$$ where $J_v$ is the Bessel function of the first kind of order $v$. From here, we get that $$E(n) = \int_0^n \int_{0}^{\infty} x^2tJ_0(xt)J_0^n(t) dt dx \tag 1$$

Equation $(1)$ can be rewritten by taking $x = u/t$ in order to get $$E(n) = \int_0^{\infty} \int_{0}^{nt} \frac{u^2}{t^2}J_0(u)J_0^n(t) du dt$$

If I could get more terms on the approximation of $E(n) \approx \frac{\sqrt{\pi}}{2} \sqrt{n}$ or perhaps even an infinite summation, that would be greatly appreciated.

Answer 1

This is far from being an answer, but it's only a simple numerical experiment.

Writing in Wolfram Mathematica 12.0:

nmax = 1000;
data = Monitor[Table[
            a = Total[Cos[Table[ToExpression[StringJoin["θ", ToString[k]]], {k, n}]]]^2;
            b = Total[Sin[Table[ToExpression[StringJoin["θ", ToString[k]]], {k, n}]]]^2;
            c = StringJoin["NIntegrate[Sqrt[", ToString[a, InputForm], "+", 
                                               ToString[b, InputForm], "] / (2π)^n,"];
            Do[If[k < n, c = StringJoin[c, "{θ", ToString[k], ",0,2π},"],
                         c = StringJoin[c, "{θ", ToString[k], ",0,2π}, Method->MonteCarlo]"]
                 ], {k, n}]; ToExpression[c], {n, nmax}],
            Row[{ProgressIndicator[n, {0, nmax}], n}, " "]];

with a little patience the first thousand integrals are obtained. So, writing:

fdata = FindFormula[data, n]

the function that best approximates these integrals is obtained:

0.885523 n^0.5

Finally, writing:

ListPlot[{data, Table[fdata, {n, nmax}]}, PlotStyle -> {Blue, Red}]

you can overlap the two graphs: