About this thread
I opened a couple of threads here and at MathOverflow regarding the following question, but always put in different contexts. Since there has been no satisfactory answer yet, I collected all accumulated results so far and put it in a more general framework.
The Problem
Let $X, Y$ be two independent, centered von Mises distributed random variables on the circle, i.e. that they can be described via the density function
$$f_X(t \mid \kappa) = \frac{e^{\kappa \cos(t)}}{2\pi I_0(\kappa)} \cdot 1_{[-\pi, \pi]}(t) \; ,$$
where
$$I_{\alpha}(z) := \sum_{m=0}^{\infty}\frac{\left(\frac{z}{2}\right)^{2m+\alpha}}{m! \Gamma(m+1+\alpha)} = \frac{1}{2\pi} \int_{-\pi}^{\pi} e^{i\alpha \tau + z \sin{\tau}} d\tau$$
denotes the modified Bessel functions of the first kind. Let further
$$\Delta := \min{\big\{ |X_1 - X_2|, \, 2\pi - |X_1 - X_2| \big\}} = \pi - \big||X_1 - X_2| - \pi \big|$$
be their geodesic distance. The pdf for $\Delta$ can be explicitly calculated and yields
$$f_{\Delta}(t \mid \kappa) = \frac{I_0 \left( 2\kappa \cos{\frac{t}{2}} \right)}{\pi I^2_0(\kappa)} \cdot 1_{[0, \pi]}(t)\; .$$
Question:
Can we find a closed expression for $$\mathbb{E}[\Delta] = \frac{1}{\pi I^2_0(\kappa)} \int_0^{\pi} t I_0 \left( 2\kappa \cos{\frac{t}{2}} \right) dt = \frac{4}{\pi I^2_0(\kappa)} \int_0^{\pi/2} t I_0 \left( 2\kappa \cos{t} \right)dt \quad ?$$
Partial results from previous threads
Tackling the integral in the series representation
Plugging in the series expansion representation of $I_0$ leads to
$$\int_0^{\pi} t \cdot I_0 \left( 2\kappa \cos{\frac{t}{2}} \right) dt = \sum_{m=0}^{\infty} \left(\frac{\kappa^m}{m!}\right)^2 \int_0^{\pi} t\cos^{2m}{\left(\frac{t}{2}\right)} dt,$$
which seemed discouraging first, but Robert Israel was able to deduce this beautiful identity
$$ \int_0^\pi t I_0(2\kappa \cos(t/2)) \; dt = \frac{\pi^2}{2} I_0(\kappa)^2 - 4 \sum_{r=0}^\infty \frac{I_{2r+1}(\kappa)^2}{(2r+1)^2} \; .$$
This led to a follow-up thread, aiming to find a closed expression for the sum appearing on the right side of the equation.
Tackling the integral in the integral representation
Plugging in the integral representation of $I_0$ leads to
$$\int_0^{\pi} t \cdot I_0 \left( 2\kappa \cos{\frac{t}{2}} \right) dt = 4 \int_{-\pi}^{\pi} \int_0^{\pi/2} t e^{2\kappa \cos{t}\sin{\tau}} dt \, d\tau \;$$
which was the actual trigger for this umbrella-thread because I didn't want to open yet another thread each time a new approach seemed somewhat promising.
Using the substitution $u = \cos{t}$ and integration by parts, one can conclude that
$$\int_0^{\pi/2} t e^{\alpha\cos{t}} dt = \frac{\pi^2}{8} + \frac{\alpha}{2} \int_0^1 e^{\alpha u} \arccos^2{u} \; du \; ,$$
which looks equally hard to solve, but with Gradshteyn and Ryzhik, 4.551 one can deduce that
$$\int_0^1 e^{\alpha u} \arccos{u} \; du = \frac{\pi}{2\alpha} \big( I_0(\alpha) + \boldsymbol{L_0}(\alpha) -\alpha\big) \; ,$$
where $\boldsymbol{L_0}$ is the modified Struve function, which makes the integral doesn't look that impossible anymore.
Tackling the integral directly
Again Gradshteyn and Ryzhik mention in 6.519.1 that
$$\int_0^{\pi/2} J_{2r}(2z \cos{t}) dt = \frac{\pi}{2} J_r^2(z) \; ,$$ where $J_v(z) = i^z I_v(-iz)$, for $v \in \mathbb{N}$. Which made me question if that information is already enough to solve the sought integral.
Tackling the integral with a Fourier transform
One can exploit the fact that
$$\mathbb{E}[\Delta] = -i\varphi'_{\Delta}(0) = \lim_{\omega \rightarrow 0} \frac{\varphi_{\Delta}(\omega) - \varphi_{\Delta}(-\omega)}{2i\omega} = \lim_{\omega \rightarrow 0} \frac{\mathcal{Im}\left(\varphi_{\Delta}(\omega)\right)}{\omega} \,$$
where
$$\varphi_{\Delta}(\omega) := \int_{-\infty}^{\infty} e^{it\omega}f_{\Delta}(t) dt ,$$
is the characteristic function of $\Delta$.
Note, however that by pulling the limit into the integral (e.g. by dominated convergence) we end up with the same integral as above. Hence, this approach can only be useful if one manages to exploit some clever properties of the Fourier transform.
