Assuming one draws two points from a von Mises distribution on a circle, I am looking for the expected distance between two such points.
Given the pdf of a centered von Mises distribution $$f_X(t \mid \kappa) = \frac{e^{\kappa \cos(t)}}{2\pi I_0(\kappa)} \cdot 1_{[-\pi, \pi]}(t) \; ,$$ one can calculate the pdf that describes the distance of two points by $$f_{\Delta}(t \mid \kappa) = \frac{I_0 \left( 2\kappa \cos{\frac{t}{2}} \right)}{\pi I^2_0(\kappa)} \cdot 1_{[0, \pi]}(t)\; ,$$ where $\Delta := \min{\big\{ |X_1 - X_2|, \, 2\pi - |X_1 - X_2| \big\}} = \pi - \big||X_1 - X_2| - \pi \big|$. Now, the expected distance corresponds to the expression $$\mathbb{E}[\Delta] = \frac{1}{\pi I^2_0(\kappa)}\int_0^{\pi} t \cdot I_0 \left( 2\kappa \cos{\frac{t}{2}} \right) dt ,$$ which I was not able to solve in a meaningful way even though there are some nice known integrals involving Bessel functions like $\int x I_0(x)\, dx = xI_1(x)$.
Trying to integrate it on its power series yields $$\int_0^{\pi} t \cdot I_0 \left( 2\kappa \cos{\frac{t}{2}} \right) dt = \sum_{m=0}^{\infty} \left(\frac{\kappa^m}{m!}\right)^2 \int_0^{\pi} t\cos^{2m}{\left(\frac{t}{2}\right)} dt,$$ which seemed equally hard to solve.
Another desperate attempt to tackle this problem with integration by parts led to this question.
TLDR:
Is there a closed form for the integral $$\frac{1}{\pi I^2_0(\kappa)}\int_0^{\pi} t \cdot I_0 \left( 2\kappa \cos{\frac{t}{2}} \right) dt \quad ?$$