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](http://web.eah-jena.de/~rsh/Forschung/Stoer/besint.pdf) 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](https://math.stackexchange.com/questions/3060103/integration-of-powers-of-trigonometric-function-with-linear-term/3060223?noredirect=1#comment6310572_3060223).


Another desperate attempt to tackle this problem with integration by parts led to [this question](https://math.stackexchange.com/questions/3060816/alternating-harmonic-series-of-modified-bessel-functions).


  

**TLDR**:   




> 
> Is there a closed form for the integral
>  $$\frac{1}{\pi I^2\_0(\kappa)}\int\_0^{\pi} t \cdot I\_0 \left( 2\kappa \cos{\frac{t}{2}} \right) dt \quad ?$$
> 
> 
> 







Closed form of integration of modified Bessel function composed with trigonometric function times a linear term
$$I\_0(2\kappa \cos(t/2)) = \sum\_{k=0}^\infty \cos(t/2)^{2k} \frac{\kappa^{2k}}{k!^2}$$
and it seems to me that
$$\int\_0^\pi \cos(t/2)^{2k} t \; dt = {2k \choose k} 2^{-2k-1} \pi^2 - \sum\_{r=0}^{\lfloor (k-1)/2 \rfloor} {2k \choose k+2r+1} \frac{2^{2-2k}}{(2r+1)^2} $$
so that, unless I've made an error,
$$ \eqalign{\int\_0^\pi & I\_0(2\kappa \cos(t/2)) t \; dt \cr & = \sum\_{k=0}^\infty \frac{\kappa^{2k}}{k!^2} {2k \choose k} 2^{-2k-1} \pi^2 - \sum\_{r=0}^\infty \sum\_{k=2r+1}^\infty {2k \choose k+2r+1} \frac{\kappa^{2k}}{k!^2} \frac{2^{2-2k}}{(2r+1)^2}\cr
&= \frac{\pi^2}{2} I\_0(\kappa)^2 - 4 \sum\_{r=0}^\infty \frac{I\_{2r+1}(\kappa)^2}{(2r+1)^2} } $$


Not closed form, of course, but still...


