Ytukyg

I was trying to find a closed form for the integral
$$4int_0^{pi/2} t , I_0(2kappacos{t}) dt ; ,$$
where

$$I_{alpha}(z) := i^{-alpha}J_{alpha}(iz) = sum_{m=0}^{infty}frac{left(frac{z}{2}right)^{2m+alpha}}{m! Gamma(m+1+alpha)} = frac{1}{2pi} int_{-pi}^{pi} e^{ialpha tau + z sin{tau}} dtau$$
are the modified Bessel functions. This integral popped up when I was trying to find the average difference of two points on a circle, where these points are assumed to be drawn independently from a von Mises distribution. It was noted by Robert Israel that this integral can be reduced to

$$ int_0^pi t I_0(2kappa cos(t/2)) ; dt = frac{pi^2}{2} I_0(kappa)^2 - 4 sum_{r=0}^infty frac{I_{2r+1}(kappa)^2}{(2r+1)^2} ; .$$
So I was wondering, if we can further simplify this expression, or to stated more clearly:

Is there a closed formula for the following sum of modified Bessel functions of the first kind?
$$sum_{r=0}^infty frac{I_{2r+1}(kappa)^2}{(2r+1)^2}$$

A lot of remarkable identities in terms of infinite sums of Bessel functions are known. E.g. Abramowitz and Stegun list in §9.6.33ff. a few of them, like:

$$begin{align}
1 &= I_0(z) + 2sum_{r=1}^{infty} (-1)^{r}I_{2r}(z) \
e^z &= I_0(z) + 2sum_{r=1}^{infty} I_{r}(z) \
cosh{z} &= I_0(z) + 2sum_{r=1}^{infty} I_{2r}(z) \
end{align}$$
WolframResearch lists another bunch of infinite series identities.
Also, Neumann's addition theorem seems to work wonders <1> <2> <3>.

Regarding the integral itself, Gradshteyn and Ryzhik mention in 6.519.1 that
$$int_0^{pi/2} J_{2r}(2kappacos{t}) = frac{pi}{2} J_r^2(kappa) ; ,$$
where $J_r(x) = i^rI_r(-ix)$. So there might be a chance to expect something along this line.

Going back to the original problem "What is the expected value of a distribution $Delta$ with the following density function"

$$f_{Delta}(t) := frac{I_0 left( 2kappa cos{frac{t}{2}} right)}{pi I^2_0(kappa)} ; ,$$
a straightforward integration leads to the integral mentioned above. Using some probability theory voodoo we can make use of the fact that

$$mathbb{E}[Delta] = -i varphi'_{Delta}(0)
= -i left[frac{d}{domega}
mathcal{F}(f_{Delta})(omega)
right] Bigg|_{omega=0}
= -i left[frac{d}{domega}
int_{-infty}^{infty} e^{itomega}f_{Delta}(t) dt
right] Bigg|_{omega=0} $$

where $varphi_{Delta}$ is the characteristic function of $f_{Delta}$ and $mathcal{F}$ the (properly scaled) Fourier transform. Now with $varphi(-omega) = overline{varphi(omega)}$, we could further rewrite

$$mathbb{E}[Delta] = -ivarphi'_{Delta}(0)
= lim_{omega rightarrow 0} frac{varphi_{Delta}(omega) - varphi_{Delta}(-omega)}{2iomega}
= lim_{omega rightarrow 0} frac{mathcal{Im}left(varphi_{Delta}(omega)right)}{omega} ,$$

to (by plugging in the integral representation of $I_0$) obtain

$$mathbb{E}[Delta]
= frac{pi}{2} - frac{4}{pi I_0^2(kappa)} sum_{r=0}^infty left( frac{I_{2r+1}(kappa)^2}{2r+1} right)^2
= frac{1}{pi^2 I_0^2{kappa}} cdot lim_{omega rightarrow 0} int_0^{pi/2} int_{-pi}^{pi} frac{sin(tomega)}{omega} e^{2kappacos{t}sin{tau}} dtau , dt ; ,$$

but this will essentially lead to the same integral we started with. The promising part about this approach is that the Fourier transform pops up, which might leave some room for the harmonic analysis people among you to do your magic.

This is not a complete answer to your question that, in the way it is stated, appears very hard. But, as said in the comments, it is easily amenable to asymptotic treatment and the approximation is not that bad. Firstly, it is known that, for $xrightarrowinfty$,
$$
I_0(x)simfrac{e^x}{sqrt{2pi x}}.
$$
So, I approximate your integral as
$$
Z(kappa)=int_0^frac{pi}{2}tI_0(2kappacos t)dtsimint_0^frac{pi}{2}tfrac{e^{2kappacos t}}{sqrt{4pikappacos t}}dt.
$$
The last integral can be managed with the Laplace method by noting that it takes the great part of contributions at $t=0$. So, I do a Taylor series for the cosine obtaining
$$
Z(kappa)sim frac{e^{2kappa}}{sqrt{4pikappa}}int_0^frac{pi}{2}te^{-kappa t^2}left(1-frac{t^2}{16pikappa}right)
$$
and we see that the next-to-leading correction can be neglected. We are left with a very easy integral and the final result will be
$$
Z(kappa)simfrac{e^{2kappa}}{sqrt{4pikappa}}frac{1}{2kappa}left(1-e^{-kappafrac{pi^2}{4}}right).
$$
Of course, this is not defined for $kappa=0$ but we know that in that case the integral has the exact value $frac{pi^2}{8}$.

So, how good is this approximation? It is fairly good indeed. Let me show some values

$Z(1)sim 0.9538227748$ the exact value is $1.658067328$.

$Z(4)sim 52.55432675$ the exact value is $61.08994014$.

$vdots$

$Z(20)sim 3.711926385cdot 10^{14}$ the exact value is $3.804956771cdot 10^{14}$.

$vdots$

$Z(10
0)sim 1.019204783cdot 10^{83}$ the exact value is $1.024131055cdot 10^{83}$.

To have a clear idea, in the range $kappa=0.01ldots 20$, I plotted the following log-log graph.

I should say that the agreement is excellent. The red curve is the exact one. I hope this will be of some help to you.