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

$$I_{\alpha}(z) := i^{-\alpha}J_{\alpha}(iz) = \sum_{m=0}^{\infty}\frac{\left(\frac{z}{2}\right)^{2m+\alpha}}{m! \Gamma(m+1+\alpha)} = \frac{1}{2\pi} \int_{-\pi}^{\pi} e^{i\alpha \tau + z \sin{\tau}} d\tau$$ 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

$$ \int_0^\pi t I_0(2\kappa \cos(t/2)) \; dt = \frac{\pi^2}{2} I_0(\kappa)^2 - 4 \sum_{r=0}^\infty \frac{I_{2r+1}(\kappa)^2}{(2r+1)^2} \; .$$ 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:

$$\begin{align} 1 &= I_0(z) + 2\sum_{r=1}^{\infty} (-1)^{r}I_{2r}(z) \\ e^z &= I_0(z) + 2\sum_{r=1}^{\infty} I_{r}(z) \\ \cosh{z} &= I_0(z) + 2\sum_{r=1}^{\infty} I_{2r}(z) \\ \end{align}$$ 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 $$\int_0^{\pi/2} J_{2r}(2\kappa\cos{t}) = \frac{\pi}{2} J_r^2(\kappa) \; ,$$ 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"

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

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

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

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

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

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

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.

  • 1
    $\begingroup$ I checked your integral numerically. It appears to diverge as $\kappa$ increases and the divergence appears to be exponential. $\endgroup$ – Jon Jan 16 '19 at 11:30
  • $\begingroup$ Thanks for checking Jon! I never thought about asymptotic, but now that you mention it, Wikipedia claims, that we have the the asymptotics $I_0(\kappa) \sim \frac{e^{\kappa}}{\sqrt{2\pi\kappa}}$, which would probably explain this phenomenon. $\endgroup$ – chickenNinja123 Jan 16 '19 at 11:55
  • $\begingroup$ Sure, I agree. Would not it be better to get some asymptotic then? $\endgroup$ – Jon Jan 16 '19 at 12:21
  • $\begingroup$ Well... I need the exact result xD The cases I am interested the most are actually where $0 \leq \kappa \leq 20$. $\endgroup$ – chickenNinja123 Jan 16 '19 at 13:14
  • $\begingroup$ Also using this exact asymptotic leads to the integral $\int t \frac{e^{2\kappa \cos{t}}}{\sqrt{4\pi\kappa\cos{t}}} dt$ which is equally, if not even harder, to solve... $\endgroup$ – chickenNinja123 Jan 16 '19 at 14:59

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 $x\rightarrow\infty$, $$ I_0(x)\sim\frac{e^x}{\sqrt{2\pi x}}. $$ So, I approximate your integral as $$ Z(\kappa)=\int_0^\frac{\pi}{2}tI_0(2\kappa\cos t)dt\sim\int_0^\frac{\pi}{2}t\frac{e^{2\kappa\cos t}}{\sqrt{4\pi\kappa\cos t}}dt. $$ 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 $$ Z(\kappa)\sim \frac{e^{2\kappa}}{\sqrt{4\pi\kappa}}\int_0^\frac{\pi}{2}te^{-\kappa t^2}\left(1-\frac{t^2}{16\pi\kappa}\right) $$ 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 $$ Z(\kappa)\sim\frac{e^{2\kappa}}{\sqrt{4\pi\kappa}}\frac{1}{2\kappa}\left(1-e^{-\kappa\frac{\pi^2}{4}}\right). $$ 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 $61.08994014$.


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


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

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

enter image description here

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.

  • $\begingroup$ Thank you Jon for your effort! As mentioned in the comments above I am actually not interested in an approximate expression, since the integral can evaluated numerically anyway. Also, the question in this thread focuses more on a closed form of the summation formula, which 'happens to coinside' with the integral, but technically this thread is not about the integral itself. $\endgroup$ – chickenNinja123 Jan 19 '19 at 12:13
  • $\begingroup$ @chickenNinja123 Sorry, but you said in the comments above "Also using this exact asymptotic leads to the integral $\int t\frac{e^{2\kappa\cos t}}{\sqrt{4\pi\kappa\cos t}}dt$ which is equally, if not even harder, to solve..." and, as a mathematician, I can't resist a challenge. Your statement is not true. $\endgroup$ – Jon Jan 19 '19 at 16:00
  • $\begingroup$ I appreciate your eager. You didn't solve the integral, though. If you count a Taylor approximation as a solution, then we could have just done that with the original integral itself. In fact, if we would be looking for a good approximation at all, simply truncating the infinite sum itself yields a rather satisfying solution; at least for $\kappa < \sqrt{2(r_{max}+1)}$ . $\endgroup$ – chickenNinja123 Jan 19 '19 at 16:41

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