I am trying to analytically evaluate

$$f(a,b)=\int_0^{2\pi} K_0(\sqrt{a^2+b^2-2ab\cos(\theta)}) d \theta$$

where $K_0$ is a modified Bessel function of the second kind and $a,b>0$. I happen to know the solution by a different method, it is

$$f(a,b)=2\pi I_0(\min \{ a,b \}) K_0(\max \{ a,b \})$$

where $I_0$ is a modified Bessel function of the first kind. But I don't know how to evaluate the integral directly, and CAS's seem to choke on it (though they can check the result above numerically just fine). It seems like it is probably useful to use the identity

$$K_0(x)=\int_0^\infty \frac{\cos(x t)}{\sqrt{t^2+1}} dt$$

and then interchange the order of integration, but then I still need to be able to compute

$$\int_0^{2\pi} \cos(\sqrt{a^2+b^2-2ab\cos(\theta)}) d \theta$$

for arbitrary $a,b>0$, before proceeding to the $dt$ integral.

Note that the problem is significantly simpler when $a=b$ because in this case $\sqrt{a^2+b^2-2ab\cos(\theta)}=2a\sin(\theta/2)$; in this case the cosine integral is $2\pi J_0(2at)$, which results in the easier problem

$$2\pi \int_0^\infty \frac{J_0(2at)}{\sqrt{t^2+1}}.$$

But in the case $a \neq b$ I don't even see how to begin.

  • $\begingroup$ Would the integral have a good geometric interpretation due to the resemblance to the cosine law? $\endgroup$ – Szeto Jun 24 '18 at 3:45
  • $\begingroup$ Just out of curiosity : could tell about the different method which leads to the beautiful $f(a,b)=2\pi \,I_0(\min \{ a,b \}))\, K_0(\max \{ a,b \})$ ? Thanks. $\endgroup$ – Claude Leibovici Jun 24 '18 at 3:57
  • $\begingroup$ @Szeto Yes, it is the same to write the integral of $K_0(|x-y|)$ as $y$ goes around the circle of radius $a$ and $x$ is any fixed vector of norm $b$. (See also my next comment.) $\endgroup$ – Ian Jun 24 '18 at 4:03
  • $\begingroup$ @ClaudeLeibovici For fixed $a$, $g(x)=f(a,|x|)$ is up to a constant factor the single layer potential associated to the equation $\Delta u = u$ in 2D. Thus it is a solution to this PDE which is bounded everywhere and has a particular jump in its normal derivative as $x$ passes through the circle of radius $a$. Given that interpretation you can solve the PDE with this jump condition and boundary conditions at the origin and infinity; this nice result comes about when you use an identity to simplify the Wronskian. $\endgroup$ – Ian Jun 24 '18 at 4:05
  • $\begingroup$ @Ian. Thanks for telling ! $\endgroup$ – Claude Leibovici Jun 24 '18 at 4:20

This is a consequence of the rather miraculous-looking Graf addition formula: if we assume that $0<b<a$ (since the symmetry is apparent), and write $w=\sqrt{a^2+b^2-2ab\cos{\theta}}$, we have $$ J_{\nu}(w) = \left(\frac{a-be^{i\theta}}{a-be^{-i\theta}}\right)^{\nu/2} \sum_{m=-\infty}^{\infty} J_{\nu+m}(a)J_{m}(b) e^{im\theta}. $$ One can prove this directly using the integral representation of $J_{\nu}$ on the right several times and swapping various integrals and sums: the proof is given in Watson's book, p. 360). Mucking about with this in the standard way gives the corresponding formulae for modified Bessel functions, (also given in Watson's book, on the next page) of which the one of interest to us is $$ K_{0}(w) = \sum_{m=-\infty}^{\infty} K_m(a) I_m(b) e^{im\theta}. $$ Integrating this from $0$ to $2\pi$ gives the result since only the $m=0$ term survives.

  • $\begingroup$ This is very helpful to me, thank you. I would not have expected an identity like this one (i.e. involving an infinite sum of Bessel functions of various orders) to be applicable here. The analogous identity for $K_1$ may also prove useful to me in a closely related problem... $\endgroup$ – Ian Jun 24 '18 at 5:15
  • $\begingroup$ Nice result. I guess you are summing over $m$ in both cases, not $n$. $\endgroup$ – Andras Vanyolos Jun 24 '18 at 8:55
  • $\begingroup$ @AndrasVanyolos Fixed, thanks. $\endgroup$ – Chappers Jun 24 '18 at 11:42
  • $\begingroup$ @Chappers Do you know if you can use the $\nu=1$ version of this formula to compute $\int_0^{2\pi} K_1(w) \frac{a-b\cos(\theta)}{w} d \theta$? The integrals under the sum seem to be more difficult in this case. $\endgroup$ – Ian Jun 24 '18 at 20:50
  • 1
    $\begingroup$ @Chappers I see now that I was being silly. What I was computing was the integral of $\frac{\partial K_0(w(a,b,\theta))}{\partial a}$ (where I drop the assumption that $a$ is the bigger one, but I do assume they are not equal). The easiest way to work with this derivative is not to use the chain rule to evaluate it in terms of $K_1(w)/w$ but just to differentiate under the Graf summation formula you already gave me for $K_0$. So thanks again. $\endgroup$ – Ian Jun 25 '18 at 3:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.