I'm searching for a suitable (hopefully simple enough) solution to the following form of integral:

$$\int_0^\infty \mathrm{d}x~x^n J_\nu(a x) J_\nu(b x) K_\mu(c x) $$

Where $n$, $\nu$, and $\mu$ are all integers, and $a$, $b$, and $c$ are all real and positive.

If not generally, a specific case would be quite helpful:

$$\int_0^\infty \mathrm{d}x~x^2 J_\nu(a x) J_\nu(b x) K_1(c x) $$

I am aware of the following:

  • Gradshteyn & Ryzhik eq. (6.522.3), which calculates the following integral as a relatively simple function:

$$\int_0^\infty \mathrm{d}x~x K_0(ax) J_\nu(b x) J_\nu(c x)$$

$$ \pi \mathrm{i} J_\nu(z) = \mathrm{e}^{-\nu\pi\mathrm{i}/2} K_\nu(-\mathrm{i} z) - \mathrm{e}^{\nu\pi\mathrm{i}/2} K_\nu(\mathrm{i} z), ~~~~|~\mathrm{arg}~z~| \leq \pi/2$$

I simply fail to see how this formula results in a simple change in his equations, especially because of the complex nature of the arguments introduced by the above formula, and the involved definition of $K_\mu$ for integer order.

Should I instead be looking for specific application of G&R eq. 6.522.17-18, which could provide the required formulas, or are there better approaches to this problem? It seems that Fabrikant in the above-linked article says the validity of these formulas is more strictly bounded than shown in G&R.

Any help here would be much appreciated.

  • $\begingroup$ If this better suits math.stackexchange.com, feel free to migrate. I couldn't decide, so I took after this question and put it here. $\endgroup$ – rubenvb Apr 25 '14 at 8:29
  • $\begingroup$ For the specific case: apply differentiation $\frac{\partial}{\partial a}$ on the formula that you cited from G&R and use $\frac{d}{dz}K_{0}(z)=-K_{1}(z)$ . $\endgroup$ – Johannes Trost Apr 28 '14 at 15:56
  • $\begingroup$ @Johannes: Yes, indeed, that hit me after writing up this question. The property $\partial_A A^\mu K_\mu(A r) = - r A^\mu K_{\mu-1}(A r)$ combined with $K_{-\mu}(x) = K_\mu(x)$ works wonders for the specific integrals I'm coming across. I'm still interested in how the relation between $J_\nu$ and $K_\nu$ transforms the integral in the Fabrikant paper. The result seems almost trivial. $\endgroup$ – rubenvb Apr 29 '14 at 8:33

Although I'm sure @Zurab's answer will indeed give me a solution (and I've heard the suggestion before in other situations), I'm not very familiar with the technique.

Following the suggestion by @Johannes in the comments, there is a much more straightforward way that, although not fully general, fits my problem exactly (i.e. it leads to expressions for the integrands I'm coming across).

Starting from G&R eq. (6.522.3), $$\int_0^\infty \mathrm{d} r~ r K_0(a r) J_\nu(b r) J_\nu(c r) = \left( \frac{r_2 - r_1}{r_2 + r_1} \right)^{|\nu|} \frac{1}{r_1 r_2}$$ where $r_1^2 = a^2 + (b-c)^2$ and $r_2^2 = a^2 + (b+c)^2$, one can use the derivative and symmetry properties of the bessel-$K$ functions to change the order from 0, incrementing the power of $r$ in the integrand at the same time. Simply calculating the derivative $c~\partial_c$ of the above righthand side gives my example from the question.


The integral should be calculable by the Mellin-transform technique. See the calculation of the similar (but different) integral in http://www.opticsinfobase.org/josaa/abstract.cfm?uri=josaa-7-7-1218 (Analysis of propagation through turbulence: evaluation of an integral involving the product of three Bessel functions, by Glenn A. Tyler).

P.S. As for the V.I. Fabrikant's article. There is a tragic story behind it. He suffered from a severe personality disorder and after being denied tenure because of his erratic behaviour, he walked into the department and shot and killed 4 people. The article about computation of integrals involving three Bessel functions he wrote in the jail. See http://xcorr.net/2013/01/17/killer-among-us-dr-fabrikant/

"O King, most high and wise Lord; How incomprehensible are thy judgments, and inscrutable thy ways!"


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