Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
    
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. –  rubenvb Apr 25 at 8:29
    
For the specific case: apply differentiation $\frac{\partial}{\partial a}$ on the formula that you cited from G&R and use $\frac{d}{dz}K_{0}(z)=-K_{1}(z)$ . –  Johannes Trost Apr 28 at 15:56
    
@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. –  rubenvb Apr 29 at 8:33
add comment

2 Answers 2

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!"

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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