An integral involving three Bessel functions

Question

I am looking for a closed form for the following integral

$$ I = \int_0^\infty \mathrm{d} x \ x \ J_0(ax) J_0(bx) J_1(cx) $$

which can be thought of as a particular case of the more general integral

$$ I(n_1,n_2,n_3) = \int_0^\infty \mathrm{d} x \ x \ J_{n_1}(ax) J_{n_2} (bx) J_{n_3} (cx) $$

i.e. $I=I(0,0,1)$. I am aware of the following related results:

• A closed form, analytic result for I(0,0,0), see for instance Ref. [1].
• A closed form, analytic result for I(0,1,1), also in Ref. [1].
• A closed form, analytic result for the general case $I(n_1,n_2,n_3)$ when $n_1 + n_2 + n_3 = 0$, in Ref. [2].
• A general expression in terms of the hypergeometric function $F_4$, which would include I(0,0,1) as a particular case, which however requires $c>a+b$. This is in Ref. [3].
• The analytic/numerical approach of Ref. [4]

Still I cannot find a result for $I=I(0,0,1)$ for every value of $a$,$b$,$c$. I have verified numerically that the integral exists finite, even when the condition $c>a+b$ is not met.

So far I have tried expressing $J_1(cx)$ using recursion relations to try and recover other known integrals, or integrating by parts one or more Bessel functions, or expressing the Bessel functions as a series. All these approaches seems to make the integral more complicated.

Similarly to this [5] other question, the motivation is coupling of momenta in Quantum Mechanics.

[1] G.N. Watson, "A Treatise on the Theory of Bessel Functions", (Cambridge University Press, Cambridge), 1966.

[2] A. D. Jackson and L. C. Maximon, "Integrals of Products of Bessel Functions", SIAM J. Math. Anal. 3, 446 (1971).

[3] W.N. Bailey, "Some Infinite Integrals Involving Bessel Functions", Proceedings of the London Mathematical Society 40, 37 (1936).

[4] http://www.mathematica-journal.com/2012/12/on-the-integral-of-the-product-of-three-bessel-functions-over-an-infinite-domain/

[5] $\mathrm{Bessel}^3$ Integral

Answer 1

Without loss of generality we can set $c=1$; Mathematica returns a closed form for $a=b$.

For $|a|<1/2$ the result is

$$I= \int_0^\infty \mathrm{d} x \ x \ J_0(ax) J_0(ax) J_1(x)=\frac{4K(\alpha)}{\pi^2\sqrt{1-4a^2}}\left[2E(\alpha)-K(\alpha)\right]\qquad\qquad(\ast)$$ where $\alpha=\tfrac{1}{2}-\tfrac{1}{2}\sqrt{1-4a^2}$, and $E$ and $K$ are elliptic integrals (as defined here and here).

For $|a|>1/2$ Mathematica returns a Meijer-G function, $$I=\frac{2}{\sqrt\pi} G_{3,3}^{2,1}\left(\frac{1}{4 a^2}\left| \begin{array}{c} 1,1,1 \\ \frac{1}{2},\frac{3}{2},\frac{1}{2} \\ \end{array}\right. \right),$$

which equals the real part of $(\ast)$, so I would conclude that the elliptic-integral expression can be used for all $a$.

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I tried to check this numerically, comparing $$A=\int_0^\infty \mathrm{d} x \ x \ J_0(ax) J_0(ax) J_1(x)$$ with $$B={\rm Re}\,\left[\frac{4K(\alpha)}{\pi^2\sqrt{1-4a^2}}\left[2E(\alpha)-K(\alpha)\right]\right],\;\;\alpha=\tfrac{1}{2}-\tfrac{1}{2}\sqrt{1-4a^2}.$$ The numerical evaluation of $A$ has rapid oscillations for $a<1/2$ as well as for $a>1/2$, and only the envelope is described by $B$. I don't know whether or not this is an artefact of the numerics.
_{[Zakk tells me they are an artefact, so it seems the elliptic-integral expression stands: A=B for $a<1/2$ and $a>1/2$.]}

_{blue: numerical evaluation of A; gold: plot of B, both as a function of $a$}

Answer 2

Gradshteyn and Ryzhik, 6.578.1, solves the integral (for c=1) in terms of Gamma factors and the Appell F4 $F4(1/2, 3/2; 1,1, a^2,b^2)$ when $a>0, b>0$, and $a+b<1.$ In certain cases the F4 can be reduced a product of Gauss hypergeometric 2F1's or, if half integer in the 3rd and 4th arguments, a single 2F1. The given arguments result in a miss of the product case by 1 unit. That 1 unit cannot be made up with recursion formulas. I think it unlikely that the F4 representation will simplify.

However, the F4 is a double series and with either a or b small enough, should converge rapidly. Alternatively, there is an integral relationship for the F4 which gets rid of the oscillations, i.e., $$F4(1/2,3/2;1,1,a^2,b^2) = (Gamma Factors) * \int_0^\infty x \,I_0(ax)\,I_0(bx)\,K_1(x) dx.$$