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Does the following integral have a closed form? $$ \mathcal{J}(2,3,5,7,11) = \int_0^\infty x J_0(x\sqrt{2})J_0(x\sqrt{3})J_0(x\sqrt{5})J_0(x\sqrt{7})J_0(x\sqrt{11})\,dx. $$

I know that some similar integrals do have a closed form $$ \mathcal{J}(2,3,5) = \frac{1}{\pi \sqrt{6}} \approx 0.12995. $$ $$ \mathcal{J}(2,3,5,7) = \frac{1}{\pi^2 210^{1/4}}K(\tfrac14\sqrt\alpha) \approx 0.110411, \qquad 43-672\alpha+42\alpha^2 = 0, \qquad \alpha\approx 15.9358. $$ (Here $K(k)$ is the complete elliptic integral of the first kind with modulus $k$.)

Numerically, the integral is approximately $$ 0.061064349908721692\ldots, $$ but I could not find a symbolic value for this constant.

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  • $\begingroup$ Just by curiosity: what was the idea behind the calculation of $\mathcal{J}(2,3,5,7)$ ? $\endgroup$ Nov 1, 2014 at 12:16
  • $\begingroup$ @O.L. I don't know, I saw it here (problem c). I'm guessing that products of Bessel functions have nice Mellin transforms (products of gamma functions), so if you integrate a bunch of Bessel functions, you get something like a Meijer G function, which I guess then simplifies. $\endgroup$
    – Kirill
    Nov 1, 2014 at 12:44
  • $\begingroup$ I think that the form of the square root coefficients in the arguments of Bessel functions should be crucial for simplification - that's why I am asking. Looking at this paper it seems unlikely that even four-argument integral $\mathcal{J}(a,b,c,d)$ can be found explicitly for generic $a,b,c,d$. $\endgroup$ Nov 1, 2014 at 13:52
  • $\begingroup$ @O.L. This identity looks relevant, so there might be a generic expression for four arguments. $\endgroup$
    – Kirill
    Nov 1, 2014 at 14:05
  • $\begingroup$ @O.L. If you know how to derive the four-argument case and can show that the same argument doesn't apply to five arguments, I'd say that's a good partial answer. $\endgroup$
    – Kirill
    Nov 1, 2014 at 14:06

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