How to prove $\int_0^\infty J_\nu(x)^3dx\stackrel?=\frac{\Gamma(1/6)\ \Gamma(1/6+\nu/2)}{2^{5/3}\ 3^{1/2}\ \pi^{3/2}\ \Gamma(5/6+\nu/2)}$? I am interested in finding a general formula for the following integral:
$$\int_0^\infty J_\nu(x)^3dx,\tag1$$
where $J_\nu(x)$ is the Bessel function of the first kind:
$$J_\nu(x)=\sum _{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+\nu+1)}\left(\frac x2\right)^{2n+\nu}.\tag2$$
Mathematica gives the following result:
$$\int_0^\infty J_\nu(x)^3dx=\frac1{\pi\,\nu}{_3F_2}\left(\frac12,\frac12-\nu,\frac12+\nu;1-\frac\nu2,1+\frac\nu2;\frac14\right)+\frac{\Gamma\left(-\frac\nu2\right)\Gamma\left(\frac{1+3\nu}2\right)\cos\frac{\pi\,\nu}2}{2^{\nu+1}\ \pi^{3/2}\ \Gamma(\nu+1)}{_3F_2}\left(\frac{1-\nu}2,\frac{1+\nu}2,\frac{1+3\nu}2;1+\frac\nu2,1+\nu;\frac14\right),\tag3$$
which can be significantly simplified for odd and half-integer values of $\nu$. The results at those points allow to conjecture another, simpler general formula:
$$\int_0^\infty J_\nu(x)^3dx\stackrel?=\frac{\Gamma\left(\frac16\right)\ \Gamma\left(\frac16+\frac\nu2\right)}{2^{5/3}\ 3^{1/2}\ \pi^{3/2}\ \Gamma\left(\frac56+\frac\nu2\right)},\tag4$$
which agrees with $(3)$ to a very high precision for many different values of $\nu$. It also has an advantage that it is defined for all $\nu>-\frac13$ whereas $(3)$ is undefined at every even $\nu$ and requires calculating a limit at those points.
Is it possible to prove the formula $(4)$?

Mathematica is able to evaluate $(1)$ for even values of $\nu$ in terms of the Meijer G-function. Plugging those expressions into $(4)$ we get another form of the conjecture:
$$G_{3,3}^{2,1}\left(\frac14\left|\begin{array}{c}2a,1,2-2a\\\frac12,1-a,a\\\end{array}
\right.\right)\stackrel?=\frac{\Gamma\left(\frac16\right)\ \Gamma\left(\frac23-a\right)}{2^{5/3}\ 3^{1/2}\ \pi\ \Gamma\left(\frac43-a\right)}.\tag5$$
Incidentally, the case $a=\frac12$ would positively resolve my another question.
 A: See section 6 in http://arxiv.org/abs/1007.0667 for that type of triple-product integrals.
A: A few years ago I tried to evaluate this integral using a generalization of Ramanujan's Master Theorem referred to as the method of brackets.
(If anyone is interested, I've previously evaluated integrals using this method here and here.)
I even had brief email exchange with Dr. Karen T. Kohl about this particular integral, someone who has authored and coauthored papers on this topic.
The case $\nu =0$ is particularly interesting and perplexing, which is what I'm going to show.

The Bessel function of the first kind of order zero has the series representation $$J_{0}(z) = {_0F_1} \left(;1;-\frac{z^{2}}{4} \right) = \sum_{m=0}^{\infty} \phi_{m} \, \frac{1}{\Gamma(m+1)} \left(\frac{z^{2}}{4} \right)^{m}.  $$
And the square of the Bessel function of the first kind has the series representation $$J_{0}(z)^{2} =   {_1F_2} \left(\frac{1}{2};1,1;-z^{2} \right) = \sum_{n=0}^{\infty} \phi_{n} \, \frac{\Gamma \left(n+ \frac{1}{2} \right)}{\Gamma \left(\frac{1}{2} \right)} \frac{1}{\Gamma(n+1)^{2}} \left(z^{2} \right)^{n}.$$
Therefore, the bracket series that corresponds to the integral $\int_{0}^{\infty} J_{0}(x)^{3} \, \mathrm dx $ is $$\sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \phi_{m,n} \,  \frac{4^{-m}}{\Gamma(m+1)} \frac{\Gamma \left(n+ \frac{1}{2} \right)}{\Gamma \left(\frac{1}{2} \right)} \frac{1}{\Gamma(n+1)^{2}} \, \langle 2m+2n+1 \rangle.  $$
Since there are two indices and only one bracket, there are two cases to look at.

If we let $m$ be free, then the bracket vanishes if $n= -m- \frac{1}{2} $,
and we have the contribution $$S_{1} = \frac{1}{2} \sum_{m=0}^{\infty} \phi_{m} \frac{4^{-m}}{\Gamma(m+1)} \frac{\Gamma(-m)}{\Gamma \left(\frac{1}{2} \right)} \frac{1}{\Gamma\left(-m + \frac{1}{2}\right)^{2}} \, \Gamma \left(m+ \frac{1}{2} \right).   $$ However, since the terms of this series are not finite, we discard it.

If we let $n$ be free, then the bracket vanishes if $m = - n - \frac{1}{2},$
and we have the contribution $$ \begin{align} S_{2} &= \frac{1}{2} \sum_{n=0}^{\infty} \phi_{n} \, \frac{4^{n+ \frac{1}{2}}}{\Gamma \left(-n+ \frac{1}{2} \right)}  \frac{\Gamma \left(n+ \frac{1}{2} \right)}{\Gamma \left(\frac{1}{2} \right)} \frac{1}{\Gamma(n+1)^{2}} \Gamma \left(n+ \frac{1}{2} \right) \\ &= \frac{1}{\pi} \sum_{n=0}^{\infty} \frac{1}{n!}    \frac{\Gamma \left(n+ \frac{1}{2} \right)^{3}}{\Gamma \left(\frac{1}{2} \right)} \frac{1}{\Gamma(n+1)^{2}} \, 4^{n}. \end{align} $$
This time the terms of the series are finite, but the series diverges.
At this point, one might also discard this contribution and conclude that the method of brackets has failed due to the fact that the difference between the number of indices and number of brackets hasn't been minimized as much as it could be.
But if the above series can be interpreted as the analytic continuation of the hypergeometric function $$   \, {_3F_2} \left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}; 1, 1 ; z \right) $$  to $z=4$, we then have, according to the Wolfram Function Site, $$S_{2} = \frac{4}{\pi^{2}} \, \Re \ \left( K \left(e^{-i \pi/3} \right)  \right) ^{2},  $$
where $K\left(e^{-i \pi/3} \right)  $ is the complete elliptic integral of the first kind with parameter $m= k^{2} =e^{- i \pi/3}$.
It would then follow that  $$\int_{0}^{\infty} J_{0}(x)^{3} \, \mathrm dx = S_{2} = \frac{4}{\pi^{2}} \, \Re \left( K \left(e^{-i \pi/3} \right) \right)^{2}, $$ which seems to check out numerically and would seem to be related to this answer.
Combining this result with the other result, we have $$\Re \left( K \left(e^{-i \pi/3} \right) \right)^{2} = \frac{\pi^{2}}{4} \frac{\Gamma \left(\frac{1}{6} \right)^{2}}{2^{5/3} 3^{1/2} \pi^{3/2} \Gamma \left(\frac{5}{6} \right)}.$$

I'll update this answer if I find out more about the role of analytic continuation with this method.
