**QUESTION**


Numerical calculation with **gp** (first to the default 38-digit
precision, then tripled) supports the conjecture that
$$
\int\_0^\infty x \, [J\_0(x)]^5 \, dx =
\frac{\Gamma(1/15) \, \Gamma(2/15) \, \Gamma(4/15) \, \Gamma(8/15)}
 {8\sqrt{5} \, \pi^4}
= \frac{2}
 {\sqrt{5} \, \Gamma(7/15) \, \Gamma(11/15) \, \Gamma(13/15) \, \Gamma(14/15)}
$$
[the two Gamma expressions are easily proved equal via the
identity $\Gamma(x) \Gamma(1-x) = \pi / \sin(\pi x)$;
the hard part is proving that either of them equals the definite integral].


*Is this a known formula?*


*If not, is it worth working out and writing up a proof?*



> 
> 
> 


**MOTIVATION: OVERVIEW**


There is a well-known analogy between
[Kloosterman sums](http://en.wikipedia.org/wiki/Kloosterman_sum)
to prime modulus,
$$
K(a,b; p) = \sum\_{x=1}^{p-1} \exp(2\pi i (ax+bx^{-1})/p)
$$
(with $x^{-1}$ being the inverse of $x \bmod p$), and the Bessel function
$$
J\_0(2\sqrt{ab}) = \frac1\pi \int\_0^{\infty} \sin(ax+bx^{-1}) \, \frac{dx}{x}
$$
[[Gradshteyn
and Ryzhik](http://www.lepp.cornell.edu/~ib38/tmp/reading/Table_of_Integrals_Series_and_Products_Tablicy_Integralov_Summ_Rjadov_I_Proizvedennij_Engl._2.pdf) 3.868 #1].
When $a,b \neq 0 \bmod p$, the Kloosterman sum is not elementary,
but is readily seen to depend only on $ab \bmod p$.


Now consider for $m=1,2,3,\ldots$ the $m$-th power moment
$$
M\_m(p) := \sum\_{c=1}^{p-1} [K(c,1;p)]^m
= \frac1{p-1} \sum\_{a=1}^{p-1} \sum\_{b=1}^{p-1} [K(a,b;p)]^m.
$$
This $M\_m(p)$ is given by an elementary formula for $m\leq 4$,
but $M\_5(p)$ involves counting points on a certain K3 surface $X(5) \bmod p$.
This suggested that the analogous Bessel integral
$\int\_0^\infty x [J\_0(x)]^5 \, dx$ might be proportional to the real
[period](http://en..ikipedia.org/wiki/Ring_of_periods) of $X(5)$.
This K3 surface has maximal Picard rank (it is "singular"), and the real period
of such a surface should in turn be proportional to a product of Gamma functions
evaluated at rationals whose denominators divide the Neron-Severi discriminant
of the surface. Here this discriminant turns out to be $-15$,
and **gp**'s function **lindep** soon came up with a candidate formula
relating the integral with the Gamma product.


**MOTIVATION: DETAILS**


We can evaluate $M\_m(p)$ by adding
to the double-sum formula for $(p-1) M\_m(p)$ also the $2p-1$ choices of $a,b$
for which $ab=0$. This increases the sum by $(p-1)^{m-1} + 2(-1)^m$.
We then expand $[K(a,b;p)]^m$ as the sum over nonzero $x\_1,\ldots,x\_m$ of
$\exp(2\pi i (\sum\_{j=1}^m ax\_j+bx\_j^{-1})/p)$ and switch the
order of summation, finding $p^2$ times the number of points
$(x\_1:x\_2:\cdots:x\_m)$ with nonzero coordinates on the projective surface
$$
X(m) := \sum\_{j=1}^m x\_j = \sum\_{j=1}^m x\_j^{-1} = 0
$$
in the projective $(m-2)$-space
$\{ (x\_1:x\_2:\cdots:x\_m) \in {\bf P}^{m-1} \mid \sum\_{j=1}^m x\_j = 0 \}$.
Therefore
$$
M\_p(m) = p^2 \#(X(m) \bmod p) - ((p-1)^{m-1} + 2(-1)^m).
$$
For example, when $m=4$ this variety is the union of three lines
$x\_1+x\_2=x\_3+x\_4=0$, $x\_1+x\_3=x\_2+x\_4=0$, and $x\_1+x\_4=x\_2+x\_3=0$,
and we find $M\_4(p) = 2p^3-3p^2-3p-1$, which incidentally also
gives an elementary proof that each $|K(c,1;p)| < (2p^3)^{1/4}$.


The next case, $m=5$, is the first one where the point count of
$X(m) \bmod p$ is not elementary.
It turns out that $X(5)$ is an open subset
(complement of $20$ rational curves) in a K3 surface studied in detail
in the paper



> 
> Christiaan Peters, Jaap Top, and Marcel van der Vlugt: The Hasse zeta function of a K3 surface related to the number of words of weight 5 in the Melas codes. *J. reine angew. Math.* (Crelle's J.) **432** (1992), 151-176.
> 
> 
> 


It follows from their analysis that
$$
M\_5(p) = (-3/p)4p^3 + 5p^2 + 4p + 1
$$
if $(-15/p) = -1$, while if $(-15/p)=+1$ then $M\_5(p)$ is given by
a more complicated formula that involves the decomposition of $4p$ as
$m^2+15n^2$ or $5m^2+3n^2$. The "magic number" $-15$ arises as the
discriminant of the intersection pairing on the Neron-Severi lattice
of the surface.


Now the K3 surface $X(5)$ is isogenous with the Kummer surface
$(E \times E) / \{\pm 1\}$ where $E$ has complex multiplication by $\sqrt{-15}$,
so the real period of $X(5)$ should be within an elementary factor of the
square of the real period of $E$,
and this square is known to be an elementary-factor
multiple of $\Gamma(1/15) \, \Gamma(2/15) \, \Gamma(4/15) \, \Gamma(8/15)$.
On the other side, a Bessel-function analogue of $M\_5$ is
$\int\_0^\infty [J\_0(2\sqrt{c})]^5 \, dc
= \frac12 \int\_0^\infty x \, [J\_0(x)]^5 \, dx$.
This integral converges slowly, but the oscillating
[asymptotic
expansion](http://en.wikipedia.org/wiki/Bessel_function#Asymptotic_forms) of $J\_0(x)$ suggests that the convergence can be accelerated
by writing it as an alternating sum
$\int\_0^\infty = \sum\_{n=0}^\infty \int\_{n\pi}^{(n+1)\pi}$
and using **gp**'s built-in routines **intnum** and **sumalt**.
Indeed the command



```
I5 = sumalt(n=0, intnum(x=n*Pi, (n+1)*Pi, x*besselj(0,x)^5))

```

takes a few seconds to return 0.32993380106006405903979065228695296470,
and a few minutes to triple the precision to 100+ digits. Then



```
lindep(log([I5, 2, 5, Pi, prod(i=0,3,gamma(2^i/15))]))

```

finds the relation with coefficients $(-2,-6,-1,-8,2)$ which yields
the first equivalent form
$$
\int\_0^\infty x \, [J\_0(x)]^5 \, dx =
\frac{\Gamma(1/15) \, \Gamma(2/15) \, \Gamma(4/15) \, \Gamma(8/15)}
 {8\sqrt{5} \, \pi^4}
$$
of our conjectured evaluation of the definite integral.


**SO WHAT'S THE QUESTION(S) AGAIN?**


I think I basically know how to prove this formula, but even once I've
honestly related the integral to the real period of $X(5)$ it would be
a nontrivial task to track the period over a sequence of transformations from
$X(5)$ to $E \times E$ and thus to obtain the relation with the Gamma product.
So I'll be happy to learn that this formula is already known, perhaps
via hypergeometric transformation formulas rather than direct manipulation
of real periods. (For what it's worth, Gradshteyn-Ryzhik has lots of
integrals of this kind with products of at most three Bessel functions,
but only a handful with four (6.579), and none with five or more.)
If the formula is not already known, is the any reason
(beyond the heuristics above) to regard it as more than a curiosity
and thus worth expending the effort to construct a complete proof?



$\int_0^\infty x \, [J_0(x)]^5 \, dx$: source and context, if any?
See *Densities of Short Uniform Random Walks* (with an appendix by Don Zagier) by
Jonathan M. Borwein, Armin Straub, James Wan, and Wadim Zudilin,
Canad. J. Math. Vol. 64 (5), 2012 pp. 961–990. <http://cms.math.ca/10.4153/CJM-2011-079-2>


Your integral is $p\_4(1)$ in Equation (2.1). They write it in terms of hypergeometric functions (doing this for $p\_4$ is one of the main results of the paper), and the OEIS reference <https://oeis.org/A244995> recapitulates in terms of a $\Gamma$-product, as given in their Theorem 5.1 (wherein they mention $\eta$-modularity and Chowla-Selberg to obtain this).


The rephrasing of the integral in terms of a probability on random walks is due to: J. C. Kluyver, *A local probability problem*. Royal Netherlands Academy of Arts and Sciences, Proceedings, 8 I, 1905, pp. 341–350. <http://www.dwc.knaw.nl/DL/publications/PU00013859.pdf>


