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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 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 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 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 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?

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    $\begingroup$ As a curiosity, here is a result from Watson's treatise on Bessel function, Eq. (8) Section 13.46: if $a_1\geq a_2\geq \cdots \geq a_m>0$ and $a_1>a_2+\cdots +a_m$, then $$\int_0^\infty x\prod_{k=1}^m J_0(a_k x) dx=0. $$ $\endgroup$ – Liviu Nicolaescu May 12 '15 at 11:58
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    $\begingroup$ When I started reading the question, my first reaction was to type a comment: "ask Noam Elkies", but then I read the entire question.... :-D $\endgroup$ – Suvrit May 12 '15 at 13:07
  • $\begingroup$ More generally, also directly relevant are pp. 419--421 of Watson's "Treatise on Bessel functions" $\endgroup$ – Suvrit May 12 '15 at 14:07
  • $\begingroup$ Thanks, but it seems that the factor of $x$ is missing there, and the integral for a product of $m$ Bessel functions is expressed as a real period of a hypersurface of dimension $m-1$, not $m-3$ as with the "curiosity" from Liviu Nicolaescu (indeed I see that in the next formula (9) Watson evaluates such integrals for $m=4$ in terms of complete elliptic integrals, i.e. real periods of elliptic curves). $\endgroup$ – Noam D. Elkies May 12 '15 at 14:30
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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

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    $\begingroup$ This was found by searching on "Gamma(1/15)" and "Gamma(2/15)" with a search engine, and then following the OEIS result. Internet mathematics at its best! If you had given the decimal expansion, maybe I could have searched on that... $\endgroup$ – GuestPoster May 12 '15 at 10:34
  • $\begingroup$ I did give the decimal expansion 0.32993380106006405903979065228695296470 and searched on that (and truncations and simple multiples) before posting the question... $\endgroup$ – Noam D. Elkies May 12 '15 at 13:49
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    $\begingroup$ Searching with Google on 0.329933801060064059 gives exactly two hits, both for "Three-Step and Four-Step Random Walk Integrals." (2010, by the first 3 authors). DuckDuckGo is similar. $\endgroup$ – GuestPoster May 12 '15 at 14:21
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    $\begingroup$ And I guess I was incredibly lucky choosing exactly that truncation (on my first try), as none others appear to work. $\endgroup$ – GuestPoster May 12 '15 at 14:43
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    $\begingroup$ ... and if I had tried 0.3299338011 then I'd have found the Borwein-Straub-Wan-Zudilin paper directly ... $\endgroup$ – Noam D. Elkies May 12 '15 at 17:28

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