ON A ZETA-FUNCTION ASSOCIATED WITH BESSEL'S EQUATION.

摘要

Let j(,n)((nu)) denote the nth positive zero of J((nu),z) = z('-(nu))J(,(nu))(z), n = 1,2,..., (nu) > -1, where J(,(nu))(z) denotes the Bessel function of the first kind of order (nu). Define the zeta-function z((nu),s) by.;It is easy to show that z((nu),s) is analytic in the half-plane (sigma) > 0 except for a simple pole at s = 1 with residue 1/(pi).;Since z(1/2,s) = (pi)('-s)(zeta)(s), where (zeta)(s) denotes the Riemann zeta-function, it is natural to ask where the zeros are in right half-planes and whether or not z((nu),s) has a meromorphic continuation to the whole s-plane.;Concerning zeros, we prove that z((nu),s) has no zeros in the half-plane (sigma) > 1/p if and only if a certain space of functions, defined in terms of a generalized "fractional part" function for the sequence of numbers j(,n)((nu))/j(,1)((nu)), n = 1,2,..., is dense in L('p)({0,1}), the space of measurable complex-valued functions on {0,1} with finite p-norm. This generalizes a theorem of Beurling for (zeta)(s). Indeed, we prove that this theorem holds for a zeta-function of the form.;(DIAGRAM, TABLE OR GRAPHIC OMITTED...PLEASE SEE DAI).;(DIAGRAM, TABLE OR GRAPHIC OMITTED...PLEASE SEE DAI).;where (lamda)(,n) = (LAMDA)n + 0(1) as n tends to (INFIN), (LAMDA) denoting a positive constant, and 0 < (lamda)(,n) < (lamda)(,n+1), n = 1,2,... .;Concerning meromorphic continuation to the s-plane, we prove that z((nu),s) has such a continuation with its only (possible) singularities being simple poles at s = 1,-3,-3,... . We also show that the residues and values of z((nu),s) at negative integers are simply expressed in terms of a sequence of polynomials (in (nu) with rational coefficients) that satisfy a quadratic recursion relation.;We also prove that z((nu),s) has an analytic continuation in (nu) to the slit plane (OMEGA) = (//C)(FDIAG)(-(INFIN),-1}, for each s in the half-plane (sigma) > 1. Since it is easy to show that z((nu),s) is analytic in the half-plane (sigma) > 1 for each (nu) in (OMEGA) from its definition, we conclude from a theorem of Hartogs that z((nu),s) is analytic in both variables (nu) and s in the obvious region of complex ((nu),s)-space.