On the real zeroes of the Hurwitz zeta-function and Bernoulli polynomials

a r X i v :m a t h /0205183v 1 [m a t h .G M ] 16 M a y 2002On the real zeroes of the Hurwitz zeta-functionand Bernoulli polynomialsFebruary 1,2008A.P.Veselov∗,∗∗and J.P.Ward∗∗Department of Mathematical Sciences,Loughborough University,Loughborough,Leicestershire,LE 113TU,UK∗∗Landau Institute for Theoretical Physics,Kosygina 2,Moscow,117940,Russiae-mails:A.P.Veselov@,J.P.Ward@Abstract.The behaviour of real zeroes of the Hurwitz zeta functionζ(s,a )=∞ r =0(a +r )−sa >0is investigated.It is shown that ζ(s,a )has no real zeroes (s =σ,a )in the regiona >−σ4πelog(−σ)+1for large negative σ.In the region 0<a <−σp=1πe.1Introduction.The classical Hurwitz zeta-function is defined for any positive real a as an analytic continuation of the seriesζ(s,a )=∞ r =0(a +r )−s .When a =1it reduces to the Riemann zeta-function.1We should note that sometimes the definitionoftheHurwitz zeta-function is re-stricted to0<a≤1(see e.g.[1,2]).From our point of view this is not natural and we follow the definition of the Hurwitz zeta-function from[3]where all positive a are allowed(cf.also the original Hurwitz paper[4]).It is known(see e.g.[2],volume1,page27)that in the special cases when s is a negative integer this function(as a function of the parameter a)reduces,up to a factor,to a Bernoulli polynomial:explicitly when s=−m,m=0,1,2,3,....ζ(−m,a)=−B m+1(a)e z−1=∞k=0B k(a)2B2(a)=a2−a+12+aπe.More precise esti-mates for N(n)have been found by Delange[6,7].In this paper we investigate the behaviour of the real zeroes of the Hurwitz zeta functionζ(σ,a)in the upper half-plane a>0.As a corollary we have a simple proof of Inkeri’s result.Our approach is different from[5,6,7]and we believe is more elementary.It is based on the remarkable Hurwitz representation of theζ(s,a)on the interval0<a≤1and Re(s)=σ<0:ζ(s,a)=2Γ(1−s)2πs)2πe.As a result we prove thatζ(σ,a)2πσ)+o(1)asσ→−∞provided0<a<−σ(2π)1−s.We show also that in the region a>−σ4πelog(−σ)+1 the Hurwitz zeta-function has no real zeroes for large negativeσ.2Figure1Our results are illustrated in Figure1which shows the behaviour of the real zeroes ofζ(σ,a).Since12they are s=0,−2,−4,...As we have mentioned above when s is a non-positiveintegerζ(s,a)reduces to a Bernoulli polynomial.We have used this fact to compute numerically the corresponding values ofζ(s,a)and to draw the picture in the regions II and IV.When s=1the Hurwitz zeta-function has a simple pole with the residue1 and forσ=Re(s)>1it is given by the convergent series with positive elements and therefore has no zeroes.2Asymptotic behaviour of the Hurwitz zeta-functionζ(σ,a)for large negativeσ.The Hurwitz zeta-function(or generalised Riemann zeta-function)is defined as a seriesζ(s,a)=∞r=0(a+r)−s a>03in the complex domain Re(s)>1and can be analytically continued to a meromorphic function in the whole complex plane with the only pole at s=1(see[1],[2],[3]).When a=1it reduces to the Riemann zeta-functionζ(s)=∞ k=1k−s.The Hurwitz zeta-function can be extended to the whole of the complex s-plane through the formulaζ(s,a)=−Γ(1−s)1−e−zd z a>0in which the integral is taken over a curve starting at‘infinity’on the real axis,encircles the origin in a positive direction and returns to the starting point(see[2]).By using an alternative integral formulation forζ(s,a)it can be shown thatζ(s,a)is analytic everywhere except for the simple pole at s=1.The Hurwitz zeta-function obviously satisfies the functional relation:ζ(s,a)=ζ(s,n+a)+n−1r=0(r+a)−s n=1,2, (1)Since each term in this relation is analytic we can assume this relation is true for the whole of the complex s plane,except for s=1.In this paper we restrict ourselves to the case when s is real:s=σ∈R.Whenσis negative Hurwitz has found the following Fourier representation forζ(σ,a)on the interval0<a≤1:ζ(σ,a)=2Γ(1−σ)2πσ)Q(σ)=sin(2πa+1(2π)1−σ.Ourfirst theorem proves that this is actually true on a muchlarger interval.As part of the theorem proofs we will use the following inequality for the function S(p,n)=1p+2p+...+n p:S(p,n)<n p 1−e−pn )p+(1−2n)pBut since1−x<e−x for x<1we have1−1n<e−2/n...1−n−1and thereforeS(p,n)<n p 1+e−p/n+e−2p/n+...+e−(n−1)p/n =n p 1−e−pr1+p.Now131+p +12p 13(24(2n)p<1r1+p <(ζ(2)−32p p e−p<Γ(1+p)<(2πp)112p p>>1Theorem1.Letσ=−p,p≥0and0<a<αp for some positiveαthen the Hurwitzzeta-function satisfies the inequalityζ(−p,a)2πp)<C1p−1/2(2πeα)p+C22−p,(3)where C1,C2are constants,which do not depend on p.In particular,on the interval 0<a<1Q(−p)=sin(2πa−1Q(σ)−ζ(σ,b)Q(σ)−ζ(σ,b)|Q(σ)|n−1r=0(r+b)−σSince0<b≤1we obviously haven−1r=0(r+b)−σ≤S(p,n)0<p=−σand,as we obtained above,S(p,n)<n p 1−e−pQ(−p)=(2π)1+pp}pπ2πp5Thus for a large pζ(−p,a )Q (−p )< 2πen√1−e −p/n<2πen√(1−e−1p <α(4)which is true since np<αby assumption.However,from Hurwitz’formula (2)it follows that ζ(−p,b )2πp )2πp )+sin(4πb −121+p+sin(6πb −131+p+....Thereforeζ(−p,b )2πp ) =ζ(−p,b )2πp ) <∞r =2r −p −1<(ζ(2)−32πe+1Q (−p )<ζ(−p,b )Q (−p ).Now assuming that p is sufficiently large we can use Stirling’s inequality Γ(1+p )<√ep e 1(2π)1+p <√π p12p However,we know from the Hurwitz formula,that when p →∞ζ(−p,b )2πp )+o (1)Therefore if(a −1)pQ (−p )>2πe (a −1)π12pwhich is greater than 1ifa −1>pπ112p 2=p2p+1Q (−p )>1it is enough to demand thata >p4πelog p +1.This implies the theorem.63The Real Zeroes of the Hurwitz Zeta-function and Bernoulli Polynomials.Let us nowfixσ=−p and considerζ(−p,a)as a function of a.It follows from theorem2that the zeroes of this function for large p are located in the interval0<a<p4πe log p+1.For given p let N(p)be the number of real zeroes ofζ(−p,a),and A(p)be the largest of these zeroes. Theorem3For p sufficiently largep−12<A(p)<p4πelog p+1(6)p−1πe +12πeare simple and close to thehalf-integer lattice:a=p2,l∈Z.Proof Let us introduce the function Z p(a)=ζ(−p,a)2πp)+o(1)on the interval I p:0<a<p∂aζ(s,a)=(−s)ζ(s+1,a).From the property of theΓ-functionΓ(p+1)=pΓ(p)it follows thatQ(−p)=2Γ(1+p)2πQ(−p+1).Thus the derivative of Z p(a)is equal toZ′p(a)=2πZ p−1(a)=2πsin(2πa−12πp)+o(1)on the interval I p−1.Similarly we have for the k th derivative of Z p(a)Z(k) p (a)=sin(k)(2πa−12πp)(and its derivative)when p→∞which ensures that for largep all the roots ofζ(−p,a)on this interval are simple and located near the pointsa=p2l∈Z.This implies the last statement of the theorem and the lowerestimates of(6)and(7).The upper estimates for A(p)follows directly from theorem2.To prove the upper estimates for N(p)we need the following simple lemma.Lemma If a function f(x)(with a continuous n th derivative)on some interval(a,b) has the property that the sign of the n th derivative is constant throughout the interval then f has no more than n roots on this interval.7Now we apply this lemma to the function Z p(a)on the interval J p:(p−12πe +12πe).p=¡Figure2Using the fact thatZ(n)p(a)=(2π)n Z p−n(a)we differentiate Z p(a)many times until we have a negative function and then applythe lemma.As one can see from Figure2if n>[y]then Z(n)p(a)will be negative in the interval J p and,as such,cannot have more than n simple roots in this interval. Now y is the solution to the equationκ (p−y)+12log(p−y)+(κ−1+1)=0We claim that the solution to this equation for large p satisfies the inequalityy<12log(p−y)+(2πe+1)is monotonically decreasingandF(12log p−2πe−1+12log p−2πe−1)+(2πe+1) =12plog p−2πe+12log p+2πe+1roots on the interval(p−12πe +12πe]we haveno more than p−1m+1(8)8Theorem3applied to these special values of p gives some estimates on the real pos-itive roots of the Bernoulli polynomials but because of their well-known symmetry properties:B m(1−a)=(−1)m B m(a)(9) we can immediately extend this result for all real roots of B m(a).In particular if N(m)is the number of all real roots of B m(a)and A(m)is the largest of these roots then from Theorem3it follows that for a large mmπe +12πe+12πe (10)2mπe+2 <N(m)<2mπe (11) Corollary(K.Inkeri).lim m→∞N(m)πe,limm→∞A(m)2πe.Remark.H.Delange in[6,7]has found sharper estimates for A(m)and N(m). In particular he showed that the additional logarithmic terms exist in both upper and lower bounds.This should be true also for the real zeroes of the Hurwitz zeta-function but it does not follow from our elementary arguments. Acknowledgements.We are grateful to John Gibbons and Peter Shiu for helpful and stimulating discussions.References[1]E.T.Whittaker and G.N.Watson,A Course of Modern Analysis.,Cam-bridge University Press,1963.[2]A.Erdelyi(Editor)Higher Transcendental Functions.,Vol.1-3.McGraw-Hill Book Company,1953.[3]G.E.Andrews,R.Askey,R.Roy Special Functions.Encyclopedia of Math-ematics and its Applications.71.Cambridge Univ.Press,1999.[4]A.Hurwitz Einige Eigenschaften der Dirichlet’schen Functionen F(s)=Σ(Dn s ,die bei der Bestimmung der Classenzahlen bin¨a rer quadratischerFormen auftreten.Zeitschrift f¨u r Math.und Phys.,XXVII(1882),86-101.[5]K.Inkeri The real roots of Bernoulli polynomials.Ann.Univ.Turku.Ser.A I,37(1959),3-19.[6]H.Delange Sur les z´e ros r´e els des polynomes de Bernoulli.C.R.Acad.Sc.Paris,s´e rie I,303(1986),539-542.[7]H.Delange Sur les z´e ros r´e els des polynomes de Bernoulli.Ann.Inst.Fourier,Grenoble,41,2(1991),267-309.9。