The Riemann Zeta Function(by David Jekel)黎曼zeta函数

TheRiemannZetaFunctionDavidJekelJune6,2013In1859,BernhardRiemannpublishedaneight-pagepaper,inwhichheestimated\thenumberofprimenumberslessthanagivenmagnitudeusingacertainmeromorphicfunctiononC.ButRiemanndidnotfullyexplainhisproofs;ittookdecadesformathematicianstoverifyhisresults,andtothisdaywehavenotprovedsomeofhisestimatesontherootsof.EvenRiemanndidnotprovethatallthezerosoflieonthelineRe(z)=12.ThisconjectureiscalledtheRiemannhypothesisandisconsideredbymanythegreatestunsolvedprobleminmathematics.H.M.Edwards'bookRiemann'sZetaFunction[1]explainsthehistor-icalcontextofRiemann'spaper,Riemann'smethodsandresults,andthesubsequentworkthathasbeendonetoverifyandextendRiemann'stheory.TherstchaptergiveshistoricalbackgroundandexplainseachsectionofRiemann'spaper.TherestofthebooktraceslaterhistoricaldevelopmentsandjustiesRiemann'sstatements.ThispaperwillsummarizetherstthreechaptersofEdwards.MypapercanserveasanintroductiontoRiemann'szetafunction,withproofsofsomeofthemainformulae,foradvancedundergraduatesfamiliarwiththerudimentsofcomplexanalysis.Iusetheterm\summarizeloosely;insomesectionsmydiscussionwillactuallyincludemoreexplanationandjustication,whileinothersIwillonlygivethemainpoints.ThepaperwillfocusonRiemann'sdenitionof,thefunctionalequation,andtherelationshipbetweenandprimes,culminatinginathoroughdiscussionofvonMangoldt'sformula.1Contents1Preliminaries32DenitionoftheZetaFunction32.1Motivation:TheDirichletSeries................42.2IntegralFormula.........................42.3Denitionof...........................53TheFunctionalEquation63.1FirstProof............................73.2SecondProof...........................84anditsProductExpansion94.1TheProductExpansion.....................94.2ProofbyHadamard'sTheorem.................105ZetaandPrimes:Euler'sProductFormula126Riemann'sMainFormula:Summary137VonMangoldt'sFormula157.1FirstEvaluationoftheIntegral.................157.2SecondEvaluationoftheIntegral................177.3TermwiseEvaluationover...................197.4VonMangoldt'sandRiemann'sFormulae...........2221PreliminariesBeforewegettothezetafunctionitself,Iwillstate,withoutproof,someresultswhichareimportantforthelaterdiscussion.Icollectthemheresosubsequentproofswillhavelessclutter.Theimpatientreadermayskipthissectionandreferbacktoitasneeded.Inordertodenethezetafunction,weneedthegammafunction,whichextendsthefactorialfunctiontoameromorphicfunctiononC.LikeEdwardsandRiemann,Iwillnotusethenowstandardnotation(s)where(n)=(n1)!,butinsteadIwillcallthefunction(s)and(n),willben!.Igiveadenitionandsomeidentitiesof,aslistedinEdwardssection1.3.Denition1.Foralls2Cexceptnegativeintegers,(s)=limN!1(N+1)sNYn=1ns+n:Theorem2.ForRe(s)1,(s)=R10exxsdx.Theorem3.satises1.(s)=Q1n=1(1+s=n)1(1+1=n)s2.(s)=s(s1)3.(s)(s)sins=s4.(s)=2p(s=2)(s=21=2).Sincewewilloftenbeinterchangingsummationandintegration,thefollowingtheoremonabsoluteconvergenceisuseful.Itisaconsequenceofthedominatedconvergencetheorem.Theorem4.Supposefnisnonnegativeandintegrableoncompactsub-setsof[0;1).IfPfnconvergesandR10Pfnconverges,thenPR10fn=R10Pfn.2DenitionoftheZetaFunctionHereIsummarizeEdwardssection1.4andgiveadditionalexplanationandjustication.32.1Motivation:TheDirichletSeriesDirichletdened(s)=P1n=1nsforRe(s)1.Riemannwantedadef-initionvalidforalls2C,whichwouldbeequivalenttoDirichlet'sforRe(s)1.HefoundanewformulafortheDirichletseriesasfollows.ForRe(s)1,byEuler'sintegralformulafor(s)2,Z10enxxs1dx=1nsZ10exxs1dx=(s1)ns:Summingovernandapplyingtheformulaforageometricseriesgives(s1)1Xn=11ns=1Xn=1Z10enxxs1dx=Z10exxs11exdx=Z10xs1ex1dx:WearejustiedininterchangingsummationandintegrationbyTheorem4becausetheintegralontherightconvergesatbothendpoints.Asx!0+,ex1behaveslikex,sothattheintegralbehaveslikeRa0xs2dxwhichcon-vergesforRe(s)1.Theintegralconvergesattherightendpointbecauseexgrowsfasterthananypowerofx.2.2IntegralFormulaToextendthisformulatoC,Riemannintegrates(z)s=(ez1)overathepathofintegrationCwhich\startsat+1,movestotheoriginalongthe\topofthepositiverealaxis,circlestheorigincounterclockwiseinasmallcircle,thenreturnsto+1alongthe\bottomofthepositiverealaxis.Thatis,forsmallpositive,ZC(z)sez1dzz=Z+1+Zjzj=+Z+1!(z)sez1dzz:Noticethatthedenitionof(z)simplicitlydependsonthedenitionoflog(z)=logjzj+iarg(z).Intherstintegral,whenzliesonthenegativerealaxis,wetakearg(z)=i.Inthesecondintegral,thepathofintegrationstartsatz=orz=,andaszproceedscounterclockwisearoundthecircle,arg(z)increasesfromitoi.(Youcanthinkoftheimaginarypartofthelogfunctionasspiralstaircase,andgoingcounterclockwisearoundtheoriginashavingbroughtusuponelevel.)Inthelastintegral,arg(z)=i.(Thus,therstandlastintegralsdonot4cancelaswewouldexpect!)Tostatethedenitionquiteprecisely,theintegraloverCisZ+1es(logzi)ez1dzz+Z20es(logi+i)eei1id+Z+1es(logz+i)ez1dzz:TheintegralsconvergebythesameargumentgivenaboveregardingR10xs1=(1ex)dx;infact,theyconvergeuniformlyoncompactsubsetsofC.Thisdenitionappearstodependon,butactuallydoesn