Recall that the series:

n=11ns$$convergesfor==1;;$s>1$bythepseriestest==.Inaddition,itdivergesfor$s=1$.Ourgoalnowistofindananalyticfunctionwhichcontinuesthisseriesfor$sC$.SuchacontinuationwillbecalledtheRiemannZetafunction.Foranyvalueof$s$andfor$n>0$,wehave

n^s \underset{def}{=} e^{x\ln(n)}e^{i\ln(n)y}

Wemayusethisasourdefinitionfor$ns$,andusethistoprovethattheseriesconvergesabsolutelyfortherealpartof$Re(s)>1$sincetheoriginalseriesconverges.Nowweseethatifwehaddefined$ζ(s)$,thenitwouldsatisfy:

\Gamma(s)\zeta(s)=\sum_{n=1}^\infty \frac{\left( \int z^{s-1}e^{-z}dz \right)}

Wecanshowthat==1;;$1ns0zs1ezdz=0zs1enz$==alsobyintegrationbyparts,sowefurthersimplifytheaboveexpression:

=\sum_{n=1}^{\infty}\int_{0}^\infty z^{s-1}e^{-nz}dz

Nowwebringthesuminsidebywishfulthinking,andusegeometricseriesformulatoget:

=\int_{0}^\infty \frac{z^{s-1}e^{-z}}{1-e^{-z}}dz=\int_{0}^\infty \frac{z^{s-1}}{e^z-1}dz

Usingtheresidueformulaagain,wecandefinethezetafunction:

\zeta(s)= \frac{1}{(e^{2\pi i s}-1)\Gamma(s)}\int_{0}^\infty \frac{z^{s-1}}{e^z-1}dz

Recallthatthisformulamakessensebecause$Γ$isentireonitsdomain,seethe[[ComplexAnalysis/Part4ResiduesandResidueTheorem/5.BuildingtheGammafunctiontrigonometricformula]].Hereweareagainusingthe[[ComplexAnalysis/Part4ResiduesandResidueTheorem/3.Theorem2.4.1ResidueTheoremResidueTheorem]].Thisisdefinedaslongas$Re(s)>1$.Butwecanalsoforcethistobethedefinitionfortheentirerighthalfplane,withpossiblesingularitieswhen$(e2πis1)Γ(s)$isnotdefined.Thatis,atthepoints$1,2,3$.Butweknowthatthezetafunctionisdefinedfor$s=2,3,$,sowedontneedtoworryaboutthesepointsWecanuseourformulatoseethatthezetafunctionhasasimplepolewhen$s=1$.Nowwecompute$ζ(1n)$for$nN$.Wecanmovesometermsaroundtoseethat:

\zeta(1-n)=\frac{(-1)^{n-1}(n-1)!}{2\pi i} \int_{C} \frac{z^{-n}}{e^{-z}-1}dz

Nowweusetheoddformulathat:

\frac{z}{e^{-z}-1}= -\frac{z}{2} +\sum_{{n=0}}^\infty \frac{b_{2n}}{(2n)!}z^

where$b2n$isthe$2n$thBernoulinumberandobservethatwecancomputetheaboveintegralwiththeresiduetheoremtofindthat:$ζ(0)=12$,$ζ(1(2m1))=ζ(2m)=0$forall$m>0$,and$ζ(12m)=(1)(2m1)!(2m)!b2n$.IwontgothroughthestepstodetermineRiemannsfamousfunctionalequationforthe$ζ$function,butitallowsustoextendthe$ζ$functiontotheentirelefthalfplane.

\zeta(1-s)=\frac{\zeta(s)}{2(2\pi)^{s-1}\Gamma(1-s)\sin\left( \frac{\pi s}{2} \right)} \hspace{1in} Re(s)<0

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