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单词 AnalyticContinuationOfRiemannZetaToCriticalStrip
释义

analytic continuation of Riemann zeta to critical strip


The   1ns=e-slogn  (see the general power) of the series

n=11ns= 1+12s+13s+14s+,(1)

defining the Riemann zeta functionDlmfDlmfMathworldPlanetmath ζ(s) for s>1,  are holomorphic in the whole s-plane and theseries converges uniformly in any closed disc of thehalf-plane  s>1 ((let  s=σ+it  withσ,t  and σ>1; then |1ns|=1nσ1n1+d for a positive d for all  n=1, 2,; the series  n=11n1+d convergessince  1+d>1;  thus the series (1) converges uniformlyin the closed half-plane  s1+d,  by theWeierstrass criterion (http://planetmath.org/WeierstrassCriterionOfUniformConvergence))). Therefore we can infer (seetheorems on complex function series (http://planetmath.org/TheoremsOnComplexFunctionSeries))that the sum ζ(s) of (1) is holomorphic in the domain s>1.

We use also the fact that the series

n=1(-1)n-1ns= 1-12s+13s-14s+-(2)

defining the Dirichlet eta functionMathworldPlanetmath η(s), a.k.a. the alternating zeta function, is convergentMathworldPlanetmathPlanetmath for s>0  and its sum is holomorphic in this half-plane.

If we multiply the series (1) by the difference1-22s, every other of the series changes its sign and we get the series (2).  So we can write

ζ(s)=η(s)1-22s,(3)

which is valid when the denominator does not vanish and  s>1.  The zeros of the denominator are obtained from  2s=2,  i.e. from

eslog2=elog2.

This gives  slog2-log2=n2iπ (see the periodicity of exponential functionDlmfDlmfMathworldPlanetmathPlanetmath), i.e.

s= 1+n2πiln2(n).(4)

Thus the zeros of the denominator of (3) are on the line  s=1.

Now the functionMathworldPlanetmath on the right hand side of (3) is holomorphic in the set

D:={ss>0}{1+n2πiln2n}

and the values of this function coincide with the values of zeta functionMathworldPlanetmath in the half-plane  s>1.

This result means that, via the equation (3), the zeta function has been analytically continued (http://planetmath.org/AnalyticContinuation) to the domain D, as far as to the imaginary axis.

Remark.  In reality, all points (4) except  s=1  are removable singularitiesMathworldPlanetmath of ζ(s) given by (3), due to the fact that they are also zeros of η(s).  The fact is considered in the entry zeros of Dirichlet eta function.

Charles Hermite has shown that the zeta function may be analytically continued to the whole s-plane except for a simple poleMathworldPlanetmathPlanetmath at  s=1,  by using the equation

ζ(s)=1Γ(s)0xs-1ex-1𝑑x.(5)

See this article (http://planetmath.org/AnalyticContinuationOfRiemannZetaUsingIntegral).

Titleanalytic continuation of Riemann zeta to critical stripMathworldPlanetmath
Canonical nameAnalyticContinuationOfRiemannZetaToCriticalStrip
Date of creation2015-08-22 13:33:33
Last modified on2015-08-22 13:33:33
Ownerpahio (2872)
Last modified bypahio (2872)
Numerical id23
Authorpahio (2872)
Entry typeExample
Classificationmsc 30D30
Classificationmsc 30B40
Classificationmsc 11M41
Related topicRiemannZetaFunction
Related topicAnalyticContinuation
Related topicMeromorphicExtension
Related topicCriticalStrip
Related topicAnalyticContinuationOfRiemannZetaUsingIntegral
Related topicFormulaeForZetaInTheCriticalStrip
Related topicGammaFunction
Definesalternating zeta function
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