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Dirichlet eta function

From Wikipedia, the free encyclopedia
Function in analytic number theory
For the modular form, seeDedekind eta function.
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Color representation of the Dirichlet eta function. It is generated as aMatplotlib plot using a version of theDomain coloring method.[1]

Inmathematics, in the area ofanalytic number theory, theDirichlet eta function is defined by the followingDirichlet series, which converges for anycomplex number having real part > 0:η(s)=n=1(1)n1ns=11s12s+13s14s+.{\displaystyle \eta (s)=\sum _{n=1}^{\infty }{(-1)^{n-1} \over n^{s}}={\frac {1}{1^{s}}}-{\frac {1}{2^{s}}}+{\frac {1}{3^{s}}}-{\frac {1}{4^{s}}}+\cdots .}

This Dirichlet series is the alternating sum corresponding to the Dirichlet series expansion of theRiemann zeta function,ζ(s) — and for this reason the Dirichlet eta function is also known as thealternating zeta function, also denotedζ*(s). The following relation holds:η(s)=(121s)ζ(s){\displaystyle \eta (s)=\left(1-2^{1-s}\right)\zeta (s)}

Both the Dirichlet eta function and the Riemann zeta function are special cases ofpolylogarithms.

While the Dirichlet series expansion for the eta function is convergent only for anycomplex numbers with real part > 0, it isAbel summable for any complex number. This serves to define the eta function as anentire function. (The above relation and the facts that the eta function is entire andη(1)0{\displaystyle \eta (1)\neq 0} together show the zeta function ismeromorphic with a simplepole ats = 1, and possibly additional poles at the other zeros of the factor121s{\displaystyle 1-2^{1-s}}, although in fact these hypothetical additional poles do not exist.)

Equivalently, we may begin by definingη(s)=1Γ(s)0xs1ex+1dx{\displaystyle \eta (s)={\frac {1}{\Gamma (s)}}\int _{0}^{\infty }{\frac {x^{s-1}}{e^{x}+1}}{dx}}which is also defined in the region of positive real part (Γ(s){\displaystyle \Gamma (s)} represents thegamma function). This gives the eta function as aMellin transform.

Hardy gave a simple proof of thefunctional equation for the eta function,[2] which isη(s)=212s112sπs1ssin(πs2)Γ(s)η(s+1).{\displaystyle \eta (-s)=2{\frac {1-2^{-s-1}}{1-2^{-s}}}\pi ^{-s-1}s\sin \left({\pi s \over 2}\right)\Gamma (s)\eta (s+1).}

From this, one immediately has the functional equation of the zeta function also, as well as another means to extend the definition of eta to the entire complex plane.

Zeros

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Thezeros of the eta function include all the zeros of the zeta function: the negative even integers (real equidistant simple zeros); the zeros along the critical line, none of which are known to be multiple and over 40% of which have been proven to be simple, and the hypothetical zeros in the critical strip but not on the critical line, which if they do exist must occur at the vertices of rectangles symmetrical around thex-axis and the critical line and whose multiplicity is unknown.[citation needed] In addition, the factor121s{\displaystyle 1-2^{1-s}} adds an infinite number of complex simple zeros, located at equidistant points on the line(s)=1{\displaystyle \Re (s)=1}, atsn=1+2nπi/ln(2){\displaystyle s_{n}=1+2n\pi i/\ln(2)} wheren is any nonzerointeger.

The zeros of the eta function are located symmetrically with respect to the real axis and under theRiemann hypothesis would be on two parallel lines(s)=1/2,(s)=1{\displaystyle \Re (s)=1/2,\Re (s)=1}, and on the perpendicular half line formed by the negative real axis.

Landau's problem withζ(s) =η(s)/(1 − (21−s)) and solutions

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In the equationη(s) = (1 − 21−s)ζ(s), "the pole ofζ(s) ats = 1 is cancelled by the zero of the other factor" (Titchmarsh, 1986, p. 17), and as a resultη(1) is neither infinite nor zero (see§ Particular values). However, in the equationζ(s)=η(s)121s,{\displaystyle \zeta (s)={\frac {\eta (s)}{1-2^{1-s}}},}η must be zero at all the pointssn=1+n2πln2i,n0,nZ{\displaystyle s_{n}=1+n{\frac {2\pi }{\ln {2}}}i,n\neq 0,n\in \mathbb {Z} }, where the denominator is zero, if the Riemann zeta function is analytic and finite there. The problem of proving this without defining the zeta function first was signaled and left open byE. Landau in his 1909 treatise on number theory: "Whether the eta series is different from zero or not at the pointssn1{\displaystyle s_{n}\neq 1}, i.e., whether these are poles of zeta or not, is not readily apparent here."

A first solution for Landau's problem was published almost 40 years later byD. V. Widder in his book TheLaplace Transform. It uses the next prime 3 instead of 2 to define a Dirichlet series similar to the eta function, which we will call theλ{\displaystyle \lambda } function, defined for(s)>0{\displaystyle \Re (s)>0} and with some zeros also on(s)=1{\displaystyle \Re (s)=1}, but not equal to those of eta.

Indirect proof ofη(sn) = 0 following Widder

λ(s)=(133s)ζ(s)=(1+12s)23s+(14s+15s)26s+{\displaystyle \lambda (s)=\left(1-{\frac {3}{3^{s}}}\right)\zeta (s)=\left(1+{\frac {1}{2^{s}}}\right)-{\frac {2}{3^{s}}}+\left({\frac {1}{4^{s}}}+{\frac {1}{5^{s}}}\right)-{\frac {2}{6^{s}}}+\cdots }

Ifs{\displaystyle s} is real and strictly positive, the series converges since the regrouped terms alternate in sign and decrease in absolute value to zero. According to a theorem on uniform convergence of Dirichlet series first proven by Cahen in 1894, theλ(s){\displaystyle \lambda (s)} function is then analytic for(s)>0{\displaystyle \Re (s)>0}, a region which includes the line(s)=1{\displaystyle \Re (s)=1}. Now we can define correctly, where the denominators are not zero,ζ(s)=η(s)122s{\displaystyle \zeta (s)={\frac {\eta (s)}{1-{\frac {2}{2^{s}}}}}}orζ(s)=λ(s)133s{\displaystyle \zeta (s)={\frac {\lambda (s)}{1-{\frac {3}{3^{s}}}}}}

Sincelog3log2{\displaystyle {\frac {\log 3}{\log 2}}} is irrational, the denominators in the two definitions are not zero at the same time except fors=1{\displaystyle s=1}, and theζ(s){\displaystyle \zeta (s)\,} function is thus well defined and analytic for(s)>0{\displaystyle \Re (s)>0} except ats=1{\displaystyle s=1}. We finally get indirectly thatη(sn)=0{\displaystyle \eta (s_{n})=0} whensn1{\displaystyle s_{n}\neq 1}:η(sn)=(122sn)ζ(sn)=122sn133snλ(sn)=0.{\displaystyle \eta (s_{n})=\left(1-{\frac {2}{2^{s_{n}}}}\right)\zeta (s_{n})={\frac {1-{\frac {2}{2^{s_{n}}}}}{1-{\frac {3}{3^{s_{n}}}}}}\lambda (s_{n})=0.}

An elementary direct andζ{\displaystyle \zeta \,}-independent proof of the vanishing of the eta function atsn1{\displaystyle s_{n}\neq 1} was published by J. Sondow in 2003. It expresses the value of the eta function as the limit of special Riemann sums associated to an integral known to be zero, using a relation between the partial sums of the Dirichlet series defining the eta and zeta functions for(s)>1{\displaystyle \Re (s)>1}.

Direct proof ofη(sn) = 0 by Sondow

With some simple algebra performed on finite sums, we can write for any complexsη2n(s)=k=12n(1)k1ks=112s+13s14s++(1)2n1(2n)s=1+12s+13s+14s++1(2n)s2(12s+14s++1(2n)s)=(122s)ζ2n(s)+22s(1(n+1)s++1(2n)s)=(122s)ζ2n(s)+2n(2n)s1n(1(1+1/n)s++1(1+n/n)s).{\displaystyle {\begin{aligned}\eta _{2n}(s)&=\sum _{k=1}^{2n}{\frac {(-1)^{k-1}}{k^{s}}}\\&=1-{\frac {1}{2^{s}}}+{\frac {1}{3^{s}}}-{\frac {1}{4^{s}}}+\dots +{\frac {(-1)^{2n-1}}{{(2n)}^{s}}}\\[2pt]&=1+{\frac {1}{2^{s}}}+{\frac {1}{3^{s}}}+{\frac {1}{4^{s}}}+\dots +{\frac {1}{{(2n)}^{s}}}-2\left({\frac {1}{2^{s}}}+{\frac {1}{4^{s}}}+\dots +{\frac {1}{{(2n)}^{s}}}\right)\\[2pt]&=\left(1-{\frac {2}{2^{s}}}\right)\zeta _{2n}(s)+{\frac {2}{2^{s}}}\left({\frac {1}{{(n+1)}^{s}}}+\dots +{\frac {1}{{(2n)}^{s}}}\right)\\[2pt]&=\left(1-{\frac {2}{2^{s}}}\right)\zeta _{2n}(s)+{\frac {2n}{{(2n)}^{s}}}\,{\frac {1}{n}}\,\left({\frac {1}{{(1+1/n)}^{s}}}+\dots +{\frac {1}{{(1+n/n)}^{s}}}\right).\end{aligned}}}

Now ifs=1+it{\displaystyle s=1+it} and2s=2{\displaystyle 2^{s}=2}, the factor multiplyingζ2n(s){\displaystyle \zeta _{2n}(s)} is zero, andη2n(s)=1nitRn(1(1+x)s,0,1),{\displaystyle \eta _{2n}(s)={\frac {1}{n^{it}}}R_{n}\left({\frac {1}{{(1+x)}^{s}}},0,1\right),}whereRn(f(x),a,b) denotes a special Riemann sum approximating the integral off(x) over[a,b].Fort = 0 i.e.,s = 1, we getη(1)=limnη2n(1)=limnRn(11+x,0,1)=01dx1+x=log20.{\displaystyle \eta (1)=\lim _{n\to \infty }\eta _{2n}(1)=\lim _{n\to \infty }R_{n}\left({\frac {1}{1+x}},0,1\right)=\int _{0}^{1}{\frac {dx}{1+x}}=\log 2\neq 0.}

Otherwise, ift0{\displaystyle t\neq 0}, then|n1s|=|nit|=1{\displaystyle |n^{1-s}|=|n^{-it}|=1}, which yields|η(s)|=limn|η2n(s)|=limn|Rn(1(1+x)s,0,1)|=|01dx(1+x)s|=|21s11s|=|11it|=0.{\displaystyle |\eta (s)|=\lim _{n\to \infty }|\eta _{2n}(s)|=\lim _{n\to \infty }\left|R_{n}\left({\frac {1}{{(1+x)}^{s}}},0,1\right)\right|=\left|\int _{0}^{1}{\frac {dx}{{(1+x)}^{s}}}\right|=\left|{\frac {2^{1-s}-1}{1-s}}\right|=\left|{\frac {1-1}{-it}}\right|=0.}

Assumingη(sn)=0{\displaystyle \eta (s_{n})=0}, for each pointsn1{\displaystyle s_{n}\neq 1} where2sn=2{\displaystyle 2^{s_{n}}=2}, we can now defineζ(sn){\displaystyle \zeta (s_{n})\,} by continuity as follows,ζ(sn)=limssnη(s)122s=limssnη(s)η(sn)22sn22s=limssnη(s)η(sn)ssnssn22sn22s=η(sn)log(2).{\displaystyle \zeta (s_{n})=\lim _{s\to s_{n}}{\frac {\eta (s)}{1-{\frac {2}{2^{s}}}}}=\lim _{s\to s_{n}}{\frac {\eta (s)-\eta (s_{n})}{{\frac {2}{2^{s_{n}}}}-{\frac {2}{2^{s}}}}}=\lim _{s\to s_{n}}{\frac {\eta (s)-\eta (s_{n})}{s-s_{n}}}\,{\frac {s-s_{n}}{{\frac {2}{2^{s_{n}}}}-{\frac {2}{2^{s}}}}}={\frac {\eta '(s_{n})}{\log(2)}}.}

The apparent singularity of zeta atsn1{\displaystyle s_{n}\neq 1} is now removed, and the zeta function is proven to be analytic everywhere ins>0{\displaystyle \Re {s}>0}, except ats=1{\displaystyle s=1} wherelims1(s1)ζ(s)=lims1η(s)121ss1=η(1)log2=1.{\displaystyle \lim _{s\to 1}(s-1)\zeta (s)=\lim _{s\to 1}{\frac {\eta (s)}{\frac {1-2^{1-s}}{s-1}}}={\frac {\eta (1)}{\log 2}}=1.}

Integral representations

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A number of integral formulas involving the eta function can be listed. The first one follows from a change of variable of the integral representation of the Gamma function (Abel, 1823), giving aMellin transform which can be expressed in different ways as adouble integral (Sondow, 2005). This is valid fors>0.{\displaystyle \Re s>0.}Γ(s)η(s)=0xs1ex+1dx=00xxs2ex+1dydx=00(t+r)s2et+r+1drdt=0101(log(xy))s21+xydxdy.{\displaystyle {\begin{aligned}\Gamma (s)\eta (s)&=\int _{0}^{\infty }{\frac {x^{s-1}}{e^{x}+1}}\,dx=\int _{0}^{\infty }\int _{0}^{x}{\frac {x^{s-2}}{e^{x}+1}}\,dy\,dx\\[8pt]&=\int _{0}^{\infty }\int _{0}^{\infty }{\frac {(t+r)^{s-2}}{e^{t+r}+1}}dr\,dt=\int _{0}^{1}\int _{0}^{1}{\frac {\left(-\log(xy)\right)^{s-2}}{1+xy}}\,dx\,dy.\end{aligned}}}

The Cauchy–Schlömilch transformation (Amdeberhan, Moll et al., 2010) can be used to prove this other representation, valid fors>1{\displaystyle \Re s>-1}.Integration by parts of the first integral above in this section yields another derivation.

21sΓ(s+1)η(s)=20x2s+1cosh2(x2)dx=0tscosh2(t)dt.{\displaystyle 2^{1-s}\,\Gamma (s+1)\,\eta (s)=2\int _{0}^{\infty }{\frac {x^{2s+1}}{\cosh ^{2}(x^{2})}}\,dx=\int _{0}^{\infty }{\frac {t^{s}}{\cosh ^{2}(t)}}\,dt.}

The next formula, due to Lindelöf (1905), is valid over the whole complex plane, when the principal value is taken for the logarithm implicit in the exponential.η(s)=(1/2+it)seπt+eπtdt.{\displaystyle \eta (s)=\int _{-\infty }^{\infty }{\frac {(1/2+it)^{-s}}{e^{\pi t}+e^{-\pi t}}}\,dt.}This corresponds to aJensen (1895) formula for the entire function(s1)ζ(s){\displaystyle (s-1)\,\zeta (s)}, valid over the whole complex plane and also proven by Lindelöf.(s1)ζ(s)=2π(1/2+it)1s(eπt+eπt)2dt.{\displaystyle (s-1)\zeta (s)=2\pi \,\int _{-\infty }^{\infty }{\frac {(1/2+it)^{1-s}}{(e^{\pi t}+e^{-\pi t})^{2}}}\,dt.}"This formula, remarquable by its simplicity, can be proven easily with the help of Cauchy's theorem, so important for the summation of series" wrote Jensen (1895). Similarly by converting the integration paths to contour integrals one can obtain other formulas for the eta function, such as this generalisation (Milgram, 2013) valid for0<c<1{\displaystyle 0<c<1} and alls{\displaystyle s}:η(s)=12(c+it)ssin(π(c+it))dt.{\displaystyle \eta (s)={\frac {1}{2}}\int _{-\infty }^{\infty }{\frac {(c+it)^{-s}}{\sin {(\pi (c+it))}}}\,dt.}The zeros on the negative real axis are factored out cleanly by makingc0+{\displaystyle c\to 0^{+}} (Milgram, 2013) to obtain a formula valid fors<0{\displaystyle \Re s<0}:η(s)=sin(sπ2)0tssinh(πt)dt.{\displaystyle \eta (s)=-\sin \left({\frac {s\pi }{2}}\right)\int _{0}^{\infty }{\frac {t^{-s}}{\sinh {(\pi t)}}}\,dt.}

Numerical algorithms

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Most of theseries acceleration techniques developed foralternating series can be profitably applied to the evaluation of the eta function. One particularly simple, yet reasonable method is to applyEuler's transformation of alternating series, to obtainη(s)=n=012n+1k=0n(1)k(nk)1(k+1)s.{\displaystyle \eta (s)=\sum _{n=0}^{\infty }{\frac {1}{2^{n+1}}}\sum _{k=0}^{n}(-1)^{k}{n \choose k}{\frac {1}{(k+1)^{s}}}.}

Note that the second, inside summation is aforward difference.

Borwein's method

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Peter Borwein used approximations involvingChebyshev polynomials to produce a method for efficient evaluation of the eta function.[3] Ifdk=n=0k(n+1)!4(n)!(2)!{\displaystyle d_{k}=n\sum _{\ell =0}^{k}{\frac {(n+\ell -1)!4^{\ell }}{(n-\ell )!(2\ell )!}}}thenη(s)=1dnk=0n1(1)k(dkdn)(k+1)s+γn(s),{\displaystyle \eta (s)=-{\frac {1}{d_{n}}}\sum _{k=0}^{n-1}{\frac {(-1)^{k}(d_{k}-d_{n})}{(k+1)^{s}}}+\gamma _{n}(s),}where for(s)12{\displaystyle \Re (s)\geq {\frac {1}{2}}} the error termγn is bounded by|γn(s)|3(3+8)n(1+2|(s)|)exp(π2|(s)|).{\displaystyle |\gamma _{n}(s)|\leq {\frac {3}{(3+{\sqrt {8}})^{n}}}(1+2|\Im (s)|)\exp \left({\frac {\pi }{2}}|\Im (s)|\right).}

The factor of3+85.8{\displaystyle 3+{\sqrt {8}}\approx 5.8} in the error bound indicates that the Borwein series converges quite rapidly asn increases.

Particular values

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Further information:Zeta constant

η(1k)=2k1kBk+,{\displaystyle \eta (1-k)={\frac {2^{k}-1}{k}}B_{k}^{+{}},} whereB+
n
is thek-thBernoulli number.

Also:

The general form for even positive integers is:η(2n)=(1)n+1B2nπ2n(22n11)(2n)!.{\displaystyle \eta (2n)=(-1)^{n+1}{{B_{2n}\pi ^{2n}\left(2^{2n-1}-1\right)} \over {(2n)!}}.}

Taking the limitn{\displaystyle n\to \infty }, one obtainsη()=1{\displaystyle \eta (\infty )=1}.

Derivatives

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The derivative with respect to the parameters is fors1{\displaystyle s\neq 1}η(s)=n=1(1)nlnnns=21sln(2)ζ(s)+(121s)ζ(s).{\displaystyle \eta '(s)=\sum _{n=1}^{\infty }{\frac {(-1)^{n}\ln n}{n^{s}}}=2^{1-s}\ln(2)\,\zeta (s)+(1-2^{1-s})\,\zeta '(s).}η(1)=ln(2)γln(2)221{\displaystyle \eta '(1)=\ln(2)\,\gamma -\ln(2)^{2}\,2^{-1}}

References

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  1. ^"Jupyter Notebook Viewer".
  2. ^Hardy, G. H. (1922). A new proof of the functional equation for the Zeta-function. Matematisk Tidsskrift. B, 71–73.http://www.jstor.org/stable/24529536
  3. ^Borwein, Peter (2000). "An efficient algorithm for the Riemann zeta function". In Théra, Michel A. (ed.).Constructive, Experimental, and Nonlinear Analysis(PDF). Conference Proceedings, Canadian Mathematical Society. Vol. 27. Providence, RI:American Mathematical Society, on behalf of theCanadian Mathematical Society. pp. 29–34.ISBN 978-0-8218-2167-1. Archived fromthe original(PDF) on 2011-07-26. Retrieved2008-09-20.
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