# zeros of Dirichlet eta function

As stated in the parent entry (http://planetmath.org/AnalyticContinuationOfRiemannZetaToCriticalStrip), the definition of the Riemann zeta function    may be analytically continued (http://planetmath.org/AnalyticContinuation) from the half-plane  $\Re{s}>1$  to the half-plane  $\Re{s}>0$  by using the Dirichlet eta function  $\eta(s)$ via the equation

 $\displaystyle\zeta(s)\;=\;\frac{\eta(s)}{1-\frac{2}{2^{s}}}.$ (1)

Then only the status of the points

 $\displaystyle s_{n}\;:=\;1\!+\!n\!\cdot\!\frac{2\pi i}{\ln{2}}\qquad(n\in% \mathbb{Z})$ (2)

which are the zeros of $1\!-\!\frac{2}{2^{s}}$, remains :  are they poles of $\zeta(s)$ or not?  E. Landau has 1909 signaled this problem, which has been elementarily solved not earlier than after 40 years, by D. V. Widder.  He proved that those numbers, except  $s=1$,  are also zeros of $\eta(s)$.  This means that they only are removable singularities of $\zeta(s)$ and that (1) in fact extends $\zeta(s)$ to every points of the half-plane  $\Re{s}>0$  except  $s=1$.

A new direct proof by J. Sondow of the vanishing of the Dirichlet eta function at the points  $s_{n}\neq 1$  was published in 2003.  It is based on a relation between the partial sums $\eta_{n}(s)$ and $\zeta_{n}(s)$ of the series defining respectively the functions  $\eta(s)$ and $\zeta(s)$ for $\Re{s}>1$, which involves the approximation of an integral  by a Riemann sum.

With some clever but not so complicated performed on finite sums, Sondow writes for any $s$ the following:

 $\displaystyle\eta_{2n}(s)$ $\displaystyle\;=\;1-\frac{1}{2^{s}}+\frac{1}{3^{s}}-\frac{1}{4^{s}}+-\ldots+% \frac{(-1)^{2n-1}}{(2n)^{s}}$ $\displaystyle\;=\;1+\frac{1}{2^{s}}+\frac{1}{3^{s}}+\frac{1}{4^{s}}+\ldots+% \frac{(-1)^{2n-1}}{(2n)^{s}}-2\left(\frac{1}{2^{s}}+\frac{1}{4^{s}}+\ldots+% \frac{1}{(2n)^{s}}\right)$ $\displaystyle\;=\;\left(1-\frac{2}{2^{s}}\right)\zeta_{2n}(s)+\frac{2}{2^{s}}% \left(\frac{1}{(n\!+\!1)^{s}}+\ldots+\frac{1}{(2n)^{s}}\right)$ $\displaystyle\;=\;\left(1-\frac{2}{2^{s}}\right)\zeta_{2n}(s)+\frac{2n}{(2n)^{% s}}\frac{1}{n}\left(\frac{1}{(1\!+\!1/n)^{s}}+\ldots+\frac{1}{(1\!+\!n/n)^{s}}\right)$

Now if $t$ is real,  $s=1\!+\!it$,  and  $2^{1-s}=2^{-it}=1$,  then the factor multiplying $\zeta_{2n}(s)$ is zero and consequently

 $\eta_{2n}(s)\;=\;\frac{1}{n^{it}}R_{n}(1/(1\!+\!x)^{s},0,1)$

where  $R_{n}(f(x),a,b)$  denotes a special Riemann sum approximating the integral of $f(x)$ over  $[a,\,b]$.  For  $s=1$,  i.e.  $t=0$, one gets

 $\eta(1)\;=\;\lim_{n\to\infty}\eta_{2n}(1)\;=\;\lim_{n\to\infty}R_{n}(1/(1\!+\!% x),0,1)\;=\;\int_{0}^{1}\frac{dx}{1\!+\!x}\;=\;\ln{2}$

and otherwise, when  $t\neq 0$,  one has  $|n^{1-s}|=|n^{-it}|=1$,  giving

 $\displaystyle|\eta(s)|$ $\displaystyle\;=\;\lim_{n\to\infty}|\eta_{2n}(s)|\;=\;\lim_{n\to\infty}|R_{n}(% 1/(1\!+\!x)^{s},0,1)|$ $\displaystyle\;=\;\left|\int_{0}^{1}\frac{dx}{(1\!+\!x)^{s}}\right|\;=\;\left|% \frac{2^{1-s}\!-\!1}{1\!-\!s}\right|\;=\;\left|\frac{1\!-\!1}{-it}\right|\;=\;0.$

Note.  By (1) the Dirichlet eta function has as zeros also the zeros of the Riemann zeta function (see Riemann hypothesis (http://planetmath.org/RiemannZetaFunction)).

## References

• 1 E. Landau: Handbuch der Lehre von der Verteilung der Primzahlen. Erster Band. Berlin (1909); p. 161, 933.
• 2
• 3 J. Sondow: “Zeros of the alternating zeta function on the line  $\Re{s}=1$”.  — Amer. Math. Monthly 110 (2003).  Also available http://arxiv.org/abs/math.NT/0209393here.
• 4 J. Sondow: “The Riemann hypothesis, simple zeros, and the asymptotic convergence degree of improper Riemann sums”.  — Proc. Amer. Math. Soc. 126 (1998).  Also available http://www.ams.org/journals/proc/1998-126-05/S0002-9939-98-04607-3/here.
Title zeros of Dirichlet eta function ZerosOfDirichletEtaFunction 2014-11-21 21:17:02 2014-11-21 21:17:02 pahio (2872) pahio (2872) 19 pahio (2872) Derivation msc 30D30 msc 30B40 msc 11M41 DirichletEtaFunction