# analytic continuation of Riemann zeta to critical strip

The   $\frac{1}{n^{s}}=e^{-s\log{n}}$  (see the general power) of the series

 $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{s}}\;=\;1+\frac{1}{2^{s}}+\frac{1}% {3^{s}}+\frac{1}{4^{s}}+\ldots,$ (1)

defining the Riemann zeta function    $\zeta(s)$ for  $\Re{s}>1$,  are holomorphic in the whole $s$-plane and the series converges uniformly in any closed disc of the half-plane  $\Re{s}>1$ ((let  $s=\sigma+it$  with $\sigma,\,t\in\mathbb{R}$  and $\sigma>1$;  then $|\frac{1}{n^{s}}|=\frac{1}{n^{\sigma}}\leq\frac{1}{n^{1+d}}$  for a positive $d$ for all  $n=1,\,2,\,\ldots$;  the series  $\sum_{n=1}^{\infty}\frac{1}{n^{1+d}}$ converges since  $1\!+\!d>1$;  thus the series (1) converges uniformly in the closed half-plane  $\Re{s}\geq 1\!+\!d$,  by the Weierstrass criterion (http://planetmath.org/WeierstrassCriterionOfUniformConvergence))).  Therefore we can infer (see theorems on complex function series (http://planetmath.org/TheoremsOnComplexFunctionSeries)) that the sum $\zeta(s)$ of (1) is holomorphic in the domain  $\Re{s}>1$.

We use also the fact that the series

 $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{s}}\;=\;1-\frac{1}{2^{s}}% +\frac{1}{3^{s}}-\frac{1}{4^{s}}+-\ldots$ (2)

If we multiply the series (1) by the difference $\displaystyle 1\!-\!\frac{2}{2^{s}}$, every other of the series changes its sign and we get the series (2).  So we can write

 $\displaystyle\zeta(s)\;=\;\frac{\eta(s)}{1-\frac{2}{2^{s}}},$ (3)

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

 $e^{s\log{2}}\;=\;e^{\log{2}}.$
 $\displaystyle s\;=\;1\!+\!n\!\cdot\!\frac{2\pi i}{\ln{2}}\quad(n\in\mathbb{Z}).$ (4)

Thus the zeros of the denominator of (3) are on the line  $\Re{s}=1$.

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

 $D\;:=\;\{s\in\mathbb{C}\,\vdots\,\,\,\Re{s}>0\}\smallsetminus\{1\!+\!n\!\cdot% \!\frac{2\pi i}{\ln{2}}\,\vdots\,\,\,n\in\mathbb{Z}\}$

and the values of this function coincide with the values of zeta function  in the half-plane  $\Re{s}>1$.

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 singularities  of $\zeta(s)$ given by (3), due to the fact that they are also zeros of $\eta(s)$.  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 pole   at  $s=1$,  by using the equation

 $\displaystyle\zeta(s)\;=\;\frac{1}{\Gamma(s)}\int_{0}^{\infty}\!\frac{x^{s-1}}% {e^{x}-1}\,dx.$ (5)

 Title analytic continuation of Riemann zeta to critical strip  Canonical name AnalyticContinuationOfRiemannZetaToCriticalStrip Date of creation 2015-08-22 13:33:33 Last modified on 2015-08-22 13:33:33 Owner pahio (2872) Last modified by pahio (2872) Numerical id 23 Author pahio (2872) Entry type Example Classification msc 30D30 Classification msc 30B40 Classification msc 11M41 Related topic RiemannZetaFunction Related topic AnalyticContinuation Related topic MeromorphicExtension Related topic CriticalStrip Related topic AnalyticContinuationOfRiemannZetaUsingIntegral Related topic FormulaeForZetaInTheCriticalStrip Related topic GammaFunction Defines alternating zeta function