# 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)

defining the Dirichlet eta function $\eta(s)$, a.k.a. the alternating zeta function, is convergent for  $\Re{s}>0$  and its sum is holomorphic in this half-plane.

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

 $e^{s\log{2}}\;=\;e^{\log{2}}.$

This gives  $s\log{2}-\log{2}=n\cdot 2i\pi$ (see the periodicity of exponential function), i.e.

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