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https://archive.lib.msu.edu/crcmath/math/math/r/r316.htm
The zeta function is defined for $\Re[s]>1$, but can be analytically continued to $\Re[s]>0$ as follows:
\begin{align*}
\sum_{n=1}^\infty (-1)^n n^{-s}+\sum_{n=1}^\infty n^{-s}& = 2\sum_{n=2, 4, \ldots}^\infty n^{-s}\\
& = 2\sum_{k=1}^\infty (2k)^{-s} = 2^{1-s}\sum_{k=1}^\infty k^{-s}\tag{27}
\end{align*}
\begin{equation}\tag{28}
\sum_{n=1}^\infty (-1)^nn^{-s}+\zeta(s)=2^{1-s}\zeta(s)
\end{equation}
\begin{equation}\tag{29}
\zeta(s) = {1\over 1-2^{1-s}} \sum_{n=1}^\infty (-1)^n n^{-s}.
\end{equation}
The Derivative of the Riemann zeta function is defined by
\begin{equation}\tag{30}
\zeta'(s)=-s\sum_{k=1}^\infty k^{-s}\ln k = -\sum_{k=2}^\infty {\ln k\over k^s}.
\end{equation}
As $s\to 0$,
\begin{equation}\tag{31}
\zeta'(0)=-{\textstyle{1\over 2}}\ln(2\pi).
\end{equation}
For Even $n\equiv 2k$,
\begin{equation}\tag{32}
\zeta(n)={2^{n-1}\vert B_n\vert\pi^n\over n!},
\end{equation}
where $B_n$ is a Bernoulli Number. Another intimate connection with the Bernoulli Numbers is provided by
\begin{equation}\tag{33}
B_n = (-1)^{n+1}n\zeta(1-n).
\end{equation}
No analytic form for $\zeta(n)$ is known for Odd $n\equiv 2k+1$, but $\zeta(2k+1)$ can be expressed as the sum limit
\begin{equation}\tag{34}
\zeta(2k+1) =({\textstyle{1\over 2}}\pi)^{2k+1}\lim_{t\to\infty}\frac1{t^{2k+1}}\sum_{i=1}^\infty \left[{\cot\left({i\over 2t+1}\right)}\right]^{2k+1}
\end{equation}
(Stark 1974). The values for the first few integral arguments are
$\displaystyle \zeta(0)$ $\textstyle \equiv$ $\displaystyle -{\textstyle{1\over 2}}$
$\displaystyle \zeta(1)$ $\textstyle =$ $\displaystyle \infty$
$\displaystyle \zeta(2)$ $\textstyle =$ $\displaystyle {\pi^2\over 6}$
$\displaystyle \zeta(3)$ $\textstyle =$ $\displaystyle 1.202 056 903 2\ldots$
$\displaystyle \zeta(4)$ $\textstyle =$ $\displaystyle {\pi^4\over 90}$
$\displaystyle \zeta(5)$ $\textstyle =$ $\displaystyle 1.036 927 755 1\ldots$
$\displaystyle \zeta(6)$ $\textstyle =$ $\displaystyle {\pi^6\over 945}$
$\displaystyle \zeta(7)$ $\textstyle =$ $\displaystyle 1.008 349 277 4\ldots$
$\displaystyle \zeta(8)$ $\textstyle =$ $\displaystyle {\pi^8\over 9450}$
$\displaystyle \zeta(9)$ $\textstyle =$ $\displaystyle 1.002 008 392 8\ldots$
$\displaystyle \zeta(10)$ $\textstyle =$ $\displaystyle {\pi^{10}\over 93{,}555}.$
Euler gave $\zeta(2)$ to $\zeta(26)$ for Even $n$, and Stieltjes (1993) determined the values of $\zeta(2)$, ..., $\zeta(70)$ to 30 digits of accuracy in 1887. The denominators of $\zeta(2n)$ for $n=1$, 2, ... are 6, 90, 945, 9450, 93555, 638512875, ... (Sloane's A002432).
https://mathworld.wolfram.com/RiemannZetaFunction.html
https://mathworld.wolfram.com/DirichletEtaFunction.html
https://artofproblemsolving.com/ ... emann_zeta_function
The most important properties of the zeta function are based on the fact that it extends to a meromorphic function on the full complex plane which is holomorphic except at $s=1$, where there is a simple pole of residue 1. Let us see how this is done.
First, we wish to extend $\zeta(s)$ to the strip $\Re(s)>0$. To do this, we introduce the alternating zeta function\[\zeta_a(s) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s} .\]For $\Re(s) > 1$, we have\[\zeta(s) = \zeta_a(s) + \frac{2}{2^s} + \frac{2}{4^s} + \frac{2}{6^s} + \cdots = \zeta_a(s) + 2^{1-s}{\zeta(s)},\]or\[\zeta(s) = \frac{1}{1- 2^{1-s}} \zeta_a(s) .\]We may thus use the alternating zeta function to extend the zeta function.
Proposition. The series $\zeta_a(s)$ converges whenever $\Re(s) \ge 0$.
Proof.
We have\[\zeta_a(s) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^s} = \sum_{n=1}^{\infty} \frac{1}{(2n-1)^s} - \frac{1}{(2n)^s} .\]Since\[\lvert d(x^{-s})/dx \rvert = \lvert s x^{-s-1} \rvert \le \left\lvert \frac{s}{(2n-1)^{s+1}} \right\rvert\]for $x \in [2n-1, 2n]$, it follows that\[\left\lvert \frac{1}{(2n-1)^s} - \frac{1}{(2n)^s} \right\rvert \le \left\lvert \frac{s}{(2n-1)^{s+1}} \right\rvert .\]Since $\Re(s+1) > 1$, the series in question converges. $\blacksquare$
Theorem 1. The function $\zeta(s)$ has a meromorphic extension to $\Re(s) > 0$, and it is holomorphic there except at $s=1$, where it has a simple pole of residue 1.
Proof. For $s \neq 1$, we have the extension\[\zeta(s) - \frac{1}{s-1} = \frac{1}{1 - 2^{1-s}}\zeta_a(s) - \frac{1}{s-1} .\]For $s= 1$, we have\[\lim_{s\to 1} \frac{(s-1) \zeta_a(s)}{1- 2^{1-s}} = \lim_{s\to1} \frac{\zeta_a(s)}{\log 2 \cdot 2^{1-s}} = \frac{\zeta_a(1)}{ \log 2} ,\]by l'Hôpital's Rule, so the pole at $s=1$ is simple, and its residue is $\zeta_a(1) / \log 2$.
Now, for all integers $n\geq 1$,\[\frac{d^n(\log t)}{(dt)^n} = \frac{(-1)^{n-1}(n-1)!}{t^n} .\]It follows that the Taylor series expansion of $\log x$ about $x=1$ is\[\sum_{k=1}^{\infty} \frac{(-1)^{k-1}(x-1)^k}{k} .\]It follows that $\zeta_a(1) = \log 2$. Thus the residue of the pole is 1. $\blacksquare$
Zeroes of the Zeta Function
Using the Euler product, it is not too difficult to show that $\zeta(s)$ has no zeros for $\Re(s)> 1$. Indeed, suppose this is the case; let $x = \Re(s)$. Then\begin{align*} \sum_{p} \bigl\lvert \log \lvert (1-p^{-s})^{-1} \rvert \bigr\rvert &= \sum_p \log \lvert p^s \rvert - \log \lvert p^s -1 \rvert = \sum_p \int\limits_{\lvert p^s - 1 \rvert}^{\lvert p^s \rvert} \frac{dt}{t} \\ &\approx \sum_p \frac{1}{\lvert p^s - 1 \rvert} \\ &< \sum_p 1/p^s, \end{align*}which converges. It follows that\[\prod_p (1 - p^{-s})^{-1} \neq 0 .\]
From the functional equation\[\zeta(1-s) = (2\pi)^{-s} 2 \cos(\pi s/2) \Gamma(s) \zeta(s),\]it is evident that the zeta function has zeroes at $s= -2n$, for $n$ a postive integer. These are called the trivial zeros. Since the gamma function has no zeros, it follows that these are the only zeros with real part less than 0.
In 1859, Georg Friedrich Bernhard Riemann, after whom the function is named, established the functional equation and proved that $\zeta(s)$ has infinitely many zeros in the strip $0 \le \Re(s) \le 1$. He conjectured that they all lie on the line $\Re(s) = 1/2$. This is the famous Riemann Hypothesis, and to this day it remains one of the great unsolved problems of mathematics. Recently it has been proven that the function's first ten trillion zeros lie on the line $\Re(s) = 1/2$[1], but proof of the Riemann hypothesis still eludes us.
In 1896, Jacque Hadamard and Charles-Jean de la Vallée Poussin independently proved that $\zeta(s)$ has no zeros on the line $\Re(s) = 1$. From this they proved the prime number theorem. We prove this result here.
We first define the phi function,\[\phi(s) = \sum_{p \text{ prime}} \frac{\log p}{p^s} .\]
Theorem 2. The function $\phi(s)$ has a meromorphic continuation to $\Re(s) > 1/2$ with simple poles at the poles and zeros of $\zeta(s)$, and with no other poles. The continuation is\[\phi(s) = - \frac{\zeta'(s)}{\zeta(s)} - \sum_p \frac{\log p}{p^s(p^s-1)} .\]
Proof.It follows from the Euler product that for $\Re(s) > 1$,\begin{align*} \frac{\zeta'(s)}{\zeta(s)} &= \sum_p \frac{d (1- p^{-s})^{-1}/ds} {(1-p^{-s})^{-1}} = -\sum_p \frac{d(1-p^{-s})/ds}{(1-p^{-s})} \\ &= -\sum_p \frac{\log p \cdot p^{-s}}{1-p^{-s}} \\ &= -\sum_p \frac{\log p}{p^s -1 } = - \phi(s) - \sum_p \frac{\log p}{p^s (p^s - 1)}. \end{align*}Since $\sum_p \frac{\log p}{p^s (p^s- 1)}$ converges when $\Re(s)> 1/2$, the theorem statement follows. $\blacksquare$
Theorem 3. The zeta function has no zeros on the line $\Re(s) = 1$.
Proof.We use the fact that $\zeta(\bar s) = \overline{ \zeta(s)}$.
Let $g(s) = 1/ \zeta(s)$. Then 1 is a zero of $g$ of order 1. Thus\[\lim_{\epsilon \to 0} \epsilon \phi(1+\epsilon) = \lim_{\epsilon \to 0} -\frac{\epsilon(1/g(1+\epsilon))'}{1/g(1+ \epsilon)} = \lim_{\epsilon \to 0} \frac{\epsilon g'(1+\epsilon)}{g(1+\epsilon)} = 1 .\]
Suppose now that $1+ki$ and $1+2ki$ are zeros of $\zeta(s)$ of $\zeta(s)$ of order $m$ and $n$, respectively. (Note that $m$ and $n$ may be zero.) Then\begin{align*} \lim_{\epsilon\to 0}\epsilon \phi(1+\epsilon\pm ki) &= \lim_{\epsilon\to 0} - \frac{\epsilon \zeta'(1+\epsilon \pm ki)}{ \zeta(1+\epsilon \pm ki)} = -m , \\ \lim_{\epsilon\to 0}\epsilon \phi(1+\epsilon\pm 2ki) &= \lim_{\epsilon\to 0} - \frac{\epsilon \zeta'(1+\epsilon \pm 2ki)}{ \zeta(1+\epsilon \pm 2ki)} = -n . \end{align*}
Now for real, positive $\epsilon$,\[\sum_{a=0}^{4} \binom{4}{a} \phi(1+\epsilon - 2ki+4kai) = \sum_p \frac{\log p}{p^{1+\epsilon}} (p^{ki/2} + p^{-ki/2})^4 \ge 0,\]since $p^{-ki/2} = \overline{p^{ki/2}}$. It follows that\[-2n - 8m + 6 \ge 0 .\]Since $m$ and $n$ must be nonnegative integers, it follows that $m=0$. Thus $\zeta(1+ki) \neq 0$. Since $k$ was arbitrary, it follows that $\zeta(s)$ has no zeros on the line $\Re(s)= 1$. $\blacksquare$Resources
Koch, Helmut (trans. David Kramer), Number Theory: Algebraic Numbers and Functions. AMS 2000, ISBN 0-8218-2054-0. |
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