Riemann zeta function

The Riemann zeta function is a function very important in number theory. In particular, the Riemann Hypothesis is a conjecture about the roots of the zeta function.

The function is defined by \[\zeta (s)=\sum_{n=1}^{\infty}\frac{1}{n^s}= 1+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\cdots\] when the real part $\Re(s)$ is greater than 1. (When $\Re(s) \le 1$ the series does not converge, but it can be extended to all complex numbers except $s = 1$—see below.)

Leonhard Euler showed that when $s=2$, the sum is equal to $\frac{\pi^2}{6}$. Euler also found that since every number is the product of a unique combination of prime numbers, the zeta function can be expressed as an infinite product: \[\zeta(s) = \left(\frac{1}{(2^0)^s} + \frac{1}{(2^1)^s}+ \frac{1}{(2^2)^s} + \cdots\right) \left(\frac{1}{(3^0)^s} + \frac{1} {(3^1)^s} + \frac{1}{(3^2)^s} + \cdots\right) \left(\frac{1}{(5^0)^s} + \frac{1}{(5^1)^s} + \frac{1}{(5^2)^s} + \cdots\right) \cdots.\] By summing up each of these geometric series in parentheses, we arrive at the following identity (the Euler Product): \[\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} = \prod_{p \text{ prime}} (1-p^{-s})^{-1}.\]

This gives a hint of why an analytic object like the zeta function could be related to number theoretic results.


Extending the 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. $\blacksquare$

Now we can extend the zeta function.

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. $\blacksquare$

The next step is the functional equation: Let \[\xi(s)=\pi^{-s/2}\Gamma(s/2)\zeta(s).\] Then $\xi(s)=\xi(1-s)$. This gives us a meromorphic continuation of $\zeta(s)$ to all of $\mathbb{C}$.

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 formula 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. $\blacksquare$

Now we proceed to the main result.

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.

See also

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