OCR Output Comparison | Datalab (Chandra & Surya)

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186 Chapter 7. THE ZETA FUNCTION AND PRIME NUMBER THEOREM

which holds for $0 \le x < 1$ , we find that

$\log \zeta(s) = \log \prod_p \frac{1}{1 - p^{-s}} = \sum_p \log \left( \frac{1}{1 - p^{-s}} \right) = \sum_{p,m} \frac{p^{-ms}}{m}.$

Since the double sum converges absolutely, we need not specify the order of summation. See the Note at the end of this chapter. The formula then holds for all $\operatorname{Re}(s) > 1$ by analytic continuation. Note that, by Theorem 6.2 in Chapter 3, $\log \zeta(s)$ is well defined in the simply connected half-plane $\operatorname{Re}(s) > 1$ , since $\zeta$ has no zeros there. Finally, it is clear that we have

$\sum_{p,m} \frac{p^{-ms}}{m} = \sum_{n=1}^{\infty} c_n n^{-s},$

where $c_n = 1/m$ if $n = p^m$ and $c_n = 0$ otherwise.

The proof of the theorem we shall give depends on a simple trick that is based on the following inequality.

Lemma 1.4 If $\theta \in \mathbb{R}$ , then $3 + 4 \cos \theta + \cos 2\theta \ge 0$ .

This follows at once from the simple observation

$3 + 4 \cos \theta + \cos 2\theta = 2(1 + \cos \theta)^2.$

Corollary 1.5 If $\sigma > 1$ and $t$ is real, then

$\log |\zeta^3(\sigma) \zeta^4(\sigma + it) \zeta(\sigma + 2it)| \ge 0.$

Proof. Let $s = \sigma + it$ and note that

$\operatorname{Re}(n^{-s}) = \operatorname{Re}(e^{-(\sigma+it)\log n}) = e^{-\sigma \log n} \cos(t \log n) = n^{-\sigma} \cos(t \log n).$

Therefore,

$\begin{aligned} \log |\zeta^3(\sigma) \zeta^4(\sigma + it) \zeta(\sigma + 2it)| \\ &= 3 \log |\zeta(\sigma)| + 4 \log |\zeta(\sigma + it)| + \log |\zeta(\sigma + 2it)| \\ &= 3 \operatorname{Re}[\log \zeta(\sigma)] + 4 \operatorname{Re}[\log \zeta(\sigma + it)] + \operatorname{Re}[\log \zeta(\sigma + 2it)] \\ &= \sum c_n n^{-\sigma} (3 + 4 \cos \theta_n + \cos 2\theta_n), \end{aligned}$

where $\theta_n = t \log n$ . The positivity now follows from Lemma 1.4, and the fact that $c_n \ge 0$ .