2.3. The series for $\log \zeta (s)$ and $\zeta'(s)/\zeta (s)$

The series for $\log \zeta (s)$

In this section, we continue to exploit the Euler product for $\zeta (s)$, deriving Dirichlet series for the functions in the title. Both these series are essential ingredients of the proof of the prime number theorem.

First, consider $\log \zeta (s)$. A logarithm of a complex number $z$ is any $w$ such that $e^w = z$. As the reader will know, there are many such $w$'s. If complex numbers $z_j$ ( $j \in \mathbb{N}$) are given, and logarithms $w_j$ of $z_j$ can be chosen so that $\sum_{j=1}^\infty w_j$ is convergent (say to $w$), then by continuity $e^w = \prod_{j=1}^\infty z_j$, so $w$ is a logarithm of $\prod_{j=1}^\infty z_j$. So, by the Euler product, we will obtain a logarithm of $\zeta (s)$ if we can choose (for each $p$) a logarithm $w_p$ of $(1- p^{-s})^{-1}$ in such a way that $\sum_{p \in P} w_p$ converges. We show now that this is achieved by using the familiar logarithmic series, and furthermore that the resulting function is holomorphic (alias differentiable).

Lemma 2.3.1. Define $h(z) = \sum_{m=1}^\infty (z^m/m)$ for $\vert z\vert < 1$. Then $h(z)$ is a logarithm of $1/(1-z)$. Also, $\vert h(z)\vert \leq 2\vert z\vert$ when $\vert z\vert \leq \frac{1}{2}$.

Proof. The series for $h(z)$ certainly converges when $\vert z\vert < 1$. By termwise differentiation (which is valid for power series), we have

$$h'(z) = \sum_{m=1}^\infty z^{m-1} = \frac{1}{1-z},$$

hence

$$\frac{d}{dz} (1-z) e^{h(z)} = -e^{h(z)} + \frac{1-z}{1-z} e^{h(z)} = 0.$$

Hence $(1-z)e^{h(z)}$ is constant (say $= c$) for $\vert z\vert < 1$. Take $z = 0$ to get $c = e^0 = 1$. Further, when $\vert z\vert \leq \frac{1}{2}$,

$$\vert h(z)\vert \leq \sum_{m=1}^\infty \vert z\vert^m = \frac{\vert z\vert}{1-\vert z\vert} \leq 2\vert z\vert. \; \Box$$

We now use the following theorem on ``double series" (for a proof, see Appendix B):

(DS) Suppose that the repeated sum  $\sum_{j=1}^\infty \sum_{k=1}^\infty \vert a_{j,k}\vert$  converges. Then the repeated sums

$$\sum_{j=1}^\infty \sum_{k=1}^\infty a_{j,k}, \qquad \sum_{k=1}^\infty \sum_{j=1}^\infty a_{j,k}$$

both converge to the same sum, say S. Further, if  $\sum_{n=1}^\infty c_n$ is any series obtained by arranging the terms $a_{j,k}$ as a single series, then  $\sum_{n=1}^\infty c_n$  converges to S.

Theorem 2.3.2. For Re $s > 1$, a holomorphic function, forming a logarithm of $\zeta (s)$, is defined by

$$H(s) = \sum_{p \in P} \sum_{m=1}^\infty \frac{1}{m p^{ms}} = \sum_{n=1}^\infty \frac{c(n)}{n^s},$$

where

$$c(n) = \left\{ \begin{array}{ll} 1/m & \mbox{if } n = p^m \mb... ...d some integer }m \\ 0 & \mbox{otherwise}. \end{array}\right.$$

Both series are absolutely convergent.

Proof. Provided that the series  $\sum_{p \in P} h(1/p^s)$ converges, its sum will be a logarithm of $\zeta (s)$. This sum is the repeated sum

$$\sum_{p \in P} \sum_{m=1}^\infty \frac{1}{m p^{ms}}.$$

Since $\vert p^{ms}\vert = p^{m\sigma }$, the required convergence, as well as the stated rearrangement as a single series, will follow if we know that

$$\sum_{p \in P} \sum_{m=1}^\infty \frac{1}{m p^{m\sigma }}$$

is convergent. But this is true, since

$$\sum_{m=1}^\infty \frac{1}{mp^{m\sigma }} = h \! \left( \frac{1}{p^\sigma } \right) \leq \frac{2}{p^\sigma },$$

and  $\sum_{p \in P} (2/p^\sigma )$  is convergent. By 1.7.9, the single series defines a holomorphic function. $\Box$

We shall use the notation $\log \zeta (s)$ (for Re $s > 1$) to mean the function $H(s)$ just defined. It does not necessarily coincide with the so-called ``principal value" of the logarithm with argument between $-\pi$ and $\pi$. However, it is clearly the real-valued logarithm when $z$ is real.

The series $\sum_{p\in P}(1/p^s)$

The double series for $\log \zeta (s)$ sheds some light on the behaviour of $\sum_{p\in P}(1/p^s)$, in which the prime terms have been picked out of the series for $\zeta (s)$. In the double series, we can reverse the order of summation (again using (DS)) and separate out the term $m = 1$, as follows:

$$\begin{eqnarray*} \log \zeta (s) & = & \sum_{m=1}^\infty \frac{1}{m} \sum_{p\in ... ... + \sum_{m=2}^\infty \frac{1}{m} \sum_{p\in P} \frac{1}{p^{ms}}. \end{eqnarray*}$$

The point is that the second sum is uniformly quite small. The next result gives the details.

Proposition 2.3.3. For real $\sigma > 1$, we have

$$\sum_{p\in P} \frac{1}{p^\sigma } < \log \zeta (\sigma ).$$

For all complex s with Re $s > 1$, we have

$$\left\vert \sum_{p\in P} \frac{1}{p^s} - \log \zeta (s) \right\vert < \mbox{$\frac{1}{2}$}.$$

Proof. The first statement follows immediately from the double series as written above. To prove the second statement, we have for all $\sigma > 1$,

$$\begin{eqnarray*} \left\vert \log \zeta(s) - \sum_{p\in P}\frac{1}{p^s} \right\v... ...sum_{n=2}^\infty \frac{1}{n(n-1)}\\ & = & \mbox{$\frac{1}{2}$}. \end{eqnarray*}$$

Hence in particular  $\sum_{p\in P}(1/p^\sigma ) \to \infty$  as $\sigma \to 1^+$. This fact is a natural companion to the divergence of $\sum_{p\in P} (1/p)$, which we proved in a similar way in 2.1.6.

The series for $\zeta'(s)/\zeta (s)$

We claim to have selected a logarithm of $\zeta (s)$ that possesses a derivative. Let us now consider this derivative. Recall that for a real function $f(x)$, the derivative of $\log f(x)$ is $f'(x)/f(x)$. The same applies to complex functions in the following sense. If $g(s)$ is differentiable and $e^{g(s)} = f(s)$, then differentiation gives $g'(s) e^{g(s)} = f'(s)$, hence $g'(s) = f'(s)/f(s)$.

In our case, with $H(s)$ defined as above, this means that

$$\begin{eqnarray*} \frac{\zeta' (s)}{\zeta (s)} & = & H'(s) \\ & = & - \sum_{n=1}^\infty \frac{c(n) \log n}{ n^s} \qquad \mbox{by 1.7.9}. \end{eqnarray*}$$

Now for $n = p^m$,

$$c(n) \log n = \frac{1}{m} m \log p = \log p.$$

In other words, the following is true:

Theorem 2.3.4. For Re $s > 1$, we have

$$\frac{\zeta'(s)}{\zeta (s)} = - \sum_{n=1}^\infty \frac{\Lambda (n)}{n^s},$$

where

$$\Lambda (n) = \left\{ \begin{array}{ll} \log p & \mbox{if $n ... ...eger $m$,}\\ 0 & \mbox{otherwise}. \end{array}\right. \; \Box$$

The function $\Lambda (n)$ is called the von Mangoldt function (and $\Lambda (n)$, again for historical reasons, is the standard notation). Exactly as with the Möbius function, the Euler product has again led us to an arithmetic function that would hardly have been obvious to guess!

Now that $\Lambda (n)$ has been identified for us, let us formulate and verify the convolution identity corresponding to this series. The series identity

$$\frac{\zeta'(s)}{\zeta (s)}\: \zeta (s) = \zeta'(s)$$

suggests the convolution identity  $\Lambda *u = \ell$, where $\ell (n) = \log n$. As before, we shall now prove this identity directly, thereby giving a second proof of Theorem 2.3.4 (note that convergence of the series for Re $s > 1$ follows from the fact that $\Lambda (n) \leq \log n$ for all $n$). At the same time, the Möbius function gives a second, equivalent identity.

Theorem 2.3.5. Write $\ell (n) = \log n$. Then

$$\Lambda * u = \ell , \qquad \ell * \mu = \Lambda .$$

Proof. The second statement follows from the first one, since $u*\mu = e_1$. To prove the first statement, note first that $\Lambda (1) = \ell (1) = 0$. Now choose $n > 1$, with prime factorization

$$n = p_1^{r_1} \ldots p_k^{r_k}.$$

The divisors $i$ of $n$ for which $\Lambda (i) \neq 0$ are the numbers $p_j^{s_j}$ for $1 \leq j \leq k$ and $1 \leq s_j \leq r_j$. Each of these has $\Lambda (i) = \log p_j$. Hence

$$\begin{eqnarray*} (\Lambda * u)(n) & = & \sum_{i\vert n} \Lambda (i)\\ & = & \sum_{j=1}^k r_j \log p_j\\ & = & \log n. \; \Box \end{eqnarray*}$$

The series for $\zeta'(s)/\zeta (s)$ plays an absolutely central part in the strategy to prove the prime number theorem. At this point, it is nearly, but not quite, clear why. Of course, our goal is to prove that $\theta (x) \sim x$. The general idea is that by knowing enough about a function defined by a Dirichlet series, we hope to derive information about the coefficients of the series - more exactly, about their partial sums. The Dirichlet series whose coefficients would deliver $\theta (x)$ in this way is, of course, the derivative of $\sum_{p \in P} 1/p^s$. However, this is an awkward function, and it is easier to establish properties of $\zeta'(s)/\zeta (s)$. The summation function of its coefficients $\Lambda (n)$ differs from $\theta (x)$ by including powers of primes as well as primes. The next task is to show that this summation function does just as well as $\theta (x)$ for our purpose. This is the subject of the next section.

Exercises

1. What convolution identity is suggested by the fact that $\zeta (1/\zeta )^\prime = - \zeta^\prime / \zeta \;$? Prove this identity by showing that $\ell (u*\mu ) = 0$ and applying a result from section 1.8.

2. Let

$$b(n) = \left\{ \begin{array}{ll} \log n & \mbox{if $n$\ is prime},\\ 0 & \mbox{otherwise} \end{array} \right.$$

Describe $(b*u)(n)$ in terms of the prime factorization of $n$.

3. Use Abel summation to show that

$$\log \zeta (s) = s \int_2^\infty \frac{\pi (x)}{x(x^s -1)}\; dx.$$

4. By termwise differentiation of the series $\sum_{p \in P} h(1/p^s)$, show that

$$- \frac{\zeta' (s)}{\zeta (s)} = \sum_{p\in P} \frac {\log p}{p^s -1},$$

and express this as an integral involving $\theta (x)$.