Complex logarithm of zeta functions and L-functions

Question

I have read multiple questions on stackexchange about the complex lograithm of $\zeta(s)$. However, I haven't seen an answer that completely addresses this problem.

Denote by $\operatorname{Log}$ the principal branch of $\log$. It is well known that $\zeta(s)$ has the Euler product representation $$ \zeta(s)=\prod_p\left(1-\frac{1}{p^s}\right)^{-1},\quad \operatorname{Re}s>1. $$ Question: is the following equation true? $$ \operatorname{Log}\zeta(s)=-\sum_{p}\operatorname{Log}\left(1-\frac{1}{p^s}\right),\quad\operatorname{Re}s>1. $$ This equation holds for $s\in(1,\infty)$. If I could show that both sides are analytic in the half plane $\operatorname{Re}s>1$, then I can conclude that they are equal by analytic continuation. Indeed, it's not hard to show that the series on the RHS converges normally in $\operatorname{Re}s>1$, so RHS is analytic. However, I do not know how whether $\zeta(s)$ passes through the branch cut $(-\infty,0]$ when $\operatorname{Re}s>1$. Is this true or not?

Similarly the Dirichlet L-function satisfies $$ L(s, \chi)=\prod_p\left(1-\frac{\chi(p)}{p^s}\right)^{-1},\quad\operatorname{Re}s>1. $$ In a note on analytic number theory, the author directly wrote

Taking logarithms on both sides (using the principal branch of the logarithm) we get, for $\operatorname{Re}s>1$, $$ \log L(s, \chi)=-\sum_p \log\left(1-\frac{\chi(p)}{p^s}\right) $$

How can I prove this identity, i.e., show that a multiple of $2\pi i$ does not appear as the difference between the two sides?

Any help would be greatly appreciated!

Answer 1

Here is a general theorem which can be used in to compute logarithms for products of functions:

Theorem. Assume that $f(s)=\prod_{n=1}^{\infty}f_n(s)$ is a product of complex-valued functions which is normally convergent on some open set $U\subseteq \mathbb{C}$ and that all the functions $f_n$ are analytic on $U$. Assume further that for each $f_n$, we have $|f_n(s)-1|<1$ for all $s\in U$. Then, the series $$g(s)=\sum_{n=1}^{\infty} \text{Log}(f_n(s))$$ is normally convergent on $U$ and $g$ is an analytic logarithm of $f$ meaning that $g$ is analytic and $e^{g(s)}=f(s)$ on $U$. Here $\text{Log}$ denotes the principal value of the logarithm.

To clarify the theorem, here is the definition of a normally convergent product: A product $\prod_{n=1}^{\infty} f_n(s)$ is called normally convergent on an open set $U\subseteq\mathbb{C}$ if the series $\sum_{n=1}^{\infty} (f_n(s)-1)$ is normally convergent on $U$, i.e. for every compact set $C\subseteq \mathbb{C}$, the series $\sum_{n=1}^{\infty} \lVert f_n-1\rVert_C$ converges where $\lVert f_n-1\rVert_C$ is the supremum of $|f_n(s)-1|$ on $C$. It can be shown that normal convergence implies pointwise convergence of the product and that the limit function is analytic if all the factors $f_n$ are. When studying analytic number theory, it is very important to learn those convergence notions for infinite products first.

Proof of the theorem: Normal convergence of the product is known to be equivalent to normal convergence of the series $\sum_{n=1}^{\infty}\text{Log}(f_n(s))$ under the assumption that all the $f_n$ satisfy $|f_n(s)-1|<1$ on $U$ (this is a basic theorem on infinite products). So what remains to show is $e^{g(s)}=f(s)$ for $s\in U$: $$\exp\bigl(\sum_{n=1}^{\infty} \text{Log}(f_n(s))\bigr)=\prod_{n=1}^{\infty}\exp\bigl(\text{Log}(f_n(s))\bigr)=\prod_{n=1}^{\infty} f_n(s)=f(s)$$ which finishes the proof.

According to the theorem, it is allowed to compute an analytic logarithm of a product by "formally" applying the logarithm to the product: $$\log f(s) = \log\prod_n f_n(s) =\sum_n \text{Log}(f_n(s))$$ whenever the conditions of the theorem are satisfied. Strictly speaking, those equations do not make sense since we do not know what $\log f(s)$ or $\log\prod_n f(s)$ should be. However, the theorem guarantees that such a computation will always result in a correct logarithm of $f(s)$.

The theorem can be applied to $\zeta(s)$ since the Euler product is normally convergent for $\text{Re }s>1$ and its factors satisfy the inequality $|f_n(s)-1|<1$ required for the theorem. Thus, the function $\log\zeta(s)$ defined by $$\log\zeta(s):=\sum_p \text{Log}\biggl(\frac{1}{1-p^{-s}}\biggr)$$ is an analytic logarithm of $\zeta(s)$ in the domain $\text{Re }s>1$. As was already pointed out in the comments, the symbol $\log \zeta$ does not mean the composition $\text{Log}\circ \zeta$ (this would not result in an analytic function since it can be shown that $\zeta(s)$ can take any value in $\mathbb{C}-\{0\}$ when considered on the domain $\text{Re }s>1$). Instead, the symbol $\log \zeta$ denotes a new function which is defined as above. The Dirichlet functions $L(s,\chi)$ can be treated similarly.

Answer 2

You ask a good question that is often misunderstood and not explained clearly in textbooks. In brief, the left side of your log equation is defined to be the right side and there is no need to refer to anything about principal logarithms or branch cuts. We still need to check the right side deserves the name "logarithm of the Euler product" on ${\rm Re}(s) > 1$, and that's what I'll do.

When $f$ is an analytic function on an open set $\Omega$ in $\mathbf C$, define a logarithm of $f$ on $\Omega$ to be any analytic function $L_f : \Omega \to \mathbf C$ such that $e^{L_f(s)} = f(s)$ for all $s$ in $\Omega$.

If $\Omega$ is connected, then two logarithms of $f$ must differ by $2\pi im$ for some integer $m$. Indeed, when $e^{g_1} = f$ and $e^{g_2} = f$ on $\Omega$, we have $e^{g_1 - g_2} = 1$ on $\Omega$, so $g_1 - g_2 \colon \Omega \to 2\pi i\mathbf Z$, where $2\pi i\mathbf Z$ has its subspace topology in $\mathbf C$, namely the discrete topology (not something crazy like the $5$-adic topology coming from $\mathbf Z$ :)). Therefore when $g_1$ and $g_2$ are continuous, $(g_1 - g_2)(\Omega)$ is connected in $2\pi i\mathbf Z$, so it is constant: there's some integer $m$ such that $g_2(s) = g_1(s) + 2\pi im$ for all $s \in \Omega$.

This shows us that if we can find any logarithm of an Euler product on a half-plane like ${\rm Re}(s) > 1$, the only logarithms of that Euler product on that half-plane (or all larger connected open sets it extends to by analytic continuation) are what you get by adding integer multiples of $2\pi i$. Thus the only thing we need to do to get a logarithm of an Euler product is (i) write down one example and (ii) find a way to characterize it that is independent of formulas, so we can think about what we built in a more conceptual way.

Example. For a positive integer $d$, let $$ L(s) := \prod_p \frac{1}{(1 - \alpha_{p,1}/p^s)\cdots(1-\alpha_{p,d}/p^s)} $$ where $|\alpha_{p,j}| \leq 1$ for all $p$ and all $j = 1, \ldots, d$. This is a degree-$d$ Euler product: the zeta-function and Dirichlet $L$-functions have $d = 1$, but important examples arise with $d > 1$ too, so let's just treat this general case in order to avoid having to go through this stuff all over again when you first meet higher-degree Euler products. For simplicity, if you wish, take $d = 1$ everywhere, but it really doesn't simplify any step (except maybe psychologically).

The product $L(s)$ converges and is analytic and nonvanishing when ${\rm Re}(s) > 1$. In practice, $|\alpha_{p,j}| = 1$ for all but finitely many $p$ and we might have $\alpha_{p,j} = 0$ for finitely many $p$ and $j$.

Our goal is to create a logarithm of $L(s)$ when ${\rm Re}(s) > 1$.

Intuitively, we want to just "take logarithms of the product by summing logarithms of the factors", so let's focus on the individual factors first. Informally, consider
$$ "\log"\left(\frac{1}{1-\alpha_{p,j}/p^s}\right) = -"\log"\left(1 - \frac{\alpha_{p,j}}{p^s}\right) = \sum_{k \geq 1} \frac{(\alpha_{p,j}/p^s)^k}{k} = \sum_{k \geq 1} \frac{\alpha_{p,j}^k}{kp^{ks}}. $$ Denote the series on the right as $\ell_{p,j}(s)$. Is this a logarithm of $1/(1-\alpha_{p,j}/p^s)$ when ${\rm Re}(s) > 1$?

On the open unit disc $|z| < 1$, $\sum_{k \geq 1} z^k/k$ is a logarithm of $1/(1-z)$, i.e., $\exp(\sum_{k \geq 1} z^k/k) = 1/(1-z)$. One way to see this is to observe that both sides are analytic on $|z| < 1$ and they agree when $z$ is in the real interval $(0,1)$ by calculus, so they are equal on all of $|z| < 1$ (there are other methods of checking that too).

Set $z = \alpha_{p,j}/p^s$, so $|z| < 1$ when ${\rm Re}(s) > 0$. For such $s$, $e^{\ell_{p,j}(s)} = \exp(\sum_{k \geq 1}\alpha_{p,j}^k/(kp^{ks})) = 1/(1 - \alpha_{p,j}/p^s)$. Thus $\ell_{p,j}(s)$ is a logarithm of $1/(1-\alpha_{p,j}/p^s)$ when ${\rm Re}(s) > 0$ (not just ${\rm Re}(s) > 1$). Now we want to "add the logarithms of the Euler factors" over all $p$ and $j = 1, \ldots, d$: define $$ \ell(s) = \sum_p\sum_{j=1}^d \ell_{p,j}(s) = \sum_p \sum_{j=1}^d \sum_{k \geq 1} \frac{\alpha_{p,j}^k}{kp^{ks}} = \sum_{p^k} \frac{\alpha_{p,1}^k + \cdots + \alpha_{p,d}^k}{kp^{ks}}, $$ which is an absolutely convergent iterated series when ${\rm Re}(s) > 1$ (not ${\rm Re}(s) > 0$ except in very unusual circumstances). That justifies rearranging and combining the terms in $\ell(s)$ however we want without changing its value as long as ${\rm Re}(s) > 1$. So $\ell$ is analytic on ${\rm Re}(s) > 1$. Notice that $\ell(s)$ as a Dirichlet series has no constant term, so $\lim_{{\rm Re}(s) \to \infty} \ell(s) = 0$. We'll come back to that point later.

Now let's take the exponential of $\ell$: when ${\rm Re}(s) > 1$, $$ e^{\ell(s)} = \exp\left(\sum_p \sum_{j=1}^d \ell_{p,j}(s)\right). $$ Since $\ell(s)$ is a limit of partial sums over $p$, by continuity of the exponential function $$ e^{\ell(s)} = \prod_p \exp\left(\sum_{j=1}^d \ell_{p,j}(s)\right). $$

Since $e^{\ell_{p,j}(s)} = 1/(1 - \alpha_{p,j}/p^s)$ for each $p$ and $j$ when ${\rm Re}(s) > 0$, when ${\rm Re}(s) > 1$ we have $$ L(s):= \prod_p \prod_{j=1}^d\frac{1}{1-\alpha_{p,j}/p^s} = \prod_p \prod_{j=1}^d e^{\ell_{p,j}(s)}, $$ and the product on the right is what we already found to be $e^{\ell(s)}$. Thus $e^{\ell(s)} = L(s)$ when ${\rm Re}(s) > 1$, so we have achieved our goal: the series $\ell(s)$ is a logarithm of $L(s)$ on ${\rm Re}(s) > 1$ because (i) it is analytic there and (ii) $e^{\ell(s)} = L(s)$ on that half-plane. Intuitively, on ${\rm Re}(s) > 1$ the sum of the logs of the factors really is a log of the Euler product because the exponential of that sum is the Euler product on that half-plane.

Now that we built one logarithm of $L(s)$ when ${\rm Re}(s) > 1$, what other options are there? Since ${\rm Re}(s) > 1$ is connected, all possible logarithms of $L(s)$ on that half-plane are the functions $\ell(s) + 2\pi im$ as $m$ runs over $\mathbf Z$. How do we single out $\ell(s)$ among all logarithms of $L(s)$ by function-theoretic properties? (I want to avoid all mumbo-jumbo about branches of logarithms.)

If $L(s) = \zeta(s)$ or $L(s) = L(s,\chi)$ where $\chi$ is a quadratic Dirichlet character, then $\ell(s)$ is real-valued wen $s > 1$ and you could say $\ell(s)$ is uniquely determined by that property because adding a nonzero integral multiple of $2\pi i$ to it wrecks that property. But if we're in the more typical case where $L(s)$ is not real-valued on $(1,\infty)$, we need another idea. And here it is: look at the constant term of $\ell(s)$: it is $0$. Meanwhile, $\ell(s) + 2\pi im$ has constant term $2\pi im$. As ${\rm Re}(s) \to \infty$, a Dirichlet series tends to its constant term, so $\ell(s) + 2\pi im \to 2\pi im$ as ${\rm Re}(s) \to \infty$. Therefore the only logarithm of $L(s)$ on ${\rm Re}(s) > 1$ that tends to $0$ as ${\rm Re}(s) \to \infty$ is $\ell(s)$. This characterizes $\ell(s)$: it is the unique analytic function on ${\rm Re}(s) > 1$ with the properties (i) its exponential is $L(s)$ and (ii) it tends to $0$ as ${\rm Re}(s) \to \infty$.