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!
Best Answer
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.