16

I know that if a function $f$ is analytic and has no zeros in a simple connected region, then we can define $\log{f}$ making it analytic in that region.

Let's assume $Re(s)>1$.

Is $\zeta(s)$ defined in a simple connected region? If not, how shall I understand $\log\zeta(s)$?

Also, is it true that $\log\zeta(s) = -\sum_p \log(1-p^{-s})$? I have no idea about this because I don't think we always have $\log{z_1z_2}=\log{z_1}+\log{z_2}$.

fan
  • 1,109
  • 1
    See http://math.stackexchange.com/questions/281252/about-the-determination-of-complex-logarithm – Siméon Mar 05 '13 at 17:24
  • 3
    You already said "Assume $Re(s)>1$." That is a simply connected region. Given any analytic function, $f$, you have to be careful about how you define $\log f(z)$, especially around zeros and poles. However, your sum formula gives a logarithm of $\zeta(s)$. It can be easily well-defined by taking the standard power series for $-log(1-z)= \sum_{k=1}^\infty \frac{z^k}{k}$. That also lets you get a nice double-sum formula for $\log \zeta(s)$. – Thomas Andrews Mar 05 '13 at 17:39

1 Answers1

3

Is $\zeta(s)$ defined in a simple connected region? If not, how shall I understand $\log\zeta(s)$?

Since you restrict $s$ to the half-plane $\Re(s) > 1$, it is indeed the case. The result that you state then tells you that there exists some analytic function $g$ defined on $\Re(s) > 1$ and such that $\exp \circ g = \zeta$ on $\Re(s) > 1$. Such a function is called a logrithm of $\zeta$ on $\Re(s) > 1$. Remark that for every $k\in\mathbb{Z}$, the function $g + i2k\pi$ is still a logarithm of $\zeta$ since $e^{i2\pi}=1$. In fact, it is the only possibility: if $g_1$ and $g_2$ are two logarithms of $\zeta$ on $\Re(s) > 1$, then there is some integer $k\in\mathbb{Z}$ such that $g_2 - g_1 = i2k\pi$.

With this understanding of the logarithm, you will only know (with the proof of user58512) that the identity $$ \log \zeta(s) = -\sum_p\log(1-p^{-s}) $$ is true modulo $i2\pi$ for every $\Re(s)>1$.

Principal value of the logarithm

The function $z\mapsto z$ is analytic on $\mathbb{C}$ but vanishes at $z = 0$. To overcome this issue, we restrict ourselves to $\mathbb{C}\setminus\mathbb{R}_-$ which is a open and simply connected domain where $z$ does not vanish. Hence, it admits logarithms and we could chose any one of them... but it would wonderfull if our complex logarithm coincides on $(0;\infty)$ with the usual real logarithm . This is possible! We call this one principal value of the complex logarithm.

Proof of the identity

Let us take $\log \zeta$ to be the logarithm of $\zeta$ on $\Re(s) > 1$ such that $\log\zeta(s)$ coincides with the real logarithm $\log(\zeta(s))$ when $s$ is real (so that it will be the principal value).

The identity $$ \log\zeta(s) = - \sum_p \log(1-p^{-s}) $$ is easy to get from Euler's product formula when $s \in (1;\infty)$ is a real number. But both sides of this equation extend analyticaly to $\Re(s) > 1$, so that according to the analytic continuation theorem, they are equal on $\Re(s)>1$.

Siméon
  • 10,664
  • 1
  • 21
  • 54
  • Thanks! But how to show $\sum \log{(1-p^{-s})}$ is analytic? – fan Mar 09 '13 at 04:47
  • 1
    This is a series of analytic functions, and it is absolutely convergent on every $\Re(s) \geq 1+\epsilon$. Hence, Morera's theorem shows that the sum is analytic on $\Re(s) > 1$. – Siméon Mar 09 '13 at 17:39
  • But since the principal branch $\operatorname{Log}$ is only analytic on $\mathbb{C}\setminus(-\infty,0]$, you would need to show that $\zeta(s)$ does not take values in $(-\infty,0]$ before concluding that $\operatorname{Log}\zeta(s)$ is analytic, right? – Sardines Nov 12 '23 at 18:06
  • @Sardines, the "$\log$" in $\log\zeta(s)$ is itself not the main branch $\text{Log} s$ of the logarithm, so this is not to be understood as a composition $\log\zeta=\text{Log}\circ\zeta$. Instead, $\log \zeta(s)$ denotes a new function $f$ such that $\exp(f(s))=\zeta(s)$ and $f(r)=\log_{\mathbb{R}} \zeta(r)$ for real $r$ numbers where $\log_{\mathbb{R}}$ is the usual real logarithm. In fact, it can be shown that $\zeta(s)$ can take any value on $\mathbb{C}$ for $\text{Re }s>1$, including the values in $(-\infty,0]$. So $\text{Log}\circ\zeta$ will not result in an analytic function. – russoo Nov 13 '23 at 04:02
  • @russoo I totally agree. That makes a lot more sense now. – Sardines Nov 13 '23 at 04:13