# how to understand $\log\zeta(s)$ (Riemann zeta function)?

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 $\log{z_1z_2}=\log{z_1}+\log{z_2}$.

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–  Siméon Mar 5 '13 at 17:24
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 5 '13 at 17:39

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 $\Re(s)>1$.

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Thanks! But how to show $\sum \log{(1-p^{-s})}$ is analytic? –  fan Mar 9 '13 at 4:47
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 9 '13 at 17:39