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I obtained a strange formula about $\zeta'(s)/\zeta(s)$

$$ \begin{split} \frac{\zeta'(s)}{\zeta(s)}-(2\pi)^s&\sum_{\Im(\rho)>0} (-i\rho)^{-s}(2\pi)^{-\rho} e^{-i\pi \rho / 2} \Gamma(\rho)\;\;\text{ converges for } \Re(s)>1 \\ &\color{red}{\text{ and it extends analytically to } \Re(s) >0} \end{split} \label{1}\tag{1} $$ In other words the generalized Dirichlet series $\sum_{\Im(\rho)>0} (2i\pi)^{-\rho}\Gamma(\rho) (\frac{\rho}{2i\pi})^{-s}$ has the same non-trivial poles as $\zeta'(s)/\zeta(s)$, it encodes the Riemann hypothesis.

I don't understand if it is obvious, expected, and how to categorize it in the zoo of Riemann zeta formulas. Would you have any insight?

The proof of \eqref{1} is as follow: for $\Re(z) >0$ let $f(z)=\sum_{n\ge 1} \Lambda(n) e^{-nz}$, so that $$\DeclareMathOperator{\Res}{Res} \frac{-\zeta'(s)}{\zeta(s)} \Gamma(s) = \int_0^\infty f(x)x^{s-1}dx,\quad\Re(s) > 1. $$ We have the Mellin inversion and explicit formula, which converges without problems $$ \begin{split} f(z) &=\frac1{2i\pi}\int_{(2)} \frac{-\zeta'(s)}{\zeta(s)} \Gamma(s) z^{-s}ds = \sum \Res\left(\frac{-\zeta'(s)}{\zeta(s)} \Gamma(s) z^{-s}\right) \\ & = z^{-1}- \sum_\rho \Gamma(\rho) z^{-\rho}+ \sum_{k=0}^\infty \Res\left(\frac{-\zeta'(s)}{\zeta(s)} \Gamma(s) z^{-s},-k\right) \end{split}$$ $\sum_\rho$ means sum over non-trivial zeros. $z^{-s}$ means $e^{-s \log z}$ with $\Im(\log z) \in (-\pi/2,\pi/2)$.

Now $\int_\epsilon^\infty f(x)x^{s-1}dx$ is entire so for the Riemann hypothesis we are interested only in the behavior of $f(x)$ as $x\to 0^+$.

$f$ is $2i\pi$-periodic so we can look instead at the asymptotic of $f(2i\pi+x)$ as $x\to 0^+$. Because $$ f'(z)= -\sum_{n\ge 1} \Lambda(n)ne^{-nz}=O\left(\sum_{n\ge 1} n e^{-n\Re(z)}\right)= O(1/\Re(z)^2) $$ we get that $$ f(2i\pi + x)=f(2i \pi e^{-i x/(2\pi)})+\int_{2i \pi e^{-i x/(2\pi)}}^{2i\pi + x}f'(z)dz= f(2i \pi e^{-i x/(2\pi)})+O(1). $$ It happens that in the sum-over-residues series for $f(2i \pi e^{-i x/(2\pi)})$ everything is $O(1)$ but $$ g(x)=- \sum_{\Im(\rho)>0} \Gamma(\rho) (2i \pi e^{-i x/(2\pi)})^{-\rho}. $$ Therefore, $\int_0^\epsilon (f(x)-g(x))x^{s-1}dx$ is analytic for $\Re(s) >0$ and due to exponential decay so is $\int_0^\infty (f(x)-g(x))x^{s-1}dx$ which is what \eqref{1} says.

Daniele Tampieri
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reuns
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  • Maybe $\frac{\zeta’(s)}{\zeta(s)}=\frac{d\ln(\zeta(s))}{ds}$, integrate to find the series of $\ln(\zeta(s))$ and take $\exp()$ on both sides to make the sum into a product? – Тyma Gaidash Dec 14 '22 at 12:52
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    Is there a typo in formula (1)? Should $\frac{\zeta'(s)}{\zeta(s)}-...$ be $\frac{\zeta'(s)}{\zeta(s)}=...$? – Steven Clark Dec 14 '22 at 19:38
  • @StevenClark No – reuns Dec 14 '22 at 19:51
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    If your answer to the question of @StevenClark is "No" then what you are asking is quite unclear: The formula for $\frac{\zeta'(s)}{\zeta(s)}$ is the expression $$\frac{\zeta'(s)}{\zeta(s)}-(2\pi)^s\sum_{\Im(\rho)>0} (-i\rho)^{-s}(2\pi)^{-\rho} e^{-i\pi \rho / 2} \Gamma(\rho)\quad\text{?}$$ Meaning that $\frac{\zeta'(s)}{\zeta(s)}$ is equal to expression? Implying that $$(2\pi)^s\sum_{\Im(\rho)>0} (-i\rho)^{-s}(2\pi)^{-\rho} e^{-i\pi \rho / 2} \Gamma(\rho) = 0 \quad\text{?}$$ – Lee Mosher Dec 14 '22 at 20:08
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    I thought perhaps there was a typo since the title implies the question is about a strange formula for $\frac{\zeta'(s)}{\zeta(s)}$. The only formula I see for $\frac{\zeta'(s)}{\zeta(s)}$ is $\frac{\zeta'(s)}{\zeta(s)}=-\frac{1}{\Gamma(s)}\int\limits_0^\infty f(x)\ x^{s-1}dx,\ \Re(s)>1$ where $f(x)=\sum\limits_{n\ge 1} \Lambda(n)\ e^{-n x}$ which doesn't seem to qualify as strange. – Steven Clark Dec 14 '22 at 20:12
  • @LeeMosher the function $\frac{\zeta'(s)}{\zeta(s)}-(2\pi)^s\sum_{\Im(\rho)>0} (-i\rho)^{-s}(2\pi)^{-\rho} e^{-i\pi \rho / 2} \Gamma(\rho)$ is analytic for $\Re(s) > 0$. ie. the generalized Dirichlet series $(2\pi)^s\sum_{\Im(\rho)>0} (-i\rho)^{-s}(2\pi)^{-\rho} e^{-i\pi \rho / 2} \Gamma(\rho)$ has the same poles as $\frac{\zeta'(s)}{\zeta(s)}$. – reuns Dec 14 '22 at 20:44
  • Ah. I wondered about the grammar in the first displayed line... : in my brand of U.S. English, I'd delete the "which" to get the sense you want. – paul garrett Dec 14 '22 at 20:54
  • Ah great, that all makes more sense now. – Lee Mosher Dec 14 '22 at 21:51

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