Logarithmic Derivative of Riemann Zeta Function
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Theorem
Let $\zeta$ be the Riemann zeta function:
- $\ds \forall s \in \C: \map \Re s > 1: \map \zeta s = \sum_{n \mathop \ge 1} n^{-s}$
Then for all $s \in \C$ with $\map \Re s > 1$:
- $\ds -\frac {\map {\zeta'} s} {\map \zeta s} = \sum_{n \mathop \ge 1} \map \Lambda n n^{-s}$
where $\Lambda$ is von Mangoldt's function.
Proof
By Sum of Reciprocals of Powers as Euler Product:
- $\ds \map \zeta s = \prod_p \frac 1 {1 - p^{-s} }$
where $p$ ranges over the primes.
From Laws of Logarithms:
- $\ds \ln \map \zeta s = - \sum_p \map \ln {1 - p^{-s} }$
This article, or a section of it, needs explaining. In particular: Is the above equality really true for complex logarithm? I believe no. Probably, we need to check the identity on $\R_{>1}$, and later extend it to $\map \Re s > 1$ You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code. |
Taking the derivative:
\(\ds \frac {\map {\zeta'} s} {\map \zeta s}\) | \(=\) | \(\ds -\sum_p \frac {\paren {\ln p} p^{-s} } {1 - p^{-s} }\) | Logarithmic Derivative of Infinite Product of Analytic Functions and Derivative of Riemann Zeta Function | |||||||||||
\(\ds \) | \(=\) | \(\ds -\sum_p \paren {\ln p} \paren {\frac 1 {1 - p^{-s} } - 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\sum_p \paren {\ln p} \sum_{n \mathop \ge 1} p^{-n s}\) | Sum of Infinite Geometric Sequence |
Also, by the definition of $\Lambda$:
- $\ds \sum_{n \mathop \ge 1} \map \Lambda n n^{-s} = \sum_p \paren {\ln p} \sum_{n \mathop \ge 1} p^{-n s}$
and the proof is complete.
$\blacksquare$