Logarithmic Derivative of Riemann Zeta Function

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Let $\zeta$ be the Riemann zeta function:

$\displaystyle \forall s \in \C: \map \Re s > 1: \map \zeta s = \sum_{n \mathop \ge 1} n^{-s}$

Then for all $s$ with $\map \Re s > 1$:

$\displaystyle -\frac {\map {\zeta'} s} {\map \zeta s} = \sum_{n \mathop \ge 1} \map \Lambda n n^{-s}$

where $\Lambda$ is von Mangoldt's function.


By Sum of Reciprocals of Powers as Euler Product:

$\displaystyle \map \zeta s = \prod_p \frac 1 {1 - p^{-s} }$

where $p$ ranges over the primes.

From Laws of Logarithms:

$\displaystyle \ln \map \zeta s = - \sum_p \map \ln {1 - p^{-s} }$

Taking the derivative:

\(\displaystyle \frac {\map {\zeta'} s} {\map \zeta s}\) \(=\) \(\displaystyle -\sum_p \frac {\paren {\ln p} p^{-s} } {1 - p^{-s} }\) Logarithmic Derivative of Infinite Product of Analytic Functions
\(\displaystyle \) \(=\) \(\displaystyle -\sum_p \paren {\ln p} \paren {\frac 1 {1 - p^{-s} } - 1}\)
\(\displaystyle \) \(=\) \(\displaystyle -\sum_p \paren {\ln p} \sum_{n \mathop \ge 1} p^{-n s}\) Sum of Infinite Geometric Sequence

Also, by the definition of $\Lambda$:

$\displaystyle \sum_{n \mathop \ge 1} \map \Lambda n n^{-s} = \sum_p \paren {\ln p} \sum_{n \mathop \ge 1} p^{-n s}$

and the proof is complete.