Derivative of Riemann Zeta Function

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Theorem

The derivative of the Riemann zeta function is:

$\displaystyle \map {\zeta'} z = \frac {\d \zeta} {\d z} = -\sum_{n \mathop = 2}^\infty \frac {\map \ln n} {n^z}$


Proof

\(\displaystyle \frac {\d \zeta} {\d z}\) \(=\) \(\displaystyle \map {\frac \d {\d z} } {\sum_{n \mathop = 1}^\infty n^{-z} }\)
\((1):\quad\) \(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \map {\frac \d {\d z} } {n^{-z} }\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \paren {-\map \ln n n^{-z} }\) Derivative of Exponential Function
\(\displaystyle \) \(=\) \(\displaystyle -\sum_{n \mathop = 1}^\infty \frac {\map \ln n} {n^z}\)
\(\displaystyle \) \(=\) \(\displaystyle -\sum_{n \mathop = 2}^\infty \frac {\map \ln n} {n^z}\) as $\ln 1 = 0$

$\blacksquare$