Riemann Zeta Function and Prime Counting Function

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Theorem

For $\operatorname{Re} \left({s}\right) > 1 $:

$\displaystyle \log \zeta \left({s}\right) = s \int_0^{\to \infty} \frac {\pi \left({x}\right)} {x \left({x^s - 1}\right)} \mathrm d x$

where $\zeta$ denotes the Riemann Zeta Function and $\pi$ denotes the Prime-Counting Function.


Proof

From the definition of the Riemann Zeta Function:

\(\displaystyle \zeta \left({s}\right)\) \(=\) \(\displaystyle \prod_p \frac 1 {1 - p^{-s} }\)
\(\displaystyle \implies \ \ \) \(\displaystyle \log \zeta \left({s}\right)\) \(=\) \(\displaystyle \log \prod_p \frac 1 {1 - p^{-s} }\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_p \log \left({\frac 1 {1 - p^{-s} } }\right)\) Sum of Logarithms
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \left({\pi \left({n}\right) - \pi \left({n - 1}\right)}\right) \log \left( \frac 1 {1 - n^{-s} } \right)\) by definition of Prime-Counting Function
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \pi \left({n}\right) \left({\log \left( \frac 1 {1 - n^{-s} } \right) - \log \left( \frac 1 {1 - \left({n + 1}\right)^{-s} }\right)}\right)\) as $\pi \left({-1}\right) = 0$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \pi \left({n}\right) \left({\log \left(1 - \left({n + 1}\right)^{-s}\right) - \log \left(1 - n^{-s} \right)}\right)\) Logarithms of Powers


By Derivative of Logarithm Function and the Chain Rule:

\(\displaystyle \frac {\mathrm d } {\mathrm d x} \log \left({1 - x^{-s} }\right)\) \(=\) \(\displaystyle \frac {s x^{- s - 1} } {1 - x^{-s} }\)
\(\displaystyle \) \(=\) \(\displaystyle \frac s {x \left({x^s - 1}\right)}\)

Hence:

\(\displaystyle \log \zeta \left({s}\right)\) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \pi \left({n}\right) \int_n^{n + 1} \frac s {x \left({x^s - 1}\right)} \mathrm d x\) by Fundamental Theorem of Calculus
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty s \int_n^{n + 1} \frac {\pi \left({x}\right)} {x \left({x^s - 1}\right)} \mathrm d x\)
\(\displaystyle \) \(=\) \(\displaystyle s \int_0^{\to \infty} \frac {\pi \left({x}\right)} {x \left({x^s - 1}\right)} \mathrm d x\)

$\blacksquare$