Product Equation for Riemann Zeta Function

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Theorem

There exists a constant $B$ such that:

$\ds \frac {\map {\zeta'} s} {\map \zeta s} = B - \frac 1 {s - 1} + \frac 1 2 \ln \pi - \frac 1 2 \frac {\map {\Gamma'} {s / 2 + 1} } {\map \Gamma {s / 2 + 1} } + \sum_\rho \paren {\frac 1 {s - \rho} + \frac 1 \rho}$

where:

$\zeta$ is the Riemann zeta function
$\rho$ runs over the nontrivial zeros of $\zeta$
$\Gamma$ is the gamma function.


Proof

Let $\xi$ be the completed Riemann zeta function:

\(\ds \map \xi s\) \(=\) \(\ds \frac 1 2 s \paren {s - 1} \pi^{-s / 2} \map \Gamma {\frac s 2} \map \zeta s\) Definition of Completed Riemann Zeta Function
\(\ds \) \(=\) \(\ds \paren {s - 1} \pi^{-s / 2} \map \Gamma {\frac s 2 + 1} \map \zeta s\) Gamma Difference Equation: $\map \Gamma {\dfrac s 2 + 1} = \dfrac s 2 \map \Gamma {\dfrac s 2}$
\(\ds \leadsto \ \ \) \(\ds \ln \map \xi s\) \(=\) \(\ds \map \ln {s - 1} - \frac s 2 \ln \pi + \ln \map \Gamma {\frac s 2 + 1} + \ln \map \zeta s\) taking the logarithm of both sides
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \quad \ds \frac {\map {\xi'} s} {\map \xi s}\) \(=\) \(\ds \frac 1 {s - 1} - \frac 1 2 \ln \pi + \dfrac 1 2 \dfrac {\map {\Gamma'} {s / 2 + 1} } {\map \Gamma {s / 2 + 1} } + \frac {\map {\zeta'} s} {\map \zeta s}\) taking the derivative of both sides


We have that the Completed Riemann Zeta Function has Order One.

So, by the Hadamard Factorisation Theorem, there exist constants $A$, $B$ such that:

$\ds \map \xi s = \map \exp {A + B s} : \prod_\rho \paren {1 - \frac s \rho} \map \exp {\frac s \rho}$

where $\rho$ runs over the zeros of $\xi$, that is, the nontrivial zeros of $\zeta$.

Therefore:

\(\ds \map \xi s\) \(=\) \(\ds \map \exp {A + B s} : \prod_\rho \paren {1 - \frac s \rho} \map \exp {\frac s \rho}\) Hadamard Factorisation Theorem
\(\ds \leadsto \ \ \) \(\ds \ln \map \xi s\) \(=\) \(\ds A + B s + \sum_\rho \paren {\map \ln {1 - \frac s \rho} + \frac s \rho}\) taking the logarithm of both sides
\(\text {(2)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \quad \ds \frac {\map {\xi'} s} {\map \xi s}\) \(=\) \(\ds B + \sum_\rho \paren {\frac 1 {s - \rho} + \frac 1 \rho}\) taking the derivative of both sides


Combining $(1)$ and $(2)$ we have:

\(\ds \frac 1 {s - 1} - \frac 1 2 \ln \pi + \dfrac 1 2 \dfrac {\map {\Gamma'} {s / 2 + 1} } {\map \Gamma {s / 2 + 1} } + \frac {\map {\zeta'} s} {\map \zeta s}\) \(=\) \(\ds B + \sum_\rho \paren {\frac 1 {s - \rho} + \frac 1 \rho}\)
\(\ds \leadsto \ \ \) \(\ds \frac {\map {\zeta'} s} {\map \zeta s}\) \(=\) \(\ds B + \sum_\rho \paren {\frac 1 {s - \rho} + \frac 1 \rho} - \paren {\frac 1 {s - 1} - \frac 1 2 \ln \pi + \dfrac 1 2 \dfrac {\map {\Gamma'} {s / 2 + 1} } {\map \Gamma {s / 2 + 1} } }\)
\(\ds \) \(=\) \(\ds B - \frac 1 {s - 1} + \frac 1 2 \ln \pi - \frac 1 2 \frac {\map {\Gamma'} {s / 2 + 1} } {\map \Gamma {s / 2 + 1} } + \sum_\rho \paren {\frac 1 {s - \rho} + \frac 1 \rho}\) rearranging terms

$\blacksquare$


Warning

The sum $\ds \sum_\rho \frac 1 {\size \rho}$ diverges, so we must be careful of the order in which we take the terms.

By the Functional Equation for Riemann Zeta Function, the zeroes occur in complex conjugate pairs, and:

$\ds \frac 1 \rho + \frac 1 {\overline \rho} = \frac {2 \map \Re \rho} {\size \rho^2} \le \frac 2 {\size \rho^2}$

and we see by the corollary to Zeroes of Functions of Finite Order that a sum of such terms does converge.

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