Product Equation for Riemann Zeta Function

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 non-trivial zeros of $\zeta$
 * $\Gamma$ is the gamma function.

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


 * $\ds \map \xi s = \frac 1 2 s \paren {s - 1} \pi^{-s / 2} \map \Gamma {\frac s 2} \map \zeta s$

By Gamma Difference Equation, this gives:


 * $\ds \map \xi s = \paren {s - 1} \pi^{-s / 2} \map \Gamma {\frac s 2 + 1} \map \zeta s$

So taking the logarithm:


 * $\ds \ln \map \xi s = \map \ln {s - 1} - \frac s 2 \ln \pi + \ln \map \Gamma {\frac s 2 + 1} + \ln \map \zeta s$

Taking the derivative:


 * $(1): \quad \ds \frac {\map {\xi'} s} {\map \xi s} = \frac 1 {s - 1} - \frac 1 2 \ln \pi + \frac {\map {\Gamma'} {s / 2 + 1} } {\map \Gamma {s / 2 + 1} } + \frac {\map {\zeta'} s} {\map \zeta s}$

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 $\zeta$.

Taking the logarithm:


 * $\ds \ln \map \xi s = A + B s + \sum_\rho \paren {\map \ln {1 - \frac s \rho} + \frac s \rho}$

which has derivative:


 * $(2): \quad \ds \frac {\map {\xi'} s} {\map \xi s} = B + \sum_\rho \paren {\frac 1 {s - \rho} + \frac 1 \rho}$

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


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

from which the result follows by rearranging the terms.

Note
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 $\zeta$, the zeros 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 Zeros of Functions of Finite Order that a sum of such terms does converge.