# Product Equation for Riemann Zeta Function

## Theorem

There exists a constant $B$ such that

- $\displaystyle \frac{\zeta'(s)}{\zeta(s)} = B - \frac 1{s-1} +\frac12 \log \pi - \frac12 \frac{\Gamma'(s/2+1)}{\Gamma(s/2+1)} + \sum_\rho \left( \frac 1{s-\rho} + \frac1\rho \right)$

where $\zeta$ is the Riemann zeta function, $\rho$ runs over the non-trivial zeros of $\zeta$, and $\Gamma$ is the gamma function.

## Proof

Let $\xi$ be the completed Riemann zeta function

- $\displaystyle \xi(s) = \frac 12 s(s-1) \pi^{-s/2} \Gamma\left(\frac s2\right) \zeta(s)$

By Gamma Difference Equation, this gives

- $\displaystyle \xi(s) = (s-1) \pi^{-s/2} \Gamma\left(\frac s2 + 1\right) \zeta(s)$

So taking the logarithm,

- $\displaystyle \log \xi(s) = \log (s - 1) - \frac s2 \log\pi + \log\Gamma\left(\frac s2 + 1\right) + \log \zeta(s)$

And taking the derivative we obtain

- $\displaystyle \frac{\xi'(s)}{\xi(s)} = \frac1{s-1} -\frac12 \log \pi + \frac{\Gamma'\left(s/2 + 1\right)}{\Gamma\left(s/2 + 1\right)} + \frac{\zeta'(s)}{\zeta(s)} \qquad (1)$

We have that the Completed Riemann Zeta Function has Order One, so by the Hadamard Factorisation Theorem, there exist constants $A$, $B$ such that

- $\displaystyle \xi(s) = \exp(A+Bs)\: \prod_\rho \left( 1 - \frac s{\rho} \right) \exp\left( \frac s{\rho} \right)$

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

Taking the logarithm, we obtain

- $\displaystyle \log \xi(s) = A+Bs + \sum_\rho \left[ \log \left( 1 - \frac s{\rho} \right) + \frac s{\rho} \right]$

which has derivative

- $\displaystyle \frac{\xi'(s)}{\xi(s)} = B + \sum_\rho \left( \frac 1{s - \rho} + \frac 1\rho\right) \qquad (2)$

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

- $\displaystyle B + \sum_\rho \left( \frac 1{s - \rho} + \frac 1\rho\right) = \frac1{s-1} -\frac12 \log \pi + \frac{\Gamma'\left(s/2 + 1\right)}{\Gamma\left(s/2 + 1\right)} + \frac{\zeta'(s)}{\zeta(s)}$

from which the result follows by rearranging the terms.

$\blacksquare$

## Note

The sum $\displaystyle \sum_\rho \frac 1{|\rho|}$ 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

- $\displaystyle \frac 1\rho + \frac1{\overline{\rho}} = \frac{2 \Re(\rho)}{|\rho|^2} \leq \frac 2{|\rho|^2}$

and we see by zeros of functions of finite order that a sum of such terms *does* converge.