Prove$\sum_{|\alpha|<1, f(\alpha)=0}\log \frac{1}{|\alpha|}= \sum_{\Re(\rho)>1/2}\log\ |\frac{\rho}{1-\rho}|$

Question

Let $$ f(z)= (s-1)\zeta(s) $$ where $s=\frac{1}{1-z}$ and $\zeta$ denotes the Riemann Zeta function.

Prove that,$$\sum_{|\alpha|<1, f(\alpha)=0}\log \frac{1}{|\alpha|}= \sum_{\Re(\rho)>1/2}\log\ \left|\frac{\rho}{1-\rho}\right|$$ I found this in this paper of Balazard Saias and Yor-.

My try- I have understood the whole paper but this one line i could not get.

Answer 1

This is very elementary math as below:

Let: $\rho=\frac{1}{1-\alpha}$

Then $1/\rho=1-\alpha$, so $\alpha=\frac{\rho-1}{\rho}$, so $1/\alpha=\frac{\rho}{\rho-1}$ so $|1/\alpha|=|\frac{\rho}{\rho-1}|=|\frac{\rho}{1-\rho}|$ Since $f(\alpha)=0$ corresponds to $\zeta(\rho)=0$ and of course $|\alpha| <1$ corresponds to $\Re \rho > 1/2$ the result should be clear now.

As an aside, the above is part of the proof of the Balazard Saias Yor criterion (which itself is part of a large number of integral criteria equivalences to RH, but it looks simpler so it has more appeal to amateurs) to prove/disprove the Riemann Hypothesis using the formula proved in the mentioned paper:

$\int_{\Re s =1/2}\frac{\log |\zeta(s)|}{|s|^2}d|s|=2 \pi\sum_{\Re(\rho)>1/2, \zeta(\rho)=0}\log\ \left|\frac{\rho}{1-\rho}\right| $

so in particular the integral is zero iff RH is true