Non trivial zeros of Riemann zeta function [closed]

Question

Question Define $f(z)=(s-1)\zeta(s)$ where $s=\frac{1}{1+z^2}$ and $\zeta$ denotes the Riemann zeta function. Prove that if $\rho$ denotes the non trivial zeros of $\zeta(s)$ then, $$\sum_{|\alpha|<1,f(\alpha)=0}\log \frac{1}{|\alpha|^2}=\sum_{\Re(\rho)>1/2} \log\left|\frac{\rho}{1-\rho}\right|$$ I am reading a paper by Balazard et al. on the zeta function where both sums converge.

My try- $\rho=\frac{1}{1+\alpha^2}$ then $\alpha^2=\frac{1-\rho}{\rho}$ so that $\alpha=\pm \sqrt{\frac{1-\rho}{\rho}}$ $$\sum_{|\alpha|<1,f(\alpha)=0}\log \frac{1}{|\alpha|^2}=\sum_{-\pi<\arg(\alpha)\leq \pi,|\alpha|<1,f(\alpha)=0}\log \frac{1}{|\alpha|^2} $$ Since the sum on the right hand side is absolutely convergent so we can write the sum in any order.$$\sum_{|\alpha|<1,f(\alpha)=0}\log \frac{1}{|\alpha|^2}=\sum_{-\pi<\arg(\alpha)\leq 0}\log \frac{1}{|\alpha|^2}+ \sum_{0<\arg(\alpha)\leq \pi}\log \frac{1}{|\alpha|^2} $$ $\rho=\frac{1}{1+\alpha^2}$ is injective on $-\pi<\arg(\alpha)\leq 0$ and also it is injective on $0<\arg(\alpha)\leq \pi$. So using $\rho=\frac{1}{1+\alpha^2}$ we get, $$\sum_{|\alpha|<1,f(\alpha)=0}\log \frac{1}{|\alpha|^2}=\sum_{-\pi/2<\arg(\rho)<\pi/2}\log\left|\frac{\rho}{1-\rho}\right|+ \sum_{-\pi/2<\arg(\rho)<\pi/2}\log\left|\frac{\rho}{1-\rho}\right| $$

Answer 1

Authors of the paper you linked actually define $f(z)$ differently. They have $$ f(z)=\left(\frac{1}{1-z}-1\right)\zeta\left(\frac{1}{1-z}\right), $$ so your $f(z)$ is their $f(-z^2)$ and every $\alpha$ in their formula corresponds to $\pm i\alpha$ from yours, so the sum on the left should actually be $\frac12$ of what is in the question, so $$ \frac{1}{2}\left(\sum_{-\pi/2<\arg(\rho)<\pi/2}\log\left|\frac{\rho}{1-\rho}\right|+ \sum_{-\pi/2<\arg(\rho)<\pi/2}\log\left|\frac{\rho}{1-\rho}\right|\right)=\sum_{-\pi/2<\arg(\rho)<\pi/2}\log\left|\frac{\rho}{1-\rho}\right| $$ and everything works just fine, no contradiction.

P.S. And yeah, unfortunately the Riemann hypothesis is still hard.

Edit for clarity: By "should actually be" I mean that to replicate the result of Balazard-Saias-Yor you should take another function. To elaborate further, let $$ f_S(z)=\left(\frac{1}{z^2+1}-1\right)\zeta\left(\frac{1}{z^2+1}\right) $$ be the Shyla's function, $$ f_{BSY}(z)=\left(\frac{1}{1-z}-1\right)\zeta\left(\frac{1}{1-z}\right) $$ be the Balazard-Saias-Yor function and three constants $A,B,C$ be $$ A=\sum_{\substack{|\alpha|<1\\ f_S(\alpha)=0}}\log\frac{1}{|\alpha|^2}, $$

$$ B=\sum_{\substack{|\alpha|<1\\ f_{BSY}(\alpha)=0}}\log\frac{1}{|\alpha|} $$ and $$ C=\sum_{-\pi/2<\mathrm{arg}\,\rho<\pi/2}\log\left|\frac{\rho}{1-\rho}\right|. $$ Then the proof of Shyla shows that $A=2C$, which is clearly equivalent to Balazard-Saias-Yor paper's $B=C$ (both are true), while the formula in question is $A=C$, which is equivalent to $C=0$, which is easily seen to be the same as the Riemann hypothesis, becuase then the sum in the definition should necessarily be empty.