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Authors of the paper you linked actually define $f(z)$ differently. They have $$ f(z)=\left(\frac{1}{1-z}-1\right)\zeta\left(\frac{1}{z-1}\right), $$$$ f(z)=\left(\frac{1}{1-z}-1\right)\zeta\left(\frac{1}{1-z}\right), $$ 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.

Authors of the paper you linked actually define $f(z)$ differently. They have $$ f(z)=\left(\frac{1}{1-z}-1\right)\zeta\left(\frac{1}{z-1}\right), $$ 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.

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.

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Authors of the paper you linked actually define $f(z)$ differently. They have $$ f(z)=\left(\frac{1}{1-z}-1\right)\zeta\left(\frac{1}{z-1}\right), $$ so your $f(z)$ is their $f(-z^2)$ and every $\alpha$ in their formula correspondcorresponds 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.

Authors of the paper you linked actually define $f(z)$ differently. They have $$ f(z)=\left(\frac{1}{1-z}-1\right)\zeta\left(\frac{1}{z-1}\right), $$ so your $f(z)$ is their $f(-z^2)$ and every $\alpha$ in their formula correspond 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.

Authors of the paper you linked actually define $f(z)$ differently. They have $$ f(z)=\left(\frac{1}{1-z}-1\right)\zeta\left(\frac{1}{z-1}\right), $$ 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.

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Authors of the paper you linked actually define $f(z)$ differently. They have $$ f(z)=\left(\frac{1}{1-z}-1\right)\zeta\left(\frac{1}{z-1}\right), $$ so your $f(z)$ is their $f(z^2)$$f(-z^2)$ and every $\alpha$ in their formula correspond to $\pm\alpha$$\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.

Authors of the paper you linked actually define $f(z)$ differently. They have $$ f(z)=\left(\frac{1}{1-z}-1\right)\zeta\left(\frac{1}{z-1}\right), $$ so your $f(z)$ is their $f(z^2)$ and every $\alpha$ in their formula correspond to $\pm\alpha$ 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.

Authors of the paper you linked actually define $f(z)$ differently. They have $$ f(z)=\left(\frac{1}{1-z}-1\right)\zeta\left(\frac{1}{z-1}\right), $$ so your $f(z)$ is their $f(-z^2)$ and every $\alpha$ in their formula correspond 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.

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