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, $$\sum_{|\alpha|<1,f(\alpha)=0}\log \frac{1}{|\alpha|^2}=\sum_{-\pi/2<\arg(\rho)<\pi/2}\log\left|\frac{\rho}{1-\rho}\right|+ \sum_{-\pi/2<\arg(\rho)<\pi/2}\log\left|\frac{\rho}{1-\rho}\right| $$

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    $\begingroup$ Why do all these sums converge? $\endgroup$ Jun 16, 2021 at 17:13
  • $\begingroup$ @მამუკა ჯიბლაძე Because they are part of Jensen's formula and Balazard Saias and Yor has proved that they converge $\endgroup$
    – user292590
    Jun 16, 2021 at 17:17
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    $\begingroup$ Why don't you add links to these? $\endgroup$ Jun 16, 2021 at 17:24
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    $\begingroup$ Thank you, but it would be better to give this link at the appropriate place in the question - for example before the first display you could say that the right hand side of the equality absolutely converges according to them, and you ask whether the left hand side does too and gives the same value. $\endgroup$ Jun 16, 2021 at 17:43
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    $\begingroup$ I’m voting to close this question as the latest iteration of the persistent efforts of a repeat offender. $\endgroup$
    – Yemon Choi
    Jun 22, 2021 at 4:14

1 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.

  • $\begingroup$ are you saying that $$\sum_{|\alpha|<1,f(\alpha)=0}\log \frac{1}{|\alpha|^2}=\frac{1}{2}\left(\sum_{-\pi/2<\arg(\rho)<\pi/2}\log\left|\frac{\rho}{1-\rho}\right|+ \sum_{-\pi/2<\arg(\rho)<\pi/2}\log\left|\frac{\rho}{1-\rho}\right| \right) $$? $\endgroup$
    – user292590
    Jun 16, 2021 at 22:50
  • $\begingroup$ @Shyla No, your final formula is correct, it's equivalent to what is stated in the article you linked. However, the formula in the Question is not true. I mean, it's truth is actually equivalent to RH. Also it is not the formula from the article. $\endgroup$ Jun 16, 2021 at 22:55
  • $\begingroup$ How is the sum in the left , $\sum_{|\alpha|<1,f(\alpha)=0} \log\frac{1}{|\alpha|^2}$ equal to $\frac{1}{2}$ times the sum $\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| $ $\endgroup$
    – user292590
    Jun 16, 2021 at 23:01
  • $\begingroup$ Which formula is correct? $\endgroup$
    – user292590
    Jun 16, 2021 at 23:02
  • $\begingroup$ The truth of which formula is equivalent to the Riemann hypothesis? $\endgroup$
    – user292590
    Jun 16, 2021 at 23:05