Define $$I= \int_{-\infty}^{\infty} \frac{\log|\zeta(\frac{1}{2}+it)|}{\frac{1}{4}+t^2} \mathrm{d}t.$$ Balazard, Saias and Yor showed that the Riemann Hypothesis is equivalent to the statement that $I=0$.

I want to prove that I is convergent.

My try:-

$$\zeta(s)= s\int_1^\infty \frac{1-x+[x]}{x^{s+1}}dx +\frac{1}{s-1}, \Re(s)>0$$

So $$\mid \zeta(1/2+it)|\leq 2\sqrt{1/4+t^2}+\frac{1}{\sqrt{1/4+t^2}}$$

$$\mid \zeta(1/2+it)|\leq \frac{2(1/4+t^2)+1}{\sqrt{1/4+t^2}}$$

$$log \mid \zeta(1/2+it)\mid \leq log(\frac{3/2+t^2}{\sqrt{1/4+t^2}})$$

$$\int_{-\infty}^{\infty}\frac{log \mid \zeta(1/2+it)\mid}{1/4+t^2}dt \leq \int_{-\infty}^{\infty} \frac{log(\frac{3/2+t^2}{\sqrt{1/4+t^2}})}{1/4+t^2}dt$$ Since $$\int_{-\infty}^{\infty}\frac{ log (1/4+t^2)}{1/4+t^2}=0$$ $$\int_{-\infty}^{\infty}\frac{log \mid \zeta(1/2+it)\mid}{1/4+t^2}dt \leq \int_{-\infty}^{\infty} \frac{log({3/2+t^2})}{1/4+t^2}dt$$

The last integral converges see this link.


1 Answer 1


Note first that the OP proof above actually shows that $I$ cannot be $\infty$ but it doesn't show that $I$ converges or it is not $-\infty$ since $I$ is an oscillating integral with $\log |\zeta|=-\infty$ infinitely many times on the critical line of course.

From general results about Hardy spaces and some changes of variable, it actually follows immediately that $I$ is finite but much more is known, namely that

$0 \le I \le 0.146$

(and much tighter bounds can be easily found by better numerics) as from the results of Balazard, Saias and Yor we know that:

$I=2\pi \sum_{\Re \rho >1/2}\log \frac{|\rho|}{|1-\rho|}=\pi \sum_{\Re \rho >1/2}\log \frac{|\rho|^2}{|1-\rho|^2}$

But now if $\rho=\sigma+it, \frac{|\rho|^2}{|1-\rho|^2}=\frac{\sigma^2+t^2}{(1-\sigma)^2+t^2}=1+\frac{\sigma^2-(1-\sigma)^2}{(1-\sigma)^2+t^2}$, so we have

$ 1 < \frac{|\rho|^2}{|1-\rho|^2} < 1+\frac{1}{(1-\sigma)^2+t^2}$ whenever $\sigma >1/2$ hence $0 < \log \frac{|\rho|^2}{|1-\rho|^2} < \frac{1}{(1-\sigma)^2+t^2}=\frac{1}{|1-\rho|^2}$

But now it is well known (see Broughan, Equivalents if The Riemann Hypothesis, vol 1, page 35) that: $\sum_{\rho} \frac{1}{|\rho|^2} \le .046191441$, where the sum is taken on all the non-trivial zeroes of zeta.

Since either $I=0$ if RH is true, or $0 <I < \pi \sum_{\Re \rho <1/2}\frac{1}{|1-\rho|^2}< \pi \times .046191441=0.145...$ if RH is false and the sum is non-void, we immediately see the claimed result; using that all the zeroes up to some huge $t$ are known to be on the critical line so not in the sum above, we can probably reduce the estimate of $I$ to a ridiculously low nonnegative number, showing the tightness of the equivalence.

  • $\begingroup$ Can you please give me a reference which shows that $ I \geq 0?$ $\endgroup$
    – user847497
    Dec 18, 2020 at 13:50
  • $\begingroup$ $I$ is either a void sum or a sum of positive numbers from the Balazard et others result since $\frac{|\rho|}{|1-\rho|}>1$ whenever $\Re \rho >1/2$ hence $\log \frac{|\rho|}{|1-\rho|} >0$ then $\endgroup$
    – Conrad
    Dec 18, 2020 at 13:54
  • $\begingroup$ How is it a void sum? $\endgroup$
    – user847497
    Dec 18, 2020 at 13:55
  • 1
    $\begingroup$ an infinite product that "converges" to zero is called divergent to zero for various reasons (essentially $0$ has bad multiplicative behavior), so the terminology can be a bit misleading but it is customary $\endgroup$
    – Conrad
    Dec 18, 2020 at 14:30
  • 1
    $\begingroup$ $I$ cannot be negative, it is either zero or positive $\endgroup$
    – Conrad
    Dec 18, 2020 at 14:39

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