# $\int_{-\infty}^{\infty} \frac{\log|\zeta(\frac{1}{2}+it)|}{\frac{1}{4}+t^2} \mathrm{d}t$ is Convergent

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.

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 $$0 if RH is false and the sum is non-void, we immediately see the claimed result; using that all the zeroes up to some huge $$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.

• Can you please give me a reference which shows that $I \geq 0?$
– user847497
Dec 18, 2020 at 13:50
• $I$ 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 Dec 18, 2020 at 13:54
• How is it a void sum?
– user847497
Dec 18, 2020 at 13:55
• 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 Dec 18, 2020 at 14:30
• $I$ cannot be negative, it is either zero or positive Dec 18, 2020 at 14:39