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 $I=0$. It seems to me that $I\leq 0$. Is this result known or significant ? Because it appears to be quite trivial to me.

Indeed, from

\begin{equation} \zeta(s) = \frac{s}{s-1}-s\int_{1}^{\infty} \lbrace x \rbrace x^{-s-1} \mathrm{d}x \end{equation} valid for $\Re(s)>0$ where $\lbrace y \rbrace$ denotes the fractional part of $y$, one easily finds that for every $\epsilon>0$, there exists some constant $\alpha$ (dependent on $\epsilon$) such that \begin{equation} |\zeta(1/2 + it)| \leq \Big(\frac{1}{4}+t^2\Big)^{\alpha+\epsilon t^2} \end{equation} for all $t\in \mathbb{R}$. Define $0<\gamma_1 < \gamma_2<\gamma_3 < \cdots $ to be the infinitely many positive zeros of $\zeta(s)$ on the line $\Re(s)=\frac{1}{2}$ and let $\delta>0$ be some real number. Consider \begin{equation} I_{\delta} = \Bigg(\int_{\delta-\gamma_1}^{0} + \int_{0}^{\gamma_{1}-\delta} \Bigg)\frac{\log |\zeta(\frac{1}{2}+it)|}{\frac{1}{4}+t^2} \mathrm{d}t + \sum_{n=1}^{\infty} \Bigg(\int_{\delta-\gamma_{n+1}}^{-\gamma_{n}-\delta} + \int_{\gamma_{n}+ \delta}^{\gamma_{n+1}-\delta} \Bigg)\frac{\log |\zeta(\frac{1}{2}+it)|}{\frac{1}{4}+t^2} \mathrm{d}t. \end{equation} Note that by our definition of $\gamma_n$, $\log|\zeta(1/2 + it)|$ is well-defined on $[0, \gamma_{1}-\delta], [\delta-\gamma_{1},0], [\gamma_{n} + \delta, \gamma_{n+1}-\delta]$ and $[\delta-\gamma_{n+1}, -\gamma_{n}-\delta]$ for every positive integer $n$, thus we can insert the inequality for $|\zeta(1/2 + it)|$ into the preceding equation and obtain \begin{equation} I_{\delta} \leq \Bigg(\int_{\delta-\gamma_1}^{0} + \int_{0}^{\gamma_{1}-\delta} \Bigg)\frac{(\alpha + \epsilon t^2)\log(\frac{1}{4}+t^2)}{\frac{1}{4}+t^2} \mathrm{d}t + \sum_{n=1}^{\infty} \Bigg(\int_{\delta-\gamma_{n+1}}^{-\gamma_{n}-\delta} + \int_{\gamma_{n}+ \delta}^{\gamma_{n+1}-\delta} \Bigg)\frac{(\alpha + \epsilon t^2)\log(\frac{1}{4}+t^2)}{\frac{1}{4}+t^2} \mathrm{d}t \end{equation} hence \begin{align} \lim_{\delta \rightarrow 0^+} I_{\delta} &\leq \Bigg(\int_{-\gamma_1}^{0} + \int_{0}^{\gamma_{1}} \Bigg)\frac{(\alpha + \epsilon t^2)\log(\frac{1}{4}+t^2)}{\frac{1}{4}+t^2} \mathrm{d}t + \sum_{n=1}^{\infty} \Bigg(\int_{-\gamma_{n}+1}^{-\gamma_{n}} + \int_{\gamma_{n}}^{\gamma_{n+1}} \Bigg)\frac{(\alpha + \epsilon t^2)\log(\frac{1}{4}+t^2)}{\frac{1}{4}+t^2} \mathrm{d}t \\ &=\int_{-\infty}^{\infty} \frac{(\alpha + \epsilon t^2) \log(\frac{1}{4}+t^2)}{\frac{1}{4}+t^2} \mathrm{d}t \\ &=0 \end{align} since $\epsilon$ is arbitrary and $\int_{-\infty}^{\infty}\frac{\log(\frac{1}{4}+t^2)}{\frac{1}{4}+t^2} \mathrm{d}t=0$. Notice that from our earlier expression for $I_\delta$ that we have $\lim_{\delta \rightarrow 0^+} I_{\delta} = \int_{-\infty}^{\infty} \frac{\log |\zeta(\frac{1}{2}+it)|}{\frac{1}{4}+t^2} \mathrm{d}t$, thus from from the last inequality we deduce that $\int_{-\infty}^{\infty} \frac{\log |\zeta(\frac{1}{2}+it)|}{\frac{1}{4}+t^2} \mathrm{d}t \leq 0$, as earlier mentioned.

  • $\begingroup$ The integral is a sum over the zeros on $\Re(s) > 1/2$ with each term of the same sign and non-zero. That's why the RH is true iff $ I = 0$ $\endgroup$ – reuns Nov 30 '18 at 18:51
  • $\begingroup$ @Reuns, you're saying ''the integral is a sum over the zeros on $\Re(s)>1/2$ with each term of the same sign.'' Which sign is this ? Positive or negative ? Because if it's positive, then notice that the result $I\leq 0$ would prove the RH... $\endgroup$ – user507152 Nov 30 '18 at 18:59
  • $\begingroup$ iml.univ-mrs.fr/~balazard/pdfdjvu/9.pdf $\endgroup$ – reuns Nov 30 '18 at 19:21
  • $\begingroup$ @reuns, i can't believe this, thanks !!!! $\endgroup$ – user507152 Nov 30 '18 at 19:27
  • $\begingroup$ @reuns, in other words you're saying the RH is equivalent to the statement that $\int_{-\infty}^{\infty} \frac{\log(0.25 + t^2)}{0.25 + t^2} \mathrm{d}t=0$. This is elementary and is known independently of the RH. See Wolfram Alpha. $\endgroup$ – user507152 Nov 30 '18 at 19:48

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