# Is the Balazard-Saias-Yor integral non-positive?

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

$$$$\zeta(s) = \frac{s}{s-1}-s\int_{1}^{\infty} \lbrace x \rbrace x^{-s-1} \mathrm{d}x$$$$ valid for $$\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$$) such that $$$$|\zeta(1/2 + it)| \leq \Big(\frac{1}{4}+t^2\Big)^{\alpha+\epsilon t^2}$$$$ for all $$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$$ be some real number. Consider $$$$I_{\delta} = \Bigg(\int_{\delta-\gamma_1}^{0} + \int_{0}^{\gamma_{1}-\delta} \Bigg)\frac{\log |\zeta(\frac{1}{2}+it)|}{\frac{1}{4}+t^2} \mathrm{d}t + \sum_{n=1}^{\infty} \Bigg(\int_{\delta-\gamma_{n+1}}^{-\gamma_{n}-\delta} + \int_{\gamma_{n}+ \delta}^{\gamma_{n+1}-\delta} \Bigg)\frac{\log |\zeta(\frac{1}{2}+it)|}{\frac{1}{4}+t^2} \mathrm{d}t.$$$$ Note that by our definition of $$\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)|$$ into the preceding equation and obtain $$$$I_{\delta} \leq \Bigg(\int_{\delta-\gamma_1}^{0} + \int_{0}^{\gamma_{1}-\delta} \Bigg)\frac{(\alpha + \epsilon t^2)\log(\frac{1}{4}+t^2)}{\frac{1}{4}+t^2} \mathrm{d}t + \sum_{n=1}^{\infty} \Bigg(\int_{\delta-\gamma_{n+1}}^{-\gamma_{n}-\delta} + \int_{\gamma_{n}+ \delta}^{\gamma_{n+1}-\delta} \Bigg)\frac{(\alpha + \epsilon t^2)\log(\frac{1}{4}+t^2)}{\frac{1}{4}+t^2} \mathrm{d}t$$$$ hence \begin{align} \lim_{\delta \rightarrow 0^+} I_{\delta} &\leq \Bigg(\int_{-\gamma_1}^{0} + \int_{0}^{\gamma_{1}} \Bigg)\frac{(\alpha + \epsilon t^2)\log(\frac{1}{4}+t^2)}{\frac{1}{4}+t^2} \mathrm{d}t + \sum_{n=1}^{\infty} \Bigg(\int_{-\gamma_{n}+1}^{-\gamma_{n}} + \int_{\gamma_{n}}^{\gamma_{n+1}} \Bigg)\frac{(\alpha + \epsilon t^2)\log(\frac{1}{4}+t^2)}{\frac{1}{4}+t^2} \mathrm{d}t \\ &=\int_{-\infty}^{\infty} \frac{(\alpha + \epsilon t^2) \log(\frac{1}{4}+t^2)}{\frac{1}{4}+t^2} \mathrm{d}t \\ &=0 \end{align} since $$\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.

• 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 $I = 0$ Nov 30 '18 at 18:51
• @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...
– user507152
Nov 30 '18 at 18:59
• iml.univ-mrs.fr/~balazard/pdfdjvu/9.pdf Nov 30 '18 at 19:21
• @reuns, i can't believe this, thanks !!!!
– user507152
Nov 30 '18 at 19:27
• @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.
– user507152
Nov 30 '18 at 19:48