According to a conjecture p.4 $|\zeta(\frac12 -\Delta + it))| > |\zeta(\frac12 + \Delta + i t|$ for $0 < \Delta < \frac12$ and $|t| > 2 \pi +1$. Since $\zeta(\overline{s}) = \overline{\zeta(s)}$, this is equvalent to $|\zeta(s)| > |\zeta(1-s)| $ for $0 < \sigma < \frac12$ and $t$ large enough.

Set $$ F(s) = {\frac {\Gamma \left( 1/2-1/2\,s \right) {\pi }^{-1/2+1/2\,s}{\pi }^{ 1/2\,s}}{\Gamma \left( 1/2\,s \right) }}$$

Then from the functional equation $$\frac{\zeta(s)}{\zeta(1-s)} = F(s) $$ and

$$\frac{|\zeta(s)|}{|\zeta(1-s)|} = |F(s)| $$

$|F(s)| > 1$ for $0 < \sigma < \frac12$ and $t$ large enough would imply the conjecture unless $s$ is a zero of zeta off the critical line.

In Maple 13 using with(MultiSeries); and assuming $0 < \sigma < 1/2$ we get:

$$ \lim_{t \to \infty} |F(\sigma+ it)| = \infty , \, 0 < \sigma < 1/2 $$

Looks like if Maple's result is correct this would mean the conjecture is true at infinity, unless $s$ is a zero off the critical line.

Is Maple's result true?

Proof that $|F(s)| > 1$ for $t$ large enough (except at zeros)?


For fixed $t>12$, let us consider for $0\leq\sigma\leq \frac{1}{2}$ the function

$$ G(\sigma):=|\pi^{-(\sigma-it)/2}\Gamma(\sigma+it)|^2 = \pi^{-\sigma}|\Gamma(\sigma+it)|^2. $$

Following the accepted answer here, we see that

$$ \frac{d}{d\sigma}\log G(\sigma)=-\sigma\log\pi + \psi(\sigma+it) + \psi(\sigma-it)$$

$$\geq -\frac{1}{2}\log\pi + 2(1 - \gamma) - \sum_{n=1}^{\infty} \frac{2}{n^2 + t^2} > 0.27-\frac{\pi}{t}>0,$$

whence $G(\sigma)$ is increasing on $[0,\frac{1}{2}]$.

It follows that $|F(s)|>1$ for $0<\Re(s)<\frac{1}{2}$ and $|\Im s|>24$.

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  • $\begingroup$ Thanks GH. Your answer means the conjecture about the inequality is true in the relevant ranges except at hypothetical zeros off the critical line? $\endgroup$ – joro Jul 16 '13 at 5:03
  • $\begingroup$ @joro: Yes, I think so. $\endgroup$ – GH from MO Jul 16 '13 at 16:51

GH has answered your question, but here is some additional remarks.

The function you are calling $F(s)$ is often called $\chi(s)$ in the theory of the Riemann zeta-function (e.g. in Titchmarsh's book The Theory of the Riemann Zeta-Function). Standard asymptotic estimates for the gamma function imply that $$ \chi(s) = \Big( \frac{2\pi}{t} \Big)^{\sigma+it-1/2} e^{i(t+\pi/4)} \Big\{ 1 + O\Big(\frac{1}{t}\Big) \Big\} $$ in any fixed strip $\alpha \le \sigma \le \beta$ as $t \to \infty$ (for $s=\sigma+it$), which also answers your question -- though not as precisely as in GH's answer.

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  • $\begingroup$ Actually this formula does not tell us what happens when $\sigma$ is very close to $\frac{1}{2}$, this is why I used the approach above. $\endgroup$ – GH from MO Jul 15 '13 at 15:49

In fact, |F(s)|>1 is more than a conjecture, and it is always satisfied, regardless of whether or not there exist non-trivial zeros off the critical line (even in case there exist some, those would result in removable singularities). In this arxiv.org/pdf/0907.2426v5.pdf I have been studying a ratio strictly related to the inverse of your F(s). Said ratio is taken between the corresponding Dirichlet eta functions. It is graphically represented in Fig. 2, which also shows the upper and lower bound "sheets".

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