I am currently reading an article that claims that the value $$\inf \{ a \in \mathbb{R}, |\zeta(\sigma + it)| = O(|t|^{a}) \}$$ is zero if $\sigma > 1$ and is $\frac{1}{2}-\sigma$ if $\sigma < 0$. I understand that the first statement is true because $$\zeta(\sigma + it)$$ is bounded for $\sigma > 1$. The claim that this value is $\frac{1}{2} - \sigma$ should follow from the functional equation $$\zeta(\sigma + it) = 2^{\sigma + it}\pi^{\sigma -1 + it} \sin(\frac{\pi(\sigma + it)}{2})\Gamma((1-\sigma)- it)\zeta((1-\sigma) - it)$$ When $\sigma < 0$, all of the above factors should be bounded from above, with the exception of $\Gamma((1-\sigma)-it)$. According to the discussion here, "Tipping point" between asymptotic behavior of gamma function along the line $z=x+mxi$, $\Gamma((1-\sigma) - it) = O(|t|^{1-\sigma - \frac{1}{2}}e^{-t})=O(|t|^{\frac{1}{2}-\sigma}e^{-t})$

However, this should mean that $\Gamma((1-\sigma) -it) = O(|t|^{b})$ for all $b \in \mathbb{R}$. This should in turn mean, from the functional equation, that $\inf \{ a \in \mathbb{R}, \zeta(\sigma + it) = O(|t|^{a}) \} = -\infty$ if $\sigma < 0$, contradicting the claim that it was $\frac{1}{2}-\sigma$. The article said it was important in Lindelorf’s hypothesis, so I want to make sure that this is correct.

Does it have to do with the sine factor that will cancel out the $e^{-|t|}$ for large $t$? Any advice on this?

  • 2
    $\begingroup$ yes of course the $\sin$ behaves like an exponential on the imaginary axis so it will cancel the exponential decay of $\Gamma$ precisely; you should look in the Titchmarsh bible for the precise asymptotic of $\chi$ the balancing factor in $\zeta(s)=\chi(s)\zeta(1-s)$ $\endgroup$
    – Conrad
    Commented Nov 7, 2023 at 1:38
  • $\begingroup$ What is the Titchmarsh bible? $\endgroup$ Commented Nov 7, 2023 at 1:58
  • $\begingroup$ Titchmarsh The Theory of the Riemann Zeta Function $\endgroup$
    – Conrad
    Commented Nov 7, 2023 at 2:25
  • $\begingroup$ Also, one other question. How do we know that $|t|^{a}\zeta(\sigma + it)$ is unbounded for large $|t|$ and $a>0$. This way we can conclude that $\zeta(\sigma + it) \neq O(\frac{1}{|t|^{a}})$ $\endgroup$ Commented Nov 7, 2023 at 3:47
  • $\begingroup$ not sure exactly what you ask - we know precise asymptotics for $\chi$ as for example $|\chi(\sigma+it)| \sim |t|^{1/2-\sigma}$ (and we know asymptotics for the imaginary part btw) so when $a \le 0$ we have $|\zeta(a+it)/\zeta(1-a+it)|\sim |t|^{1/2-a}$ and we know the distribution of $|\zeta(1+\epsilon +it)|$ (it goes quasi periodically from $(0, \zeta(1+\epsilon)]$ if $\epsilon$ is small enough and from $(2-\zeta(1+\epsilon), \zeta(1+\epsilon]$ if $\epsilon$ larger than the value where $2-\zeta(1+\epsilon)=0$, while on the line $1$ we know that $\zeta$ is unbounded, so the result follows $\endgroup$
    – Conrad
    Commented Nov 7, 2023 at 4:07

1 Answer 1


When $x$ lies in a fixed interval and $y\to+\infty$, Stirling's approximation gives

$$ \log\Gamma(x+iy)=\left(x+iy-\frac12\right)\log(iy)-iy+\frac12\log2\pi+O(y^{-1}), $$

so taking real components gives

$$ \log|\Gamma(x+iy)|=\left(x-\frac12\right)\log y-{\pi y\over2}+\frac12\log2\pi+O(y^{-1}). $$

Exponentiating gives

$$ |\Gamma(x+iy)|\sim y^{x-\frac12}e^{-\pi y/2}. $$

Because $\Gamma(\overline z)=\overline{\Gamma(z)}$, we have the following general asymptotic formula:

$$ |\Gamma(x+iy)|\sim|y|^{x-\frac12}e^{-\pi|y|/2},\quad(y\to\pm\infty).\tag1 $$

From Euler's formula, we have for $x$ lying in a fixed interval and $y\to+\infty$ that

$$ \sin(x+iy)={1\over2i}(e^{ix-y}-e^{-ix+y})=-{e^{-ix+y}\over2i}+O(e^{-y}), $$

so generalizing gives

$$ |\sin(x+iy)|\sim\frac12e^y,\quad(y\to\pm\infty).\tag2 $$

By the functional equation, we have for $s=\sigma+it$ that

$$ \zeta(s)=2(2\pi)^{s-1}\Gamma(1-s)\sin\left(\pi s\over2\right)\zeta(1-s). $$

For fixed $\sigma<0$ and $t\to\pm\infty$, (1) and (2) gives

\begin{aligned} |\zeta(\sigma+it)| &\sim2(2\pi)^{\sigma-1}|t|^{\frac12-\sigma}e^{-\pi|t|/2}\cdot e^{\pi|t|/2}|\zeta(1-\sigma-it)| \\ &=2(2\pi)^{\sigma-1}|t|^{\frac12-\sigma}\left|\sum_{n=1}^\infty n^{\sigma+it-1}\right| \\ &\le=2(2\pi)^{\sigma-1}|t|^{\frac12-\sigma}\sum_{n=1}^\infty n^{\sigma-1}=O(|t|^{\frac12-\sigma}). \end{aligned}

  • $\begingroup$ I think inside the sum in line two at the end, it should be $$|\sum_{n = 1}^{\infty} n^{(1-\sigma) -it}| $$ $\endgroup$ Commented Nov 8, 2023 at 3:29
  • $\begingroup$ I mean because it is $|\zeta((1-\sigma)-it)|$ $\endgroup$ Commented Nov 8, 2023 at 3:54
  • $\begingroup$ And the upper bound should be $\sum_{n = 1}^{\infty} n^{1-\sigma}$. $\endgroup$ Commented Nov 8, 2023 at 3:56
  • $\begingroup$ $\zeta(s)=\sum_{n=1}^\infty n^{-s}$ when $\Re(s)>1$ $\endgroup$
    – TravorLZH
    Commented Nov 8, 2023 at 13:32
  • $\begingroup$ okay. I see. I was thinking the exponents were positive. $\endgroup$ Commented Nov 8, 2023 at 18:53

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