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?
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}