Correct growth rate of logarithmic derivative of zeta, outside critical strip

Question

Let $\zeta$ be the Riemann zeta-function, and let $t> 0$. I'm interested in the growth rate of $$ \left|\frac{\zeta'}{\zeta}\left(-\frac{1}{2}+it\right)\right| $$ as $t\to\infty$. It is easy to find references in the literature to estimations like $O(\log t)$. What is the actual growth rate?

Answer 1

It is $\log t + O(1)$. Without the absolute value, $-(\zeta'/\zeta)(-1/2 + it) = \log |t| + O(1)$ as $|t| \to \infty$.

First, by taking the logarithmic derivative of the functional equation $$ \zeta(1-s) = 2(2\pi)^{-s}\Gamma(s)\cos\left(\frac{\pi}{2}s\right)\zeta(s) $$ we have $$ -\frac{\zeta'(1-s)}{\zeta(1-s)} = -\log(2\pi) + \frac{\Gamma'(s)}{\Gamma(s)} - \frac{\sin((\pi/2)s)}{\cos((\pi/2)s)}\frac{\pi}{2} + \frac{\zeta'(s)}{\zeta(s)}. $$ Let $s = 3/2 - it$: $$ -\frac{\zeta'(-1/2 + it)}{\zeta(-1/2+it)} = -\log(2\pi) + \frac{\Gamma'(3/2-it)}{\Gamma(3/2-it)} - \frac{\sin((\pi/2)(3/2-it))}{\cos((\pi/2)(3/2-it))}\frac{\pi}{2} + \frac{\zeta'(3/2-it)}{\zeta(3/2-it)}. $$ All terms on the right side, as functions of $t$, are bounded except for the $\Gamma'/\Gamma$ term. The last term on the right side bounded since $|\zeta'(s)/\zeta(s)| \leq |\zeta'(\sigma)/\zeta(\sigma)|$ for $\sigma = {\rm Re}(s) > 1$ by the Dirichlet series for $\zeta'(s)/\zeta(s)$ when ${\rm Re}(s) > 1$. The trigonometric term is bounded since on each vertical line in $\mathbf C$ we have $|\sin(\sigma + it)|, |\cos(\sigma + it)| \sim (1/2)e^{|t|}$ as $|t| \to \infty$, so $|\tan(\sigma + it)| \sim 1$ as $|t| \to \infty$. (That growth estimate on $|\sin(\sigma + it)|$ and $|\cos(\sigma + it)|$ as $|t| \to \infty$ is true not just on each vertical line in $\mathbf C$, but also in each vertical strip $a \leq \sigma \leq b$ in $\mathbf C$ and is uniform in $\sigma$, so $|\tan(\sigma + it)| \to 1$ as $|t| \to \infty$ in each vertical strip. Thus $\tan(\sigma + it)$ is bounded as $|t| \to \infty$ in each vertical strip.)

It remains to see what $\Gamma'(s)/\Gamma(s)$ looks like on the vertical line ${\rm Re}(s) = 3/2$. I'll show for $s = 3/2 - it$ that it is $\log|t| + O(1)$. The proof of the complex Stirling's formula for $\log \Gamma(s)$ in Lang's Complex Analysis (3rd edtion, Chapter XV, Sect. 2) gives an exact error term (see equation $\Gamma$ 13 on the top of p. 405) and differentiation of that formula gives a formula for $\Gamma'(s)/\Gamma(s)$ with an exact error term (see equation $\Gamma$ 15 on the top of p. 410) that implies $$ \frac{\Gamma'(s)}{\Gamma(s)} = \log s + O\left(\frac{1}{|s|}\right) $$ on the right half-plane ${\rm Re}(s) > 0$ (more precisely, on the complex plane outside of an angular sector symmetric with respect to the negative real axis, exactly the kind of region where the complex Stirling's formula is proved). Here $\log s$ means the usual logarithm defined on the complement of the real axis: if $s = re^{i\theta}$ where $-\pi < \theta < \pi$, then $\log s := \log r + i\theta$. So $\log s = \log r + O(1) = \log |s| + O(1)$. Taking $s = 3/2 - it$, $$ \frac{\Gamma'(3/2-it)}{\Gamma(3/2-it)} = \log\left|\frac{3}{2}-it\right| + O(1) $$ as $|t| \to \infty$. Since $|3/2 - it| = \sqrt{9/4 + t^2} \sim |t|$ as $|t| \to \infty$, $\log|3/2 - it| = \log |t| + o(1)$, so $$ \frac{\Gamma'(3/2-it)}{\Gamma(3/2-it)} = \log|t| + O(1) $$ as $|t| \to \infty$. Thus $-(\zeta'/\zeta)(-1/2 + it) = \log|t| + O(1)$ as $|t| \to \infty$.

The same reasoning shows in each vertical strip $a \leq \sigma \leq b < 0$ to the left of the imaginary axis, $-(\zeta'/\zeta)(\sigma + it) = \log |t| + O(1)$ as $|t| \to \infty$, uniformly in $\sigma$.

More generally, suppose $L(s)$ is an Euler product of degree $d$ with an absolutely convergent Dirichlet series for ${\rm Re}(s) > 1$ and $\Lambda(s) := cA^{s/2}\prod_{j=1}^d \Gamma((\mu_j/2)s + \nu_j/2)L(s)$ has a meromorphic continuation to $\mathbf C$ that satisfies a functional equation $\Lambda(s) = w\overline{\Lambda(1-\overline{s})}$, where $c \not= 0$, $A > 0$, $\mu_j > 0$, and $w \not= 0$. Then for each $\sigma < 0$ (or in fact uniformly in each vertical strip to the left of the imaginary axis), we have $-(L'/L)(\sigma + it) = (\sum_{j=1}^d \mu_j)\log |t| + O(1)$ as $|t| \to \infty$. (In the terminology of the Selberg class, $\sum_{j=1}^d \mu_j$ is the "degree" of $L(s)$. In practice each $\mu_j$ is $1$, so that sum is $d$.) The $\Gamma$-factor in the definition of $\Lambda(s)$ can be written in many ways by identities for the $\Gamma$-function, so no individual $\mu_j$ is uniquely determined, but the estimate on $-(L'/L)(s)$ on vertical lines in the left half-plane, coming from a functional equation, shows $\sum_{j=1}^d \mu_j$ is well-defined.