Let $\chi$ be a primitive Dirichlet character $\mod q$, $q>1$. Is there a neat, simple way to give a good bound on $L'(1,\chi)/L(1,\chi)$?
Assuming no zeroes $s=\sigma+it$ of $L(s,\chi)$ satisfy $\sigma>1/2$ and $|t|\leq 5/8$ (note: much more is known for $q\leq 200000$ or so), I can give a bound of the form
$$\left|\frac{L'(1,\chi)}{L(1,\chi)}\right| \leq
\frac{5}{2} \log M(q) + 15.1$$
(constants not optimal) using Landau/Borel-Carathéodory, where $M(q) = \max_n |\sum_{m\leq n} \chi(m)|$, and then of course I can bound $M(q)$ using Pólya-Vinogradov (in its original form or one of its stronger, more recent variants), but I was wondering whether there was a simpler and/or more standard way. (Or, perhaps, who knows, even a closed expression I ought to know but don't.)
Thank you very much for all the very good answers – I've left comments below. Here is a remark making reference to the accepted answer (Lucia's).
Lucia says: "the constant $B(\chi)$ is a little tricky to bound". In fact, Lucia's answer, which avoids using $B(\chi)$, gives a very good bound on $|L'(1,\chi)/L(1,\chi)|$… and thus on $B(\chi)$. Let me explain the implication. Write $b(\chi)$ for the constant coefficient of the Laurent expansion of $L'(s,\chi)/L(s,\chi)$. Using the functional equation, one can easily prove that, for $q>1$, $$b(\chi) = \log \frac{2\pi}{q} + \gamma – \frac{L'(1,\overline{\chi})}{L(1,\overline{\chi})}.$$
It is immediate from Lucia's equation (1) and the Laurent expansion
$\Gamma'(s)/\Gamma(s) = -1/s – \gamma + (\dotsc) s$ that $$b(\chi) = – \frac{1}{2} \log \frac{q}{\pi} + \frac{\gamma}{2} + B(\chi).$$ Hence $$B(\chi) = \frac{1}{2} \log \frac{4 \pi}{q} + \frac{\gamma}{2} – \frac{L'(1,\overline{\chi})}{L(1,\overline{\chi})}.$$
Thus, Lucia's bound implies that $B(\chi)\leq \frac{3}{2} \log q$, up to a check for small $q$ (and should give $B(\chi)\leq (1+\epsilon) \log q + c_\epsilon$ with $c_\epsilon$ explicit in general. Moreover, since $L'(1,\chi)/L(1,\chi) = o(\log q)$ in reality (conditionally on GRH), it must actually be the case that $B(\chi) = (1/2 + o(1)) \log q$.
I take these bounds on $B(\chi)$ must be known?
Best Answer
Suppose that $\chi(-1)=1$ and that all non-trivial zeros $\beta+i\gamma$of $L(s,\chi)$ with $|\gamma|\le 1/2$ are on the critical line $\beta=1/2$. Recall the Hadamard factorization formula (see Davenport Chapter 12) which gives $$ \frac{L^{\prime}}{L}(s,\chi) = -\frac 12 \log \frac q\pi - \frac 12 \frac{\Gamma^{\prime}}{\Gamma}(s/2) + B(\chi) + \sum_{\rho} \Big( \frac{1}{s-\rho} +\frac{1}{\rho}\Big). \tag{1} $$ The constant $B(\chi)$ is a little tricky to bound, but its real part is well known to equal $-\sum_{\rho} \text{Re} (1/\rho)$. Thus we also have $$ \text{Re} \frac{L^{\prime}}{L}(s,\chi) = -\frac 12\log \frac{q}{\pi} -\frac 12 \frac{\Gamma^{\prime}}{\Gamma}(s/2) + \sum_{\rho} \text{Re} \Big(\frac{1}{s-\rho}\Big). \tag{2} $$
Apply (1) with $s=1$ and $s=3/2$ and subtract. This gives $$ \frac{L'}{L}(1,\chi) - \frac{L'}{L}(3/2,\chi) = \frac 12\Big(\frac{\Gamma^{\prime}}{\Gamma}(3/4) -\frac{\Gamma'}{\Gamma}(1/2) \Big) + \sum_{\rho} \frac{1/2}{(1-\rho)(3/2-\rho)}. $$ Therefore, by the triangle inequality, and a trivial bound for $|L'/L(3/2,\chi)|$ we find $$ \Big|\frac{L'}{L}(1,\chi)\Big| \le -\frac{\zeta'}{\zeta}(3/2) +\frac 12\Big| \frac{\Gamma'}{\Gamma}(3/4) -\frac{\Gamma'}{\Gamma}(1/2)\Big| +\sum_{\rho} \frac{1/2}{|(1-\rho)(3/2-\rho)|}. \tag{3} $$ By assumption $|\gamma|\le 1/2$ implies that $\beta=1/2$. This means that $|1-\rho|\ge 1/2$ always and that $$ |3/2-\rho| \le 1/2 + |1-\rho| \le 2|1-\rho|. $$ Therefore, the sum over zeros in (3) is bounded above by $$ \le \sum_{\rho} \frac{1}{|(3/2-\rho)|^2} \le 2 \sum_{\rho} \text{Re} \frac{1}{3/2-\rho} = \log \frac q{\pi} +\frac{\Gamma'}{\Gamma}(3/4) +2 \text{Re}\frac{L^{\prime}}{L}(3/2,\chi), $$ upon using (2) in the last estimate. Inserting this in (3), and again bounding $L'/L(3/2,\chi)$ trivially, we conclude that $$ \Big|\frac{L'}{L}(1,\chi)\Big| \le \log \frac{q}{\pi} -3 \frac{\zeta'}{\zeta}(3/2) + \frac{\Gamma'}{\Gamma}(3/4) + \frac 12 \Big|\frac{\Gamma'}{\Gamma}(3/4)-\frac{\Gamma'}{\Gamma}(1/2)\Big|. $$ Calculating these constants gave a bound $\le \log q + 2.75$ in this case.
The case when $\chi(-1)=-1$ is similar -- you only need to modify the $\Gamma$-factors. Obviously one can play with the argument with a different $\sigma$ than $3/2$ (chosen more or less arbitrarily). If you don't want to make an assumption on the low lying zeros, you could isolate the contribution of zeros near $1$, and then bound the rest of the zeros as above. Obviously some condition on zeros very near $1$ is necessary to give bounds for $L'/L(1,\chi)$, but as can be seen from (2), a general one-sided bound is given by $$ -\text{Re} \frac{L'}{L}(1,\chi) \le \frac 12 \log \frac{q}{\pi} +\frac 12\frac{\Gamma'}{\Gamma}(1/2). $$