$|L'(1,\chi)/L(1,\chi)|$

Question

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.)

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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?

Answer 1

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 $$ -\text{Re} \frac{L'}{L}(1,\chi) \le \frac 12 \log \frac{q}{\pi} +\frac 12\frac{\Gamma'}{\Gamma}(1/2). $$

Answer 2

You may use the local method of Landau with some bounds for L(s,chi) (expressing L'/L in terms of the local zeros, the approximation being controlled by an upper bound for |L(s,chi)| in a slightly larger region). Then, the convexity bound, as given by Rademacher for instance, should give you about what you get here. I would say, this would remove the +15.1, but both results should be rather close.

Tim (Trudgian) is the one who has the most precise explicit Landau formula (well, this is a combination of Caratheodory's inequality [bounds for Log f] with Koebe's inequality for derivatives).

That's all I have in my purse! Pintz-II (series of papers "Elementary methods in L-functions theory"] 1976 may be a good read.

Best, Olivier

Answer 3

Few remarks I already wrote privately to the OP.

First, my joint paper with Alessandro Zaccagnini mentioned by Pace Nielsen was published in a slightly different form in 2009 on Experimental Mathematics 19 (3), 279-284 (with an interesting final section by Karl K. Norton). The goal there was to get an extremely good estimate (at least 100 digits) for the Meissel-Mertens constants in arithmetic progressions p\equiv a \mod q, for every (a,q)=1 and q up to a certain bound (q\le 100, in this case). The results up to 20 digits and for every q\le 300 are available at my web page dedicated to this paper (the link is the one written in the paper). Please remark that such computations were performed about ten years ago and it is clear that now we can beat such results using the same algorithm on the more efficient PCs we have now at our disposal. Possible improvements there can be obtained using the more efficient implementations we have now to compute the needed values of Dirichlet L- functions at positive integers (for example the one we have now in Pari/gp). All these remarks essentially apply also to the computation performed for another paper of Zaccagnini and myself on the Mertens’ product in arithmetic progressions published in 2007 on Mathematics of Computation 78 (265), 315-326.

Second: I think that the faster approach we have to compute the value at 1 of the logarithmic derivative of Dirichlet L-functions mod q is, at least for q prime, the one described in a paper by Ford, Luca and Moree and improved in my arxiv preprint https://arxiv.org/abs/1903.05487 (see the references there for the paper of Ford, Luca and Moree). In such a preprint of mine I also tried to analyze the cost of directly compute the mentioned quantity using the Pari/gp function lfun; it seems that this approach is slower if compared with the other ones presented there (I never tried to compare the speediness of ARB and Pari/gp, though). The algorithms used to compute the log-derivative at 1 are based on classical formulae that link L(1,\chi) and L’(1, chi) to some special functions evaluated at some rational numbers in (0,1). For more details please refer to my preprint mentioned above. Final remark on this point: I have already precomputed and stored the special functions values needed to get the Euler-Kronecker constants for every q prime up to 5*10^5 and I am now performing the needed computations for reaching 10^6. This means that with suitable modifications to my programs I can get the data about the logarithmic derivative evaluated at 1 for every Dirichlet character mod q, q prime up to 5*10^5, by now, and in a month or so (I hope...) for every prime q up to 10^6. This clearly means to divert some computational power (and time to write and test the new programs) from the project I am now working on; if there is no hurry in getting such data, I will insert them in the next arxiv version of my work on the Euler-Kronecker constants (hoping that such a version will be the final one...)

Third: about computing the values of L(1,\chi), chi mod q, q prime: a similar approach used in my preprint mentioned in the second item of this list was used for performing the needed computation in an ongoing research project I am now working on (it is a collaboration with Pieter Moree, Sumaia Saad Eddin ad Alisa Sedunova). It is about the Kummer ratio for the class number of cyclotomic fields (please see our preprint https://arxiv.org/abs/1908.01152). Directly using the Pari/gp function lfun seems to be slower than computing L(1,\chi) using some classical formulae that make use of some special functions values. In this case too I did not compare the speediness of Pari/gp and ARB. From a computational point of view, this problem is simpler than the one on the Euler-Kronecker constants because the special function involved in the needed formulae here is the Psi function and, using a decimation in frequency strategy, we can in fact use the values of the cotangent function. An alternative algorithm uses the first chi-Bernoulli number instead. Again, please refer to our preprint mentioned above for more details.

Answer 4

One way to improve the explicit bound mentioned in the question is simply to compute $L'(1,\chi)/L(1,\chi)$ for whatever characters $\chi$ are needed. The bound in the question depends on a GRH verification anyhow (up to a trivial height), and, as was correctly pointed out in the comments, you can't hope for a good two-sided bound for $\chi$ quadratic (good = better than $O(\sqrt{q})$ or so) without checking that there is no exceptional zero. Thus, there is no avoiding calculations, so we might as well try a direct one.

We have, with non-absolute convergence, $$L(1,\chi) = \sum_n \frac{\chi(n)}{n},\;\;\;\; L'(1,\chi) = - \sum_n (\log n) \frac{\chi(n)}{n}.$$ Let me focus on the first sum; it should be possible to deal with the second one in much the same way.

Define $c_{a,q}$ by

$$\mathop{\sum_{n\leq N}}_{n\equiv a\mod q} \frac{1}{n} = \frac{\log N/q}{q} + c_{a,q} + o(1).$$

Then, for $\chi$ non-principal,

$$\begin{aligned}L(1,\chi) &= \sum_n \frac{\chi(n)}{n} = \lim_{N\to \infty} \sum_{n\leq N} \frac{\chi(n)}{n}\\ &= \lim_{N\to \infty} \sum_{a \mod q} \chi(a) \mathop{\sum_{n\leq N}}_{n\equiv a \mod q} \frac{1}{n} = \sum_{a \mod q} \chi(a) c_{a,q}. \end{aligned}$$

It is clear that $c_{0,q} = \gamma$. For $a\not\equiv 0 \mod q$, $$\mathop{\sum_{n\leq N}}_{n\equiv a \mod q} \frac{1}{n} = \frac{1}{q} \sum_{0\leq n\leq N/q} \frac{1}{n + a/q}$$

We can then use Euler-Maclaurin to compute $c_{a,q}$ to any level of accuracy in time logarithmic on the size of the tolerated error $\epsilon$. (Consider the terms with $n\leq \log(1/\epsilon)$ (say) apart from the rest.) The total time taken for all $a \mod q$ is thus about $O(q)$ (for constant tolerance). We then use FFT to compute $L(1,\chi)$ for all $\chi \mod q$ in time $O(\phi(q) \log \phi(q))$.

We compute $L'(1,\chi)$ similarly, and thus obtain $L'(1,\chi)/L(1,\chi)$ for all $\chi \mod q$ in time $O(q \log q)$. Hence, doing all $q\leq 10^6$ would seem to be very much within the range of semi-amateur programming -- it is surprising it hasn't been done. Or am I missing something?

Answer 5

Just a remark on Lucia's answer, generalizing it and improving it a little.

As Lucia says, one choose an arbitrary $\sigma>1$ instead of $3/2$. What is more, one can halve the coefficient of $q$ for a given $\sigma$ by taking into the account the symmetry $\sigma \mapsto 1-\overline{\sigma}$ of the roots of $L(s,\chi)$ and doing a little additional work. Thus instance, for $\sigma=3/2$, one gets $$\left|\frac{L'(1,\chi)}{L(1,\chi)}\right| \leq \frac{1}{2} \log q + \begin{cases} 2.334 &\text{if $\chi(-1)=1$,}\\ 2.5 &\text{if $\chi(-1)=-1$,}\end{cases}$$ under the assumption that all zeros $\rho$ with $|\Im \rho|\leq 3/2$ satisfy GRH (let us call that assumption GRH($3/2$)).

More generally, for any $\sigma>1$, $$ \left|\frac{L'(1,\chi)}{L(1,\chi)}\right| \leq (\sigma-1) \log q + c_0(\sigma,\kappa),$$ where $$\begin{aligned} c_0(\sigma,\kappa) &= (1 + 2 (\sigma-1)) \left| \frac{\zeta'(\sigma)}{\zeta(\sigma)}\right| + \frac{1}{2} \left(\digamma\left(\frac{\sigma+\kappa}{2} \right) - \digamma\left(\frac{1+\kappa}{2} \right)\right)\\ &+ (\sigma-1) \left(\digamma\left(\frac{\sigma+\kappa}{2}\right) - \log \pi\right), \end{aligned}$$ $\kappa = 1$ if $\chi(-1)=-1$, $\kappa=0$ if $\chi(-1)=1$, and $\digamma(s)$ is the digamma function.

I can include the proof here if needed.