# Bateman-Horn, continued even further

As before, consider the "singular series", which shows up in the Bateman-Horn conjecture: for an irreducible polynomial $f,$ this is equal to

$$s(f) = \prod_p \frac{1-\frac{n_f(p)}p}{1-\frac1p},$$ which is obviously the value at $1$ of the function

$$L_f(s) = \prod_p \frac{1-\frac{n_f(p)}{p^s}}{1-\frac1{p^s}}.$$

The question is: is there some alternative way to evaluate these objects (without multiplying out the Euler product) - an integral formula, or some such?

• By "evaluate" do you mean get more rapidly convergent expressions for them? – KConrad Aug 27 '15 at 14:41
• Look at the page guests.mpim-bonn.mpg.de/moree/Moree.en.html and the links there. – KConrad Aug 27 '15 at 14:44
• @KConrad (responding to your first comment): yes, faster is good, but also not giving the first several primes preferential treatment is good! – Igor Rivin Aug 27 '15 at 14:56
• @KConrad I looked at Moree's thing, but I don't immediately see how it applies, since the function here is not a rational function of $p.$ – Igor Rivin Aug 27 '15 at 15:07
• Replace the role of the zeta-function there with the zeta-function of the number field $K$ cut out by a root of $f$ or some of its "irreducible parts" (Dirichlet $L$-functions or more general Artin $L$-functions). For instance, the linear part of the $p$-factor is $-(n_f(p)-1)/p^s$, which is the linear part of $\zeta_K(s)/\zeta(s)$. So multiply $L_f(s)$ by this ratio and its reciprocal, and insert the $p$-factors for one of them into the Euler product for $L_f(s)$ to speed up convergence at $s=1$ (if you can compute $\zeta_K(s)/\zeta(s)$ rapidly by other methods). Does that make sense? – KConrad Aug 27 '15 at 15:51

See the following two papers of Nobushige Kurokawa, both appearing in Proc. Japan Acad. Ser. A:

"On Some Euler Products II" (volume 60, 1984, 365-368, esp. Proposition 1)

"Special Values of Euler Products and Hardy-Littlewood Constants" (volume 62, 1986, 25-28), where he writes $Z(s,f)$ for your $L_f(s)$ and $C(f)$ for your $s(f)$, allowing reducible $f$. In this paper, Theorem A2 rewrites $C(f)$ as rapidly convergent product involving Artin $L$-functions associated to the number field cut out by roots of $f$ and having a cutoff parameter $M$ that can be adjusted.

• Are Artin $L$-functions fun and easy to compute? – Igor Rivin Aug 27 '15 at 16:56
• Henri Cohen has a "draft of a preprint" on the subject, I don't think in principle he does anything that Moree or Kurokawa do, but it seems more explicit and might be useful (Section 4 and 5 particularly). math.u-bordeaux1.fr/~cohen/hardylw.dvi – kantelope Aug 27 '15 at 18:27
• @kantelope, from the way your wrote your comment, do you mean you don't think in principle that Cohen does anything that Moree and Kurokawa don't do? – KConrad Aug 27 '15 at 19:49
• Yes, I don't. I realized I made a typo, but figured the context was clear enough. – kantelope Aug 27 '15 at 19:54
• – kantelope Aug 27 '15 at 20:31

I record this here in the manner of usefulness.

The preprint draft http://www.math.u-bordeaux1.fr/~cohen/hardylw.dvi of Cohen gives a method for computing such constants.

The idea is to turn an Euler product into an Euler sum by taking logarithms, transform by Möbius inversion to turn sums over prime powers into logs of $L$-functions at positive integers, and then evaluate the $L$-functions by "standard" methods. To speed the convergence, one can consider the primes up to $A$ (say 30) separately.

In Sections 1 and 3, Cohen considers how to evaluate said $L$-functions (such methods are standard following Hecke/Lavrik, and are implemented by Dokchitser in GP/PARI or Magma), and in sections 4 and 5 he describes the cases of quadratic and cubic fields more fully. Section 2 gives an overview about the manipulations with sums over prime powers.

As a model case to illustrate, consider (Cohen's Section 4) the case of a quadratic Dirichlet character $\chi$. The base objects are the sums $S_m(\chi)=\sum_p \chi(p)/p^m$, and by Möbius inversion these are equal to

$$S_m(\chi)=\sum_{k\ge 1}^\infty {\mu(k)\over k}\log L(km,\chi^k).$$

However, by splitting the $p$-sum at $A$, one in fact by a similar inversion has

$$S_m(\chi)=\sum_{p\lt A}{\chi(p)\over p^m}+\sum_{k\ge 1}^\infty {\mu(k)\over k}\log L_{p\ge A}(km,\chi^k),$$

and this $\log L_{p\ge A}(km,\chi^k)$ is bounded by $O(1/A^{km})$ (Cohen does this in detail in Section 2.1).

To use the above formula, one needs a method to compute $L(km,\chi^k)$, and as was said, for any $L$-function this is standard (by an approximate functional equation with weights given by an incomplete Mellin transform of the $\Gamma$-factors), and takes time proportional to approximately the square root of the conductor of $\chi^k$ (with mild constants dependent on the desired precision, and also the size of $km$).

In order to obtain the Hardy-Littlewood or Bateman-Horn constants from the base objects $S_m(\chi)$, one needs to piece them together in the proper manner, which is largely bookkeeping (see 4.1 and 4.2 of Cohen), involving Dirichlet convolution of arithmetic sequences.

For higher degree polynomials, instead of Dirichlet $L$-functions one will also need Artin $L$-functions, which Cohen points out could be computed by $\zeta$-quotients, though this would be rather inefficient. Instead, a Magma package of Dokchitser allows one to compute with Artin representations rather easily. The time is still proportional to approximately square root of the conductor, though with higher degree $L$-functions the constant factor(s) are not quite so nice.

I am not sure this really answers the question. In this method, one does not "multiply out" the Euler product, except for the small primes (though one could take $A=2$, if treating small primes differently is not pleasing). The bulk contribution is determined by an $L$-function computation, which I suppose could be seen as an "integral" (in terms of a Mellin transform?) if desired.

• Yes, I read (ok, skimmed) the Cohen preprint, and it looks very interesting. The question is: if you DON'T want to compute the HL constants to 1000 decimal places, is this method much better than the stupid multiplying out of the Euler product? Which raises the question: if I approximate the constant by the first N terms of the Euler product, do I have some rigorous error estimate ( - it seems that the formula in Kurokawa's paper can be used for this, but I am not entirely sure. – Igor Rivin Aug 28 '15 at 21:03
• From my understanding, the convergence of the singular series in the first place is equivalent to non-vanishing of $L$-functions at 1 (for instance with quadratic polynomials, one needs that the quadratic character has $L(1,\chi)\neq 0$, so that the mean of $\eta(p)$ is indeed 1 with natural density, as opposed to merely Dirichlet density), and so probably the convergence rate depends on zero-free regions. – kantelope Aug 28 '15 at 21:06
• In fact, I think Cohen might be hiding the issue of convergence, where I think one needs to consider the issue of $S_1(\chi)=\sum_p \chi(p)/p$ more carefully. In the final line of page 14, you can note that the appearance (with $N=1$) of $\log L(1,\chi)$ in his formula. – kantelope Aug 28 '15 at 21:18
• As to the first question, I would guess that by multiplying out terms up to $X$ in the Euler product, under GRH one would be within $\tilde O(1/\sqrt X)$ of the answer, this coming from a (weighted and swiftly convergent) sum over zeros on the half-line, each giving a $X^{\rho-1}$ contribution. Any arithmetic correction from fiddling with higher prime powers should only induce $\sum_{p\ge X} 1/p^2=O(1/X)$. For $X$ less than the square root of the conductor, this is likely the way to go (as Cohen implies in his last sentence of section 4). – kantelope Aug 28 '15 at 21:28
• That is very interesting! Is the constant in the $O$ independent of $f?$ – Igor Rivin Aug 31 '15 at 9:51

As it seems to be of some interest, let me record some comments on the convergence rate of the Euler product (which are probably standard).

We consider $$C(f)=\prod_p \frac{1-\eta_f(p)/p}{1-1/p},$$ and as was pointed out, this is kind of like an $L$-function at $s=1$, namely that at good primes the Euler factor $(1-\eta_f(p)/p^s)^{-1}$ gives the same Dirichlet series coefficient at $p$ (though not powers of $p$) as the Dedekind $\zeta$-function of the field associated to $f$.

Everything below will be in terms of relative precision, which should not matter much as $C(f)$ itself is not particularly big nor small. Consider the contribution in the above product from $p\ge X$ and take logs, getting $$\sum_{p\ge X}\log(1-\eta_f(p)/p)-\log(1-1/p)=\sum_{p^k\ge X}\frac{1-\eta_f(p)^k}{p^k}=\sum_{p\ge X}\frac{1-\eta_f(p)}{p}+O(1/X),$$ where the big-Oh constant should be no worse than about the square of the degree of $f$. Under GRH and the Artin conjecture, we will show this $p$-sum tail is $O((\log |\Delta_f|)/\sqrt X)$.

As indicated above, $\eta_f(p)$ relates to Dedekind $\zeta$-function for $f$. Let us write $A(s)=\zeta_f(s)/\zeta(s)$, and this Artin $L$-function will be "nice" if we assume enough hypotheses. In particular, by Perron's formula we have $$\int_{(1)}X^s\frac{A'}{A}(s+1){ds\over 2\pi is}= -\sum_{\|{\frak p}\|^k\le X}{\log\|{\frak p}\|\over \|{\frak p}\|^k} +\sum_{p^k\le X}{\log p\over p^k}.$$ The terms with $k\ge 2$ contribute $B_1+B_2+O(1/X)$, where $B_i$ are the relevant sums extended to infinity (not cut off at $X$), and the prime ideals with $\|{\frak p}\|\neq p$ contribute $O(1/\sqrt X)$.

The prime ideals with $\|{\frak p}\|=p$ are those counted by $\eta_f(p)$ (at least for good primes), so we have $$\int_{(1)}X^s\frac{A'}{A}(s+1){ds\over 2\pi is}=B_1+B_2+\sum_{p\le X}(1-\eta_f(p))\frac{\log p}{p}+O(1/\sqrt X).$$ We then truncate the line of integration at say height $X^3$, with error $O(X/X^3)$ (again the big-Oh depends on the degree of $f$). We then move to contour to the left, say $\sigma=-3/4$. Under GRH the zeros are on $\sigma=-1/2$, and the contribution from $\sigma=-3/4$ is $O((\log N)(\log X)/X^{3/4})$, where $N$ is the conductor, thus the absolute value of the discriminant of $f$. Similarly the zeros themselves contribute $O(\log N)(\log X)/\sqrt X$, by zero-density estimates (namely $\log N$) and a harmonic sum up to height $X^3$ from the $1/s$.

So we get $$\sum_{p\le X}(1-\eta_f(p))\frac{\log p}{p}=B+O((\log N)(\log X)/\sqrt X)$$ and by partial summation $$\sum_{p\le X}\frac{1-\eta_f(p)}{p}=\tilde B+O((\log N)/\sqrt X).$$ In particular we have $$\sum_{p\ge X}\frac{1-\eta_f(p)}{p}=O((\log N)/\sqrt X),$$ which as above gives the convergence rate for the desired constant $C(f)$.

Maybe some analytic number theorist will come around and correct anything (and fix any sign errors I made). Also, the Artin conjecture might not be necessary, just the GRH for $\zeta_f$ and $\zeta$ (poles on the half-line are not bothersome).

• Thanks! There is another form of the remainder term in Kurokawa's paper - I assume that does not make things easier?! – Igor Rivin Sep 1 '15 at 9:09
• In Kurokawa's Theorem A2, he writes $C(f)=\prod_{p\le M}(...)\times \prod_{(\rho,n)\neq (Id,1)} L^M(n,E,\rho)^{...}$, where the first part is an Euler product up to $M$ as wanted, and the second (as in Cohen's writeup) is a product over (partial) $L$-function values, with only $p\ge M$ considered in their Euler products. These partial $L$-functions converge sufficiently rapidly to 1 for $n\gt 1$. However, for $n=1$ (and all nontriv irreps of the Galois group), we again need to estimate how fast the $p\ge M$ part of the Euler product converges to $L^M(1)$, and involves zero-free regions I fear. – kantelope Sep 1 '15 at 9:56
• On the other hand, one can calculate these (partial) $L$-values to any precision desired, viewing it as a Dirichlet series (rather than an Euler product) and using an approximate functional equation with about $\sqrt N$ terms, as sketched in Cohen's Section 2 (and gone over in more detail and generality in 10.3 of his second GTM book). – kantelope Sep 1 '15 at 10:01
• Ah, I see, thanks! If you want to chat off-line, perhaps you could drop me an email (ir40@st-andrews.ac.uk)? – Igor Rivin Sep 1 '15 at 10:42