An explicit upper bound for L (1 , χ ) when χ is quadratic

We consider Dirichlet L -functions L ( s, χ ) where χ is a non-principal quadratic character to the modulus q . We make explicit a result due to Pintz and Stephens by showing that | L (1 , χ ) | (cid:2) 12 log q for all q (cid:3) 2 · 10 23 and | L (1 , χ ) | (cid:2) 920 log q for all q (cid:3) 5 · 10 50 .


Introduction and results
A central problem in number theory concerns estimates on L (1, χ), where χ is a nonprincipal Dirichlet character to the modulus q, and where L(s, χ) is its associated Dirichlet L-function.Bounding sums of χ(n) trivially leads to the bound |L(1, χ)| log q + O(1).The Pólya-Vinogradov inequality allows one to improve this to (1/2) log q + O(1).An interesting history of these developments is given by Pintz [28].
Explicit versions of the above results date back to Hua [10].See also work by Louboutin [22] and the second author [32,33] for finding small pairs c, q 0 such that |L(1, χ)| (1/2) log q + c for all q q 0 .It appears difficult to improve on these bounds for generic q.
Theorem 1 Let χ be a quadratic odd primitive Dirichlet character modulo q 2 • 10 23 .We have L(1, χ) (log q)/2.For all even characters (not just quadratic ones) it is known that |L(1, χ)| (log q)/2, for all q 2. This is proved in [32] after several papers by Louboutin, the last of which is [22].Bounds relying on additional constraints on the characters at the small primes have been investigated by Louboutin in [23], by the second author in [33], by Saad Eddin in [36] and by Platt and Saad Eddin in [30].On taking q to be larger, we can improve on the factor 1/2 in Theorem 1.
We note that, on the Generalized Riemann hypothesis much more is known.Littlewood [21] showed that L (1, χ) log log q.This has been made explicit for large q in [15] by Lamzouri, Li and Soundararajan, and then for all q in [17] by Languasco and the third author.Finally, although we do not consider lower bounds on L(1, χ), we direct the reader to a survey of explicit and inexplicit bounds of Mossinghoff, Starichkova and the third author in [25], and to the recent work [16].
The outline of this paper is follows.In § 2 we assemble some bounds on prime counting functions and on related inequalities.In § 3 we collect the necessary explicit results on character sums.In § 4 we prepare the technical preliminaries to Stephens' approach, and analyse these in § 5. Our § 6 is purely centred on the optimization in (an improved version of) Stephens' method, and contains no number-theoretic input.Finally, in § 7 we prove Theorems 1 and 2.
We use the notation for the range of x considered.Note that this differs from the usual big-O notation, as it takes the implied constant to be 1.We also let for some prime p and integer k 1, 0, otherwise be the usual von Mangoldt function.Throughout this paper we shall let H > 1 be a parameter, which ultimately we shall optimise.With that in mind, we define In addition, for x 0 define We note that the second inequality in (2) follows from the first and partial summation.Namely, which rearranges to the desired result.
Our aim is to majorize F (1). Define It is also convenient to introduce the points

Preliminary results
We now list a trivial result that follows immediately from partial summation.
The following result is slightly more subtle.
Proof First note that we only need to prove the result for y an integer.If y / ∈ Z, then Similarly it suffices to prove the result for integral x.Assume, therefore, that y x 1 where both x, y ∈ Z.We therefore have as required.
We now list some bounds related to the prime number theorem.In particular, we give a range of bounds for the Chebyshev function The first is a classical result from Rosser and Schoenfeld, see [35,Thm 12].
We note that the result of Rosser and Schoenfeld gives 1.03883 in Lemma 5, which is an approximation to ψ(113)/113.To improve the bound in Lemma 5 it would be necessary to take x x 0 > 113, which, while possible, would complicate greatly the ensuing analysis for only a marginal improvement.
The second is an explicit bound of the form ψ(x) − x = o(x) coming from [3, Table 15] by Broadbent, Kadiri, Lumley, Ng, and Wilk.We remark that slightly weaker versions of Lemma 7, but ones that hold in a longer range of x have been provided by the first author in [12].We require the following result to be used in tandem with Lemma 7.This is obtained directly from [3,Table 8].The key feature here is that e 40 < 10 19 so that Lemma 7 and Lemma 8 between them cover all values of x > 11.Better results are available when x is very large, say log x 1000 -see [31] by Platt and Trudgian,and [13] by the first author and Yang -but Lemmas 7 and 8 suffice for our needs.
We now turn to estimates on to aid in the evaluation of h(χ, y) and h (1, y) in (1).To obtain such estimates we correct a result of the second author in [34].
Lemma 9 For x 71 we have n x where γ is the Euler-Mascheroni constant, and where Proof As discussed by Chirre, Hagen, and Simonič in [6], by fixing a couple of small typos, Lemma 2.2 in [34] can be replaced by where ρ runs over the non-trivial zeros of the Riemann zeta-function, counting multiplicity.Following [34, §5], we have Finally, since x 71, Proof Using Lemma 9 with the bounds from Lemmas 6 and 7, we obtain that, for x 10 5 , and for 10 5 x 10 19 .We then extend these estimates to smaller values of x by direct computation, giving ( 6) and ( 7).
An immediate consequence of this result is as follows.
We now examine the weighted average of n u (n)/n.

Lemma 12
We have This integral may be of interest in its own right.While the true value of this integral seems close to 0.41, we have no idea of the conjectured limiting value of the integral.To this end, see a similar problem discussed in [2].
When the variable u is small, we compute directly by using the fact that ψ(u) is constant on [n, n + 1) and that, with The second case is treated by splitting the integral at u = e τ .We compute in this manner that We now use Lemma 8 and Lemma 9 to show that, for some x 1 10 19 , To handle the integration beyond x 1 we use ( 6) in Lemma 10, whence the total integral is Choosing x 1 = exp(500) gives the result.
We remark that we could further divide the range to use more entries in the tables in [3], but the above result is sufficient for our purposes.

Character sum estimates
The work of Stephens and Pintz relied on the Burgess bound from [4].Explicit versions of this are known but are still numerically rather weak.When the modulus is prime, such bounds have been provided by Francis [8] improving on work by Treviño [38] and McGown [24].If we restrict our attention here to quadratic characters to prime modulus congruent to 1 modulo 4, we may rely on the slightly stronger bounds of Booker in [1].Recently, Jain-Sharma, Khale and Liu have produced in [11] an explicit version of the Burgess inequality for a composite modulus, but only for q primitive and q exp(exp(9.6)).
Instead of the Burgess bound we shall rely on versions of the Pólya-Vinogradov inequality.We first require an explicit version of the Pólya-Vinogradov inequality.
The following is from [18,19] by Lapkova, which makes a small improvement on the earlier result from [9, Theorem 2] by Frolenkov and Soundararajan.
When A = 0 and χ is even, we may divide this bound by 2.
In what follows, we write V = V (χ) for brevity, and apply the appropriate bound from Lemma 13 according to the parity of the character χ.
Here is a smoothed version of the Pólya-Vinogradov inequality that we take from Levin, Pomerance and Soundararajan in [20].
Lemma 14 Let χ be a primitive Dirichlet character modulo q > 1.Let M and N be real numbers with 0 < N q.With W (t) = max(0, 1 − |t − 1|), we have Lemma 15 Let χ be a primitive Dirichlet character modulo q > 1.Let M and N be real numbers with 0 < N q.When χ is odd,we have When χ is even, we have Proof We may assume that M is an integer.Notice first that the lemma is trivial when N √ q, so we may assume N > √ q.Let K 1 be an integer and let A = N /K .Keeping the notation of Lemma 14, we first notice that Therefore a which is readily seen to be of size at most A 2 + 1.On using Lemma 14, we get By computing the derivative with respect to c, we check that this quantity is maximised at c = 1.The lemma follows readily.
Before introducing the next lemma, we recall the functions F, f defined in (2).
, where V is defined in Lemma 13.
Proof By summation by parts, we find that

Preliminaries to Stephens' approach
Using the definition of f (x) from ( 2) in § 1, we have Now, (m) is only non-zero when m is a prime power, so We now write s = i p i α i as the prime decomposition for each s H x , so that (8) becomes 1 using the definition of (x) from (3).We therefore have We now recast this for greater ease of use in what follows.

Lemma 17
We have, for x 0, If H V > 1 we also have where We note that the proofs of Theorems 1 and 2 only require the upper bound in the O * error term in (10): see (17) for where this is used.Nevertheless, it is easy enough to prove (10) as it is written.
Proof We find that and the first part of the lemma follows readily.Concerning the upper bound for | 1 0 f (u)H u du|/H, we proceed as follows.

Case of even characters
By Lemma 13 and 15, we have three upper bounds for |f (u)|: either 1, We momentarily set Ṽ = V /2.We define Define the real parameter a by 1 2 (1 − a) log H = log( √ qH/ Ṽ ).We get

Resuming the proof
Inequality (10) follows: indeed, by ( 9), the left-hand side is (1)/ log H which we compute with the first formula of the present lemma.We complete the proof by using the bound above for 1 0 |f (u)|H u du.

Lemma 18
We have, for x 0, Proof On joining (9) and Lemma 17, we get This is the equivalent of [37, (55)] by Stephens.The next step is to integrate the above relation: As for the left-hand side, we first check that And finally by bounding |f (u)| by 1.

Lemma 19
We have, when x 0 Proof We start from the right-hand side: Recall ψ from (5).We approximate ψ(H x−t ) by (x − t) log H − γ , getting the main term and this is γ log H and treat the error term by bounding |f (t)| by 1: We then majorize this last term by Lemma 12.If follows from the definition of F in (2) that this last term is not more than 0.411/ log 2 H.

Lemma 20
We have, when x 0, Proof This follows by combining Lemmas 18 and 19.

A comparison and the main inequality
This section is devoted to the comparison between and F (1), where x m is defined in (4).The important observation, essentially due to Stephens, is that since f has tame variations, both should be about equal.In § 7, where we prove Theorems 1 and 2, we need only to bound (12) from below by F (1) plus some error term.
We first connect F (1) with the bounds on character sums, that is, with the V from Lemma 13.

Lemma 21
We have, for any D 1, Proof We have Using that Lemma 22 Let D 0 1, and A 2 be such that We then have, for any D D 0 , mn H Please notice that we would need only the lower estimate in the last bound.
Proof We write By Lemma 3, the last sum over n is bounded in absolute value by V .It then follows by Lemma 5 that the second summand of ( 14) satisfies Concerning the first summand of ( 14), we use three steps.For the first step, we restrict to the range H/10 19 < n H/D and use (13).Note that Lemma 7 tells us that we can take A 2 = 0.94 provided D 0 > 11.A quick calculation also shows that A 2 = √ 2 works for D 0 1, or A 2 = 0.956 works for D 0 7. Now, where for the second equality we used Lemma 3.
For the second step, we use Lemma 8 and consider the range H/A < n H/10 19 , where A exp(40) is to be chosen later.That is, where for the second equality we used Lemma 4.
For the third step we consider the sum over n H/A.First, if H < A then there is nothing to add.On the other hand, if H A we use Lemma 6 to get Since n log 2 (H/n) is increasing when n H/e 2 , we have that Since the above is decreasing in H, and H A, we can set A = exp(574) to bound the O * terms in ( 15) and ( 16) by 1.6 • 10 −5 H.

Lemma 23 For any D D
where A 2 is as in Lemma 22. Proof By using the definition of f , we get H log H and we appeal to Lemma 22.This leads to Note further that, by Lemma 21 (we need only the upper estimate), we have We are now in a position to prove the following crucial lemma.Recall the definitions of h(1, x) and h(χ, x) from (1), and of R χ (H, V, q) from ( 11).
Lemma 24 Let H 10 6 and 1/2.Then we have where A 2 is as in Lemma 22 when D is taken to be A 2 2.04 (see (19)).
This is the equivalent of [37, Lemma 2] by Stephens.
Proof The first inequality follows by Lemma 11.Concerning the second one, we proceed as follows.Define Since |χ(m)|, |f (x m )| 1, we have S 0. Furthermore, on expanding and using the second part of Lemma 17, we find that Now we use Lemma 23.Since S 0, this leads to the inequality We select (19) so that the expression in (( 18)) involving D is minimised.This then gives Let us extend this inequality to h(χ, x).We simply write

A result in optimization
This section contains a refined version of a result by Stephens: see Theorem 4 in [37].No further arithmetical material is being introduced.We start with a technical lemma.

Lemma 25
We have, for θ fixed and θ ∈ (0, x), Proof Notice that 2 u log udu = u 2 log u − (u 2 /2) and thus Next, and thus whence the lemma follows after some simple algebraic rearrangement.that for some parameters a and ε 2 , we have, when x 1/2, and that, for some parameter ε 1 we have, when x 1/2, The function ϕ is increasing (its derivative is −2 log y) on (0, 1] and takes the positive value −2θ log θ at y = θ.Note that θ 1/2 since G(1) 1, and that when θ 1/ √ e, our result is immediate.Let us assume that θ < 1/ √ e so that θ + 2θ log θ < 0. Assume that, when θ y Z, we have G(y) ϕ(y).This latter inequality translates into Our initial remark is that θ is such a number.
Proof Indeed, if it where not, we would have since G(x) x.We notice next that G(1) = 2 − 2θ , so that the above inequality can be rewritten as G( 1) G(1) + θ + 2θ log θ < G(1) by the inequality assumed for θ , leading to a contradiction.We define for this proof We find that, for Z x θ, by bounding above G(y) by y by 0 ) when y θ.We study separately the two right-hand side sums, say S 1 and S 2 .First we note that, on recalling the definition (20) of g: and this amounts to In the first integral, we bound above g(x − u) by 2(x − u) by (H 2 ).We split the second integral at u = θ; between x − θ and θ , we bound above g(u) again by 2u while in the later range, we bound above g(u) by 2θ + ε 1 by (H − 3) (valid since u θ 1/2).We infer in this manner that By Lemma 25 and noticing that ϕ(θ ) − θ = −θ (1 + 2 log θ ), we get (again bounding above g(x − u) by 2(x − u) by (H 2 )) By (H 1 ) and the above, we infer that (x + a)G(x) xϕ(x) + θ 2 (1 + 2 log θ ) + ε 1 + ε 2 .
We will further majorize f (1) by V /2 when χ is even and by V /H when χ is odd.

Using Lemma 26
So we infer that Since a < 0, if this inequality is satisfied for some H 0 > 1 then it remains true for H H 0 .
We select H = BV , for some parameter B, and bound |f (1)| by V /H.Therefore, if q q 0 , for some q 0 > 1 which we shall specify later, we have that So, here are possible choices: B 39.6 → A 2 = 0.956, B 79.5 → A 2 = 0.94.

Setting the numerics
We can now prove Theorems 1 and 2. We use the expression for V given in Lemma 13.We take H = BV which we assume to be 10 6 , we also assume that q q 0 so that V V 0 .Given a choice of B, we select 2 B −1/3 − B −1 + R χ (BV 0 , V 0 , q 0 ) log(BV 0 ) .
Hypothesis (H 0 ) is granted by the bound |f (u)| 1.By Lemma 20 we can then set and this gives us Hypothesis (H 1 ).Lemma 10 is enough to grant Hypothesis (H 2 ).Finally, by Lemma 24 and provided that H max(V, 10 6 ), Hypothesis (H 3 ) is satisfied with