On the generalised Dirichlet divisor problem

We improve unconditional estimates on $\Delta_k(x)$, the remainder term of the generalised divisor function, for large $k$. In particular, we show that $\Delta_k(x) \ll x^{1 - 1.889k^{-2/3}}$ for all sufficiently large fixed $k$.


Introduction
For integer k ≥ 2, let d k (n) denote the number of ways that n can be written as the product of exactly k factors.Partial sums of d k (n) are known to satisfy the asymptotic formula where P k−1 (t) is a degree k − 1 polynomial, and ∆ k (x) is a remainder term.The generalised Dirichlet divisor problem, concerning the order of the quantity ∆ k (x) as x → ∞, is an open problem that has attracted much interest in analytic number theory.It has been conjectured that ∆ k (x) ≪ ε x 1/2−1/(2k)+ε for any ε > 0, a result that implies the well-known Lindelöf Hypothesis [Tit86,Ch. 13].
While the true order of ∆ k (x) is currently unknown, substantial partial progress has been made.We briefly review two types of results, of the form ∆ k (x) ≪ ε x α k +ε , (1.1) and For the first type of bound, it is known for instance that α 2 ≤ 517/1648 [BW17], and various bounds are also known for small k > 2 [Kol81; IO89; Ivi03].Much less is known for large k, with the current best-known bounds taking the form α k ≤ 1 − Dk −2/3 for some constant D > 0, known also as the Karatsuba constant [Kol11].In particular, it has been shown that if the Riemann zeta-function ζ(s) satisfies Richert's bound [Ric67], of the form uniformly for 1/2 ≤ σ ≤ 1 and some constant B > 0, then there exists c 0 , c 1 > 0 for which α k ≤ 1 − c 0 (Bk) −2/3 and β k ≤ 1 − c 1 (Bk) −2/3 , for sufficiently large fixed k.
In this article we make two contributions.First, we adapt the method of [Kol11] to use Heath-Brown's estimate (1.4) and thus benefit from the improved B constant.In doing so, we obtain a new Karatsuba constant of D = (2/3) 2/3 B −2/3 ≈ 1.224.Second, we develop a new method of estimating ∆ k (x) that directly uses an exponential sum estimate instead of a bound on ζ(s).Combined with an improved estimate of such exponential sums, we obtain a Karatsuba constant of D ≈ 1.889 for sufficiently large k.
We remark that although the second estimate is sharper, the first estimate has two benefits.First, the explicit dependence on B means any improvement in this constant directly translates to an improved Karatsuba constant.Second, the result holds for all fixed k ≥ 30 instead of for sufficiently large k.
Theorem 1.1.Let k be a fixed positive integer.Then, for k ≥ 30

More generally, we show that
1 In [Kol11] it was claimed that the same result holds for all k ≥ 93, however since [Kol11, Thm.5] is ultimately used to bound σ k/2 instead of σ k , we in fact require k ≥ 186.
Theorem 1.3.Let δ > 0. For all fixed integers k ≥ Aδ −3 , where A is an absolute constant, we have In particular, for sufficiently large k.
1.1.Omega results and sign changes of ∆ k (x).For completeness we briefly compare Theorem 1.1 to the best possible results.It has been conjectured [IO89] that The conjecture involving β k is equivalent to the Lindelöf Hypothesis, while the conjecture involving α k implies the Lindelöf Hypothesis.Currently, the best-known Ω-results concerning ∆ k (x) are much closer to the conjectured truth than O-results.[Har17;Haf81;Sou03].Recent developments have focused on lower-order factors, in particular Soundararajan [Sou03] has shown that for all k ≥ 2. Here, log j x represents the jth iterated logarithm.Another type of result of interest is the frequency of sign changes of ∆ k (x).In 1955, Tong [Ton55] proved that for k ≥ 2, ∆ k (x) changes sign at least once in the interval [X, X + h k ] for h k ≫ k X 1−1/k .In the case k = 2, Heath-Brown and Tsang [HBT94] has shown that Tong's theorem is best possible up to factors of log X.Recently, Baluyot and Castillo [BC23] showed that, assuming the Riemann Hypothesis, Tong's theorem is also sharp for k = 3 (up to factors of log X).

Background and useful lemmas
The primary tool leading to improvements in [Kol11] over [IO89] are lower bounds on the Carlson exponent m(σ).This is defined as the supremum of all numbers m ≥ 4 such that for any ε > 0. We first recall the following classical result of Ivić [Ivi03, Thm.8.4] This is an estimate of order (1 − σ) −1 for σ close to 1.The main result of [Kol11] depends on a sharper estimate, of order (1 − σ) −3/2 .In particular, it was shown 3 that We improve this estimate in Lemma 3.2.The argument we use relies on estimates of the Carlson abscissa σ k .For k > 0, define σ k as the infimum of numbers σ for which As mentioned in [Tit86, p. 153], an equivalent definition for σ k is the infimum of numbers σ for which for any fixed ǫ > 0. Furthermore, throughout let We can obtain upper bounds on σ k using upper bounds on µ k , via the following result, originally due to Carlson [Car22;Car26].Lemma 2.1.For any 0 < η < 1, we have Proof.See e.g.Titchmarsh [Tit86, Thm.7.9].
We make use of this result in the proof of Theorem 1.2.Lastly, for the proof of Theorem 1.2, we also require the following upper bound on β k , due to Titchmarsh.
Lemma 2.2.[Tit86, Thm.12.5] For any integer k ≥ 2, β k is equal to the lower bound of positive numbers σ for which

Proof of Theorem 1.2
The proof of Theorem 1.2 relies primarily on an inductive argument used in [Kol11], which produces an upper bound on σ k , and hence by extension a lower bound on m(σ).Such a bound is crucial to the improvement of the α k estimate in [Kol11] over that in [IO89].
Assume now that P (r) holds for some k 0 ≤ r < k.In particular, we have say.We will show that this implies P (r + ∆) where ∆ is a fixed positive quantity.From the definition of σ r , we have This implies, for any fixed δ > 0 and fixed ε 1 > 0 (in particular, independent of r) (3.4) Hence, using the definition of µ r+δ (η r ), we have However, from Lemma 2.1, if and only if where We take which is a fixed positive quantity.This gives (for Since c < 2/3 for any ε 0 > 0, (3.5) is true for sufficiently small x > 0. In particular, we can choose a fixed ∆ = ∆(k 0 , k 2 ) such that (3.5) is true for all 0 < δ ≤ ∆ and r ≥ k 0 > k 2 .The result follows.
Proof.From Lemma 3.1, we have Since g is bijective, letting σ = g(m/2) we have 3.1.Bounds on α k .We now proceed to the proof of the first part of Theorem 1.2.Throughout, assume that x is half an odd integer.Applying Perron's formula (see e.g.Titchmarsh [Tit86, Lem.3.12]), we obtain, for any c > 0, σ + c > 1 and x half an odd integer, where we have used d k (n) ≪ ε n ε .We choose c = 1 + 1/ log x and σ > 0, so that the error term is where ε > 0 is arbitrarily small.Applying the residue theorem to the integral, we obtain where xP k−1 (log x) is the residue of the integrand ζ k (s)x s /s at the point s = 1 and E 3 is an arc of radius 1/2 centered at the point z 0 = 1 from the point β − ih to the point β + ih, with 0 < h < 1/2, in the clockwise direction.Figure 1 displays the contour used.It follows that We estimate separately the moduli of the integrals J i .Using the relation (1.4)4 to estimate |ζ(σ + iT )| for σ ∈ [β, 1], and the estimate |ζ(σ + iT )| ≪ log 2/3 T for σ ∈ (1, c], we obtain where and, to estimate the second integral, we used the fact that x c = x 1+1/ log x = ex.Since } for all β ≤ σ ≤ 1.It follows that (3.7) Now, we estimate the integral J 3 .Using the analytic continuation of ζ(s) for σ > −1 (see [Tit86, (2.1.4)])and the fact that |s − 1| = 1/2 and |s| ≥ 1/2 on E 3 , we have since the integral is convergent.Finally, we estimate the integrals J 2 and J 4 the same way.Let (3.9) so that, by Lemma 3.2, Then, for all β such that m 0 (β) ≤ k, we have, for 1 ≪ Z ≪ T , 5 Using a dyadic division and summing over O(log T ) intervals, we get for any ε > 0, and similarly for J 2 .At this point, combining all the estimates, we have In order to optimise the above estimate, we choose β so as to balance the main terms The two terms are equal if x f (β) = T , where .
3.2.Bounds on β k .We now turn our attention to the constant c 1 .By Lemma 2.2, it suffices to prove that the integral By symmetry, it suffices to show that the integral converges, since the integral converges on [−1, 1] and σ < 1.
For any 1 ≪ Z ≪ T , we have where m 0 (σ) is defined in (3.9)6 , and since m 0 (σ) ≤ m(σ) by Lemma 3.2, for all ε > 0 arbitrarily small.Using a dyadic division, we thus have The exponent of the parameter T of the RHS is which is the same as (3.13) since ε 0 is arbitrarily small.Therefore, the exponent on the RHS of (3.16) is non-positive for ε sufficiently small.Taking T → ∞, the second part of Theorem 1.2 then follows from Lemma 2.2.

Proof of Theorem 1.3
We begin by refining an exponential sum estimate due to Heath-Brown [HB17], which are in turn based on the bounds on the Vinogradov mean value integral proved in [Woo16;BDG16].
Observe that for k ≥ 2, However, for ρ k ≤ ρ ≤ ρ k+1 and k ≥ 2, where the first inequality is verified via a routine calculation.Therefore Lemma 4.1 leads to the following improved estimate of Carlson's abscissa, σ k .
Lemma 4.2.Let δ > 0 be fixed and sufficiently small.Then, there exists an absolute constant A such that for all integers k ≥ Aδ −3 , we have since the desired result then follows by summing the integrals taken over [T /2, T ], [T /4, T /2] and so on.Note that the parameter δ used here is different to that in the proof of Theorem 1.2.Using Minkowski's inequality, for T ≤ t ≤ 2T we have For σ > 0, (4.1) where in the last inequality we have used d k (n) ≪ ε n ε and the mean value theorem for Dirichlet polynomials. 7To bound the second integral, consider We have for any ε > 0. Furthermore, by counting the maximum number of terms in the product expansion.Combining (4.3) and (4.4) and applying the mean-value theorem for Dirichlet polynomials, for any ε > 0. Additionally, by Lemma 4.1 if ρ ≥ 3 and the trivial bound if 2 ≤ ρ < 3, uniformly for T ≤ t ≤ 2T .Note that in this step we have used ℓ ≤ k, a bound that originates from using of the mean value theorem for Dirichlet polynomials to cover the range ρ ∈ [k, ∞).Hence, combining (4.5) and (4.6), and using N = T 1/ρ and partial summation, However, since ρ − 1 < ℓ ≤ ρ by (4.2), the exponent of T is majorised by for some constant α > 0 and substituting, we find, via a direct evaluation The function h(ρ) is maximised on [2, k] by the choice ρ satisfying (αk 1/3 + 1)ρ 3 + 2ρ 2 − 3(k + 4)ρ + 12(k + 1) = 0.
Next, we require a more precise estimate of ζ(s) close to σ = 1.Since the proof shares some similarities with Lemma 4.2, we shall keep the exposition terse where possible.
Lemma 4.3.If δ > 0 is fixed, then for all k ≥ Aδ −3 , where A is an absolute constant, we have for any ε > 0.
The result follows from replacing δ with δ/3.