Beyond the LSD method for the partial sums of multiplicative functions

The Landau–Selberg–Delange method gives an asymptotic formula for the partial sums of a multiplicative function f whose prime values are \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\alpha $$\end{document}α on average. In the literature, the average is usually taken to be \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\alpha $$\end{document}α with a very strong error term, leading to an asymptotic formula for the partial sums with a very strong error term. In practice, the average at the prime values may only be known with a fairly weak error term, and so we explore here how good an estimate this will imply for the partial sums of f, developing new techniques to do so.

However, if α / ∈ Z, then the function ζ(s) α has an essential singularity at s = 1, so the usual method of shifting the contour of integration to the left and using Cauchy's residue theorem is not applicable.
A very similar integral in the special case when α = 1/2 was encountered by Landau in his work on integers that are representable as the sum of two squares [4], as well as on his work counting the number of integers all of whose prime factors lie in a given residue class [5]. Landau discovered a way to circumvent this problem by deforming the contour of integration around the singularity at s = 1, and then evaluating the resulting integral using Hankel's formula for the Gamma function. His technique was further developed by Selberg [7] and then by Delange [1,2]. In its modern form, it permits us to establish a precise asymptotic expansion for the partial sums of τ α and for more general multiplicative functions. These ideas collectively form what we call the Landau-Selberg-Delange method or, more simply, the LSD method. 1 Tenenbaum's book [8] contains a detailed description of the LSD method along with a general theorem that evaluates the partial sums of multiplicative functions f satisfying a certain set of axioms. Loosely, if F(s) is the Dirichlet series of f with the usual notation s = σ + it, then the axioms can be rephrased as: (a) | f | does not grow too fast; (b) there are constants α ∈ C and c > 0 such that F(s)(s − 1) α is analytic for σ > 1 − c/ log(2 + |t|). Ifc 0 ,c 1 , . . . are the Taylor coefficients of the function F(s)(s − 1) α /s about 1, then Theorem II.5.2 in [8, p. 281] implies that for each fixed J . Our goal in this paper is to prove an appropriate version of the above asymptotic formula under the weaker condition for some α ∈ C and some A > 0. In particular, this assumption does not guarantee that F(s)(s − 1) α has an analytic continuation to the left of the line Re(s) = 1. It does guarantee however that F(s)(s − 1) α can be extended to a function that is J times continuously differentiable in the half-plane Re(s) ≥ 1, where J is the largest integer < A. We then say that F(s)(s − 1) α has a C J -continuation to the half-plane Re(s) ≥ 1, and we set for j ≤ J , the first J + 1 Taylor coefficients about 1 of the functions (s − 1) α F(s) and (s − 1) α F(s)/s, respectively. Since s = 1 + (s − 1) and, as a consequence, 1/s = 1 − (s − 1) + (s − 1) 2 + · · · for |s − 1| < 1, these coefficients are linked by the relationsc j = j a=0 (−1) a c j−a and c j =c j +c j−1 (0 ≤ j ≤ J ) with the convention thatc −1 = 0. Since ζ(s) ∼ 1/(s − 1) and f is multiplicative, we also have that Theorem 1 Let f be a multiplicative function satisfying (1.2) and such that | f | ≤ τ k for some positive real number k. If J is the largest integer < A, and the coefficients c j andc j are defined by (1.3), then The implied constants depend at most on k, A, and the implicit constant in (1.2). The dependence on A comes from both its size, and its distance from the nearest integer.
We will demonstrate Theorem 1 in three successive steps, each one improving upon the previous one, carried out in Sects. 3, 4 and 5, respectively. Section 2 contains some preliminary results.
In Sect. 6, we will show that there are examples of such f with a term of size x(log x) Re(α)−1−A in their asymptotic expansion, for arbitrary α ∈ C \ Z ≤0 and arbitrary positive non-integer A > |α| − Re(α). We deduce in Corollary 8 that the error term in (1.5) is therefore best possible when α = k is a positive real number, and A is not an integer.
The condition | f | ≤ τ k can be relaxed significantly, but at the cost of various technical complications. We discuss such an improvement in Sect. 7.
Theorem 1 is of interest to better appreciate what ingredients go in to proving LSDtype results, which fits well with the recent development of the "pretentious" approach to analytic number theory in which one does not assume the analytic continuation of F(s). In certain cases, conditions of the form (1.2) are the best we can hope for. This is the case when F(s) = L(s) 1/2 , where L(s) is an L-function for which we only know a zero-free region of the form {s = σ + it : σ > 1 − 1/(|t| + 2) 1/A+o(1) }. Examples in which this is the best result known can be found, for instance, in the paper of Gelbart and Lapid [3], and in the appendix by Brumley [6].
Wirsing, in the series [9,10], obtained estimates for the partial sums of f under the weaker hypothesis p≤x ( f ( p) − α) = o(x/ log x) as x → ∞, together with various technical conditions ensuring that the values of f ( p)/α are restricted in an appropriate part of the complex plane (these conditions are automatically met if f ≥ 0, for example). Since Wirsing's hypothesis is weaker than (1.2), his estimate on the partial sums of f is weaker than Theorem 1. The methods of Sects. 4 and 5 bear some similarity with Wirsing's arguments.

Initial preparations
Let f be as in the statement of Theorem 1. Note that |α| ≤ k. All implied constants here and for the rest of the paper might depend without further notice on k, A, and the implicit constant in (1.2). The dependence on A comes from both its size, and its distance from the nearest integer.
The first thing we prove is our claim that F(s)(s − 1) α has a C J -continuation to the half-plane Re(s) ≥ 1. To see this, we introduce the function τ f whose Dirichlet series is given by for all primes p and all ν ≥ 1. We also write f = τ f * R f and note that R f is supported on square-full integers and satisfies the bound |R f | = | f * τ − f | ≤ τ 2k . If F 1 and F 2 denote the Dirichlet series of τ f and R f , respectively, then F 2 (s) is analytic for Re(s) > 1/2. Hence our claim that F(s)(s − 1) α has a C J -continuation to the half-plane Re(s) ≥ 1 is reduced to the same claim for the function F 1 (s)(s − 1) α . This readily follows by (1.2) and partial summation, since Next, we simplify the functions f we will work with. Define the function f by the convolution formula We claim that we may assume that f = τ f . Indeed, for the function τ f introduced above, we have that τ f ( p ν ) = f ( p) log p ; in particular, | τ f | ≤ k . Moreover, if we assume that Theorem 1 is true for τ f , then we may easily deduce it for f : since R f is supported on square-full integers and satisfies the bound |R f | ≤ τ 2k , we have for C big enough. Now, if Theorem 1 is true for τ f , then it also follows for f , since if C is large enough. From now on, we therefore assume, without loss of generality, that f = τ f so that the values of f at f ( p k ) is determined by its value at f ( p), and in particular | f | ≤ k . Consider, now, the functions Q(s) := F(s)(s −1) α andQ(s) = Q(s)/s. As we saw above, they both have a C J -continuation to the half-plane Re(s) ≥ 1. In particular, if c j andc j are given by (1.3), then for each ≤ J we have To this end, we introduce the notations as well as the "error terms" We have the following lemma: Lemma 2 Let f be a multiplicative function such that f = τ f and for which (1.2) holds. Let also s = σ + it with σ > 1.
(a) Let ≤ J , m ≥ 0, and |s − 1| ≤ 2. Then (d) Let |t| ≥ 1, ≤ J , and m ≥ 0. Then All implied constants depend at most on k, A and the implicit constant in (1.2). The dependence on A comes from both its size, and its distance from the nearest integer.
(b) As in part (a), we focus on the claimed bound on E Estimating E J +1 (s) is trickier than estimating E (s) with ≤ J , because we can longer use Taylor's expansion for Q, as we only know that Q is J times differentiable. Instead, we will show that there are coefficients c 0 , c 1 , . . . , c J independent of s such that Notice that for s as in the hypotheses of part (b), the error term is 1, so that the claimed estimate for E J +1 (s) readily follows when m = 0 by exponentiating (2.5) and multiplying the resulting asymptotic formula by (s − 1) −α .
By our assumption that Using Taylor's theorem, we find that The last term is as needed, since 0 < A − J ≤ 1. This completes the proof of part (b) by taking Since |α| ≤ k and |s − 1| ≤ 2, we find that |s − 1| k−Re(a) ≤ 2 2k , whence . We argue similarly for the bound onẼ ( +m) (s).
(d) Let |t| ≥ 1, and j ≤ J , and fix for the moment some N ≥ 1. Summation by parts implies and (1.2) imply that for |t| ≥ 1. Taking log N = |t| 1/A yields the estimate This can be proven by induction on and by noticing that We thus conclude that Finally, in order to calculate the main term in Theorem 1, we need Hankel's formula for 1/ (z): for Re(s) < 0. By Mellin inversion we then have that f (x) = 1 2πi Re(s)=c F(s)x −s ds for c < 0. Making the change of variables s → 1 − s completes the proof.
Alternatively, we may give a proof when x > 1 that avoids the general Mellin inversion theorem. We note that it suffices to prove that since the claimed formula will then follow by differentiating with respect to x and then with respect to u, which can be justified by the absolute convergence of the integrals under consideration.
Using the formula valid for Re(s) > 1, we find that Since x − y = x y du, relation (2.6) follows.

Using Perron's formula
In this section, we prove a weak version of Theorem 1 using Perron's formula: Theorem 4 Let f be a multiplicative function satisfying (1.2) and such that | f | ≤ τ k for some positive real number k. If is the largest integer < A/2, and the coefficients c j andc j are defined by (1.3), then The implied constants depend at most on k, A, and the implicit constant in (1.2).
Proof As we discussed in Sect. 2, we may assume that f = τ f . We may also assume that A > 2, so that ≥ 1; otherwise, the theorem is trivially true. We fix T ∈ [ √ log x, e √ log x ] to be chosen later as an appropriate power of log x, and we let ψ be a smooth function supported on [0, 1 and whose derivatives satisfy for each fixed j the growth condition ψ ( j) (y) j T j uniformly for y ≥ 0. For its Mellin transform, we have the estimate This estimate is useful for small values of t. We also show another estimate to treat larger values of t. Integrating by parts, we find that Iterating and using the bound ψ ( j) (y) j T j , we find that We thus conclude that Now, let r denote an auxiliary large integer. Then For s = 1+1/ log x +it with 1 ≤ |t| ≤ (log x) ε T , we use the bounds (s) 1/|t| and F (r +2 ) (s) |t| 2 /A (log x) k+r , with the second one following from Lemma 2(d) with m = r and 2 in place of . Thus In the remaining part of the integral, we use the formula We then choose T = (log x) and use Lemma 2(c) with m = r + to write Since (x s − 1)/s = x 1 y s−1 dy and we find that Partial summation the completes the proof of (3.1). To deduce (3.2), we integrate by parts in (3.1). Alternatively, we may use a modification of the argument leading to (3.1), starting with the formula that is obtained by integrating by parts r + 2 times. We then bound the above integral as before: in the portion with |t| ≥ 1, we estimate F and its derivatives by Lemma 2(d), and we use the bound ( j) (s) j |t| −1 /(1 + |t|/T ) j−1 ; in the portion with |t| ≤ 1, we use the bound d j ds j ( (s) − 1/s) 1/T j+1 and we approximate (F(s)/s) (r +2 ) byG (r +2 ) (s) using Lemma 2(c).
Evidently, Theorem 4 is weaker than Theorem 1. On the other hand, if f = τ α , then (1.2) holds for arbitrarily large A, so that we can take to be arbitrarily large in (3.1) and (3.2). For general f , we may write f = τ α * f 0 . The partial sums of τ α can be estimated to arbitrary precision using (3.1) with as large as we want. On the other hand, f 0 satisfies (1.2) with α = 0. So if we knew Theorem 1 in the special case when α = 0, we would deduce it in the case of α = 0 too (with a slightly weaker error term, as we will see). The next section fills in the missing step.

The case˛= 0 of Theorem 1
Theorem 5 Let f be a multiplicative function with | f | ≤ τ k and The implied constant depend at most on k, A and the implicit constant in (4.1).
Proof As we discussed in Sect. 2, we may assume that f = τ f . Our goal is to show the existence of an absolute constant M such that We argue by induction on the dyadic interval on which x lies: if x ≤ 2 j 0 , where j 0 is a large integer to be selected later, then (4.2) holds by taking M large enough in terms of j 0 (and k). Assume now that (4.2) holds for all x ≤ 2 j with j ≥ j 0 , and consider where the restriction a ≥ 2 is automatic by the fact that f is supported on prime powers. We may thus estimate the first sum in (4.3) by the induction hypothesis, and the second sum by (4.1). Hence The implied constant here and below depends on k, A and the implied constant in (4.1), but not on our choice of M. Since | f | ≤ k (and thus | f | ≤ τ k ), as well as uniformly for x ∈ [2 j/2 , 2 j+1 ]. By partial summation, we thus conclude that To complete the inductive step, we take j 0 = 2/ε to be large enough so as to make the εM part of the upper bound ≤ M/2, and then M to be large enough in terms of j 0 so that the ε −A part of the upper bound is also ≤ M/2. The theorem is thus proven.
By Theorem 5 and the discussion in the last paragraph of Sect. 3, we obtain Theorem 1 with the error term being O(x(log x) k+2|α|−A−1 log log x). The reason for this weaker error term is that for the function f 0 = f * τ −α we only know that | f 0 | ≤ (k + |α|) . To deduce Theorem 1 in the stated form, we will modify the proof of Theorem 5 to handle functions f satisfying (1.2) for general α. This is accomplished in the next section.

Proof of Theorem 1
We introduce the auxiliary functions Our goal is to show that Theorem 1 then readily follows, since partial summation implies that We start by showing a weak version of (5.1) for smoothened averages of d: Lemma 6 Let f be a multiplicative function such that f = τ f and for which (1.2) holds. Let ψ : R → R be a function in the class C ∞ (R) supported in [γ, δ] with 0 < γ < δ < ∞. There are integers J 1 and J 2 depending at most on A and k such that for x ≥ 2, with the implied constant depending on A, k and the implicit constant in (1.2), but not on ψ.
Proof All implied constants might depend on A, k and the implicit constant in (1.2) without further notice. We will prove the lemma with J 2 = 1 + k + (A + 2k)(J + 2)/A and J 1 = J 2 + m, where m = J + k + 1. Set ϕ(y) = ψ(y)/y m and note that It thus suffices to prove that We consider the Mellin transform of the function y → ϕ(log y/ log x), that is to say the function We then have that Moreover, if we integrate by parts j times in δ γ ϕ(u)x su du, we deduce that we used here our assumption that supp(ϕ) ⊂ [γ, δ], which implies that ϕ ( j) (u) = 0 for all j and all u / ∈ (γ , δ). Putting together the above estimates, we conclude that Finally, we bound the portion of the integral in (5.3) with |t| ≤ (log x) A J +2 −1 . Note that Since we have assumed that m = J + k + 1 ≥ J + |α| + 1, and here we have that |t| ≤ 1 and σ = 1/ log x, partial summation implies that in the notation of Sect. 2, where we used (2.7) with s replaced by s + 1 to obtain the second equality. We will apply Lemma 2(b) with s = 1 + 1/ log x + it. Notice that we have |t| ≤ (log x) A J +2 −1

(log x)
A J +1 −1 / log log x, so that the hypotheses of Lemma 2(b) are met. Consequently, We conclude that the portion of the integral with |t| ≤ (log x) This completes the proof of the lemma.
We have that f log = f * f . Since ∞ n=1 g(n)/n s approximates the analytic behaviour of F, we might expect that the function is small on average. In reality, its asymptotic behaviour is a bit more complicated: Lemma 7 Let f be a multiplicative function such that f = τ f and for which (1.2) holds. There is a constant κ ∈ R such that n≤x

The implied constant depend at most on k, A and the implicit constant in (1.2). The dependence on A comes from both its size, and its distance from the nearest integer.
Proof Set h = f * g − g log. We begin by showing that there are coefficients We will later show by a different argument that the coefficients κ j / (α − j + 1) with j = α must vanish.

Partial summation implies that
as well as that where the terms with j = α can be trivially excluded because 1/ (0) = 0, and we used that J + 1 ≥ A and Re(α) ≤ k.
We apply Dirichlet's hyperbola method to the partial sums of f * g to find that We then insert relations (1.2) and (5.9) to deduce that Consequently, For the sum of g(b)/b, we use the Euler-Maclaurin summation formula to find that It remains to estimate the sum over a. By partial summation and (1.2), we find that (log(x/y)) y dy + αg(log x) where R(y) := n≤y f (n) − α y y(log y) −A and using the fact that c j =c j +c j−1 . In the main term, we make the change of variables t = log(x/y). In the error term, we develop q into Taylor series about log x: we have that The first two terms on the right hand side of this last displayed equation can be computed exactly: they equal since c j =c j +c j−1 . Using the estimates q ( j) (log x) Putting together the above estimates yields the formula where κ andκ j are some constants that can be explicitly computed in terms of the constants c, c j and I j . We may then write the above formula in the form (5.7) using the fact that thus completing the proof of (5.7).
To complete the proof of the lemma, we will show that κ j / (α − j + 1) = 0 for all j = α with j < A − k + Re(α). To see this, let ψ be a smooth test function such that We calculate L(x) in two different ways. On the one hand, partial summation and (5.7) imply that On the other hand, we have that h = d log − f * d by (5.6). An application of Lemma 6 yields that Then we observe that, for each fixed b ≤ √ x, the function u → ψ(u + log b/ log x) is smooth and supported in [1/6, 1]. We re-apply Lemma 6 to find that Finally, for fixed a ≤ √ x, we use relation (1.2) to find that We thus conclude that The function is a smooth function supported in [0, 1/2] and that is constant for u ≤ 1/6. Hence the function ϕ(u) := (2u) − (u) is supported on [1/12, 1/2]. Lemma 6 then implies that By our choice of ψ, comparing the above estimate with (5.10) proves that κ j / (α − j + 1) = 0 for all j = α with j < A − k + Re(α), and the lemma follows.
We are finally ready to prove our main result: Proof of Theorem 1 We will prove that there is some constant M such that for all x ≥ 2. Together with (5.8) and (5.9), this will immediately imply Theorem 1.
As in the proof of Theorem 5, we induct on the dyadic interval in which x lies. We fix some large integer j 0 and note that (5.11) is trivially true when 2 ≤ x ≤ 2 j 0 by adjusting the constant M. Fix now some integer j ≥ j 0 and assume that (5.11) holds when 2 ≤ x ≤ 2 j . We want to prove that (5.11) also holds for x ∈ [2, 2 j+1 ]. Whenever we use a big-Oh symbol, the implied constant will be independent of the constant M in (5.11).
Let x ∈ [2 j(1−ε) , 2 j+1 ] and ε = 2/ j 0 . We have that We estimate the sum a≤x/b f (a) by (1.2), and the sum b≤x/a d(b) by the induction hypothesis, since a ≥ 2 here. As in the proof of Theorem 5, and using the bound for all x ∈ [2 j(1−ε) , 2 j+1 ]. If we could show that the main terms cancel each other, then the induction would be completed as in Theorem 5. To show this, we will use Lemma 6. Firstly, note that when x ∈ [2 j(1−ε) , 2 j+1 ], we have that x ε ≥ 2, so that x 1−ε ≤ x/2 ≤ 2 j . Re-applying the induction hypothesis yields the bound for all x ∈ [2 j(1−ε) , 2 j+1 ]. Set X = 2 j and let ψ be a smooth function that is nonnegative, supported on [1 − ε, 1], assumes the value 1 on [1 − ε/2, 1 − ε/3], and for which ψ ( j) ∞ j ε − j for all j. Then Lemma 6 gives us that On the other hand, if we set ϕ(u) = ψ(u)/u and R(x) = n≤x d(n) log n − λ j x, then partial summation and (5.12) yield that Noticing that we also have that ∞ 0 ϕ(u)du ε by our choice of ϕ, we deduce that We then apply partial summation to find that Choosing ε to be small enough, and then M to be large enough in terms of ε, similarly to the proof of Theorem 5, completes the inductive step. Theorem 1 then follows.

The error term in Theorem 1 is necessary
To obtain the specific shape of the error term in Theorem 1.2, we had to use increasingly complicated arguments. A natural question is whether one can produce a sharper error term. We will show that the error term in Theorem 1.2 is optimal, when α is a nonnegative real number and A is not an integer. Precisely, we have the following result: Corollary 8 Let α = k and A be given real numbers with α ≥ 1, where A > 0 is not an integer and let J be the largest integer < A. There exists a multiplicative function f satisfying (1.2) and the inequality | f | ≤ τ k , and coefficientsc j defined by (1.3), as well as γ = 0, such that This follows easily from the following theorem: Let α ∈ C and A > |α| − Re(α). There exists a multiplicative function f satisfying (1.2) and the inequality | f | ≤ max{|α|, 1} , and for which there exist coefficients β j , j < A, and γ = 0 such that Remark 1 We say a few words to explain our hypotheses in Corollary 8. Comparing the result in Theorem 9 with Theorem 1, we see that each β j =c j . To ensure that the error term is as big as desired we need that k = Re(α) and so, since |α| ≤ k, this implies that α = k is a non-negative real number. To obtain that | f | ≤ τ k we need that | f | ≤ k and so k ≥ 1. The term with exponent α − 1 − A is only not part of the series of terms with exponents α − j − 1 if A is not an integer. This explains the assumptions in Corollary 8.
To construct f in the proof of Theorem 9, we let θ = arg(α), fix a parameter ε ∈ [0, 1] that will be chosen later, and set that is to say f is the multiplicative function with Dirichlet series We have selected α p so that it is a real scalar multiple of α, with |α p | ≤ max{|α|, 1}. Therefore f satisfies (1.2), as well as the inequality | f | ≤ max{|α|, 1} . We have the following key estimate: Proof The Dirichlet series of g is given by Since |α − α p | ≤ 1, we have that |g| ≤ 1. Note also that g( p) 1/(log p) A , so that p, ν≥1 |g( p ν )|/ p ν = O(1). By multiplicativity, we conclude that In particular, this proves that λ 0 is well-defined.
To estimate the partial sums of g, we take y := x 1/ log log x and decompose n as n = ab, with a having all its prime factors ≤ y and b having all its prime factors > y. Since |g| ≤ 1, the n's with b not being square-free contribute x/y to the sum n≤x g(n), and the n's with b = 1 contribute (cf. Corollary III.5.19 in [8]). Similarly, the number of n's with a > √ x contribute x/(log x) 2 A+1 . Finally, if b is square-free with ω(b) ≥ 2, then we write n = mpq with p being the largest prime factor of n and q being its second largest prime factor, for which we know that p, q > y. We thus find that the contribution of such n is Before continuing, we note for future reference that the exact same argument can be applied with |g| in place of g and yield the estimate When 3/2 > Re(α) + j > 0, the lemma follows by partial summation and (6.4), whereas when Re(α) + j > 3/2, we use (6.5). Next, when Re(α) + j < 0, we note that the sum ∞ m=1 τ α (m)(log m) j /m converges amd is equal to 0. Indeed, it equals (−1) j times the j-th derivative of ζ(s) α evaluated at s → 1 + , which tends to 0 in virtue of our hypothesis that Re(α) + j < 0. Estimating the right-hand side using (6.5) and partial summation proves the lemma in this case too. It remains to consider the lemma when Re(α) = − j. We then simply observe that We are now ready to estimate the partial sums of f :

Proof of Theorem 9
For the summatory function of τ α , we already know an asymptotic series expansion: there exist constants κ 0 , κ 1 , . . . such that for any fixed ≥ 1, where θ ∈ [1/3, 2/3] is a parameter to be chosen in the end of the proof. Letting, as usual, J to be the largest integer < A, and using relation (6.6), we find that In the notation of Theorem 1, we have β v =c v , so that the sum over v constitutes the main term in (1.5). For the second term in (6.7), we have Since A > |α| − Re(α), we conclude that Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.