Sary Drappeau

ON THE AVERAGE DISTRIBUTION OF DIVISORS OF FRIABLE NUMBERS

A number is said to be y-friable if it has no prime factor greater than y. In this paper, we prove a central limit theorem on average for the distribution of divisors of y-friable numbers less than x, for all (x, y) satisfying 2 ≤ y ≤ e (log x)/(log log x) 1+ε . This was previously known under the additional constraint y ≥ e (log log x) 5/3+ε , by work of Basquin. Our argument relies on the two-variable saddle-point method.

, Basquin studies the slight variant where e v is replaced by n v / log x . This change does not affect the estimate of Theorem A.

Introduction

An integer n ≥ 1 is said to be y-friable, or y-smooth, if its greatest prime factor P (n) is less than or equal to y, with the convention P (1) = 1. We denote S(x, y) := n ≤ x : P (n) ≤ y , Ψ(x, y) := card S(x, y). Friable integers are a recurrent object in analytic number theory: we refer the reader to the surveys [START_REF]Integers without large prime factors[END_REF][START_REF] Granville | Smooth numbers: computational number theory and beyond, Algorithmic number theory: lattices, number fields, curves and cryptography[END_REF][START_REF]Integers without large prime factors: from Ramanujan to de Bruijn, Integers 14A[END_REF] for an overview of recent results and applications. An important aspect of results about friable numbers is their uniformity with respect to y. The difficulty in this context is the fact that y-friable numbers tend to rarefy very rapidly -much more so than what would be expected from sieve heuristics, for instance. In this respect, analytic methods have proven to be very effective. The object of this paper is to study, using these analytic methods, the distribution of divisors of friable numbers on average.

For any n ≥ 1, define D n to be the random variable taking the value log d where d is chosen among the τ (n) divisors of n with uniform probability. It was shown by Tenenbaum [Ten80] that D n / log n does not converge in law on any sequence of integers n of positive upper density in N. However it can be expected that the discrepancies arising from the erratic behaviour of the multiplicative structure are smoothed out upon averaging over n. When one averages over all the integers, this question was settled by Deshouillers, Dress and Tenenbaum [START_REF] Deshouillers | Lois de répartition des diviseurs. I[END_REF] who established

(1.1) 1 x n≤x Prob(D n ≤ t log n) = 2 π arcsin √ t + O 1 √ log x , (t ∈ [0, 1]).
The error term here is optimal if one seeks full uniformity with respect to t. See also [START_REF] Bareikis | On the DDT theorem[END_REF] for a generalization (where one changes the probability measure one puts on the divisors). Expectedly, the analog problem for friable numbers has a different structure. Choosing a divisor of n at random is equivalent to choosing an integer k ∈ {0, . . . , ν} uniformly at random for every factor p ν n appearing in the decomposition of n. We may thus write (1.2)

D n = p ν n D p ν
Date: January 27, 2016.

1
where the variables D p ν on the right-hand side are independent. As is known from work of Alladi [START_REF] Alladi | An Erdős-Kac theorem for integers without large prime factors[END_REF] and Hildebrand [START_REF] Hildebrand | On the number of prime factors of integers without large prime divisors[END_REF], the number ω(n) of prime factors of n ∈ S(x, y) typically tends to grow with n, in such a way that we may expect the sum (1.2) to satisfy the central limit theorem. We are therefore led to the prediction that

(1.3) Prob D n ≥ 1 2 log n + v n ≈ Φ(v) := ˆ∞ v e -z 2 /2 dz √ 2π
for all fixed v ∈ R and almost all integers n ∈ S(x, y), where n denotes the standard deviation of D n , given by (1.4)

2 n = p ν n ν(ν + 2) 12 (log p) 2 .
La Bretèche and Tenenbaum [dlBT02, Corollaire 2.2] consider the case of the primorial number N 1 (y) := p≤y p (which is the largest square-free y-friable number). They obtain

(1.5) Prob D N 1 ≥ 1 2 log N 1 + v N 1 (y) = Φ(v) 1 + O 1 + v 4
y/ log y for 0 ≤ v (y/ log y) 1/4 . Note that 2 N 1 (y) ∼ (y log y)/4. We emphasize that there is no average over the integers under study. Another related example considered recently by Tenenbaum [Ten14, Corollaire 1.4], is the case of N 2 (y) := p≤y p (log y)/ log p . There, Tenenbaum obtains an analogous result to (1.5).

Such a law obviously does not hold for all y-friable numbers, as illustrated by the example of N 3 (y) = 2 y/ log 2 (which is roughly of the same size as N 1 and N 2 , but for which D N 3 converges to the uniform law). It is therefore natural to ask what the output is, if we on average over friable numbers. One option would be to study the average 1 Ψ(x, y) n∈S(x,y)

Prob(D n ≥ 1 2 log n + v n ).

However, a more interesting variant is deduced from observing that an additive function of n naturally appears in the formula (1.4). A fundamental result in probabilistic number theory, the Turán-Kubilius inequality, developped in the context of friable numbers by La Bretèche and Tenenbaum [START_REF]Entiers friables: inégalité de Turán-Kubilius et applications[END_REF], ensures the existence of a quantity (x, y) independent of n such that (1.6) n ∼ (x, y) for a relative proportion 1+o(1) of integers n ∈ S(x, y), when y → ∞ and y = x o (1) . The exact definition of (x, y) involves the saddle-point α(x, y), defined as the only positive solution to the equation p≤y log p p α -1 = log x.

Then the approximation (1.6) holds with = (x, y) := 1 4 p≤y p α -1 3 (p α -1) 2 (log p) 2 1/2 .

We will prove below that (x, y) 2 ∼ (log x)(log y) 1 4 + log x 6y (y → ∞, y = x o(1) ).

In view of the above, we consider for v ∈ R the quantity D(x, y; v) := 1 Ψ(x, y) n∈S(x,y)

Prob(D n ≥ 1 2 log n + v ) (1.7) = 1 Ψ(x, y) n∈S(x,y) 1 τ (n) d|n d≥n 1/2 e v 1 (2 ≤ y ≤ x, v ∈ R).
The asymptotic behaviour of D(x, y; v) was studied previously by Basquin [START_REF] Basquin | Loi de répartition moyenne des diviseurs des entiers friables[END_REF] 1 for relatively large values of y. There, Basquin quantifies the shift from the arcsine law (1.1) to a contracted normal law similar to (1.5): we refer the reader to [Bas14, Théorème 1.1] for more details about this transition. We shall focus on the gaussian behaviour for small values of y: let u := (log x)/ log y, ū := min{u, π(y)},

(1.8) H(u) := exp{u/(log 2u) 2 }, where π(y) denotes the counting function of primes. Then Theorem 1.1 and Corollary 1.3 of [START_REF] Basquin | Loi de répartition moyenne des diviseurs des entiers friables[END_REF] (along with [START_REF] Hildebrand | On integers free of large prime factors[END_REF]equation (7.19)] to relate with the quantity ξ (u) involved there) imply the following.

Theorem A. Then for all ε > 0 and all x and y satisfying

(H ε ) exp{(log log x) 5/3+ε } ≤ y ≤ x,
we have

(1.9) D(x, y; v) = Φ(v) + O ε 1 u + 1 √ log y + log(u + 1) log y (v ∈ R).
The range of validity in x and y here is inherent to the method used, which is based on the "indirect" saddle-point method (see also [START_REF] Saias | Sur le nombre des entiers sans grand facteur premier[END_REF]). The purpose of the present work is to introduce a variant of the two-variable (direct) saddle-point method which allows us to obtain a significant improvement of the range of validity and of the error term in Theorem A.

Theorem 1. Let ε > 0. Whenever (G ε )
x ≥ 3, 2 ≤ y ≤ e (log x)/(log log x) 1+ε , and 0 ≤ v (ū) 1/4 , we have

(1.10) D(x, y; v) = Φ(v) 1 + O ε 1 + v 4 ū .
The condition y ≤ e (log x)/(log log x) 1+ε is purely technical. For x and y in the complementary range u ≤ (log log y) 1+ε , the Gaussian approximation is less relevant and the methods of [START_REF] Basquin | Loi de répartition moyenne des diviseurs des entiers friables[END_REF] are better suited.

The range v (ū) 1/4 is the natural range of validity of the Gaussian approximation. As is typically the case in large deviation theory, one could expect an asymptotic formula to hold in the range v (ū) 1/2-ε by adding correction terms to the exponent z 2 /2 in the definition (1.3) of Φ(v). We prove that such is indeed the case.

Theorem 2. Let (x, y) ∈ (G ε ). There exists a sequence of numbers (b j (x, y)) j≥0 satisfying b 0 (x, y) = -1/2, b j (x, y) j (ū) -j , such that the following holds. Let k ≥ 1 and

v max (ū) k/(2k+2)
be given, and assume that 0

≤ v ≤ v max . Letting (1.11) R k (z) = R k (x, y; z) := k-1 j=0 b j (x, y)z 2(j+1) (z ≥ 0),
we have

(1.12) D x, y; v = 1 + O k 1 + v 2 ū + v 2(k+1) (ū) k ˆ2vmax v e R k (z) dz √ 2π . Remark. Note that R k (z) = -z 2 /2 + O(z 4 /ū)
, which explains the shape of the error term in (1.10).

The coefficients b j (x, y) for j ≥ 1 could be expressed, if one wished, as an explicit but complicated expression involving sums over primes less than y and the saddlepoint α(x, y) defined below. As ū → ∞, they can be approximated by elementary expressions involving x and y, in the same shape as formula (1.18) below. We refrain to do so here.

1.1. The saddle-point method. We now recall the explanation for the limitation on y in the estimate of Basquin [START_REF] Basquin | Loi de répartition moyenne des diviseurs des entiers friables[END_REF]. The range (H ε ) is classical in the study of friable numbers: it is typically linked to the approximation of Ψ(x, y) by Dickman's function2 ρ:

(1.13) Ψ(x, y) = xρ(u) 1 + O ε log(u + 1) log y ((x, y) ∈ (H ε )).
This estimate is a theorem of Hildebrand [START_REF] Hildebrand | On the number of positive integers ≤ x and free of prime factors >[END_REF], improving in particular De Bruijn's work [dB51] (see also [START_REF] Moree | the enchanter of friable integers[END_REF]). The range (H ε ) is tighly linked to the best known error term in the prime number theorem: it was shown by Hildebrand [Hil84] that if one could prove the weaker estimate Ψ(x, y) = xρ(u) exp{O(y ε )} for y ≥ (log x) 2+ε , for all fixed ε > 0, then the Riemann hypothesis would follow. In many applications however, including that of interest here, one seeks a control on the local variations of Ψ(x, y) with respect to x, rather than a control of Ψ(x, y) itself. By "local variations" we mean, for instance, quantities of the shape Ψ(x/d, y)/Ψ(x, y) for relatively small d ≥ 1. The saddle-point approach to estimating Ψ(x, y), developped by Hildebrand and Tenenbaum [START_REF] Hildebrand | On integers free of large prime factors[END_REF], is very suitable for such applications: it enabled very substantial progress to be made in the last decades regarding the uniformity with respect to y, for example in friable analogs of the Turán-Kubilius inequality [START_REF]Entiers friables: inégalité de Turán-Kubilius et applications[END_REF] or distribution of friable numbers in arithmetic progressions [START_REF] Soundararajan | The distribution of smooth numbers in arithmetic progressions, Anatomy of integers[END_REF].

We now recall Hildebrand and Tenenbaum's result. When 2 ≤ y ≤ x, the saddlepoint α(x, y) is defined as the positive real number satisfying (1.14)

p≤y log p p α -1 = log x.
It is therefore the positive number optimizing Rankin's simple but remarkably efficient upper bound

(1.15) Ψ(x, y) ≤ min σ>0 ζ(σ, y)x σ , where ζ(s, y) := p≤y (1 -p -s ) -1 = P (n)≤y n -s (Re(s) > 0).
Here and in what follows, the letter p always denotes a prime number. As was pointed out in [START_REF] Hildebrand | On integers free of large prime factors[END_REF], the point s = α(x, y) is a saddle point for the Mellin transform x s ζ(s, y) relevant to Ψ(x, y):

Ψ(x, y) = 1 2πi ˆσ+i∞ σ-i∞ x s ζ(s, y) ds s , (x ∈ N, σ > 0).
Letting

φ 2 (s, y) = p≤y (log p) 2 p s (p s -1) 2 (Re(s) > 0),
they obtain for 2 ≤ y ≤ x the following estimate [HT86, Theorem 1] :

(1.16) Ψ(x, y) = ζ(α, y)x α α 2πφ 2 (α, y) 1 + O 1 ū .
The denominator α 2πφ 2 (α, y) in (1.16) may be estimated using [HT86, Theorem 2.(ii)].

We have

(1.17) α(x, y) = log(1 + y/(log x)) log y 1 + O log log(1 + y) log y , (1.18) φ 2 (x, y) = 1 + log x y (log x) log y 1 + O 1 log(u + 1) + 1 log y .
However, the question of approximating ζ(α, y)x α up to an factor (1 + o(1)) by a smooth and explicit function of x and y -for instance, in terms of the Dickman function ρ, is tightly related to the error term in the prime number theorem. In a way, α encodes the irregularities in the distribution of prime numbers that prevent us from having a smooth, explicit estimate for Ψ(x, y) when (x, y) ∈ (H ε ) for all ε > 0.

On the other hand, the local variations of α(x, y) with respect to x are relatively well controlled : such local estimates were obtained by La Bretèche-Tenenbaum [START_REF]Propriétés statistiques des entiers friables[END_REF]. We note however that at the current state of knowledge, when (x, y) ∈ (H ε ), we are not able to deduce from them an equivalent e.g. of the quantity Ψ(x 2 , y)/Ψ(x, y), or the quantity

(1.19) 1 Ψ(x, y) n∈S(x,y) 1 τ (n) .
This is hinted, for instance, by the fact that the error terms of [dlBT05b, Théorème 2.4], which result from the estimation of Ψ(x/d, y)/Ψ(x, y), are of the same size as the main term if d = √

x. Note that if y ≥ (log x) 3 , say, the saddle-point relevant to the sum in (1.19) is roughly of the same size as α(x 2 , y) (because 1/τ (p) = 1/2 for prime p). The issue at hand when studying D(x, y; v) is precisely the estimation of such sums as the one in (1.19); in our case however, as will be apparent, the upper bound on n will be roughly of size x 1/2+o(1) , and the relevant saddle-point will indeed be well-approximated, to some extent, by α(x 1+o(1) , y).

A truncated convolution and the two-variable saddle-point method.

We now sketch our proof of Theorem 1. Inverting summations yields D(x, y; v) = S Ψ(x, y) ,

where

S := P (nd)≤y nd≤x, d 1/2 ≥n 1/2 e v 1 τ (nd)
.

The obvious approach here consists in first approximating the sum over n by a "nice" function of d, and then estimating the remaining sum over d. This is the method followed e.g. in [START_REF] Basquin | Loi de répartition moyenne des diviseurs des entiers friables[END_REF]. There, one relies on estimates for friable sums of multiplicative functions from [START_REF] Smida | Valeur moyenne des fonctions de Piltz sur les entiers sans grand facteur premier[END_REF], which are a generalization of (1.13). These however are still subject to the limitation (x, y) ∈ (H ε ). One could presumably follow the same strategy by using the estimate (1.16) along with local estimates for the saddle-point. The need for uniformity in d for the estimation of the inner sum, however, is likely to produce significant technical complications due to the dependence of the summand on the multiplicative structure of d. Here instead we study the double sum as a whole by applying the Perron formula twice, which yields x (s+w)/2 e γ(w-s) F y (s, w) dwds (s -w)(s + w) .

(1.20) S = 1 (2πi) 2 ˆσ+i∞ σ-i∞ ˆκ+i∞ κ-i∞ x s e -γw F y (s + w/2, s -w/2) dw w ds s , ( 2σ
The effect of the factor 1/(s -w) cannot be fully neglected; although a direct analysis would likely be possible (as in [dlBT02, Corollary 2.2]), we circumvent this issue by truncating off values of s and w with large imaginary parts, and differentiating with respect to v. Therefore, for some T > 0 of a suitable size and for some optimal choice of (σ, κ) (depending on x, y and γ), one wishes to estimate

2 (2πi) 2 ˆσ+iT σ-iT ˆκ+σ+iT κ+σ-iT
x (s+w)/2 e γ(w-s) F y (s, w) dwds s + w .

The integrals there can be analyzed by the saddle-point method, which eventually yields the expected approximation Ψ(x, y)e -v 2 /2 / √ 2π. Finally, we note that very recently Robert and Tenenbaum [START_REF] Robert | Sur la répartition du noyau d'un entier[END_REF] used a variant of the two-variable saddle-point method to study the distribution of integers with small square-free kernel. Compared with theirs, our setting is simplified by the fact that the series F y (s, w) is symmetric and to some extent comparable to ζ(s, y) 1/2 ζ(w, y) 1/2 (for the study of which we can use the work of Hildebrand and Tenenbaum [START_REF] Hildebrand | On integers free of large prime factors[END_REF]). 1.3. Acknowledgments. The author was supported by a CRM-ISM Postdoctoral fellowship. The author is grateful to Régis de la Bretèche, Andrew Granville, Gérald Tenenbaum and an anonymous referee for helpful comments on an earlier version of this manuscript.

Preliminary remarks and notations

We will keep throughout the notation

s = σ + iτ, w = κ + it, ((σ, τ, κ, t) ∈ R 4 ).
We write A B or A = O(B) whenever A and B are expressions where B assumes nonnegative values, and there exists a positive constant C such that |A| ≤ CB uniformly. The constant C may depend on various parameters, which are then displayed in subscript (e.g. A ε B if the constant depends on ε). Moreover, the letters c 1 , c 2 , . . . designate positive constants, which are tacitly assumed to be absolute, unless otherwise specified.

At various places in our arguments, functions such as z → 1/(log z) -1/(z -1) are involved, which are regular at some particular point of their domain of definition, where the explicit expression diverges (here z = 1). It will be implicit that one should consider the holomorphic extension at said point.

Finally, every instance of the complex logarithm function we consider is, unless otherwise specified, the principal determination defined on C R -. For all r > 0 and any function f defined on C R -, we denote f (-r + 0i) := lim ε→0+ f (-r + iε) and similarly f (-r -0i) := lim ε→0+ f (-r -iε), whenever those limits exist.

Saddle-point estimates for ζ(s, y)

For all k ∈ N, s ∈ C with Re(s) > 0 and y ≥ 2, we define

φ 0 (s, y) := log ζ(s, y) = - p≤y log(1-p -s ), φ k (s, y) := ∂ k φ 0 ∂s k (s, y), φ k (s, y) := p≤y (log p) k (p s -1) k , σ k := φ k (α, y), σ k := φ k (α, y).
Bear in mind that σ k and σ k depend on x and y, the values of which will be clear from the context. In particular, by the definition of α,

(3.1) σ 1 = -log x, σ 2 = p≤y (log p) 2 p α (p α -1) 2 , σ 2 = p≤y (log p) 2 (p α -1) 2 .
We quote the following useful estimates on α(x, y) and φ k (α, y) from Theorem 2 and Lemmas 2, 3 and 4 of [START_REF] Hildebrand | On integers free of large prime factors[END_REF]. They will be implicitly used thoughout our argument. Uniformly for 2 ≤ y ≤ x, we have

σ k (u log y) k (ū) 1-k , α ū u log y (y log x), α 1 log y (y log x), (1 -α) log y log ū, √ ū α √ σ 2 min{ √
ū log y, y/ log y}. We will also require the following two bounds, which are corollaries of the calculations of [HT86, page 281]. We have

(3.2) ˆ∞ (ū) 2/3 /(log x) 1 + τ 2 σ 2 y/(log y) -cy/(log y) dτ 1 ū√ σ 2 , ˆ∞ 0 1 + t 2 σ 2 y/(log y) -cy/(log y) dt 1 √ σ 2 .
Regarding φ k (x, y), using prime number sums in the same way as [HT86, Lemma 4 and 13], we deduce that

φ k (σ, y) k |φ k (σ, y)| (k ≥ 2, σ > 0, y ≥ 2).
Note that we trivially have σ 2 ≤ σ 2 . The next lemma relates more precisely the two quantities.

Lemma 1. As y, u → ∞,

(3.3) σ 2 σ 2 = 1 1 + (y/ log x) + o(1).
Proof. When α ≥ 0.6, we certainly have y/(log x) → ∞ as well as σ 2 = O(1) and σ 2 → ∞, so that the desired estimate holds. We may thus assume that α < 0.6. Let ε ∈ (0, 1/10]. By [HT86, Lemma 3], we have φ 2 (α, y) y 1-α log y whenever 1/(log y) α ≤ 0.6. The same conditions are satisfied when one replaces y by y 1/2 ; we deduce

(3.4) φ 2 (α, y 1/2 ) y -(1-α)/2 φ 2 (α, y) ≤ y -α/3 φ 2 (α, y).
Suppose first that y ≥ (1/ε) log x. Then log(1/ε)/ log y α < 0.6, and we have

φ 2 (α, y) = p≤y 1/2 (log p) 2 (p α -1) 2 + y 1/2 <p≤y (log p) 2 (p α -1) 2 ≤ φ 2 (α, y 1/2 )+y -α/2 φ 2 (α, y) ε c φ 2 (α, y)
for some absolute constant c > 0, because of our assumption on α. Assume next that y ≤ ε log x. Then α ε/ log y and p α = 1 + O(ε) uniformly for p ≤ y, so that φ 2 (α, y) = {1 + O(ε)}φ 2 (α, y). Finally assume that y = t log x where t varies inside (ε, 1/ε) and let ū → ∞. Then we have α ∼ log(1 + t)/ log y, so that y α ∼ ε (1 + t) (the decay of the implied o(1) there may depend on ε). Evaluating the sum over primes defining φ 2 (α, y) using [HT86, Lemma 13], we have

φ 2 (α, y) = 1 + o(1) (1 -y -α ) 2 ˆy 2 (log z)dz z α + O(1) ∼ ε y 1-α log y (1 -y -α ) 2 ∼ ε (t -1 + t -2 )y log y.
The same set of calculations show that, on the other hand,

φ 2 (α, y) = 1 + o(1) (1 -y -α ) 2 ˆy 2 (log z)dz z 2α + O(1) ∼ ε y 1-2α log y (1 -y -α ) 2 ∼ ε t -2 y log y. We deduce φ 2 (α, y) ∼ ε (1 + t) -1 φ 2 (α, y).
Grouping our estimates, we have in any case

lim sup ū→∞ σ 2 σ 2 - 1 1 + (y/ log x) ε c
for some absolute c > 0 and all ε > 0, and we conclude by letting ε → 0.

Having the above facts at hand, we let = (x, y) be defined for 2 ≤ y ≤ x by

(3.5) := 1 2 (σ 2 -σ 2 /3) 1/2 (log x)/ √ ū.
As ū → ∞, we therefore have

2 ∼ (log x) log y 1 4 + log x 6y .

Lemmas

The following lemma is a truncated Perron formula suited for sparse sequences, cf. [Ten07, Exercices II.2.2 and II.

2.3]. Let K(τ ) := max{0, 1 -|τ |} (τ ∈ R).
Lemma 2. Let (a n ) be any sequence of complex numbers, and assume that the series

F (s) := n≥1 a n n s
is absolutely convergent on the half-plane Re(s) > σ 0 for some σ 0 > 0. For all such s, let F 0 (s) := n≥1 |a n |n -s . Then for all x ≥ 2, σ > σ 0 and T ≥ 2, we have

n≤x a n = 1 2πi ˆσ+iT σ-iT F (s) x s ds s + O x σ √ T F 0 (σ) + ˆ√T - √ T x iτ F 0 (σ + iτ )K(τ / √ T )dτ .
Remark. The integral in the error term is a non-negative real number, as is apparent from the proof.

Proof. The estimate follows classically from the formula, valid for all z > 0,

(4.1) 1 2πi ˆσ+iT σ-iT z s ds s = 1 z≥1 + O z σ min{1, (T | log z|) -1 } z σ . Indeed the error term is O z σ {T -1/2 + 1 | log z|≤T -1/2
} , and we have

(4.2) 1 | log z|≤T -1/2 sin( √ T (log z)/2) √ T (log z)/2 2 = ˆ1 -1 z iτ √ T K(τ )dτ.
We then specialize at z = x/n and sum over n against the coefficients a n .

Basic properties of F y (s, w).

Let

H := {s ∈ C : Re(s) > 0}, U := {z ∈ C : |z| < 1}.
For all (s, w) ∈ H 2 , we write

F y (s, w) := P (nd)≤y 1 τ (nd)n s d w .
Note that we have the Euler product expansion

F y (s, w) = p≤y k, ≥0 p -ks-w k + + 1 = p≤y log(1 -p -s ) -log(1 -p -w ) p -w -p -s .
In what follows, the letters a, b and z shall denote complex numbers. Whenever z ∈ C R -, taking principal determinations of the logarithms, we have

Re z 1/2 -z -1/2 log z > 0
where the fraction is analytically extended with value 1 at z = 1. It follows that the function

(4.3) g(z) := log z 1/2 -z -1/2 log z is a well-defined analytic function of z ∈ C R -. Since we have (1-a)/(1-b) ∈ C R - for all (a, b) ∈ U 2 , it follows that the function (4.4) Ξ(a, b) := - 1 2 log(1 -a) - 1 2 log(1 -b) -g 1 -a 1 -b is an analytic function of (a, b) ∈ U 2 . When a, b ∈ (-1, 1), we have exp{Ξ(a, b)} = log(1 -a) -log(1 -b) b -a .
This identity therefore holds on U 2 by analytic continuation. Putting (4.5)

f y (s, w) := p≤y Ξ(p -s , p -w ),
we obtain that f y (s, w) is an analytic function of (s, w) ∈ H 2 , and

F y (s, w) = exp{f y (s, w)}.
For any (k, ) ∈ N 2 and function f (a, b) of class C k+ , we shall use the notation

∂ k f := ∂ k+ f ∂a k ∂b .
The hessian will play an important role: for a class C 2 function f of two variables, we denote Hess

[f ] := (∂ 20 f )(∂ 02 f ) -(∂ 11 f ) 2 .
In the rest of the paper, Ξ(a, b) will always denote the function defined by equations (4.4) and (4.3) in the proof of the previous lemma. The next lemma regroups some useful facts concerning the power series expansion of Ξ(a, b). Lemma 3. (i) For some sequence of positive coefficients (d k, ) k+ ≥1 with d 1,0 = d 0,1 = 1/2, the power series expansion of Ξ(a, b) at (0, 0) is

(4.6) Ξ(a, b) = k+ ≥1 d k, a k b ((a, b) ∈ U 2 ).
(ii) For some analytic function ξ(a, b) of (a, b) ∈ U 2 , we may write

(4.7) g 1 -a 1 -b = (a -b) 2 ξ(a, b) ((a, b) ∈ U 2 ).
(iii) For some sequences (d k, ) k, ≥0 and (d k, ) k, ≥0 of positive numbers with d 0,0 = d 0,0 = 1/24, we have

(4.8) ξ(a, b) = k, ≥0 d k, a k b ((a, b) ∈ U 2 ), (4.9) ∂ k ξ(a, a) = d k, (1 -a) k+ +2
(a ∈ U).

(iv) For all (a, b) ∈ (0, 1), we have

(4.10) [(∂ 20 Ξ)(∂ 02 Ξ) -(∂ 11 Ξ) 2 ](a, b) > 0.
The useful feature in points (i) and (iii) is the positivity of the coefficients, which will provide a neat way to establish bounds on F y (s, w).

Proof. Recall that the function g is defined by (4.3). Note that g(z) = O(log(|z|+|z| -1 )) uniformly for z ∈ C R -, and g(1) = 0. Thus, whenever z ∈ R -and Γ is an oriented circle inside C R -circling around z counter-clockwise, the Cauchy formula yields

g(z) = z -1 2πi ˛Γ g(w)dw (w -z)(w -1) = z -1 2πi ˆ0 -∞ g(w + 0i) -g(w -0i) (w -z)(w -1) dw,
where the last equality follows from modifying the contour of integration into a Hankel contour, first from -∞ to 0 with argument π, then from 0 to -∞ with argument -π.

Setting t = 1/(1 -w), we obtain

g(z) = (1-z) ˆ1 0 K(t)dt 1 -t(1 -z) ,
where

K(t) = 1 π arctan 1 π log t 1 -t (t ∈ (0, 1))
which we extend by continuity at t = 0 and 1. Letting z = (1 -a)/(1 -b), we deduce that

(4.11) g 1 -a 1 -b = (a -b) ˆ1 0 K(t)dt 1 -(ta + (1 -t)b) .
Note that the function K is differentiable in (0, 1) and

K (t) = 1 t(1 -t) π 2 + log(t/(1 -t)) 2 > 0 (0 < t < 1).
Expanding the rational fraction in the RHS of (4.11) as a power series, and taking into account the factor (b -a), we obtain for some coefficients ( d j ) j≥0 the expression (4.12)

g 1 -a 1 -b = j≥0 d j {a j + b j } + ˆ1 0 K(t) k, ≥1 a k b + k k 1 k + kt k-1 (1 -t) -t k (1 -t) -1 dt = j≥0 d j {a j + b j } - k, ≥1 a k b k + k 1 k + ˆ1 0 K (t)t k (1 -t) dt
by an integration by parts. The point here is that the coefficients of terms a k b with positive exponents are negative. We return now to Ξ(a, b). Setting b = 0, we have

Ξ(a, 0) = log - log(1 -a) a (a ∈ U).
By considering the derivative of this expression, it is easily obtained that the coefficients (d j,0 ) j≥1 in the expansion Ξ(a, 0) = j≥1 d j,0 a j are positive, and d 1,0 = 1/2. Using this expansion, the expression (4.12) for g and equation (4.4) (as well as the symmetry between a and b), we finally get

Ξ(a, b) = j≥1 d j,0 a j + b j ) + k, ≥1 a k b k + k 1 k + ˆ1 0 K (t)t k (1 -t) dt = k+ ≥1 d k, a k b
say, where the coefficients (d k, ) k+ ≥1 are positive and d 1,0 = d 0,1 = 1/2. This yields (4.6).

We continue with the expression (4.11). Since K(1 -t) = -K(t), we deduce

g 1 -a 1 -b = (a -b) 2 ˆ1 0 (t -1/2)K(t)dt (1 -(tb + (1 -t)a))(1 -(ta + (1 -t)b))
from which we deduce the existence of the function ξ(a, b) satisfying (4.7) and its analyticity. Note that (t -1/2)K(t) ≥ 0 for t ∈ [0, 1]. For all t ∈ [0, 1], we let

R t (a, b) := 1 (1 -(tb + (1 -t)a))(1 -(ta + (1 -t)b)) = k, ≥0 r k, (t)a k b ,
for some numbers r k, (t), by expanding the rational fraction as a power series in ta + (1 -t)b and tb + (1 -t)a, which in turn is a power series in a and b whose coefficients are polynomial combinations of t and 1 -t with positive coefficients. Therefore, for all k, ≥ 0, r k, (t) is a non-zero polynomial in t with r k, (t) ≥ 0 (t ∈ [0, 1]). Setting

d k, := ˆ1 0 (t -1/2)K(t)r k, (t)dt,
the expansion (4.8), along with the positivity of the coefficients, follows at once. Furthermore, it is easily seen by induction that for all k, ≥ 0,

∂ k, R t (a, b) = 1≤j≤k+ +1 P (j) k, (t) (1 -(tb + (1 -t)a)) j (1 -(ta + (1 -t)b)) k+ +2-j
for some non-zero polynomials P (j) k, (t) ≥ 0 (t ∈ [0, 1]). This yields the equation (4.9). The fact that d 0,0 = 1/24 is a simple calculation; it implies that d 0,0 = 1/24 by specialization at a = 0.

The inequality (4.10) is proved by a direct computation. Let

z := (1 -a)/(1 -b) > 0. Then [(∂ 20 h)(∂ 02 h) -(∂ 11 h) 2 ](a, b) = 1+z z-1 log z -2 (1 -a) 2 (1 -b) 2 (log z) 2
which is extended by continuity as 1/(6(1 -a) 4 ) when a = b. When z = 1, the positivity of the numerator is easy to establish.

We introduce for δ ≥ 0 the subset

D δ (α; y) := (σ, κ) ∈ (0, 1] 2 : (σ-α)(α-κ) ≥ 0, 1 1 + δ ≤ 1 -2 -σ
1 -2 -κ ≤ 1+δ and |σ-κ| log y ≤ δ . The first condition simply means that α is between is σ and κ. The other guarantee that σ and κ are adequately close to each other. Note that D δ (α; y) ⊂ D δ (α; y) whenever δ ≤ δ, and that if (σ, κ) ∈ D δ (α; y), then uniformly for p ≤ y, we have (4.13)

1 -p -σ = (1 -p -α ){1 + O(δ)}, σ = α{1 + O(δ + (log y) -1 )}, p σ = p α {1 + O(δ)},
and similarly for κ.

In the next lemma, we deduce from the properties of the series Ξ(a, b) some information about the function f y (s, w) defined in (4.5). Recall that σ 2 and σ 2 were defined by (3.1). Lemma 4. For some absolute constant δ 0 > 0 and all 2 ≤ y ≤ x, the following assertions hold. (i) For all (s, w) ∈ H 2 and k, ≥ 0, we have Re

∂ k f y (s, w) ≤ ∂ k f y (σ, κ).
(ii) For all σ, κ > 0, we have

[∂ 20 f y + ∂ 11 f y ](σ, κ) > 0 and Hess[f y ](σ, κ) > 0.
(iii) For all non negative integers k, with (k, ) = (0, 0), we have

∂ k f y (σ, κ) k, |φ k+ (α, y)| ((σ, κ) ∈ D δ 0 (α; y)).
(iv) Whenever 0 ≤ δ ≤ δ 0 and (σ, κ) ∈ D δ (α; y), we have

∂ 20 f y (σ, κ) + ∂ 11 f y (σ, κ) = σ 2 2 + O(δσ 2 ), Hess[f y ](σ, κ) = σ 2 4 σ 2 - σ 2 3 + O(δσ 2 2 ).
Proof. Part (i) follows immediately from part (i) of Lemma 3. Indeed, for all fixed indices k, ≥ 0, we can write ∂ k Ξ(a, b) = j 1 ,j 2 ≥0 d j 1 ,j 2 a j 1 b j 2 for some non-negative coefficients d j 1 ,j 2 depending on k and . Then

Re ∂ k f y (s, w) -∂ k f y (σ, κ) = - j 1 ,j 2 ≥0 d j 1 ,j 2 p≤y 1 -cos((j 1 τ + j 2 t) log p) p j 1 σ+j 2 κ ≤ 0
by positivity. Regarding part (ii), the inequality [∂ 20 f y + ∂ 11 f y ](σ, κ) > 0 also follows immediately by linearity from the equality

[∂ 20 f y + ∂ 11 f y ](σ, κ) = p≤y (log p) 2 a∂ 10 Ξ(a, b) + a 2 ∂ 20 Ξ(a, b) + ab∂ 11 Ξ(a, b) a=p -σ b=p -κ
and the positivity of the coefficients in the expansion (4.6). Concerning the hessian, we apply the Cauchy-Schwarz inequality, getting

[(∂ 20 f y )(∂ 02 f y )](σ, κ) ≥ p≤y (log p) 2 (a∂ 10 Ξ(a, b) + a 2 ∂ 20 Ξ(a, b))(b∂ 01 Ξ(a, b) + b 2 ∂ 02 Ξ(a, b)) a=p -σ b=p -κ 2 ≥ p≤y (log p) 2 a 2 b 2 ∂ 20 Ξ(a, b)∂ 02 Ξ(a, b) a=p -σ b=p -κ 2 .
By (4.10), the last sum over p is strictly greater than p≤y

(log p) 2 p -σ-κ ∂ 11 Ξ(p -σ , p -κ ) = ∂ 11 f y (σ, κ)
as required.

We now turn to estimating the derivatives of f y . Assume that (σ, κ) ∈ D δ (α; y) for some small δ. Recall that

Ξ(a, b) = - 1 2 log(1 -a) - 1 2 log(1 -b) -(a -b) 2 ξ(a, b).
Let k, ≥ 0 be fixed with k + ≥ 1. Then the derivative ∂ k Ξ(a, b) can be written as a linear combination with bounded coefficients of terms assuming one of the following four shapes:

(4.14)              (1 -a) -k if = 0, or (1 -b) -if k = 0, (a -b) 2 ∂ k ξ(a, b), (a -b)∂ j 1 j 2 ξ(a, b) with j 1 + j 2 = k + -1 and j i ≥ 0, ∂ j 1 j 2 ξ(a, b) with j 1 + j 2 = k + -2 and j i ≥ 0.
Suppose for simplification that a ≤ b. Then for any j 1 , j 2 ≥ 0, we have (4.15)

∂ j 1 j 2 ξ(a, a) ≤ ∂ j 1 j 2 ξ(a, b) ≤ ∂ j 1 j 2 ξ(b, b) j 1 j 2 (1 -b) -j 1 -j 2 -2
by virtue of (4.8) and (4.9). Noting that |a -b| ≤ 1 -a, It follows that each of the four expressions given in (4.14) is bounded by

O k, ((1 -a) 2 (1 -b) -k--2 ), so that ∂ k Ξ(a, b) k, (1 -a) 2 (1 -b) -k--2 (k + ≥ 1, a ≤ b).
By symmetry, when b ≤ a the same estimate holds if we swap a and b in the right-hand side. Next, we specialize a = p -σ , b = p -κ . By the property (4.13), if δ is small enough, we have for all k, ≥ 0 with k + ≥ 1 (4.16)

∂ k Ξ(p -σ , p -κ ) k, (1 -p -α ) -k-.
Differentiating the function (σ, κ) → Ξ(p -σ , p -κ ), k times with respect to σ and times with respect to κ yields a linear combination of terms of the shape

(log p) k+ p -j 1 σ-j 2 κ ∂ j 1 j 2 Ξ(p -σ , p -κ ) (1 ≤ j 1 + j 2 ≤ k + )
each of which is bounded by O((log p) k+ (p α -1) -j 1 -j 2 ) (here we used (4.13) and (4.16)). Summing over p ≤ y, we obtain

∂ k f y (σ, κ) k, p≤y (log p) k+ 1 (p α -1) k+ + 1 p α -1 ≤ φ k+ (α, y) + u(log y) k+ .
Each of the last two terms is bounded from above by O(|φ k+ (α, y)|) = O((u log y) k+ (ū) 1-k-) and this proves part (iii). We now estimate the hessian. A direct calculation reveals that

∂ 20 f y (σ, κ) = 1 2 φ 2 (σ, y)- p≤y (log p) 2 2a(a-b)ξ+a(a-b) 2 ∂ 10 ξ+2a 2 ξ+4a 2 (a-b)∂ 10 ξ+a 2 (a-b) 2 ∂ 20 ξ a=p -σ b=p -κ , ∂ 11 f y (σ, κ) = - p≤y (log p) 2 ab -2ξ + 2(a -b)(∂ 10 ξ -∂ 01 ξ) + (a -b) 2 ∂ 11 ξ a=p -σ b=p -κ
, where we abbreviated for simplicity

∂ k ξ = ∂ k ξ(a, b
). Using (4.13) and the properties (4.9) and (4.15), we obtain

∂ 20 f y (σ, κ) = 1 2 φ 2 (α, y) -2 p≤y (log p) 2 p -2α ξ(p -σ , p -κ ) + O(δφ 2 (α, y)), ∂ 11 f y (σ, κ) = 2 p≤y (log p) 2 p -2α ξ(p -σ , p -κ ) + O(δφ 2 (α, y)).
Using once more the properties (4.8) and (4.9), along with the value d 0,0 = 1/24, we get

∂ 20 f y (σ, κ) = σ 2 2 - σ 2 12 + O(δσ 2 ), ∂ 11 f y (σ, κ) = σ 2 12 + O(δσ 2 ).
Using the symmetry of f y with respect to σ ↔ κ, we finally obtain

Hess[f y ](σ, κ) = σ 2 2 - σ 2 12 2 - σ 2 12 2 + O(δσ 2 2 ) = σ 2 4 σ 2 - σ 2 3 + O(δσ 2 2 )
which gives part (iv) of the lemma.

Decay estimates along vertical lines.

For the saddle-point method to succeed, it is required that the tails of the integrals in (1.20) contribute a negligible quantity.

The following lemma, which provides sufficient information for this purpose, states that the decay of F y (s, w) away from the to-be saddle-points is reasonnably good compared with what a Taylor formula at order 2 would predict, even in a range where the Taylor formula turns out not to be relevant. It is an analog of [HT86, Lemma 8]. We recall our notation that c 1 , c 2 , . . . denote constants, which are absolute unless otherwise specified. (iii) For all ε > 0, there exists c 2 (ε) > 0 depending only on ε such that whenever

min{σ, κ} ≥ ε/ log y, max{σ, κ} ≤ 0.6, max{|τ |, |t|} ≤ c 2 (ε)/ log y, we have Re f y (s, w) -f y (σ, κ) ≤ -c 2 (ε) τ 2 φ 2 (σ, y) + t 2 φ 2 (κ, y) .
Although it is elementary, the proof of this lemma is somewhat lengthy and otherwise unrelated to the rest of the argument: it is postponed to the appendix. We deduce the following estimate for F y (s, w).

Corollary 1. Let |τ |, |t| ≤ exp{(log y) 4/3 }. For some absolute constants δ, c 3 > 0, whenever (σ, κ) ∈ D δ (α; y), the following holds.

(i) For max{|τ |, |t|} 1/ log y, we have

(4.17) F y (s, w) F y (σ, κ) exp -c 3 ū τ 2 (1 -α) 2 + τ 2 + t 2 (1 -α) 2 + t 2 . (ii) For max{|τ |, |t|} ≤ c 3 / log y,
-w ) = -1 2 log(1 -p -s ) -1 2 log(1 -p -w ) + O(p -2σ + p -2κ
), from which it follows that F y (s, w) = ζ(s, y) 1/2 ζ(w, y) 1/2 exp{O(1)}. In this case, Corollary 1 is a direct consequence of Lemma 8 of [START_REF] Hildebrand | On integers free of large prime factors[END_REF].

Next let c 1 be the constant in Lemma 5.(ii), and suppose that |τ |, |t| ≤ c 1 /(2 log y) and α ≤ c 1 /(4 log y). If δ is sufficiently small, this implies σ, κ ≤ c 1 /(2 log y). If moreover δ is sufficiently small in terms of c 1 , then the conditions of Lemma 5.(i) are fulfilled and for some constant c > 0, we have

|F y (s, w)| ≤ F y (σ, κ) exp -c y log y log 1 + τ σ 2 + log 1 + t κ 2 .
Since under our hypotheses σ 2 α 2 y/(φ 2 (α, y) log y), and similarly for κ, we have the required estimate Assume next that η/(4 log y) < α < 0.55, and |τ |, |t| ≤ c 1 /(2 log y). In this case, assuming δ is small enough, we deduce c 1 /(5 log y) ≤ σ ≤ 0.6 and similarly for κ, so that the conditions of Lemma 5.(iii) (with ε < c 1 /5 being absolute) are satisfied: for some absolute constant c > 0, we have

|F y (s, w)| ≤ F y (σ, κ) exp -c τ 2 φ 2 (σ, y) + t 2 φ 2 (κ, y) .
Note that φ 2 (σ, y) φ 2 (α, y) and similarly for κ. Furthermore, we have under the current hypotheses τ 2 φ 2 (α, y) log y/y u(log y)/y 1. Therefore

y log y log 1 + τ 2 φ 2 (α, y) y/(log y) τ 2 φ 2 (α, y)
and similarly for t. This yields the required estimate Suppose next that α < 0.55, |t| ≤ |τ | and |τ | 1/ log y (as we may without loss of generality). Then from part (i) of Lemma 3 we deduce

Re f y (s, w)-f y (σ, κ) = - k+ ≥1 d k p≤y 1 -cos((kτ + t) log p) p kσ+ κ ≤ - p≤y 1 -cos(τ log p) 2p σ
dropping all but one term by positivity. Note that p σ p α . It follows from Lemma 8 of [START_REF] Hildebrand | On integers free of large prime factors[END_REF] that for some c > 0, we have

p≤y 1 -cos(τ log p) p α cūτ 2 (1 -α) 2 + τ 2 .
(Note that the condition |τ | ≥ 1/(log y) in the statement of [START_REF] Hildebrand | On integers free of large prime factors[END_REF] We shall use the first estimate of the previous corollary in the form of the following bounds. We recall the definition (1.8) of H(u).

Corollary 2. Suppose (σ, κ) ∈ D δ (α; y) for some sufficiently small δ ≥ 0. Then the following assertions hold.

(i) For all 1 ≤ X ≤ exp{(log y) 4/3 } and λ ∈ R, ˆX -X |F y (σ + iλτ, κ + iτ )|dτ F y (σ, κ) 1 + Xe -c 5 ū .
(By symmetry the same bound holds for the analogous integral over t). (ii) For all 0 < δ ≤ 1, (µ 1 , µ 2 ) ∈ (0, 2] 2 , (λ 1 , λ 2 ) ∈ R 2 and 5 ≤ X ≤ exp{(log y) 4/3 }, we have

(4.20) ¨( * ) |F y (σ+i(τ + λ 2 t), κ + i(τ + λ 1 t))| dτ µ 1 + |τ | dt µ 2 + |t| δ F y (σ, κ) log(X/(µ 1 µ 2 )) 2 H(ū) -c 5 δ 2 (log x) -1 ,
where the integration domain is

( * ) = (τ, t) : max{|τ |, |t|} ≤ X, max{|τ + λ 1 t|, |τ + λ 2 t|} ≥ δ/ log y .
Proof. The proof is very similar to the calculations of [HT86, pages 277 and 279]. We only sketch the details. Part (i) follows in a straightforward way from bounding trivially by O(F y (σ, κ)) the contribution to the integral of |τ | ≤ 1, and bounding by O(Xe -cū F y (σ, κ)), for some c > 0, the contribution of |τ | ≥ 1 using Corollary 1.(i).

Regarding part (ii), first note that log x = H(ū) o(1) as ū → ∞ as soon as the two conditions y ≥ (log log x) 1+ε and u ≤ (log log y) 1+ε . The second was assumed from the outset of our argument. Let us first assume that y > (log log x) 2 . Then Corollary 1.(ii) yields ¨( * )

|F y (σ + i(τ + λ 1 t), κ + i(τ + λ 2 t))| dτ µ 1 + |τ | dt µ 2 + |t| F y (σ, κ)   sup ν∈R ˆX δ/ log y exp -cū τ 2 (1 -α) 2 + τ 2 dτ min{µ 1 , µ 2 } + |τ -ν|   2
for some c > 0. Since for any |τ | ≥ δ/ log y, we have

τ 2 (1 -α) 2 + τ 2 ≥ δ 2 δ 2 + ((1 -α) log y) 2 δ 2 (log ū) 2 , we obtain the bound F y (σ, κ)H(ū) -c δ 2 log(X/(µ 1 µ 2 )) 2
for some c > 0. Since (log x) δ,ε H(ū) c δ 2 /2 , the above is an acceptable error term.

If on the contrary y ≤ (log log x) 2 , then by Corollary 1.(iii), we have

¨( * ) |F y (σ + i(τ + λ 1 t), κ + i(τ + λ 2 t))| dτ µ 1 + |τ | dt µ 2 + |t| e O(y/ log y) F y (σ, κ)   sup ν∈R ˆX δ/ log y 1 + (τ log x) 2 (1 + τ 2 )y 2 -cy/ log y dτ min{µ 1 , µ 2 } + |τ -ν|   2
for some c > 0. We certainly have, for τ ≥ δ/ log y,

1 + (τ log x) 2 (1 + τ 2 )y 2 δ 2 log(u/y) δ 2 log x,
so that for some c > 0, we have a bound δ F y (σ, κ)(log x) -c δ 2 y/ log y log(X/(µ 1 µ 2 )) 2 . This is clearly acceptable since y ≤ (log log x) 2 . 4.3. The saddle-points. Let 2 ≤ y ≤ x, and γ = v with v ∈ R, |v| ≤ (log x)/ . We are interested in the properties of the pair of positive abscissae satisfying

(β 1 , β 2 ) = argmin (σ,κ)∈(R * + ) 2
x (σ+κ)/2 e γ(w-s) F y (σ, κ) .

This pair will be more easily dealt with if defined by extrapolation from the case v = 0. We let β : (-v 0 , v 0 ) → R be the maximal solutions (here v 0 ∈ R * + ∪ {∞}) to the differential equation

(4.21) β (v) = [∂ 02 f y + ∂ 11 f y ](β(v), β(-v)) Hess[f y ](β(v), β(-v))
satisfying the initial condition β(0) = α(x, y).

Lemma 6. For all |v| < v 0 , the couple (β(v), β(-v)) satisfies the saddle-point equation

(4.22)          ∂ 10 f y (β(v), β(-v)) + log x 2 -v = 0, ∂ 01 f y (β(v), β(-v)) + log x 2 + v = 0.
Moreover, the function β is defined in the interval

(4.23) V = V(x, y) = [-c 6 √ ū, c 6 √ ū],
and for v ≥ 0, v ∈ V, we have

(β(v), β(-v)) ∈ D δ 0 (α; y), 0 ≤ β(v) -β(-v) v √ σ 2
where δ 0 > 0 is an absolute constant such that Lemma 4.(iii)-(iv) hold.

Proof. That (β(v), β(-v)) satisfies the saddle-point equation can be seen by differentiating the system (4.22) with respect to v, granted that it is satisfied at v = 0. To check this last fact, we remark that from the definitions (4.5), (4.4) and Lemma 3.(ii) (more specifically, using the fact that g((1 -a)/(1 -b)) vanishes at order 2 when a = b), we have

∂ 10 f y (σ, σ) = ∂ 01 f y (σ, σ) = 1 2 φ 1 (σ, y) (σ > 0).
Then, when v = 0, both equations in (4.22) reduce to the definition of α(x, y). From Lemma 4.(ii), we have that β (v) > 0 for |v| < v 0 . Let δ 0 > 0 be, as in the statement, an absolute constant such that Lemma 4.(iii)-(iv) hold, and let

(4.24) v m := inf v ∈ (0, v 0 ) : β(v) -β(-v) = c 7 ū u log y or 2 -β(-v) -2 -β(v)
1 -2 -β(-v) = δ 0 > 0 where c 7 > 0 is absolute and such that 0 < c 7 ≤ δ 0 and α ≥ 2c 7 ū/(u log y). Since ū ≤ u, we have (β(v), β(-v)) ∈ D δ 0 (α; y), (0 ≤ v < v m ). In particular, β 1 , β 2 and their derivatives have a limit at v = v m and the theorem of Cauchy-Lipschitz yields v m < v 0 . Lemma 4.(iv) ensures that for 0

≤ v ≤ v m we have β (v) (log x)/( σ 2 ) 1/ √ σ 2 . It follows that (4.25) β(v) -β(-v) v √ σ 2 (0 ≤ v ≤ v m ).
We claim that v m √ ū. Indeed, assume first that the limiting condition in (4.24

) is β(v m )-β(-v m ) = c 7 ū/(u log y). Then from (4.25) it follows that v m / √ ū 1. If on the contrary the limiting condition in (4.24) is (2 -β(-vm) -2 -β(vm) )/(1 -2 -β(-vm) ) = δ 0 , then we write this condition as f (v m ) = 0, where f (v) := δ 0 + 2 -β(v) -(1 + δ 0 )2 -β(-v) (0 ≤ v ≤ v m ). We have f (0) = δ 0 (1 -2 -α ) α, f (v) |β (v)| 1/ √ σ 2 (0 ≤ v ≤ v m )
from which we deduce that v m α √ σ 2 √ ū. This proves the lemma.

The next lemma describes more precisely the variations of β and of the quantities x (β 1 +β 2 )/2 e γ(β 2 -β 1 ) F y (β 1 , β 2 ) and Hess[f y ](β 1 , β 2 ) (where (β 1 , β 2 ) = (β(v), β(-v))) with respect to v. Lemma 7. Define for all 2 ≤ y ≤ x and v ∈ V (4.26)

E(v) = E(v; x, y) := log(x (β 1 +β 2 )/2 e γ(β 2 -β 1 ) F y (β 1 , β 2 )) ((β 1 , β 2 ) = (β(v), β(-v)).
Then for some sequence of functions (b j (x, y)) j≥0 satisfying

b 0 (x, y) = -1/2, b j (x, y) j (ū) -j ,
and for each fixed k ≥ 1, we have the Taylor expansion

(4.27) E(v) = E(0) + k-1 j=0 b j (x, y)v 2j+2 + O v 2k+2 (ū) k (v ∈ V).
Moreover, we have

(4.28) Hess[f y ](β 1 (v), β 2 (v)) = Hess[f y ](α, α) 1 + O v 2 ū , (4.29) β 1 (v) = α - v + O v 2 u log y .
Proof. First we note that

E(v) = f y (β(v), β(-v)) + β(v) + β(-v) 2 log x -v(β(v) -β(-v)) .
By the saddle-point property (4.22), we have

E (v) = -β(v) -β(-v) (v ∈ V).
We wish to differentiate this expression further. In order to simplify the presentation, we introduce the following temporary notation. For m ≥ 2, let D m be the set of linear combinations with coefficients independent of x and y of functions of the shape v →

∂ k f y (β 1 (v), β 2 (v)) defined for v ∈ V,
where k + = m. We also denote by

D m 1 • • • D m k the set of products f 1 • • • f k
H : v → Hess[f y ](β(v), β(-v)), H ∈ D 2 2 we have from (4.21) (4.30) β ∈ H • D 2 . (which reads "the function v → (H(v)/ )β (v) is in D 2 ", etc.) It follows that E ∈ 2 H • D 2 .
By differentiating further with respect to v, we obtain

E ∈ 3 H 3 • D 3 2 D 3 , E (4) ∈ 4 H 5 • D 6 2 D 4 + D 5 2 D 2 3 .
More generally, an induction over j readily yields

E (j) ∈ j H 2j-3 • mrm=7j-12 rm=3j-5 m D rm m (j ≥ 2)
where the summation is over sequences of non-negative integers (r m ) m≥2 satisfying

m≥2 mr m = 7j -12, m≥2 r m = 3j -5.
Recall that σ m := |φ m (α, y)|. The definition of and Lemma 4 imply that for v ∈ V,

σ 2 2 , H(v) σ 2 2 , f ∞ m |σ m | m (u log y) m (ū) 1-m (m ≥ 2, f ∈ D m ).
It follows that for all j ≥ 2,

E (j) (v) j σ 6-7j/2 2 mrm=7j-12 rm=3j-5 m σ rm m j (ū) 1-j/2 .
Since the function E is even, the estimate (4.27) and the bound b j (x, y) (ū) -j are a consequence of the Taylor formula, and there remains to compute b 0 (x, y). Lemma 4.(iv) applied with the parameters (σ, κ) = (α, α) ∈ D 0 (α; y) yields

E (0) = -2 β (0) = - 4 2 σ 2 -σ 2 /3 = -1
by definition of . This proves that b 0 (x, y) = -1/2. Estimate (4.28) follows on the same lines. Indeed, since H ∈ D 2 2 , we have

H ∈ H • D 2 2 D 3 , H ∈ 2 H 3 • D 5 2 D 4 + D 4 2 D 2 3 .
We deduce

H ∞ σ 4 + σ 2 3 σ -1 2 H(0)/ū. Since H is even, H (0)
= 0 and the estimate (4.28) follows from a Taylor formula at order 2.

Finally, from (4.30) we obtain

(4.31) β ∈ 2 H 3 • D 3 2 D 3
so that a Taylor formula at order 2 yields

β(v) = α + v + O v 2 σ 3 σ 2 2 = α + v + O v 2 u log y as claimed.

Proof of Theorem 1

Let 2 ≤ y ≤ x be large numbers, v ≥ 0 such that v ∈ V (which we recall was defined in (4.23)), and

γ := v ∈ [-(log x)/2, (log x)/2] if the constant c 6 in the definition of V was chosen small enough. Recall that β = β(v)
is defined by (4.21). By the definition (1.7), swapping the sums over n and d, we recall that

D(x, y; v) = S(x, y; v) Ψ(x, y) ,
where S(x, y; v) :=

P (nd)≤y ne 2v ≤d≤x/n 1 τ (nd) . Let R(v) = R(x, y; v) := E(v) -E(0), (v ∈ V),
where E(v) is the quantity defined in (4.26).

Proposition 1. Let v m ∈ V, v m ≥ 1 be fixed. Assume 2 ≤ y ≤ x and 0 ≤ v ≤ v m .
Then we have

(5.1) S(x, y; v) Ψ(x, y) = ˆvm v 1 + O 1 + z 2 ū e R(z) dz √ 2π + O e R(vm) v m + e R(v) ū .
Proof that Proposition 1 implies Theorems 1 and 2. Let v m = v max (ū) k/(2k+2) be given.

Recall that R k (v) is defined by (1.11). By Lemma 7, we have

R(z) = R k (z) + O z 2k+2 (ū) k (0 ≤ z ≤ v m ).
The error term is absolutely bounded. Let

I(v) := ˆ2vm v e R k (z) dz √ 2π , (0 ≤ v ≤ v m ).
Then it is easily verified that for 0

≤ v ≤ v m , e R(v) (1 + v)I(v), I(v m ) 1 + v 2k+2 (ū) k I(v), ˆvm v z e R k (z) dz (1 + v )I(v) ( ∈ {2, 2k + 2}).
We deduce that ˆvm

v 1 + O 1 + z 2 ū e R(z) dz √ 2π = 1 + O 1 + v 2 ū + v 2k+2 (ū) k I(v), e R(vm) v m 1 + v 2k+2 (ū) k I(v)
. This implies Theorem 2. Theorem 1 follows by specialization at k = 1.

We define

T := min{H(u), exp((log y) 5/4 )}. Note that α √ σ 2 T o(1) as ū → ∞.

Small values of v. Let

v 1 := 1 (log x)ūα √ σ 2
and consider first the case when 0 ≤ v ≤ v 1 . Note that log(1/v 1 ) log log x. The right-hand side of (5.1) varies by an amount at most O(Ψ(x, y)/ū). The left-hand side of (5.1) varies by at most

P (nd)≤y nd≤x 1≤ d n ≤x 2v 1 1 τ (nd)
where we used the rough bound (log x) 2 . By Rankin's trick and (4.2) for z = d/n at the height T = (ūα √ σ 2 ) 2 ≤ exp((log y) 4/3 ) (for y large enough), the quantity above is bounded by

P (nd)≤y 1 τ (nd) x nd α ˆ1 -1 d n iτ /(v 1 log x) K(τ )dτ ≤ x α ūα √ σ 2 ˆūα √ σ 2 -ūα √ σ 2 |F y (α + iτ, α -iτ )|dτ.
By (1.16), Corollary 2.(i) and the fact that

F y (α, α) = ζ(α, y), this is Ψ(x, y)α √ σ 2 1 ūα √ σ 2 + e -c 8 ū Ψ(x, y) ū .
Therefore, the estimate (5.1) for 0 ≤ v ≤ v 1 is implied by the trivial case v = 0, and we can suppose from now on that v ≥ v 1 .

Applying the Perron formula. For all n ∈ S(x 1/2 e -γ , y), we apply Lemma 2 at the abscissa

σ = 1 2 (β(v) + β(-v
x σ τ (nd)(nd) σ + O 1 √ T ˆ√T - √ T P (d)≤y ne 2γ ≤d x σ+iτ τ (nd)(nd) σ+iτ K(τ / √ T )dτ .
Let κ := β(v) -β(-v) > 0. We sum the previous estimate over all y-friable d ≥ 1. By using Rankin's trick 1 ne 2γ ≤d ≤ e -κγ (d/n) κ/2 on the error terms, we obtain S(x, y; v) = 1 2πi ˆσ+iT σ-iT P (d)≤y n∈S(de -2γ ,y)

x s τ (nd)(nd) s ds s + O x (β 1 +β 2 )/2 e γ(β 1 -β 1 ) F y (β 1 , β 2 ) √ T + O x (β 1 +β 2 )/2 √ T ˆ√T - √ T F y (β 1 + iτ, β 2 + iτ )K(τ / √ T )dτ
where, here and in what follows, we abbreviate

(β 1 , β 2 ) = (β(v), β(-v)).
Next we express the inner sum over n in the main term using again Lemma 2, at the abscissa κ/2 and the height T /2. By similar calculations and using

S 2 = M + O(R 1 + R 2 + R 3 ),
where

M := 1 (2πi) 2 ˆσ+iT σ-iT ˆκ+iT κ-iT
x s e -γw F y (s + w/2, s -w/2) dw w ds s , (5.3)

R 1 := x (β 1 +β 2 )/2 e γ(β 2 -β 1 ) F y (β 1 , β 2 ) √ T x (β 1 +β 2 )/2 e γ(β 2 -β 1 ) F y (β 1 , β 2 )T -c 9 , R 2 := x (β 1 +β 2 )/2 e γ(β 2 -β 1 ) √ T ˆ√T - √ T |F y (β 1 + iτ, β 2 + iτ )|dτ, R 3 := x (β 1 +β 2 )/2 e γ(β 2 -β 1 ) √ T ˆ√T - √ T |F y (β 1 -it/2, β 2 + it/2)|dt.
First truncation. By Corollary 2.(i), we have

R 1 +R 2 +R 3 x (β 1 +β 2 )/2 e γ(β 2 -β 1 ) F y (β 1 , β 2 ) T -1/2 +e -c 10 ū x (β 1 +β 2 )/2 e γ(β 2 -β 1 ) F y (β 1 , β 2 )T -c 11 .
By (4.27), we obtain

(5.4)

R 1 + R 2 + R 3 x α ζ(α, y)e R(v) T -c 11 Ψ(x, y) e R(v) ū .
Second truncation. We now consider M , defined at (5.3). Let c 3 be the absolute constant given by Corollary 1, and put T 1 := c 3 /(2 log y). We write the integration domain in the double integral (5.3) as the disjoint union

[-T, T ] 2 = D 1 D 2 ,
where D 1 := {(τ, t) : |τ -t/2| ≤ T 1 and |τ + t/2| ≤ T 1 }, D 2 := {(τ, t) : max{|τ |, |t|} ≤ T, and |τ -t/2| > T 1 or |τ + t/2| > T 1 )}. Accordingly, we call I 1 the contribution of (τ, t) ∈ D 1 to (5.3), and I 2 the contribution of D 2 , so that M = I 1 + I 2 . The hypotheses of Corollary 2 are satisfied for I 2 , with the parameters

(X, δ, λ 1 , λ 2 , µ 1 , µ 2 ) = (T, c 3 /2, -1/2, 1/2, β 1 + β 2 , β 1 -β 2 ).
We deduce

I 2 x (β 1 +β 2 )/2 e γ(β 2 -β 1 ) F y (β 1 , β 2 )(log(T (log x)/(α 2 v 1 )) 2 (log x) -1 H(ū) -c 12 ε x (β 1 +β 2 )/2 e γ(β 2 -β 1 ) F y (β 1 , β 2 )T -c 13 since β 1 -β 2 v 1 √ ū/ log
x and log(1/v 1 ) log log x. As for (5.4), we conclude that (5.5)

I 2 Ψ(x, y) e R(v) ū .
Bounds for large v. By the change of variables (s, w) ← (s + w/2, s -w/2), we write

I 1 (v) := 2 (2πi) 2 ˆβ1 +iT 1 β 1 -iT 1 ˆβ2 +iT 1 β 2 -iT 1 x (s+w)/2 e γ(w-s) F y (s, w) dwds (s -w)(s + w) .
We first give a rough bound for I(v m ). Consider first, then, that v = v m . By the triangle inequality,

I 1 (v m ) x (β 1 +β 2 )/2 e γ(β 2 -β 1 ) ˆT1 -T 1 ˆT1 -T 1 |F y (β 1 + iτ, β 2 + it)| dτ dt (β 1 -β 2 )α (v = v m ).
From Corollary 1 and (3.2), we have ˆT1

-T 1 ˆT1 -T 1 |F y (β 1 + iτ, β 2 + it)| dτ dt (β 1 -β 2 )α F y (β 1 , β 2 ) ακσ 2 (v = v m ).
Since we have κ v m / √ σ 2 , we conclude that (5.6)

I 1 (v m ) x α ζ(α, y)e R(vm) v m α √ σ 2 Ψ(x, y) e R(vm) v m .
Differentiation, third truncation. Now we let v vary inside [v 1 , v m ]. For all such v, the quantity I 1 is differentiable at v and

I 1 (v) = - 2 (2πi) 2 ˆβ1 +iT β 1 -iT ˆβ2 +iT β 2 -iT
x (s+w)/2 e γ(w-s) F y (s, w) dwds s + w .

Let T 0 := (ū) 2/3 /(u log y). We split the previous integrals as

I 1 (v) = J 0 (v) + J 0 (v),
where J 0 is the integral over the box D 0 := {(τ, t) : |τ |, |t| ≤ T 0 }, and J 0 is the complementary contribution.

Taylor range. When (τ, t) ∈ D 0 , we Taylor expand the integrand : the calculations are very much similar to [HT86, page 280]. Letting

Q(τ, t) := τ 2 2 ∂ 20 f y (β 1 , β 2 ) + τ t∂ 11 f y (β 1 , β 2 ) + t 2 2 ∂ 02 f y (β 1 , β 2 ),
we have by a Taylor expansion at order 4, using Lemma 4.(i) and (iv),

f y (s, w) = f y (β 1 , β 2 )+iτ ∂ 10 f y (β 1 , β 2 )+it∂ 01 f y (β 2 , β 2 )-Q(τ, t)+ 3 j=0 λ j τ j t 3-j +O((|τ |+|t|) 4 σ 4 )
for some coefficients λ j σ 3 . Since T 4 0 σ 4 and T 3 0 σ 3 are O(1), we have exp

3 j=0 λ j τ j t 3-j +O((|τ |+|t|) 4 φ 4 (α, t)) = 1+ 3 j=0 λ j τ j t 3-j +O (|τ |+|t|) 6 σ 2 3 +(|τ |+|t|) 4 σ 4 . Moreover, 1 s + w = 1 β 1 + β 2 1 -i t + τ β 1 + β 2 + O |τ | + |t| α 2 .
Since we have σ 3 α -1 σ 4 , we obtain for some numbers µ 1 , µ 2 independent of τ and t, (5.7) x (s+w)/2 e γ(w-s) F y (s, w) s

+ w = x (β 1 +β 2 )/2 e γ(β 2 -β 1 ) F y (β 1 , β 2 ) β 1 + β 2 e -Q(τ,t) 1 + 3 j=0 λ j τ j t 3-j + µ 1 τ + µ 2 t + O (|τ | + |t|) 6 σ 2 3 + (|τ | + |t|) 4 σ 4 + (|τ | + |t|) 2 α -2 .
Note that we have the formulae (see also [START_REF] Robert | Sur la répartition du noyau d'un entier[END_REF]formula (11.13)]):

¨(τ,t)∈R 2 e -Q(τ,t) dτ dt = 2π Hess[f y ](β 1 , β 2 ) 1 σ 2 , (5.8) ¨(τ,t)∈R 2 |τ | k |t| e -Q(τ,t) dτ dt k, 1 (σ 2 ) 1+(k+ )/2 , (k, ∈ N).
We integrate the quantity (5.7) over the square D 0 . By the symmetry of D 0 , the contribution of terms involving λ j and µ j vanishes. Therefore, using (5.8), we have

J 0 = -2 1 + O 1 ū x (β 1 +β 2 )/2 e γ(β 2 -β 1 ) F y (β 1 , β 2 ) (2π) 2 (β 1 + β 2 ) ¨(τ,t)∈D 0 e -Q(τ,t) dτ dt.
On the other hand, following again [RT13, section 11.3], we have (5.9) ¨max{|t|,|τ|}≥T 0 e -Q(τ,t) dτ dt

≤ ¨max{|t|,|τ|}≥T 0 exp - Hess[f y ] 4 τ 2 ∂ 02 f y + t 2 ∂ 20 f y dτ dt
where the partial derivatives of f y are evaluated at (β 1 , β 2 ). By Lemma 4, we have that Hess[f y ](β 1 , β 2 ) σ 2 2 and ∂ 20 (β 1 , β 2 ) σ 2 (similarly for ∂ 02 f y ). Therefore, for some c > 0, the right-hand side of (5.9) is

≤ ¨max{|t|,|τ|}≥T 0 e -cσ 2 (τ 2 +t 2 ) dτ dt e -cσ 2 T 2 0 T 0 σ 3/2 2 .
Since T 2 0 σ 2 ū1/3 , the above is certainly O((σ 2 ū) -1 ). We conclude that

J 0 = -2 1 + O 1 ū x (β 1 +β 2 )/2 e γ(β 2 -β 1 ) F y (β 1 , β 2 ) 2π(β 1 + β 2 ) Hess[f y ](β 1 , β 2 )
. Now by equations (4.28) and (4.29), we have

Hess[f y ](β 1 , β 2 ) = Hess[f y ](α, α) 1 + O v 2 ū , β 1 + β 2 = 2α + O v 2 log x ,
and by definition of we have Hess[f y ](α, α) = 2 φ 2 (α, y). We obtain (5.10)

J 0 = -1 + O 1 + v 2 ū x (β 1 +β 2 )/2 e γ(β 2 -β 1 ) F y (β 1 , β 2 ) 2π √ σ 2 α = -Ψ(x, y) 1 + O 1 + v 2 ū e R(v) √ 2π
by (4.27) and (1.16). This is our expected main term. We note that (5.11) J 0 Ψ(x, y)e R(v) .

Bounds away from the Taylor range. It remains to estimate J 0 , which is the contribution to I 1 (v) of those (τ, t) which satisfy max{|τ |, |t|} ≥ T 0 . By Corollary 1.(ii) and symmetry, we have

J 0 √ σ 2 x (β 1 +β 2 )/2 e γ(β 2 -β 1 ) F y (β 1 , β 2 ) α ˆT1 0 ˆT1 T 0 1+ τ 2 σ 2 y/ log y 1+ t 2 σ 2 y/ log y -cy/ log y dtdτ.
By (3.2), we have (5.12)

J 0 x (β 1 +β 2 )/2 e γ(β 2 -β 1 ) F y (β 1 , β 2 ) ūα √ σ 2 Ψ(x, y) e R(v) ū J 0 ū
On the other hand, we have

| arg(z)| = arg 1 -a 1 -b = arctan sin(µ 1 τ ) e µ 1 -cos(µ 1 τ ) -arctan sin(µ 2 t) e µ 2 -cos(µ 2 t) ≤ arctan(|τ |) + arctan(|t|).
Using (A.7) and Lemma 8.(e), we have

S 2 ≤ 0.15 + O(η) arctan(|τ |) + arctan(|t|) |τ | 1 + τ 2 + |t| 1 + t 2 ≤ 0.3 + O(η) τ 2 1 + τ 2 + t 2 1 + t 2 .
We finally obtain in this case

S ≤ -0.1 + O(η) τ 2 1 + τ 2 + t 2 1 + t 2 which is once again acceptable.
To conclude the proof of part (i) of Lemma 5, it remains to justify Lemma 8.

Proof of Lemma 8. Since r(1/z) = -r(z) and r(z) = r(z), and similarly for s(z), it suffices to consider the case when |z| ≥ 1 and arg(z

) ≥ 0. For all ω ∈ C, | Im ω| ≤ π/2, let (A.8) L(ω) := 1 (sinh ω) 2 - 1 ω 2 = k∈Z {0} 1 (ω + ikπ) 2 .
We shall prove that (A.9) -0.6 ≤ Re L(ω) ≤ 0,

Im L(ω) ≥ 0, (0 ≤ Re ω, 0 ≤ Im ω ≤ π/2.)
Let us first prove that (A.9) implies Lemma 8. Let z 1 , z 2 ∈ C R -. We have

r(z 2 ) -r(z 1 ) = 1 2 Re ˆ(log z 2 )/2 (log z 1 )/2 L(ω)dω = 1 2 ˆ(log |z 2 |)/2 (log |z 1 |)/2 Re L(t + i arg(z 1 )/2)dt - 1 2 ˆ(arg z 2 )/2 (arg z 1 )/2 Im L (log |z 2 |)/2 + it dt.
Assuming 1 ≤ |z 1 | ≤ |z 2 | and 0 ≤ arg z 1 ≤ arg z 2 , the integrals are respectively nonpositive and non-negative in view of (A.9), and we get in this case r(z 1 ) ≥ r(z 2 ). By setting z 1 = 1 and z 2 = z, it follows that r(z) ≤ 0. By setting z 1 = z and z 2 = -X + i0 and letting X → ∞, it follows that r(z) ≥ lim sup X→∞ r(-X + i0) = -1/2. Finally, in the case where |z| ≤ 2, setting z 1 = z and z 2 = -2 + i0, we obtain r(z) ≥ r(-2 + i0), which evaluates numerically to -0.0997 ± 10 -5 ≥ -1/10. This proves parts (a), (b) and (c) of the lemma. On the other hand, noting that L(ω) ∈ R when ω ∈ R, we obtain

s(z) = 1 2 ˆ(arg z)/2 0 Re L (log |z|)/2 + it dt.
The integrand being non-positive, we obtain 0 ≥ s(z) ≥ s(-|z| + i0) = - It remains to prove (A.9). We recall that L was defined in (A.8). Let ω = a + ib ∈ C be fixed with a ≥ 0 and 0 ≤ b ≤ π/2. We have

Re L(ω) ≤ L(ω) ≤ k∈Z {0} 1 |ω + ikπ| 2 ≤ k∈Z {0} 1 (b + kπ) 2 = 1 (sin b) 2 - 1 b 2 .
The last expression is easily seen to be maximal when b = π/2; its value at this point is 1 -4/π 2 ≤ 0.6.

Next, suppose a ≤ π/2. Then using the series representation (A.8), we have

Re L(ω) = k∈Z {0} a 2 -(b + kπ) 2 (a 2 + (b + kπ) 2 ) 2 ≤ 0.
If on the contrary a > π/2, then we have

Re L(ω) = Re 1 (sinh ω) 2 - 1 ω 2 = sinh 2 a cos(2b) -sin 2 b (sinh 2 a + sin 2 b) 2 - a 2 -b 2 (a 2 + b 2 ) 2 . This is obviously non-positive if π/4 ≤ b ≤ π/2. If b < π/4, then the above is ≤ 1 sinh 2 a - 1 a 2 φ(b/a),
where φ(t) := (1 -t 2 )/(1 + t 2 ) 2 . It is easily verified that φ(t) ≥ 0.48 ≥ (a/ sinh a) 2 for |t| ≤ 1/2 and a > π/2. Since indeed 0 ≤ b/a ≤ 1/2, it follows that Re L(ω) ≤ 0 as required.

We turn now to Im L(ω). Consider first the case a ≤ 4.9. Using the series representation (A.8) and grouping indices with same absolute values, we obtain Given that b ≤ π/2, the numerator is less than a 4 + (b 2 -π 2 )(2a 2 + b 2 + 3π 2 ); this last expression is maximal when b = π/2. At this point, it equals a 4 -3π 2 /2a 2 -39π 4 /16 which is negative by our assumption that 0 ≤ a ≤ 4.9; in view of (A.10), we have thus obtained Im L(ω) ≥ 0 when a ≤ 4.9. Suppose now on the contrary that a > 4.9. Then This concludes the proof of (A.9), hence of part (i) of Lemma 5. From the definition (4.4), we deduce that it will suffice to prove (A.11) Re(g(z)) ≥ O(1) (z ∈ C R -).

Proof of part (ii) of

Because g(z) = g(z) and g(1/z) = g(z), we may assume |z| ≥ 1 and arg z ≥ 0. In terms of w = 2 log z, this means Re w ≥ 0, Im w ∈ [0, π/2) and we have g(w) := g(e 2w ) = log sinh w w .

Notice that, with the definition (A.8), g (w) = -L(w).

From (A.9), we deduce Therefore, (A.11) holds and so does part (ii) of Lemma 5.

Proof of part (iii) of Lemma 5. First we note that since σ ≤ 0.6, by the previously done calculation (3.4), we have φ 2 (σ, y 1/2 ) y -0.2 φ 2 (σ, y) and similarly for κ. For large enough y, it will therefore be sufficient to show that under the stated conditions, there exists c(ε) > 0 such that Re Ξ(p -s , p -w ) -Ξ(p -σ , p -κ ) ≤ -c(ε) p σ (τ log p) 2 (p σ -1) 2 + p κ (t log p) 2 (p κ -1) 2

for y 1/2 ≤ p ≤ y (so that log p log y). We shall once again depart slightly from the notation used up to now. Relabelling a ← p -s , b = p -w and τ ← arg a, t ← arg b, we wish to show that for all ε > 0, there exists c = c(ε) > 0 such that for all (a, b) ∈ U 2 satisfying (A.12) max{|a|, |b|} ≤ 1 -ε, max{τ, t} ≤ c, (1 -|b|) 2 . We consider ε as being fixed, and let implicit constants depend of ε throughout the rest of the proof. Replacing τ by λτ and t by λt and differentiating with respect to λ as before, we see that it will be sufficient to show under the same assumptions (A.12), we have Suppose, then that the conditions (A.12) hold for some c > 0. In the same way as before, we write the left-hand side as S 1 + S 2 , where Here we used the fact that the quantities r(|z|) and s(z) are uniformly bounded. Note that ψ 2 (|z|) < 0. By the upper bound (λ -µ) 2 ≤ 2(λ 2 + µ 2 ) (λ, µ ∈ R), we get

S ≤ - |a|τ 2 (1 -|a|) 2 1 2 -ψ 1 (|z|)+2|a|ψ 2 (|z|) - |b|t 2 (1 -|b|) 2 1 2 +ψ 1 (|z|)+2|b|ψ 2 (|z|) +O(c(|a|τ 2 +|b|t 2 )).
It is easily seen that 1/2 -|ψ 1 (ρ)| + 2ψ 2 (ρ) > 0 for ρ > 0. Since |z| is bounded in terms of ε, it follows that both parentheses in the above expression are greater than some c = c (ε) > 0, so that

S ≤ -c + O(c) |a|τ 2 (1 -|a|) 2 + |b|t 2 (1 -|b|) 2 .
Then, choosing c sufficiently small relative to c , we have

S ≤ -c |a|τ 2 (1 -|a|) 2 + |b|t 2 (1 -|b|) 2
as required.

  > κ > 0), provided x ∈ N and e 2γ ∈ Q. Here F y (s, w) is the Dirichlet series relevant to our problem F y (s, w) := P (nd)≤y 1 τ (nd)n s d w , (Re(s), Re(w) > 0), and γ = v . One wishes to apply the saddle-point method for the double-integral in (1.20). A linear change of variables yields

Lemma 5 .

 5 (i) Whenever |σ -κ| ≤ c 1 σ, max{σ + |τ |, κ + |t|} ≤ c 1 /(log y),we have Re f y (s, w) -f y (σ, κ) ≤ -c 1 When σ ≤ c 1 / log y and |τ | ≤ e (log y) 3/2-ε , we have Re f y (s, w) -f y (σ, κ) ≤ -c 1

  may be relaxed to |τ | 1/ log y without changing the proof). Since the fraction in the right-hand side is an increasing function of |τ | and |τ | ≥ |t|, we obtain the required result. Finally, if α ≤ c 3 / log y and |τ | ≤ e (log y) 3/2-ε , exponentiating the upper bound of Lemma 5.(ii) immediately yields the desired result.

  where for each j, f j ∈ D m j ; and we write D r m = D m • • • D m (r times). Using the shorthand

,

  as well as the triangle inequality, we obtain (5.2)

π(

  log |z|) 2 +π 2 ≥ -1/π and this proves part (d) of the lemma. Moreover, by the triangle inequality, we obtain |s(z)| ≤ (arg z) 4 sup Re ω≥0 0≤Im ω≤π/2 | Re L(ω)| ≤ 0.15 × (arg z) and this proves part (e).

(A. 10 )

 10 Im L(ω) = -4ab k≥1 a 4 + (b 2 -(kπ) 2 )(2a 2 + b 2 + 3(kπ) 2 ) (a 2 + (b + kπ) 2 ) 2 (a 2 + (b -kπ) 2 ) 2 .

Im L(ω) = - 2

 2 sinh a cosh a sin b cos b (sinh 2 a + sin 2 b) 2 + 2ab (a 2 + b 2 ) 2 = -2ψ sin b sinh a cosh a cos b sinh 2 a + 2 a 2 ψ(b/a)where ψ(t) := t/(1 + t 2 ) 2 . It is easily established that ψ(t)/t ≥ 0.81 for 0 ≤ t ≤ 1/3, and ψ(t) ≤ t for t > 0. Since b/a ≤ 1/3, we obtainIm L(ω) ≥ -2 sin b cos b cosh a sinh 3 a + 1.62 × b a 3 .Using a > 4.9, we have a 3 (cosh a)/ sinh 3 a ≤ 0.03. Consequently, Im L(ω) ≥ b a 3 1.62 -0.06 ≥ 0.

  Lemma 5. We quote from [dlBT15, equation (5.2)] the boundζ(s, y) ζ(σ, y) 1 + τ σ log y -cεπ(y) (|τ | ≤ e (log y) 3/2-ε , 0 < σ 1/ log y).

000

  Re g (w) = -ˆRe w 0 Re L(t)dt + ˆIm w 0 Im L(Re w + it)dt ≥ 0, Im g (w) = -ˆIm w Re L(Re w + it)dt ≤ 0.6π/2 ≤ 1, so that Re g(z) = Re g(w) = ˆRe w Re g (t)dt -ˆIm w Im g (Re w + it)dt ≥ -π/2.

  a, b) -Ξ(|a|, |b|) ≤ -c |a|τ 2 (1 -|a|) 2 + |b|t 2

  Im τ a∂ 10 Ξ(a, b) + tb∂ 01 Ξ(a, b) ≤ -c |a|τ 2 (1 -|a|) 2 + |b|t 2 (1 -|b|) 2 .

  and z = (1 -a)/(1 -b). Here the function r(z) and s(z) are again defined by(A.4). Now we note thatIm -τ a 1 -a = -|a|τ sin τ (1 -|a|) 2 = -{1 + O(c)} |a|τ 2 (1 -|a|) 2 , Re τ a 1 -a = |a|τ (cos τ -|a|) (1 -|a|) 2 = {1 + O(c)} |a|τ 1 -|a| granted that c is small enough in terms of ε (note that |1 -a| = (1 + O(c))(1 -|a|)). Moreover, we have arg(1 -a) = arctan |a| sin τ 1 -|a| cos(τ ) = {1 + O(c)} |a|τ 1 -|a| = O(|a|τ ).By symmetry the same estimates hold for the analogous quantities for b. Note that by our hypotheses we have |z| = O(1), which impliesr(z) = r(|z|) + O(c), s(z) = (arg z) ∂ s(z ) ∂(arg z ) z =|z| + O(c) .Letψ 1 (ρ) := r(ρ) = -(ρ) := ∂s(z ) ∂(arg z ) z =ρ = ρ (ρ -1) 2 -1 (log ρ) 2 .We regroup the estimates above to obtainS 1 = |a|τ 2 (1 -|a|) 2 -1 2 + ψ 1 (|z|) + |b|t 2 (1 -|b|) 2 -1 2 -ψ 1 (|z|) + O c(|a|τ 2 + |b|t 2 ) , S 2 = -ψ 2 (|z|) -|a|τ 1 -|a| + |b|t 1 -|b| 2 + O(c(|a|τ + |b|t) 2 ).

Dickman's function ρ is the unique continuous function on R + which is differentiable on (1, ∞), satisfies ρ(u) = 1 for u ∈ [0, 1], and uρ (u)+ρ(u-1) = 0 for u > 1. We have ρ(u) = u -u+o(u) as u → ∞.

by (4.27) and (1.16). Regrouping our estimates (5.12) and (5.10), we conclude that (5.13)

Integration. Since I 1 (v) = I 1 (v m ) -´vm v I 1 (z)dz, estimates (5.13) and (5.6) imply

(5.14)

We regroup estimate (5.14) with (5.5) and (5.4) to obtain the required result

Proof of part (i) of Lemma 5. We shall actually prove that for all p ≤ y,

from which the lemma follows by summing over p ≤ y. For this proof, it will be more convenient to depart slightly from the notation used in the rest of this paper. We put

under the hypotheses

We abbreviate further

The equation (A.1) is trivially satisfied at τ = t = 0. Replacing τ by λτ and t by λt with λ varying in [0, 1], and then differentiating with respect to λ, we have that it will be sufficient to establish that under the same hypotheses,

The quantity on the LHS can be written as

Note that r(1/z) = -r(z) and s(1/z) = -s(z). We can write S = S 1 + S 2 , where

Given the hypotheses (A.2), we remark that we have

By symmetry we have the analog estimates for µ 2 tb/(1 -b). We shall use the following inequalities concerning the functions r and s.

On the other hand, using (A.7), Lemma 8.(d) and |τ | ≥ 3/2, we have

Adding the bounds for S 1 and S 2 , we obtain

which is acceptable granted that η is sufficiently small. Assume now |τ |, |t| ≤ 3/2. Then |z| ≤ {1 + O(η)} 1 + (3/2) 2 ≤ 2 if η is small enough; similarly |z| ≥ 1/2. Then proceeding as before, but this time using Lemma 8.(c), we obtain