Explicit estimates for the distribution of numbers free of large prime factors
aa r X i v : . [ m a t h . N T ] M a y EXPLICIT ESTIMATES FOR THE DISTRIBUTION OFNUMBERS FREE OF LARGE PRIME FACTORS
JARED D. LICHTMAN AND CARL POMERANCE
Abstract.
There is a large literature on the asymptotic distribution of num-bers free of large prime factors, so-called smooth or friable numbers. But thereis very little known about this distribution that is numerically explicit. Inthis paper we follow the general plan for the saddle point argument of Hilde-brand and Tenenbaum, giving explicit and fairly tight intervals in which thetrue count lies. We give two numerical examples of our method, and withthe larger one, our interval is so tight we can exclude the famous Dickman–deBruijn asymptotic estimate as too small and the Hildebrand–Tenenbaum mainterm as too large. Introduction
For a positive integer n >
1, denote by P ( n ) the largest prime factor of n , andlet P (1) = 1. Let Ψ( x, y ) denote the number of n ≤ x with P ( n ) ≤ y . Such integers n are known as y -smooth, or y -friable. Asymptotic estimates for Ψ( x, y ) are quiteuseful in many applications, not least of which is in the analysis of factorizationand discrete logarithm algorithms.One of the earliest results is due to Dickman [7] in 1930, who gave an asympoticformula for Ψ( x, y ) in the case that x is a fixed power of y . Dickman showed thatΨ( x, y ) ∼ xρ ( u ) ( y → ∞ , x = y u )(1.1)for every fixed u ≥
1, where ρ ( u ) is the “Dickman–de Bruijn” function, defined tobe the continuous solution of the delay differential equation uρ ′ ( u ) + ρ ( u −
1) = 0 ( u > ,ρ ( u ) = 1 (0 ≤ u ≤ . There remain the questions of the error in the approximation (1.1), and also thecase when u = log x/ log y is allowed to grow with x and y . In 1951, de Bruijn [4]proved that Ψ( x, y ) = xρ ( u ) (cid:16) O ε (cid:16) log(1 + u )log y (cid:17)(cid:17) holds uniformly for x ≥
2, exp { (log x ) / ε } < y ≤ x , for any fixed ε >
0. Afterimprovements in the range of this result by Maier and Hensley, Hildebrand [12]showed that the de Bruijn estimate holds when exp( { (log log x ) / ε } ) ≤ y ≤ x .In 1986, Hildebrand and Tenenbaum [13] provided a uniform estimate for Ψ( x, y )for all x ≥ y ≥
2, yielding an asymptotic formula when y and u tend to infinity. Date : October 5, 2018.
The starting point for their method is an elementary argument of Rankin [17] from1938, commonly known now as Rankin’s “trick”. For complex s , define ζ ( s, y ) = X n ≥ P ( n ) ≤ y n − s = Y p ≤ y (1 − p − s ) − (where p runs over primes) as the partial Euler product of the Riemann zeta function ζ ( s ). Then for 0 < σ <
1, we have(1.2) Ψ( x, y ) = X n ≤ xP ( n ) ≤ y ≤ X P ( n ) ≤ y ( x/n ) σ = x σ ζ ( σ, y ) . Then σ can be chosen optimally to minimize x σ ζ ( σ, y ).Let φ j ( s, y ) = ∂ j ∂s j log ζ ( s, y ) . The function φ ( s, y ) = − X p ≤ y log pp s − α = α ( x, y ) to φ ( α, y ) + log x = 0 gives theoptimal σ in (1.2). We also denote σ j ( x, y ) = | φ j ( α ( x, y ) , y ) | .In this language, Hildebrand and Tenenbaum [13] proved that the estimateΨ( x, y ) = x α ζ ( α, y ) α p πσ ( x, y ) (cid:16) O (cid:16) u + log yy (cid:17)(cid:17) holds uniformly for x ≥ y ≥
2. As suggested by this formula, quantities α ( x, y ) and σ ( x, y ) are of interest in their own right, and were given uniform estimates whichimply the formulae α ( x, y ) ∼ log(1 + y/ log x )log y and σ ( x, y ) ∼ (cid:16) xy (cid:17) log x log y, together which implyΨ( x, y ) ∼ x α ζ ( α, y ) √ πu log( y/ log x ) (if y/ log x → ∞ ) , Ψ( x, y ) ∼ x α ζ ( α, y ) p πy/ log y (if y/ log x → . These formulae indicate that Ψ( x, y ) undergoes a “phase change” when y is oforder log x , see [3]. This paper concentrates on the range where y is considerablylarger, say y > (log x ) .The primary aim of this paper is to make the Hildebrand–Tenenbaum methodexplicit and so effectively construct an algorithm for obtaining good bounds forΨ( x, y ). UMBERS FREE OF LARGE PRIMES, EXPLICITLY 3
Explicit Results.
Beyond the Rankin upper bound Ψ( x, y ) ≤ x α ζ ( α, y ), wehave the explicit lower boundΨ( x, y ) ≥ x − log log x/ log y = x (log x ) u due to Konyagin and Pomerance [11]. Recently Granville and Soundararajan [10]found an elementary improvement of Rankin’s upper bound, which they have gra-ciously permitted us to include in an appendix in this paper. In particular, theyshow that Ψ( x, y ) ≤ . y − σ x σ ζ ( σ, y ) / log x for every value of σ ∈ [1 / log y, x, y ) for given x, y asinputs. For example, using an accuracy parameter c , Bernstein [2] created analgorithm to generate bounds B − ( x, y ) ≤ Ψ( x, y ) ≤ B + ( x, y ) with B − Ψ ≥ − log xc log 3 / log 2 and B + Ψ ≤ xc log 3 / log 2 , running in O (cid:16) y log y + y log x log y + c log x log c (cid:17) time. Parsell and Sorenson [15] refined this algorithm to run in O (cid:16) c y / log y + c log x log c (cid:17) time, as well as obtaining faster and tighter bounds assuming the Riemann Hy-pothesis. The largest example computed by this method was an approximation ofΨ(2 , ).As seen in Figure 1, the bounds presented in this paper far outshine best-knownupper and lower bounds for the two examples presented. We also provide themain term estimates x α ζ ( α, y ) /α √ πσ from [13] and ρ ( u ) x from [7] as points ofreference. It is interesting that our estimates in the second example are closerto the truth than are the Dickman–de Bruijn and Hildebrand–Tenenbaum mainterms. The second-named author has asked if Ψ( x, y ) ≥ xρ ( u ) holds in generalfor x ≥ y ≥
2, see [9, (1.25)]. This inequality is known for u bounded and x sufficiently large, see the discussion in [14, Section 9]. JARED D. LICHTMAN AND CARL POMERANCE
Figure 1.
Examples. x y KP 1 . · . · R 4 . · . · GS 5 . · . · DD 2 . · . · HT 2 . · . · Ψ − . · . · Ψ + . · . · Here,KP is the Konyagin–Pomerance lower bound x/ (log x ) u , R is the Rankin upper bound x α ζ ( α, y ) , GS is the Granville–Soundararajan upper bound 1 . y − α x α ζ ( α, y ) / log x, DD is the Dickman–de Bruijn main term ρ ( u ) x, andHT is the Hildebrand–Tenenbaum main term x α ζ ( α, y ) / ( α √ πσ ) . Our principal result, which benefits from some notation developed over the courseof the paper, is Theorem 3.11. It is via this theorem that we were able to estimateΨ(10 , ) and Ψ(10 , ) as in the table above.2. Plan for the paper
The basic strategy of the saddle-point method relies on Perron’s formula, whichimplies the identity Ψ( x, y ) = 12 πi Z σ + i ∞ σ − i ∞ ζ ( s, y ) x s s ds, for any σ >
0. It turns out that the best value of σ to use is α = α ( x, y ) discussedin the Introduction. We are interested in abridging the integral at a certain height T and then approximating the contribution given by the tail. To this end, we have(2.1) Ψ( x, y ) = 12 πi Z α + iTα − iT ζ ( s, y ) x s s ds + Error . There is a change in behavior occurring in ζ ( s, y ) when t is on the order 1 / log y .In [13] it is shown that (cid:12)(cid:12)(cid:12) ζ ( s, y ) ζ ( α, y ) (cid:12)(cid:12)(cid:12) = Y p ≤ y (cid:12)(cid:12)(cid:12) − p − α − p − s (cid:12)(cid:12)(cid:12) = Y p ≤ y (cid:16) − cos( t log p )) p α (1 − p − α ) (cid:17) − / ≤ exp n − X p ≤ y − cos( t log p ) p α o . (2.2)Thus when t is small (compared to 1 / log y ) the oscillatory terms are in resonance,and when t is large the oscillatory terms should exhibit cancellation. This behavior UMBERS FREE OF LARGE PRIMES, EXPLICITLY 5 suggests we should divide our range of integration into | t | ≤ T and T < | t | < T ,where T ≈ / log y is a parameter to be optimized.The contribution for | t | ≤ T will constitute a “main terrm”, and so we will tryto estimate this part very carefully. In this range we forgo (2.2) and attack theintegrand ζ ( s, y ) x s /s directly. The basic idea is to expand φ ( s, y ) = log ζ ( s, y ) as aTaylor series in t . This approach, when carefully done, gives us fairly close upperand lower bounds for the integral. In our smaller example, the upper bound is lessthan 1% higher than the lower bound, and in the larger example, this is better bya factor of 20. Considerably more noise is encountered beyond T and in the Errorin (2.1).For the second range T < | t | < T , we focus on obtaining a satisfactory lowerbound on the sum over primes, X p ≤ y − cos( t log p ) p α . Our strategy is to sum the first L terms directly, and then obtain an analyticformula W ( y, w ) to lower bound the remaining terms starting at some w ≥ L ,where essentially W ( y, w ) = y − α − w − α − α + error . With an explicit version of Perron’s formula, the Error in (2.1) may be handledby (cid:12)(cid:12)
Error (cid:12)(cid:12) ≤ x α X P ( n ) ≤ y n α min (cid:16) , πT | log( x/n ) | (cid:17) ≤ x α X P ( n ) ≤ yT | log( x/n ) | >T d n α πT | log( x/n ) | + X P ( n ) ≤ yT | log( x/n ) |≤ T d (cid:16) xn (cid:17) α ≤ x α ζ ( α, y ) πT d + e αT d − h Ψ( xe T − d , y ) − Ψ( xe − T − d , y ) i . Here d ≈ is a parameter of our choosing, which we set to balance the two termsabove. Thus the problem of bounding | Error | is reduced to estimating the numberof y -smooth integers in the “short” interval (cid:0) xe − T − d , xe T − d (cid:3) .This latter portion is better handled when T is large, but the earlier portion inthe range [ T , T ] is better handled when T is small. Thus, T is numerically set tobalance these two forces.In our proofs we take full advantage of some recent calculations involving theprime-counting function π ( x ) and the Chebyshev functions ψ ( x ) = X p m ≤ x log p, ϑ ( x ) = X p ≤ x log p, with p running over primes and m running over positive integers. As a corollary ofthe papers [5], [6] of B¨uthe we have the following excellent result. Proposition 2.1.
For ≤ x ≤ we have . √ x ≤ x − ϑ ( x ) ≤ . √ x. JARED D. LICHTMAN AND CARL POMERANCE
We have | ϑ ( x ) − x | x < . · − , when x > , . · − , when x > e , . · − , when x > e , . · − , when x > e . Proof.
The first assertion is one of the main results in B¨uthe [6]. Let H be a numbersuch that all zeros of the Riemann zeta-function with imaginary parts in [0 , H ] lieon the 1 / x/ log x ≤ H / . and x ≥ | ϑ ( x ) − x | x < (log x −
2) log x π √ x . We can take H = 3 · , see Platt [16]. Thus, we have the result in the range10 ≤ x ≤ e . For x ≥ e we have from B¨uthe [5] that | ψ ( x ) − x | /x ≤ . · − .Further, we have (see [18, (3.39)]) for x > ψ ( x ) ≥ ϑ ( x ) > ψ ( x ) − . x / − x / . (This result can be improved, but it is not important to us.) Thus, for x ≥ e we have | ϑ ( x ) − x | /x ≤ . · − , establishing our result in this range. For thelatter two ranges we argue similarly, using | ψ ( x ) − x | ≤ . · − when x ≥ e and | ψ ( x ) − x | ≤ . · − for x ≥ e , both of these inequalities coming from[5]. (cid:3) We remark that there are improved inequalities at higher values of x , found in[5] and [8], which one would want to use if estimating Ψ( x, y ) for larger values of y than we have done here. 3. The main argument
As in the Introduction, for complex s , define ζ ( s, y ) = X n ≥ P ( n ) ≤ y n − s = Y p ≤ y (1 − p − s ) − , which is the Riemann zeta function restricted to y -smooth numbers, and for j ≥ φ j ( s, y ) = ∂ j ∂s j log ζ ( s, y ) . UMBERS FREE OF LARGE PRIMES, EXPLICITLY 7
We have the explicit formulae, φ ( s, y ) = − X p ≤ y log pp s − ,φ ( s, y ) = X p ≤ y p s log p ( p s − ,φ ( s, y ) = − X p ≤ y ( p s + p s ) log p ( p s − ,φ ( s, y ) = X p ≤ y ( p s + 4 p s + p s ) log p ( p s − ,φ ( s, y ) = − X p ≤ y ( p s + 11 p s + 11 p s + p s ) log p ( p s − . Note that for y ≥ σ > φ ( σ, y ) is strictly increasing from 0, so there is a uniquesolution α = α ( x, y ) > x + φ ( α, y ) = 0 . Since we cannot exactly solve this equation, we shall assume any choice of α thatwe use is a reasonable approximation to the exact solution, and we must take intoaccount an upper bound for the difference between our value and the exact value.We denote φ j = φ j ( α, y ) , σ j = | φ j | = ( − j φ j , B j = B j ( t ) = σ j t j /j !so that the Taylor series of φ ( s, y ) = log ζ ( s, y ) about s = α is φ ( α + it, y ) = X j ≥ σ j j ! ( − it ) j = X j ≥ ( − i ) j B j . Our first result, which is analogous to Lemma 10 in [13], sets the stage for ourestimates.
Lemma 3.1.
Let < d < and T > . We have that (cid:12)(cid:12)(cid:12) Ψ( x, y ) − πi Z α + iTα − iT ζ ( s, y ) x s s ds (cid:12)(cid:12)(cid:12) ≤ x α ζ ( α, y ) πT d + e αT d − h Ψ( xe T d − , y ) − Ψ( xe − T d − , y ) i . Proof.
We have12 πi Z α + iTα − iT ζ ( s, y ) x s s ds = 12 πi Z α + iTα − iT X P ( n ) ≤ y (cid:16) xn (cid:17) s dss = X P ( n ) ≤ y πi Z α + iTα − iT (cid:16) xn (cid:17) s dss , where the interchange of sum and integral is justified since ζ ( s, y ) is a finite product,hence uniformly convergent as a sum. JARED D. LICHTMAN AND CARL POMERANCE
By Perron’s formula (see [1, § (cid:12)(cid:12)(cid:12) πi Z α + iTα − iT (cid:16) xn (cid:17) s dss (cid:12)(cid:12)(cid:12) ≤ ( x/n ) α min (cid:16) , πT | log( x/n ) | (cid:17) if n > x, (cid:12)(cid:12)(cid:12) − πi Z α + iTα − iT (cid:16) xn (cid:17) s dss (cid:12)(cid:12)(cid:12) ≤ ( x/n ) α min (cid:16) , πT | log( x/n ) | (cid:17) if n ≤ x. Together these imply (cid:12)(cid:12)(cid:12) Ψ( x, y ) − πi Z α + iTα − iT ζ ( s, y ) x s s ds (cid:12)(cid:12)(cid:12) ≤ x α X P ( n ) ≤ y n α min (cid:16) , πT | log( x/n ) | (cid:17) ≤ x α X P ( n ) ≤ y | log( x/n ) | >T d − n α πT | log( x/n ) | + x α X P ( n ) ≤ y | log( x/n ) |≤ T d − n α ≤ x α ζ ( α, y ) πT d + e αT d − h Ψ( xe T d − , y ) − Ψ( xe − T d − , y ) i . This completes the proof. (cid:3)
In using this result we have the problems of performing the integration from α − iT to α + iT and estimating the number of y -smooth integers in the interval (cid:0) xe − T d − , xe T − d (cid:3) . We turn first to the integral evaluation.Recall that B j = B j ( t ) = σ j ( x, y ) t j /j ! and let B ∗ = B ∗ ( t ) = t log x − B ( t ).Note that B ∗ = 0 if α is chosen perfectly. Lemma 3.2.
For s = α + it , we have Re n ζ ( s, y ) x s s o = x α ζ ( α, y ) α + t (cid:0) α cos( B + B ∗ + b ) + t sin( B + B ∗ + b ) (cid:1) exp (cid:8) − B + B + a (cid:9) , where a , b are real numbers, depending on the choice of t , with | a + ib | ≤ B ( t ) .Proof. We expand φ ( α + it, y ) = log ζ ( α + it, y ) in a Taylor series around t = 0.There exists some real ξ between 0 and t such that φ ( α + it, y ) = φ ( α, y ) + itφ − t φ − it φ + t φ − i t
5! ( α + iξ, y )= B − iB − B + iB + B − i t φ ( α + iξ, y ) . Since ζ ( s, y ) = exp( φ ( s, y )), we obtain ζ ( s, y ) x s s = ζ ( α, y ) x α α + it exp n it log x − iB − B + iB + B + i t φ ( α + iξ, y ) o = x α ζ ( α, y ) α + it exp n − B + B + i ( B ∗ + B ) + i t φ ( α + iξ, y ) o . Letting iφ ( α + iξ ) t /
5! = a + b i , we have ζ ( s, y ) x s s = x α ζ ( α, y ) α + t ( α − it ) (cid:0) cos( B ∗ + B + b ) + i sin( B ∗ + B + b ) (cid:1) exp (cid:8) − B + B + a (cid:9) , UMBERS FREE OF LARGE PRIMES, EXPLICITLY 9 and taking the real part gives the result. (cid:3)
The main contribution to the integral in Lemma 3.1 turns out to come from theinterval [ − T , T ], where T is fairly small. We have12 πi Z α + iT α − iT ζ ( s, y ) x s s ds = 12 π Z T − T ζ ( α + it, y ) x α + it α + it dt. Note that the integrand, written as a Taylor series around s = α , has real coeffi-cients, so the real part is an even function of t and the imaginary part is an oddfunction. Thus, the integral is real, and its value is double the value of the integralon [0 , T ].Consider the cosine, sine combination in Lemma 3.2: f ( t, v ) := α cos( B ( t ) + v ) + t sin( B ( t ) + v ) , and let v ( t ) = | B ∗ ( t ) | + B ( t ) . We have, for each value of t , the constraint that | v | ≤ v ( t ). The partial derivativeof f ( t, v ) with respect to v is zero when arctan( t/α ) − B ( t ) ≡ π ). Let u ( t ) = arctan( t/α ) − B ( t ) . If u ( t ) [ − v ( t ) , v ( t )], then f ( t, v ) is monotone in v on that interval; otherwise ithas a min or max at u ( t ). Let T , T , T , T be defined, respectively, as the leastpositive solutions of the equations u ( t ) = v ( t ) , u ( t ) = − v ( t ) , u ( t ) + π = v ( t ) , u ( t ) + π = − v ( t ) . Then 0 < T < T < T < T . We have the following properties for f ( t, v ):(1) For t in the interval [0 , T ] we have f ( t, v ) increasing for v ∈ [ − v ( t ) , v ( t )],so that f ( t, − v ( t )) ≤ f ( t, v ) ≤ f ( t, v ( t )) . (2) For t in the interval [ T , T ], we have f ( t, v ) increasing for − v ( t ) ≤ v ≤ u ( t )and then decreasing for u ( t ) ≤ v ≤ v ( t ). Thus,min { f ( t, − v ( t )) , f ( t, v ( t )) } ≤ f ( t, v ) ≤ f ( t, u ( t )) . (3) For t ∈ [ T , T ], f ( t, v ) is decreasing for v ∈ [ − v ( t ) , v ( t )], so that f ( t, v ( t )) ≤ f ( t, v ) ≤ f ( t, − v ( t )) . (4) For t ∈ [ T , T ], we have f ( t, v ) decreasing for v ∈ [ − v ( t ) , u ( t ) + π ] andincreasing for v ∈ [ u ( t ) + π, v ( t )]; that is, f ( t, u ( t ) + π ) ≤ f ( t, v ) ≤ max { f ( t, − v ( t )) , f ( t, v ( t )) } . Note too that f ( t, v ) has a sign change from positive to negative in the interval[ T , T ]. Let Z − , Z + be, respectively, the least positive roots of f ( t, v ( t )) = 0, f ( t, − v ( t )) = 0.Let I +0 be an upper bound for the function appearing in Lemma 3.2 on [0 , T ]using | a | , | b | ≤ B and the above facts about f ( t, v ), and let I − be the corre-sponding lower bound. We choose a = B in I +0 when the cos, sin combination ispositive, and a = − B when it is negative. For I − , we choose a in the reverseway. Let(3.1) J +0 = Z T I +0 ( t ) dt, J − = Z T I − ( t ) dt. We thus have the following result, which is our analogue of Lemma 11 in [13].
Lemma 3.3.
We have x α ζ ( α, y ) π J − ≤ πi Z α + iT α − iT ζ ( s, y ) x s s ds ≤ x α ζ ( α, y ) π J +0 . In order to estimate the integral in Lemma 3.1 when | t | > T we must knowsomething about prime sums to y . Lemma 3.4.
We have (cid:12)(cid:12)(cid:12) Z α + iTα + iT ζ ( s, y ) x s s ds (cid:12)(cid:12)(cid:12) ≤ x α ζ ( α, y ) J , where J := Z TT exp (cid:0) − W ( y, , t ) (cid:1) dt √ α + t and (3.2) W ( v, w, t ) := X w
For 0 ≤ v ≤ < t , equation (3.14) in [13] states that(1 + 4 vt/ ( t − ) − ≤ exp {− v/t } . Applied to (3.17) in [13] with v = (1 − cos( t log p )) /
2, we have that (cid:12)(cid:12)(cid:12) ζ ( s, y ) ζ ( α, y ) (cid:12)(cid:12)(cid:12) = Y p ≤ y (cid:12)(cid:12)(cid:12) − p − α − p − s (cid:12)(cid:12)(cid:12) = Y p ≤ y (cid:16) − cos( t log p )) p α (1 − p − α ) (cid:17) − / ≤ exp n − X p ≤ y − cos( t log p ) p α o . (3.3)This completes the proof. (cid:3) Our goal now is to find a way to estimate W ( v, w, t ). The following result isanalogous to Lemma 6 in [13]. Lemma 3.5.
Let s be a complex number, let < w < v , and define F s ( v, w ) := X w
e . Proof. (i) By partial summation, X w
Lemma 3.6.
For t ∈ R , z > , and β = 1 − α , let δ z := t log z − arctan( t/β ) . (i) For ≤ w < v ≤ we have that W ( v, w, t ) ≥ W ( v, w, t ) , where W ( v, w, t ) log v = v β − w β β − v β cos δ v − w β cos δ w p β + t − v / − α + w / − α ) − α + | s | ) w / − α − v / − α α − / . (ii) For ≤ w < v we have that W ( v, w, t ) ≥ W ( v, w, t ) , where W ( v, w, t ) log v = v β − w β β − v β cos δ v − w β cos δ w p β + t − ε w (cid:0) v β + w β (cid:1) − ε w ( α + | s | ) (cid:16) v β − w β β (cid:17) . Proof.
We apply Lemma 3.5 with s = 1 − β and s = 1 − β + it , and take the realpart of the difference. Letting the difference of the sums be S , we have that S : = X w
Suppose that w, L satisfy , L ≤ w . If y ≤ , then J ≤ Z TT exp (cid:0) − W ∗ ( y, w, t ) − W ( L, , t ) (cid:1) dt √ α + t . If y > e and , L ≤ w ≤ , let W = W ∗ (10 , w, t ) , W = W ∗ ( e , , t ) , W = W ∗ ( e , e , t ) ,W = W ∗ ( e , e , t ) , W = W ∗ ( y, e , t ) . Then J ≤ Z TT exp (cid:0) − W − W − W − W − W − W ( L, , t ) (cid:1) dt √ α + t . We remark that if 10 < y ≤ e , then there is an appropriate inequality for J involving fewer W j ’s. If y is much larger than our largest example of y = 10 , onemight wish to use better approximations to ϑ ( y ) than were used in Proposition 2.1. Proof.
If 1427 ≤ w < v and [ w, v ] satisfy the hypotheses of Lemma 3.5, we have W ( v, w, t ) = W ( v/e ⌊ log( v/w ) ⌋ , w, t ) + ⌊ log( v/w ) ⌋− X j =0 W ( v/e j , v/e j +1 , t ) ≥ W ( v/e ⌊ log( v/w ) ⌋ , w, t ) + ⌊ log( v/w ) ⌋− X j =0 W ( v/e j , v/e j +1 , t ) . The result then follows from Lemma 3.4. (cid:3)
Remark . We implement Lemma 3.7 by choosing L as large as possible so as notto interfere overly with numerical integration. We have found that L = 10 workswell. The ratio e in the definition of W ∗ is convenient, but might be tweaked forslightly better results. The individual terms in the sum W ( L, , t ) are as in (3.2),except for the first 30 primes, where instead we forgo using the inequality in (3.3),using instead the slightly larger expression12 log (cid:16) − cos( t log p )) p α (1 − p − α ) (cid:17) . We choose w as a function w ( t ) in such a way that the bound in Lemma 3.6 isminimized. For simplicity, we ignore the oscillating terms, i.e., we set ∂∂w h − w β /β − w / − α + 2( α + | s | ) w / − α / (1 / − α ) i = − w β − − w − / − α / (1 / − α ) + 2( α + | s | ) w − / − α equal to 0. Multiplying by w / α and solving for w gives w ( α, t ) := (cid:16) α − / α + 2 p α + t (cid:17) . We let w ( t ) := max { L, w ( α, t ) } . Our next result, based on [13, Lemma 9], gives a bound on the number of y -smooth integers in a short interval. Lemma 3.9.
Let < d < , T > be such that z := ( e T d − − − > . We have Ψ( xe T − d , y ) − Ψ( xe − T − d , y ) ≤ e α / z − αT d − x α ζ ( α, y ) r eπ J z . where, with W ( y, w, t ) as in Lemma 3.6, J := Z ∞ exp n − t z − W ( y, , t ) o dt. Proof.
Let ξ = xe − T d − , so that(3.5) Ψ( xe T d − , y ) − Ψ( xe − T d − , y ) = Ψ( ξ + ξ/z, y ) − Ψ( ξ, y ) . For ξ < n ≤ ξ + ξz , we have that1 > ξn ≥ (cid:16) z (cid:17) − , so 0 > log( ξ/n ) ≥ − log(1 + 1 /z ) ≥ − z , which implies that 0 < [ z log( ξ/n )] ≤ . Thus,Ψ( ξ + ξ/z, y ) − Ψ( ξ, y ) = X P ( n ) ≤ yξ 12 [ z log( ξ/n )] o . For σ, v ∈ R , we have the formula e − v / = 1 √ π e σ / − σv Z + ∞−∞ exp n − t + it ( σ − v ) o dt Letting σ = α/z , v = − z log( ξ/n ), we obtainΨ( ξ + ξ/z, y ) − Ψ( ξ, y ) ≤ r e π X P ( n ) ≤ yξ We have Ψ( ξ + ξ/z, y ) − Ψ( ξ, y ) ≤ ξ α ζ ( α, y ) e α / z r eπ J z , and the lemma now follows from (3.5) and the definition of ξ . (cid:3) Remark . For t large, say t > z log z , we can ignore the term W ( y, , t ) in J , getting a suitably tiny numerical estimate for the tail of this rapidly convergingintegral. The part for t small may be integrated numerically with w ( t ) , L as inRemark 3.8.With these lemmas, we now have our principal result. Theorem 3.11. Let d, T, z be as in Lemma 3.9, let J ± be as in (3.1) , J as inLemma 3.4, and J as in Lemma 3.9. We have Ψ( x, y ) ≥ x α ζ ( α, y ) π (cid:16) J − − J − T − d − e α / z √ πe J z (cid:17) and Ψ( x, y ) ≤ x α ζ ( α, y ) π (cid:16) J +0 + J + T − d + e α / z √ πe J z (cid:17) . Computations In this section we give some guidance on how, for a given pair x, y , the numbers α , ζ ( α, y ), and σ j for j ≤ x, y ) via Theorem 3.11.4.1. Computing α . Given a number a ∈ (0 , 1) and a large number y we mayobtain upper and lower bounds for the sum σ ( a, y ) = X p ≤ y log pp a − . First, we choose a moderate bound w ≤ y where we can compute the sum σ ( a, w )relatively easily, such as w = 179 , , X w
Suppose f ( t ) is positive and f ′ ( t ) is negative on [ w , w ] . Supposetoo that t − √ t < ϑ ( t ) ≤ t on [ w , w ] . Then Z w w (1 − / √ t ) f ( t ) dt + ( w − ϑ ( w ) − √ w ) f ( w ) ≤ X w
We record our calculations of α and the numbers σ j for two examples.Note that we obtain bounds for ζ via σ = log ζ . Figure 2. Data. x y α . . ζ , ± 16 2 . · ± · σ ∗ . · − . · − σ , . ± . 03 71 , . ± . σ , . ± . , , . ± . σ , , ± , , ± σ +5 . · . · Note that σ ∗ is an upper bound for | σ − log x | , and σ +5 is an upper bound for σ .The functions α ( x, y ) and σ j ( x, y ) are of interest in their own right. A simpleobservation from their definitions allows for more general bounds on α and σ j usingthe data in Figure 2, as described in the following remark. Remark . For pairs x, y and x ′ , y ′ , if x ≥ x ′ and y ≤ y ′ then α ( x, y ) ≤ α ( x ′ , y ′ ).Similarly, if α ( x, y ) ≥ α ( x ′ , y ′ ) and y ≤ y ′ then σ j ( x, y ) ≤ σ j ( x ′ , y ′ ). UMBERS FREE OF LARGE PRIMES, EXPLICITLY 17 A word on numerical integration. The numerical integration needed toestimate J , J is difficult, especially when we choose a large value of L , like L =10 . We performed these integrals independently on both Mathematica and Sageplatforms. It helps to segment the range of integration, but even so, the softwarecan report an error bound in addition to the main estimate. In such cases we havealways added on this error bound and then rounded up, since we seek upper boundsfor these integrals. In a case where one wants to be assured of a rigorous estimate,there are several options, each carrying some costs. One can use a Simpson ormidpoint quadrature with a mesh say of 0 . . 1, where on each interval and for each separate cosineterm appearing, the maximum contribution is calculated. If this is done with T = 4 · and L = 10 , there would be magnitude 10 of these calculations.The extreme value of the cosine contribution would either be at an endpoint of aninterval or − π mod 2 π . We have donea mild form of this method in our estimation of the integrals J ± .4.5. Example estimates. We list some example values of x, y and the correspond-ing estimates in the figure below. Figure 3. Results. x y T . . T . . Z − . . Z + . . T . . T . . T · d . 57 0 . J − . · − . · − J +0 . · − . · − J . · − . · − J . · − . · − Ψ − . · . · Ψ + . · . · Appendix We prove the following theorem. Theorem 5.1 (Granville and Soundararajan) . If ≤ y ≤ x and / log y ≤ σ ≤ ,then Ψ( x, y ) ≤ . y − σ log x x σ ζ ( σ, y ) . Proof. By the identity log n = P d | n Λ( d ), we have X n ≤ xP ( n ) ≤ y log n = X m ≤ xP ( m ) ≤ y X d ≤ x/mP ( d ) ≤ y Λ( d ) = X m ≤ xP ( m ) ≤ y X p ≤ min { y,x/m } log p j log( x/m )log p k ≤ X m ≤ xP ( m ) ≤ y π (cid:0) min { y, x/m } (cid:1) log( x/m ) . Thus,Ψ( x, y ) log x = X n ≤ xP ( n ) ≤ y (log n + log( x/n )) ≤ X n ≤ xP ( n ) ≤ y (cid:0) π (min { y, x/n } ) (cid:1) log( x/n ) . Using the estimates in [18] we see that the maximum of (1 + π ( t )) / ( t/ log t ) occursat t = 7, so that 1 + π ( t ) < . t/ log t for all t > 1. The above estimate then givesΨ( x, y ) log x < . X x/y 1, then y − σ ( x/n ) σ ≥ ( x/n, if x/y < n ≤ x,y log( x/n ) / log y, if n ≤ x/y. Indeed, in the first case, since t − σ is non-decreasing in t , we have ( x/n ) − σ ≤ y − σ . And in the second case, since t − σ log t is decreasing in t for t ≥ y , we have( x/n ) − σ log( x/n ) ≤ y − σ log y .We thus haveΨ( x, y ) log x < . X n ≤ xP ( n ) ≤ y y − σ ( x/n ) σ < . y − σ x σ ζ ( σ, y ) . This completes the proof. (cid:3) Acknowledgments We warmly thank Jan B¨uthe, Anne Gelb, Habiba Kadiri, Dave Platt, BradRodgers, Jon Sorenson, Tim Trudgian, and John Voight for their interest and help.We are also very appreciative of Andrew Granville and Kannan Soundararajan forallowing us to include their elementary upper bound prior to the publication oftheir book. The first author was partially supported by a Byrne Scholarship atDartmouth. 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