Chapter 10. The singular integral Introducing S(n) and J(n)

Size: px

Start display at page:

Download "Chapter 10. The singular integral Introducing S(n) and J(n)"

Joella Goodman
5 years ago
Views:

1 Chaptr Th singular intgral Our aim in this chaptr is to rplac th functions S (n) and J (n) by mor convnint xprssions; ths will b calld th singular sris S(n) and th singular intgral J(n). This will b don in sction.. W shall show that th ordr of magnitud of th singular intgral is n 2 in sction.2.. Introducing S(n) and J(n) Dfin whr Thus, S(q) := a q gcd(a,q)= S(q, a) = S(q, a) 9 ( an/q), q (am 3 /q). m= S (n) = S(q) q. 9 q n /3 Lmma.. Lt a, q b coprim intgrs and ɛ any positiv ral. Thn S(a, q) ɛ q 3 4 +ɛ. 65

2 Proof. Th argumnt is an analogu of th diffrncing procss (s pags 5-53 of Chaptr 8). W hav S(q, a) 2 = ( ) a q (m3 m 3 2). h (mod q) m 2 (mod q) m (mod q) Th transformation m h givn by m h + m 2 (mod q) shows that S(q, a) 2 = ( ) a ( ) a q h3 q (3h (h m 2 + m 2 2)), hnc th triangl inquality givs S(q, a) 2 h (mod q) m 2 (mod q) Now Cauchy s inquality rvals that S(q, a) 4 q h (mod q) m 2 (mod q) m 2 (mod q) Th innr trm is ( ) a q (3h (h m 2 + m 2 2 2)) = m 2 (mod q) ( ) a. q (3h (h m 2 + m 2 2)) ( ) a q (3h (h m 2 + m 2 2 2)). m 2,m 3 (mod q) ( ) a q (3h (h (m 2 m 3 ) + (m 2 2 m 3 3))) and th substitution m 2 h 2 givn by m 2 hm 3 + h 2 (mod q) lads to ( 3 a ) q (h2 h 2 + h h 2 2) (6 aq ) h h 2 m 3. h 2 (mod q) m 3 (mod q) Not that th coprimality of a, q shows that th sum ovr m 3 quals q whn q divids 6h h 2 and vanishs othrwis. W obtain that S(q, a) 4 q 2 #{ h, h 2 q : q 6h h 2 }. Th intgrs 6h h 2 li in th rang [, 6q 2 ] and ar divisibl by q. Hnc thr xists i [, 6q] such that 6h h 2 = iq. Thrfor 6q S(q, a) 4 q 2 #{ h, h 2 q : 6h h 2 = iq}. i= 66

3 In ordr to hav 6h h 2 = iq both intgrs h, h 2 must divid iq and thr ar only τ(iq) 2 ɛ (iq) ɛ/2 q ɛ such pairs, whr ɛ is any positiv ral. This concluds our proof. Th last lmma shows that S(q) q 9 ɛ q + 4 +ɛ, thrfor th following sris, usually rfrrd to as th singular sris, (.) S(n) := q= S(q) q 9 convrgs absolutly and satisfis S(n) S (n) ɛ q + q>n /3 4 +ɛ n /3 dt t + 4 +ɛ ɛ n 2 +ɛ. This shows that (.2) S(n) and by (9.6) w obtain (.3) R (n) = (S(n) + O ɛ (n 2 +ɛ ))J (n). W nxt rplac J (n) by a mor suitabl intgral. For this w shall nd to nd th bhaviour of v(β) = 3 n m= (βm)/m2/3 in th rang β 2. Lmma.2. Lt β R with β 2. Thn v(β) min{n /3, β /3 }. Proof. If β is clos to thn th trms (βm) in th dfinition of v(β) rmain clos to. Hnc using th triangl inquality on dos not loos much information, v(β) 3 m n m n dt + O() n /3. t 2 3

4 If β /n thn β /3 > n /3, hnc th claim of our lmma is vidnt. In th rmaining cas β > /n w s that m / β (βm) m 2/3 m / β m 2/3 (/ β )/3, which is accptabl. W us partial summation to stimat th rmaining sum For this purpos w dfin for t R, A(t) := m t / β <m n (βm) m 2/3. (βm) = (βm) (β[t]) (β) and obsrv that th inquality (β) β, valid for β < /2, yilds A(t) β, with an implid constant that is indpndnt of t. Partial summation now givs / β <m n (βm) m 2/3 = A(n) n 2/3 A(/ β ) β 2/3 + n / β A(t) dt t 5/3, which is / β n 2/3 + β 2/3 β + β β 2/3 β /3. Dfin th following intgral (which is usually calld singular intgral), (.4) J(n) := /2 /2 and obsrv that Lmma.2 shows that J(n) /n v(β) 9 ( βn)dβ n 9/3 dβ + 68 /2 /n dβ β 3,

5 hnc (.5) J(n) n 2. Now rcall th dfinition of J (n) in (9.8). W hav J(n) J (n) = v(β) 9 ( βn)dβ, which according to Lmma.2 is /2 n +/3 β /2 n /3 /n β 3 dβ n 2 5. Using (.2), (.3) and (.5) w find an absolut constant δ > such that R (n) = S(n)J(n) + O(n 2 δ ), which whn combind with (8.8), (8.) and (9.5) yilds th following thorm. Thorm.3. W hav lim n + R(n) n 2 S(n)J(n) n 2 =..2 Th singular intgral Th Bta function is dfind as B(x, y) := t x ( t) y dt, for x, y >. Bfor rlating th singular intgral J(n) to th Bta function w nd som information on sums of monoton arithmtic functions. Lt f : Z R b any monotonic function. Comparing th sum y<n x f(n) with th intgral x f(t)dt y w s that x f(n) = f(t)dt + O( + f(y) + f(x) ). y<n x Thrfor if f : Z R is monotonic in ach intrval y (, x ), (x, x 2 ),..., (x k, x) 69

6 thn f(m) = n m n f(t)dt + O( + f() + f(n) + Th function f(x) := x β (n x) α, dfind for k f(x i ) ). i= β (, ], α β, has drivativ x β 2 (n x) α 2 ((β )n x(α + β 2)), which vanishs at and n and X =. If X (, n) thn β α+β 2 m n f(m) = n f(t)dt + O( + f(x) ) and if X / (, n) thn th rror trm is O(). Th substitution t y givn by t = ny shows that n and th rror trm is n α, thrfor (.6) n m= f(t)dt = n α+β B(β, α) m β (n m) α = n α+β ( B(β, α) + O(n β ) ). Bfor procding w nd to rcall a fw standard facts about th Gamma function. It is dfind as and satisfis Γ(t) := (.7) Γ() =, t x t dt for x > (.8) Γ(t + ) = tγ(t) for t >, (.9) B(x, y) = Γ(x)Γ(y) Γ(x + y) for x, y >. Obsrv that Γ(x) = (x )! for vry positiv intgr x. So B(x, y) = x + y xy for all positiv intgrs x, y. W hav th following thorm. ( x + y x ) 7

7 Thorm.4. W hav J(n) = Γ ( ) 9 4 n 2 ( + O(n /3 ) ). 3 2 Proof. W bgin by proving by induction that for vry intgr s 2, that on has (.) 3 s m,...,m s n m + +m s=n (m m s ) 2 3 = Γ ( ) s 4 ( s ) Γ ( s n 3 + O(n /3 ) ). 3 3 For s = 2 this is valid du to (.6) with α = β = /3, as wll as (.8) and (.9). Assuming that (.) is valid for som intgr s 2 thn quals which is Γ ( ) s 4 ( s ) Γ m s+ n 3m 2 3 s+ 3 s+ m,...,m s+ n m + +m s+ =n m n 3 s (m m s+ ) 2 3 m,...,m s n m + +m s=n m s+ m 3 m (n m) s 3 + O ( (m m s ) 2 3 m n, m /3 (n m) (s ) 3 du to th induction hypothsis. Using (.6) with β = and α = s, (s ) rspctivly for th main and th rror trm, w conclud th proof of (.). givs Combining (.4) and th dfinition of v, v(β) = 3 m n (βm) m 2 3 ) J(n) = 3 9 m,...,m 9 n (m m 9 ) (β(n m m 9 ))dβ. 7

8 Th intgral vanishs xcpt whn n m m 9 =, thus obtaining J(n) = 3 9 m,...,m 9 n m + +m 9 =n and according to (.) our thorm is valid. (m m 9 )

Recall that by Theorems 10.3 and 10.4 together provide us the estimate o(n2 ), S(q) q 9, q=1

Recall that by Theorems 10.3 and 10.4 together provide us the estimate o(n2 ), S(q) q 9, q=1 Chaptr 11 Th singular sris Rcall that by Thorms 10 and 104 togthr provid us th stimat 9 4 n 2 111 Rn = SnΓ 2 + on2, whr th singular sris Sn was dfind in Chaptr 10 as Sn = q=1 Sq q 9, with Sq = 1 a q gcda,q=1