On exponential sum over primes and application in Waring-Goldbach problem REN Xiumin
Department of Mathematics, The university of Hong Kong, Pokfulam road, Hong Kong (email:
[email protected])
Abstract In this paper, we prove the following estimate on exponential sums over primes: Let k ≥ 1, βk = 1/2 + log k/ log 2, x ≥ 1 and α = a/q + λ subject to (a, q) = 1, 1 ≤ a ≤ q, and λ ∈ R. Then à ! X p x Λ(m)e(αmk ) ¿ (d(q))βk (log x)c x1/2 q(1 + |λ|xk ) + x4/5 + p . q(1 + |λ|xk ) x σa . Let A(x) =
max
x/2≤n≤2x
|an |, x ≥ 1;
B(σ) =
∞ X |an | n=1
nσ
,
P∞
an n=1 ns
be absolutely
σ > σa .
Then for T ≥ 1 and for any s0 = σ0 + it0 and b > 0 with σ0 + b > σa , one has Z b+iT X an 1 xs = F (s + s) ds + R(x, T ), 0 s n0 2πi b−iT s n≤x
where, on writing kxk for the distance from x to the nearest integer N , ½ ¾ xb B(b + σ0 ) log x 1−σ0 R(x, T ) ¿ +x A(x) min 1, T T ¾ ½ x . +x−σ0 |aN | min 1, T kxk
(2.5)
This is Perron’s formula, a proof can be found, for example, in ref. [15]. Lemma 2.5. If there exist Mi , Mj with 1 ≤ i < j ≤ 5 such that Mi Mj > M 2/5 , then Lemma 2.1 is true. Proof. Without loss of generality, we may assume that i = 1 and j = 2. By applying Lemma 2.4 with T = M β , s0 = 1/2 + it and b = 1/2 + 1/ log M , we get µ ¶ 1 f1 + it, χ 2 ¶ Z 1/2+1/L+iM β µ 1 1 (2M1 )w − M1w =− L0 + it + w, χ dw + O(1), (2.6) 2πi 1/2+1/L−iM β 2 w where we have written L = log M . Now we move the integral leftward along the rectangular contour with vertices ±iM β , 1/2 + 1/L ± iM β to the line Re(w) = 0. Note that the integrand is regular inside the contour except for the simple pole at w = 1/2 − it when χ = χ0 . We have Y L(s, χ0 ) = ζ(s) (1 − p−s ), (2.7) p|q
and L0 (s, χ0 ) = ζ 0 (s)
Y X log p1 Y ¡ ¢ 1 − p−s . (1 − p−s ) + ζ(s) s p1 p|q p|q
p1 |q
p6=p1
So the residue at the simple pole is ) ( 1/2−it X log p1 Y ¡ ¢ (2M1 )1/2−it − M1 1 − p−1 , p1 1/2 − it p|q p1 |q
p6=p1
4
(2.8)
which can be estimated as ½X ¾½Y ¾ 1/2 1/2 M1 M1 (log log q)2 log p −1 ¿ (1 − p ) ¿ , p 1 + |t| 1 + |t| p|q
p|q
by elementary estimates X log p p|q
p
¿ log log q,
Y (1 − p−1 ) ¿ log log q. p|q
Thus (2.6) becomes (Z µ ¶ Z iM β Z 1/2+1/L+iM β ) −iM β 1 1 f1 + it, χ = − + + 2 2πi 1/2+1/L−iM β −iM β iM β Ã ! 1/2 δχ M1 L +O + O(1), 1 + |t|
(2.9)
where δχ = 1 or 0 according as χ = χ0 or not. Now we recall the following bound: For σ ≥ 1/2 and |t| ≥ 2, L(k) (σ + it, χ) ¿ (logk+2 q(|t| + 2)) max{1, q (1−σ)/2 (|t| + 2)1−σ }.
(2.10)
When χ 6= χ0 , this can be found, for example, in ref. [15], p.269, (13) and p.271, Exercise 6. When χ = χ0 , the above inequality can be derived from (2.7) and (2.8), by applying the well-known bound (see for example ref. [15], p.140, Theorem 2): ζ (k) (σ + it) ¿ (logk+1 |t|) max{1, |t|1−σ },
for σ ≥ 1/2 and |t| ≥ 2,
and the elementary estimates: For 0 < σ < 1, X X Y r−σ ¿ (d(q))1−σ , d−σ ≤ (1 − p−s ) ¿ d|q
p|q
(2.11)
r≤d(q)
and X log p1 Y ¡ ¢ 1 − p−s s p1 p|q p1 |q
p6=p1
¯ ¯ ¯ ¯ X Y X ¯ log p ¯¯ 1−σ −s ¯ ¿ (d(q)) (log q) r−σ (1 − p ) ¿¯ ¯ σ p ¯ ¯ p|q r≤log q p|q ¿ (d(q))1−σ (log q)2−σ .
(2.12)
Applying (2.10) with σ = 1/2 + u, we see that the contribution from the two horizontal segments in (2.9) is ¿ L3
max
0≤u≤1/2+1/L
q (1−(1/2+u))/2 M β(1−(1/2+u))
M1u ¿ M u(1−2β) L3 ¿ 1, Mβ
since q ≤ M 2β and β ≥ 1. Moreover, on the vertical segment from −iM β to iM β , one has 1 (2M1 )iv − M1iv ¿ . iv 1 + |v| 5
Therefore on writing Z g1 (t, χ) =
¯ µ ¶¯ ¯ dv ¯ 0 1 ¯L + it + iv, χ ¯¯ + 1, ¯ 2 1 + |v| β
Mβ
−M
(2.13)
we have µ f1
¶ 1/2 δχ M1 L 1 + it, χ ¿ g1 (t, χ) + . 2 1 + |t|
(2.14)
¶ 1/2 δχ M2 L 1 + it, χ ¿ g2 (t, χ) + , 2 1 + |t|
(2.15)
Similarly we have µ f2 where ¯ µ ¶¯ ¯ ¯ dv 1 ¯ L g2 (t, χ) = + it + iv, χ ¯¯ + 1. ¯ 2 1 + |v| −M β Z
Mβ
Write g3 (t, χ) =
10 Y
µ fj
j=3
1 + it, χ 2
¶
X
=
M3 ···M10
