Feb 1, 2008 - Correlation Conjecture on the zeros of the Riemann zeta function under ... Assuming Riemann Hypothesis, he proved in [8] that, as T â â,.
arXiv:math/0206293v1 [math.NT] 27 Jun 2002
More precise Pair Correlation Conjecture Tsz Ho Chan February 1, 2008 Abstract In this paper, we derive a more precise version of the Strong Pair Correlation Conjecture on the zeros of the Riemann zeta function under Riemann Hypothesis and Twin Prime Conjecture.
1
Introduction
In the early 1970s, H. Montgomery studied the distribution of the difference γ − γ ′ between the imaginary parts of the non-trivial zeros of the Riemann zeta function. Let X ′ 4 F (x, T ) = xi(γ−γ ) w(γ − γ ′ ) and w(u) = . (1) 4 + u2 0≤γ≤T 0≤γ ′ ≤T
Assuming Riemann Hypothesis, he proved in [8] that, as T → ∞, F (x, T ) ∼
T T (log T )2 log x + 2π 2πx2
for 1 ≤ x ≤ T (actually he only proved for 1 ≤ x ≤ o(T ) and the full range was done by Goldston [4]). He conjectured that F (x, T ) ∼
T log T 2π
for T ≤ x which is known as the Strong Pair Correlation Conjecture. From this, one has the (Weak) Pair Correlation Conjecture: Z sin πu 2 X T log T α du. 1− 1∼ 2π πu 0 ′ 0 0 and A ≥ 1 + ǫ, F (x, T ) =
T T T log − + O(T 1−ǫ1 ) 2π 2π 2π
(3)
holds uniformly for T 1+ǫ ≤ x ≤ T A with some ǫ1 > 0. It would be interesting to know how F (x, T ) changes from (2) to (3) when x is close to T . We have the following Theorem 1.1. Assume Riemann Hypothesis and Twin Prime Conjecture. For any small ǫ > 0 and any integer M > 2, Z T T 4x T /x sin v x2 X S(h) 1 − cos log x − dv + 2π 3π 0 v πT h2 x h≤H ∗ Z ∞ Z Z Ty B sin Txy x 4T ∞ f (y) sin Tx y 11 x ∞ sin x − dy + dy − dy + T 2π 1 y2 2 12 π 1 y4 π 1 y2 xy Ry Z Z Z ∞ 6T ∞ f (u) sin Tx y 2T ∞ 1 f (u)du sin Tx y dy + du T dy y + T π 1 y3 π 1 u4 y xy xy x1+6ǫ T + O(x1/2+7ǫ ) + O . + O T (log T )M−2
F (x, T ) =
for (logTT )M ≤ x ≤ T 2−ǫ. B = −C0 − log 2π and C0 is Euler’s constant 0.5772156649... H ∗ , S(h) and f (u) are defined as in the next section. The implicit constants may depend on ǫ and M . As corollaries of Theorem 1.1, we have Corollary 1.1. Assume Riemann Hypothesis and Twin Prime Conjecture. For any integer M > 2, F (x, T ) = for
T (log T )M
T T log x + O(x) + OM M−2 2π (log T )
≤ x ≤ T.
Corollary 1.2. Assume Riemann Hypothesis and Twin Prime Conjecture. For any small ǫ > 0 and any integer M > 2, F (x, T ) =
T T T T T 1/2−ǫ + Oǫ,M log − + Oǫ T M−2 2π 2π 2π x (log T )
for T ≤ x ≤ T 2−29ǫ .
2
2
Preparations
We mentioned Twin Prime Conjecture in the previous section. The form needed is the following: For any ǫ > 0, N X
Λ(n)Λ(n + d) = S(d)N + O(N 1/2+ǫ )
n=1
uniformly in |d| ≤ Q N . Λ(n) is the von Mangoldt lambda function. S(d) = Q p−1 1 2 p>2 1 − (p−1) 2 p|d,p>2 p−2 if d is even, and S(d) = 0 if d is odd. We also need a lemma concerning S(d). Lemma 2.1. For any ǫ > 0, h X
(h − k)S(k) =
k=1
1 2 1 h − h log h + Ah + O(h1/2+ǫ ) 2 2
where A = 12 (1 − C0 − log 2π) and C0 is Euler’s constant. Proof: This is a theorem in Montgomery and Soundararajan [9]. Borrowing from [6], Sα (y) :=
X
S(h)hα −
h≤y
and Tα (y) :=
y α+1 for α ≥ 0, α+1
X S(h) for α > 1. hα
h>y
Then from [3], 1 S0 (y) = − log y + O((log y)2/3 ). 2
(4)
Suppose S0 (y) = − 12 log y + ǫ(y). By partial summation, Z y 1 α yα , + ǫ(y)y α − α ǫ(u)uα−1 du + + Sα (y) = − 2α 2α α + 1 1
(5)
and 1 ǫ(y) 1 Tα (y) = − α − +α (α − 1)y α−1 y 2αy α
Z
Lemma 2.2. For any ǫ > 0, Z y B ǫ(u)du = y + O(y 1/2+ǫ ) 2 1 where B = −C0 − log 2π as in the previous section. 3
∞ y
ǫ(u) du. uα+1
(6)
Proof: By Lemma 2.1, Z y Z y X 1 S(h) − u + log u du ǫ(u)du = 2 1 1 h≤u
=
X
h≤y
= =
1 1 1 (y − h)S(h) − y 2 + y log y − y + 1 2 2 2
1 Ay − y + O(y 1/2+ǫ ) 2 B y + O(y 1/2+ǫ ). 2
Now, let us define f (y) :=
Z
y
ǫ(u) −
1
B du. 2
By integration by parts and Lemma 2.2, one has Z y Z y B 2 B B ǫ(u)udu = y + yf (y) − f (u)du − = y 2 + O(y 3/2+ǫ ), 4 4 4 1 1
(7)
and Z
y
∞
B f (y) ǫ(u) du = 2 − 3 + 3 u3 4y y
Z
y
∞
B f (u) du = 2 + O(y −5/2+ǫ ). u4 4y
(8)
Next, we are going to define a smooth weight ΨU (t). Fix a small positive real number ǫ and let K be a large integer depending on ǫ. Let M be an integer greater than 2 and U = (log T )M . We want ΨU (t) to have support in [−1/U, 1 + 1/U ], 0 ≤ ΨU (t) ≤ 1, ΨU (t) = 1 for 1/U ≤ t ≤ 1 − 1/U , and (j) ΨU (t) ≪ U j for j = 1, 2, ..., K. Let ∆ = 1/(2K U ). We define a sequence of functions as follow (which is Vinogradov’s construction) : 1, if 0 ≤ t ≤ 1, χ0 (t) = 0, else. Z ∆ 1 χi (t) = χi−1 (t + x)dx for i = 1, 2, ..., K + 1. 2∆ −∆ Clearly, 0 ≤ χi (t) ≤ 1 for 1 ≤ i ≤ K + 1. One can easily check by induction that χi (t) = 1 for 2i−1 ∆ ≤ t ≤ 1 − 2i−1 ∆, and χi (t) = 0 for t < −2i−1 ∆ or t > 1 + 2i−1 ∆ for i = 1, 2, ..., K + 1. (j)
(j)
Lemma 2.3. χi (t) exist and are continuous, and χi (t) ≤ ∆−j for 0 ≤ j ≤ i − 1 and 2 ≤ i ≤ K + 1.
4
Proof: Induction on i. First note that χ1 (t) is continuous because |χ1 (t + δ) − χ1 (t)|
= = = ≤
Z ∆ 1 Z ∆ 1 χ0 (t + δ + x)dx − χ0 (t + x)dx 2∆ −∆ 2∆ −∆ Z ∆ 1 Z ∆+δ 1 χ0 (t + x)dx − χ0 (t + x)dx 2∆ −∆+δ 2∆ −∆ Z −∆+δ 1 Z ∆+δ 1 χ (t + x)dx − χ0 (t + x)dx 0 2∆ ∆ 2∆ −∆ δ . ∆
Similarly, χ2 (t + h) − χ2 (t) h
= =
Z ∆+h Z −∆+h i 1h 1 1 χ1 (t + x)dx − χ1 (t + x)dx h 2∆ ∆ 2∆ −∆ 1 [χ1 (t + ∆ + ξ1 ) − χ1 (t − ∆ + ξ2 )] 2∆
for some 0 ≤ ξ1 , ξ2 ≤ h by mean-value theorem. So χ′2 (t) exists and equals to (j) 1 1 2∆ [χ1 (t + ∆) − χ1 (t − ∆)] which is continuous and ≤ ∆ . Assume that χi (t) (j) are continuous and satisfy χi ≪ ∆−j for some 2 ≤ i ≤ K and all 0 ≤ j ≤ i − 1. R ∆ (j) (j) 1 χ (t + x)dx ≤ ∆−j by induction Now, for 0 ≤ j ≤ i − 1, χi+1 (t) = 2∆ −∆ i hypothesis. For j = i, (i−1)
(i)
χi+1 (t) = = =
(i−1)
χi+1 (t + h) − χi+1 (t) h→0 h Z ∆+h Z −∆+h i 1 1h 1 (i−1) (i−1) (t + x)dx − (t + x)dx lim χi χi h→0 h 2∆ ∆ 2∆ −∆ 1 (i−1) (i−1) (t − ∆)] (t + ∆) − χi [χ 2∆ i lim
which is continuous and ≤ ∆−i by induction hypothesis. 2π∆y Lemma 2.4. χ ˆ0 (y) = eπiy sinπyπy and χ ˆi+1 (y) = χ ˆi (y) sin2π∆y for 0 ≤ i ≤ K. R∞ ˆ ˆ Here f (y) denotes the inverse Fourier transform of f (t), f (y) = −∞ f (t)e(yt)dt.
Note: We use inverse Fourier transform so that the notation matches with [5] and [6].
5
Proof: χ ˆ0 (y) =
R1
χ ˆi+1 (y)
0
e(yt)dt = =
Z
e2πiy −1 2πiy
= eπiy sinπyπy .
∞
χi+1 (t)e(yt)dt
−∞
= = =
1 2∆
Z
∆
−∆ Z ∆
Z
∞
χi (t + x)e(yt)dtdx
−∞
1 χ ˆi (y)e(−yx)dx 2∆ −∆ sin 2π∆y χ ˆi (y) e(−y∆) − e(y∆) =χ ˆi (y) . 2∆ −2πiy 2π∆y
Now we take ΨU (t) = χK+1 (t), then ΨU (t) has the required properties by the above discussion and Lemma 2.3. From Lemma 2.4, we know that ˆ U (y) = eπiy sin πy ( sin 2π∆y )K+1 . It follows that Ψ πy 2π∆y ˆ U (y) = ReΨ
sin 2πy sin 2π∆y K+1 , 2πy 2π∆y
ˆ U (y) ≪ y −K for y ≫ T ǫ , Ψ
(9)
ˆ U (T y) ≪ T −Kǫ for y ≫ τ −1 where τ = T 1−ǫ . and Ψ These are similar to (18) and (19) in [5]. Also, by Lemma 2.3, it follows from the discussion in [5] that ˆ U (y), Ψ ˆ ′U (y) ≪ min 1, ( U )K Ψ 2πy which is (17) in [5]. Consequently, the results in [5] are true with our choice of ΨU (t). Moreover, if one follows their arguments carefully, one has their Corollaries 1 & 2 (except that the error term may need to be modified by a factor of N ǫ ) and Theorem 3 as long as τ = T 1−ǫ ≤ x. We shall need the following lemmas concerning our weight function ΨU (t). Here we assume T ∆ ≤ x. Lemma 2.5. For any integer n ≥ 1, Z
∞ 1
1 ˆ U T y dy = x Re Ψ yn 2πx T
Z
1
6
∞
sin Txy 1 . dy + O K∆ log y n+1 ∆
Proof: By a change of variable v = Txy and (9), the left hand side T n−1 Z ∞ 1 sin v sin ∆v K+1 = dv n v x ∆v T /x v T n−1 Z ∞ 1 T n−1 Z 1/∆ sin v 2 2 (1 + O(K∆ v ))dv + O dv = n+1 n+1 x x T /x v 1/∆ v Z 1/∆ T n−1 Z 1/∆ sin v T n−1 T n−1 1 2 = dv + O dv + O K∆ ∆n n+1 n−1 x v x v x T /x T /x T n−1 Z ∞ sin v 1 dv + O K∆ log = n+1 x ∆ T /x v Z ∞ Ty sin x 1 x dy + O K∆ log = n+1 T 1 y ∆ because T ∆ ≤ x. Note that the error term comes from the case n = 2. If n 6= 2, then we can replace the error term by O(K∆). Lemma 2.6. If F (y) ≪ y −3/2+ǫ for y ≥ 1, then Z ∞ Z ∞ Ty sin T y ˆ F (y)ReΨU dy = F (y) T x dy + O(K∆). 2πx 1 1 xy Proof: By a change of variables v = Txy and (9), the left hand side Z x ∞ x sin v sin ∆v K+1 = dv F ( v) T T /x T v ∆v Z Z ∞ |F ( Tx v)| x x sin v x 1/∆ 1 + O(K∆2 v 2 ) dv + O dv F ( v) = T T /x T v T ∆K+1 1/∆ v K+2 Z x sin v x 1/∆ T 1/2−ǫ 3/2−ǫ F ( v) ∆ = dv + O K T T /x T v x Z x Z ∞ |F ( x v)| x ∞ x sin v T = dv + O dv + O(K∆) F ( v) T T /x T v T 1/∆ v Z ∞ sin T y F (y) T x dy + O(K∆). = 1 xy Finally, we need the following Lemma 2.7. Assume Riemann Hypothesis. For any ǫ > 0, X 1 1 Λ(n)2 n = x2 log x − x2 + O(x3/2+ǫ ) 2 4 n≤x
X Λ(n)2 1 log x 1 1 1 = + + O( 5/2−ǫ ) 3 2 2 n 2 x 4 x x n>x where the implicit constants may depend on ǫ. 7
Proof: By partial summation and the form of prime number theorem under Riemann Hypothesis.
3
Proof of main results
Throughout this section, we assume τ = T 1−ǫ ≤
T (log T )M
≤ x, U = (log T )M
for M > 2, H ∗ = τ −2 x2/(1−ǫ) , and ΨU (t) is defined as in the previous section. Keep in mind the ǫ and M dependency in the error terms. Proof of Theorem 1.1: Our method is that of Goldston and Gonek [5]. Let s = σ + it, X Λ(n) X Λ(n) and A∗ (s) := . A(s) := s n ns n>x n≤x
Assume Riemann Hypothesis, it follows from Theorem 3.1 of [1] with slight modification that F (x, T ) = Z x Z ∞ Z T 2 1 1 3 1 u1/2−it du + x A∗ ( + it) − u−3/2−it du dt A(− + it) − 2π 0 x 2 2 1 x +O((log T )3 ).
Inserting ΨU (t/T ) into the integral and extending the range of integration to the whole real line, we can get F (x, T ) = where
and
x1+6ǫ T (log T )2 x2 1 + O I (x, T ) + I (x, T ) + O 1 2 2πx2 2π U T
(10)
∞
Z x 2 t 1 ΨU u1/2−it du dt, I1 (x, T ) = A(− + it) − T 2 −∞ 1 Z
∞
Z ∞ 2 t ∗ 3 ΨU I2 (x, T ) = u−3/2−it du dt. A ( + it) − T 2 −∞ x Z
T T and T − U , This is essentially by Lemma 1 of [6] with modification that V = − U 2T and W = U . Riemann Hypothesis is assumed here so that the contribution from the cross term is estimated via Theorem 3 of [5].
Now, we assume the Twin Prime Conjecture in the previous section. By Corollary 1 of [5] (see also the calculations at the end of [5] and [6]) and Lemma
8
2.7, one has, I1 (x, T ) =
ˆ U (0)T Ψ
X
Λ2 (n)n
n≤x
T 3 Z ∞ +4π 2π T /2πx
X
h≤2πxv/T
=
ˆ U (v) dv S(h)h2 ReΨ v3
Z 2πxv/T T 3 Z ∞ ˆ U (v) dv −4π u2 du ReΨ 2π v3 T /2πτ x 0 x3+6ǫ +O + O(x5/2+7ǫ ) T 1 2 1 T x log x − T x2 2 4 Z 2πxv/T T 3 Z ∞ X ˆ U (v) dv u2 du ReΨ S(h)h2 − +4π 2π v3 0 T /2πx h≤2πxv/T
− =
4π 3 x 3
Z
T /2πx
ˆ U (v)dv + O ReΨ
T /2πτ x
1 1 2 T x log x − T x2 2 4 T 3 Z ∞ +4π 2π T /2πx
X
h≤2πxv/T
2 − x3 3
Z
T /x
0
x3+6ǫ T
2
S(h)h −
Z
+ O(x5/2+7ǫ )
0
2πxv/T
ˆ U (v) dv u2 du ReΨ v3
x3+6ǫ KT x2 sin v + O(x5/2+7ǫ ) + O dv + O v (log T )M T
because, from (9), Z
T /2πx
ˆ U (v)dv ReΨ
=
T /2πτ x
Z
T /2πx
0
=
1 2π
=
1 2π
T sin 2πv sin 2π∆v K+1 dv + O 2πv 2π∆v τx Z T /x T sin u (1 + O(K∆2 u2 ))du + O u τx 0 Z T /x K∆2 T 2 T sin u du + O . +O u x2 τx 0
9
Similarly, by Corollary 2 of [5] and Lemma 2.7, ˆ U (0)T I2 (x, T ) = Ψ
X Λ2 (n) n3 x