More precise Pair Correlation Conjecture

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Feb 1, 2008 - Correlation Conjecture on the zeros of the Riemann zeta function under ... Assuming Riemann Hypothesis, he proved in [8] that, as T → ∞,.
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

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