Pair correlation and distribution of prime
numbers
by
Tszho Chan
A dissertation submitted in partial fulfillment
of the requirements for the degree of
Doctor of Philosophy
(Mathematics)
in The University of Michigan
2002
Doctoral Committee:
Professor Hugh L. Montgomery, Chair
Professor David E. Barrett
Professor Emeritus Donald J. Lewis
Professor Romesh Saigal
Associate Professor Kannan Soundararajan
ACKNOWLEDGEMENTS
Firstly, I would like to thank my advisor, Professor Montgomery, for his guidance
and help on my thesis work. Secondly, I would like to thank all the Professors who
teach me and guide me in both mathematics and teaching. Thirdly, I want to thank
my fellow classmates and friends for their care and support. Lastly, I would like to
express my gratitude towards my dissertation committee for their time and effort.
ii
TABLE OF CONTENTS
ACKNOWLEDGEMENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
ii
LIST OF FIGURES
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v
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vi
NOTATION
CHAPTER
I. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.1
1.2
1.3
1.4
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1
2
3
4
II. “Equivalence” of three formulas . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
2.1
2.2
2.3
2.4
Pair correlation . . . . . . . . . . . . . . .
Primes in short interval . . . . . . . . . . .
Beyond Pair Correlation . . . . . . . . . .
More precise Pair Correlation Conjectures
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Pair Correlation Conjecture
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V. “Equivalence” of Odd Moments . . . . . . . . . . . . . . . . . . . . . . . . . . .
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37
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Introduction . . . . . . . . . . .
Results on λ-th moments . . . .
“Equivalence” for odd moments
Numerical evidence . . . . . . .
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5.1
5.2
5.3
5.4
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IV. “Equivalence” of Higher Even Moments . . . . . . . . . . . . . . . . . . . . .
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A brief introduction . . . . . . . . . . . . . . . . .
Some preparations . . . . . . . . . . . . . . . . . .
Main result for k-th moments . . . . . . . . . . .
Its converse . . . . . . . . . . . . . . . . . . . . . .
Concluding remarks and some numerical evidence
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4.1
4.2
4.3
4.4
4.5
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III. Beyond Pair Correlation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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when k is even
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5
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26
Introduction . . . . . . . . . . . . .
An asymptotic formula . . . . . . .
A more precise formula for µk (N, h)
Numerical evidence . . . . . . . . .
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3.1
3.2
3.3
3.4
The starting point . . . . . . . .
Preparation for the more precise
Proof of main result . . . . . . .
Conclusion . . . . . . . . . . . .
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54
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VI. More precise Pair Correlation Conjectures . . . . . . . . . . . . . . . . . . . .
6.1
6.2
6.3
6.4
6.5
6.6
6.7
Introduction . . . . . . . . . . . . . . . . . . . . . . . .
Some preparations . . . . . . . . . . . . . . . . . . . . .
A formula for F (x, T ) . . . . . . . . . . . . . . . . . . .
Improvement on asymptotic formula when 1 ≤ x ≤ logT T
Preparation for the case (logTT )M ≤ x . . . . . . . . . .
Asymptotic formula of F (x, T ) when (logTT )M ≤ x . . .
Conclusion . . . . . . . . . . . . . . . . . . . . . . . . .
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65
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APPENDIX . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
91
BIBLIOGRAPHY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
99
iv
LIST OF FIGURES
Figure
6.1
graph of rα,δ
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88
6.2
graph of rα,2δ − rα−δ,2δ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
90
v
NOTATION
A, B, B 0 , C, ...
absolute constants
k, n
integers
C0
Euler’s constant 0.5772156649...
², ²0 , ²1 , ²2
small positive real numbers
p
prime number
ζ(s)
Riemann zeta function
γ, γ 0
imaginary parts of non-trivial zeros of ζ(s)
Λ(n)
log p if n is some power of prime p; 0 otherwise
P
n≤x Λ(n)
ψ(x)
£ ¤
x
greatest integer less than x
a|b
a divides b
(a, b)
the greatest common divisor of a and b
µ(n)
Mobius function
φ(n)
Euler’s phi function
e(x)
e2πix
f (x) ∼ g(x)
limx→∞
f (x)
g(x)
=1
f (x) = Oθ (g(x)) |f (x)| ≤ Cθ g(x), where Cθ is some constant depending on θ
f (x) ¿θ g(x)
same as f (x) = Oθ (g(x))
f (x) = o(g(x))
limx→∞
a≈b
C1 b ≤ a ≤ C2 b for some constants C1 , C2 > 0
f (x)
g(x)
=0
vi
CHAPTER I
Introduction
1.1
Pair correlation
Riemann Hypothesis tells us that every non-trivial zero of the Riemann zeta
function is of the form ρ =
1
2
+ iγ. In the early 1970s, H. Montgomery studied the
distribution of the difference γ − γ 0 between the zeros. Let
F (X, T ) =
X
0
X i(γ−γ ) w(γ − γ 0 ) and w(u) =
0≤γ≤T
0≤γ 0 ≤T
4
.
4 + u2
He proved in [13] that, as T → ∞,
F (X, T ) =
³
´
T
T
T
2
2
log X +
(log
T
)
+
o
T
log
X
+
(log
T
)
2π
2πX 2
X2
for 1 ≤ X ≤ T , and conjectured that
F (X, T ) =
T
log T + o(T log T )
2π
for T ≤ X which is known as the Strong Pair Correlation Conjecture.
Rescale the imaginary parts γ1 ≤ γ2 ≤ . . . of the zeros { 12 + iγ} of ζ(s) by
γ̃j =
1
γ
2π j
log γj so that the mean spacing is 1. Thus,
T → ∞. Let R(a, b) = limN →∞
1
#{(j1 , j2 )
N
1
#{j
T
≥ 1 : γ̃j < T } → 1 as
: 1 ≤ j1 , j2 ≤ N, γ̃j1 − γ̃j2 ∈ (a, b)} be
the limiting distribution of pairs of zeros provided that the limit exists. The Strong
1
2
Pair Correlation Conjecture implies that
Z b³
R(a, b) =
1−
a
¡ sin 2πu ¢2 ´
du + δ(a, b)
2πu
where δ(a, b) = 1 if 0 ∈ [a, b], δ(a, b) = 0 otherwise. This is known as the Weak Pair
Correlation Conjecture.
F. J. Dyson remarked that this is what one would expect if the zeros of the
Riemann zeta function behaved like the eigenvalues of a random matrix from the
Gaussian Unitary Ensemble (GUE). H. Montgomery suggested that one can extend
the study to the k-tuple correlation of the zeros of the zeta function. In 1994, Hejhal
[12] computed the three-level correlation function for zeros of ζ(s). Later, Rudnick
and Sarnak [19] extended the result to n-level correlation of the non-trivial zeros of
primitive L-functions.
1.2
Primes in short interval
Goldston and Montgomery [9] showed that, under Riemann Hypothesis, the Strong
Pair Correlation Conjecture and the following assertions are equivalent to each other:
As X → ∞,
1.
Z
X¡
1
2.
Z
X¡
1
¢2
X
ψ(x + h) − ψ(x) − h dx ∼ hX log
for X ² ≤ h ≤ X 1−² .
h
¢2
1
1
ψ(x + δx) − ψ(x) − δx dx ∼ δX 2 log for X −1+² ≤ δ ≤ X −² .
2
δ
This suggests that ψ(x + h) − ψ(x) has mean h and variance h log X/h , as x
varies 1 ≤ x ≤ X with h in the range X ² ≤ h ≤ X 1−² . Meanwhile, Cramér’s model
predicts that the variance should be h log X instead. Numerical evidence suggests
that the Strong Pair Correlation Conjecture is the more accurate model when h is
3
big. Moreover, Odlyzko [18] and Forrester & Odlyzko [2] analyzed the distribution
of the zeros of the zeta function and found that the data is in close agreement with
the Weak Pair Correlation Conjecture.
Recently, Montgomery and Soundararajan [14] proposed a more precise asymptotic formula:
Z X
¡
¢2
X
(1.1)
ψ(x + h) − ψ(x) − h dx = hX log + BhX + smaller terms
h
1
for X ² ≤ h ≤ X 1−² where B = −C0 − log 2π and C0 is Euler’s constant. On the
other hand, Goldston and Yildirim [10] proved more generally that
Z 2x ³
q
X
h ´2
ψ(y + h; q, a) − ψ(y; q, a) −
dy
φ(q)
x
a=1
(a,q)=1
³
xq
π X log p ´
− hx C0 + log +
+ O² (h2 + h1−² x)
h
2
q−1
p|q
P
≤ x1/2−2² , where ψ(x; q, a) = n≤x,n≡a(mod q) Λ(n). They assumed the
= hx log
for q 4² ≤
h
q
usual quantitative form of Twin Prime Conjecture (see Hypothesis HL in Chapter
III with r = 2).
Our first goal in this thesis is using (1.1) to get more precise asymptotic formulas
RX
for F (X, T ) and 1 (ψ(x + δx) − ψ(x) − δx)2 dx. Goldston mentioned something
about these in [6] already. This is the content of Chapter II.
1.3
Beyond Pair Correlation
Montgomery and Soundararajan [15] considered the higher moments:
Z X
¡
¢k
ψ(x + h) − ψ(x) − h dx.
µk (X, h) =
1
They showed that, under Hardy-Littlewood Prime k-tuple Conjecture (same as Hypothesis HL in Chapter III),
³
¡
¢
X ´k/2
µk (X, h) = ck + o(1) Xhk/2 log
as X → ∞,
h
4
for X ² ≤ h ≤ X 1/k−² . Here ck are the moments of the normalized normal variable:


k!

, if k is even,
(k/2)!2k/2
ck =

 0,
if k is odd.
This leads us to expect that the distribution of ψ(x + h) − ψ(x), for 0 ≤ x ≤ X,
is approximately normal with mean h and variance h log X/h. Another goal of this
thesis is to get a more precise understanding of µk (X, h), as well as obtaining its
RX
“equivalence” form for 1 (ψ(x+δx)−ψ(x)−δx)k dx. These are achieved in Chapters
III, IV and V.
1.4
More precise Pair Correlation Conjectures
Finally, based on the result of Montgomery [13] and our work in Chapter II,


 T log X + O(X log X) + O(T ), if T ² ≤ X ≤ T ,
2π
F (X, T ) =

 T log T − T + O(T 1−²0 ),
if T 1+² ≤ X.
2π
2π
2π
This arouses our curiosity on how F (X, T ) behaves when X is near T . The final
goal of this thesis is to study the second main terms for F (X, T ) which bridge the
transition when X is near 1 and T . Using these, one can then derive a more precise
Weak Pair Correlation Conjecture. These are done in Chapter VI.
CHAPTER II
“Equivalence” of three formulas
2.1
The starting point
Montgomery and Soundararajan [14] considered possible cancellations among the
error terms in Twin Prime Conjecture. They propose a more precise asymptotic
formula: As X → ∞,
Z
X¡
1
¢2
X
ψ(x + h) − ψ(x) − h dx = hX log + BhX + o(hX)
h
for X ² ≤ h ≤ X 1−² , where B = −C0 − log 2π and C0 is Euler’s constant. This
more precise form suggests that one may be able to get a more precise form of the
Strong Pair Correlation Conjecture of Montgomery [13] by employing the method of
Goldston and Montgomery [9].
First, we would like to quote the following results by Saffari and Vaughan [20],
under Riemann Hypothesis:
Z
X¡
(2.1)
¢2
ψ(x + δx) − ψ(x) − δx dx ¿ δX
1
2
³
2 ´2
log
δ
for 0 < δ ≤ 1, and
Z
X¡
(2.2)
1
2X ´2
ψ(x + h) − ψ(x) − h dx ¿ hX log
h
¢2
for 0 < h ≤ X.
5
³
6
Theorem II.1. Assume Riemann Hypothesis. For every 0 < ² < 1/2, if
Z
X¡
(2.3)
1
´
³
¢2
X
hX
ψ(x + h) − ψ(x) − h dx = hX log + BhX + O
h
(log X/h)5
holds uniformly for X ² ≤ h ≤ X 1−² , then for every 0 < ² < 1/2,
Z
X¡
(2.4)
1
³ δX 2 ´
¢2
1
1
ψ(x + δx) − ψ(x) − δx dx = δX 2 log + CδX 2 + O
2
δ
log 1/δ
holds uniformly for X −1+² ≤ δ ≤ X −² . Here C = (1 + B)/2.
Proof: The method is that of Saffari and Vaughan [20] employed in Goldston and
Montgomery [9]. For ²1 > 0, we want to deduce from (2.3) that
Z ∆Z X
¡
¢2
ψ(x + δx) − ψ(x) − δx dxdδ
0
1
(2.5)
³ ∆2 X 2 ´
1
1
= ∆2 X 2 log + K∆2 X 2 + O
4
∆
(log 1/∆)5
for X −1+²+²1 ≤ ∆ ≤ X −²−²1 . Here K = 3/8 + B/4. One can accomplish this by
showing
Z
∆
0
(2.6)
=
3³1
4 4
Z
X
¡
¢2
ψ(x + δx) − ψ(x) − δx dxdδ
X/2
∆2 X 2 log
´
³ ∆2 X 2 ´
1
+ K∆2 X 2 + O
∆
(log 1/∆)5
for X −1+² ≤ ∆ ≤ X −² . We substitute h = δx. Then the left hand side of (2.6)
becomes
(2.7)
Z ∆X Z
∆X/2
X
h/∆
f (x, h)2
dxdh +
x
Z
∆X/2
X ²/2
Z
X
X/2
f (x, h)2
dxdh +
x
Z
X ²/2
0
Z
X
X/2
f (x, h)2
dxdh
x
where f (x, h) = ψ(x + h) − ψ(x) − h. By integration by parts, we have from (2.3)
that
Z
V
U
¢
h¡
f (x, h)2
dx =
(log V )2 − (log U )2 + h(1 + B)(log V − log U )
x
2
´
³
h
−h log h(log V − log U ) + O
(log X/h)5
7
provided that V ²/2 ≤ h ≤ U 1−²/2 . Here we need (2.3) to hold uniformly for X ² ≤
h ≤ X 1−² for every ². Using this, the first term of (2.7) becomes
³ log 2
8
−
3´ 2 2
(log 2)2 2 2 (3 + 2B) log 2 2 2
∆ X log ∆ −
∆X −
∆X
16
16
16
³9
³ ∆2 X 2 ´
3B ´ 2 2
+
+
∆ X +O
32
16
(log 1/∆)5
h 1−²/2
if X ²/2 ≤ h ≤ ( ∆
)
when
∆X
2
≤ h ≤ ∆X. One can easily check that this is okay
when X −1+² ≤ ∆ ≤ X −² . The second term of (2.7) becomes
³ ∆2 X 2 ´
log 2 2 2
(log 2)2 2 2 (3 + 2B) log 2 2 2
−
∆ X log ∆ +
∆X +
∆ X +O
8
16
16
(log 1/∆)5
if X ²/2 ≤ h ≤ ( X2 )1−²/2 when X ²/2 ≤ h ≤
∆X
.
2
Again, one can check that this is okay
when X −1+² ≤ ∆ ≤ X −² . By (2.2), the third term of (2.7) is
¿ X ²/2 X ² (log X)2 ¿
∆2 X 2
.
(log 1/∆)5
Combining these, we have (2.6) and hence (2.5) by replacing X by X2−k in (2.6),
summing over 0 ≤ k ≤ M = [ 7 logloglog2 X ] and by appealing to (2.1) with X replaced by
X2−M −1 . Note that when X −1+²+²1 ≤ ∆ ≤ X −²−²1 , (X2−k )−1+² ≤ ∆ ≤ (X2−k )−²
for 0 ≤ k ≤ M . That is why we can use (2.6) repeatedly.
We now deduce (2.4) from (2.5). For X −1+²+2²1 ≤ ∆ ≤ X −²−2²1 , we set η =
1
.
(log 1/∆)4
By (2.5) and Taylor’s expansion for log (1 + η), we have
Z
(2.8)
(1+η)∆
∆
Z
X
1
1 2 2
1 ³
1 ´ 2 2
f (x, δx) dxdδ = η∆ X log + 2Kη − η ∆ X
2
∆
4
³
³ ∆2 X 2 ´
1´
2 2 2
+
O
η
∆
X
log
.
+O
(log 1/∆)5
∆
2
Let g(x, δx) = f (x, ∆x). By the identity
f 2 − g 2 = 2f (f − g) − (f − g)2
8
and Cauchy’s inequality, we find that
Z (1+η)∆ Z X
f (x, δx)2 − g(x, δx)2 dxdδ
∆
1
³Z Z
´1/2 ³Z Z ¡
´1/2
¢2
2
(2.9)
¿
f (x, δx) dxdδ
f (x, δx) − g(x, δx) dxdδ
Z Z
¡
¢2
+
f (x, δx) − g(x, δx) dxdδ.
But f (x, δx) − g(x, δx) = f ((1 + ∆)x, (δ − ∆)x), so
Z (1+η)∆ Z X
¡
¢2
f (x, δx) − g(x, δx) dxdδ
∆
1
(2.10)
Z η∆ Z (1+∆)X
1+∆
=
0
f (x, δx)2 dxdδ ¿ η 2 ∆2 X 2 log
1+∆
1
∆η
by (2.5), our choice for η and the range for ∆. Hence by (2.8), (2.9) and (2.10), we
have
Z
X¡
¢2
Z
(1+η)∆
Z
X
g(x, δx)2 dxdδ
ψ(x + ∆x) − ψ(x) − ∆x dx =
η∆
1
∆
1
Z (1+η)∆ Z X
³
1´
=
f (x, δx)2 dxdδ + O η 3/2 ∆2 X 2 log
∆
∆
1
´
³
³ ∆2 X 2 ´
³
1 2 2
1´
1
1
=
η∆2 X 2 + O η 3/2 ∆2 X 2 log
+O
η∆ X log + 2K −
.
2
∆
4
∆
(log 1/∆)5
We divide both sides by η∆, and obtain
Z
X¡
1
1
1 ³
1´
2
ψ(x + ∆x) − ψ(x) − ∆x dx = ∆X log + 2K −
∆X 2
2
∆
4
¢2
³
³
´
1´
∆X 2
1/2
2
+O η ∆X log
+O
.
∆
η(log 1/∆)5
Recall η =
Z
X¡
1
as 2K −
1
,
log (1/∆)4
we have, for X −1+²+2²1 ≤ ∆ ≤ X −²−2²1 ,
³ ∆X 2 ´
¢2
1
1
ψ((1 + ∆)x) − ψ(x) − ∆x dx = ∆X 2 log + C∆X 2 + O
2
∆
log 1/∆
1
4
=
1+B
2
= C. Since, ²1 is arbitrary, we have the theorem.
One can also carry out a similar calculation from (2.4) to (2.3). Then one obtains
the following converse with the help of (2.2):
9
Theorem II.2. Assume Riemann Hypothesis. For every 0 < ² < 1/2, if
Z
X¡
(2.11)
1
³ δX 2 ´
¢2
1
1
ψ(x + δx) − ψ(x) − δx dx = δX 2 log + CδX 2 + O
2
δ
(log 1/δ)5
holds uniformly for X −1+² ≤ δ ≤ X −² , then for every 0 < ² < 1/2,
Z
X¡
(2.12)
1
³ hX ´
¢2
X
ψ(x + h) − ψ(x) − h dx = hX log + BhX + O
h
log X/h
holds uniformly for X ² ≤ h ≤ X 1−² . Again C = (1 + B)/2.
The error terms for (2.3), (2.4), (2.11) and (2.12) are reasonable because, for
instance, from Montgomery and Soundararajan [14], one expects that the error term
0
for (2.3) and (2.12) to be O(hX 1−² ). Similarly, one would expect the error term for
0
(2.4) and (2.11) to be O(δX 2−² ). Thus, it would be nice if we can have variations
of Theorem II.1 and Theorem II.2 with these (smaller) error terms. But, in order to
do that, one has to be more careful about the argument and one has to shorten the
range for δ and h a little bit, respectively.
Theorem II.3. Assume Riemann Hypothesis. For every 0 < ² < 1/2, if, for some
²1 > 0 (²1 may depend on ²),
Z
X¡
(2.13)
1
¢2
X
ψ(x + h) − ψ(x) − h dx = hX log + BhX + O(hX 1−²1 )
h
holds uniformly for X ² ≤ h ≤ X 1−² , then
Z
X¡
(2.14)
1
¢2
1
1
ψ(x + δx) − ψ(x) − δx dx = δX 2 log + CδX 2 + O(δX 2−²2 )
2
δ
holds uniformly for X −1+²+2²1 ≤ δ ≤ X −² /2 for some ²2 > 0. Here C = (1 + B)/2.
Proof: Let f (x, h) = ψ(x + h) − ψ(x) − h. Let X −1+²+²1 ≤ ∆ ≤ X −² . By
integration by parts, we have from (2.13) that
Z
V
U
¢
f (x, h)2
h¡
dx =
(log V )2 − (log U )2 + h(1 + B)(log V − log U )
x
2
10
−h log h(log V − log U ) + O(hU −²1 )
as long as V ² ≤ h ≤ U 1−² with U ≤ V ≤ 2U . Then, by substituting h = δx,
Z
∆
Z
0
V
f (x, δx)2 dxdδ
V /2
Z ∆V /2 Z V
Z V² Z V
f (x, h)2
f (x, h)2
f (x, h)2
dxdh +
dxdh +
dxdh
=
x
x
x
∆V /2 h/∆
V²
V /2
0
V /2
¡
¢
3³1 2 2
1 ¡ 3 B ¢ 2 2´
=
∆ V log +
+
∆ V + O(∆2 V 2−²1 ) + O V 2² (log 2V )2
4 4
∆
8
4
Z
∆V
Z
V
where the last error term comes from evaluating the second integral at V ² and estih 1−²
mating the third integral by (2.2). The above is okay as long as V ² ≤ h ≤ ( ∆
)
∆V
2
when
≤ h ≤ ∆V , and V ² ≤ h ≤ ( V2 )1−² when V ² ≤ h ≤
∆V
2
Now, suppose X 1−.6²1 ≤ V ≤ X. One has V ² ≤ X ² ≤ X
³ ´1−²
V −² V
V
≤
. One only needs to check h ≤ ( ∆h )1−² when
2
2
.
−1+²+²1 X 1−.6²1
2
∆V
2
≤
∆V
2
≤
≤ h ≤ ∆V . But
∆1−² h² ≤ ∆1−² (∆V )² ≤ ∆X ² ≤ 1. Thus
Z
Z
V
V /2
∆¡
0
´
¢2
3³1 2 2
1
ψ(x + δx) − ψ(x) − δx dδdx =
∆ V log + K∆2 V 2 + O(∆2 V 2−²1 )
4 4
∆
where K =
3
8
+
B
4
as long as X 1−.6²1 ≤ V ≤ X. Replacing X by X2−k in above,
1 log X
] and appealing to (2.1) with V = X2−M −1 ,
summing over 0 ≤ k ≤ M = [ .6²log
2
Z
∆
Z
0
X¡
1
¢2
1
1
ψ(x + δx) − ψ(x) − δx dxdδ = ∆2 X 2 log + K∆2 X 2 + O(∆2 X 2−²1 ).
4
∆
Consider X −1+²+2²1 ≤ ∆ ≤ X −² /2 and set η = X −2²1 /3 . One just mimics the
reduction in Theorem II.1 and obtain
Z
¢2
1
1 ³
1´
ψ(x + ∆x) − ψ(x) − ∆x dx =
∆X 2 log + 2K −
∆X 2
2
∆
4
³
³ ∆X 2−²1 ´
1´
+O η 1/2 ∆X 2 log
+O
∆
η
1
1
=
∆X 2 log + C∆X 2 + O(∆X 2−²2 )
2
∆
X¡
1
for X −1+²+2²1 ≤ ∆ ≤ X −² /2, whenever 0 < ²2 < ²1 /3.
11
Conversely (see Theorem IV.5 in Chapter IV for details), one has
Theorem II.4. Assume Riemann Hypothesis. For every 0 < ² < 1/2, if, for some
²1 > 0,
Z
X¡
(2.15)
1
¢2
1
1
ψ(x + δx) − ψ(x) − δx dx = δX 2 log + CδX 2 + O(δX 2−²1 )
2
δ
holds uniformly for X −1+² ≤ δ ≤ X −² , then
Z X
¡
¢2
X
(2.16)
ψ(x + h) − ψ(x) − h dx = hX log + BhX + O(hX 1−²2 )
h
1
holds uniformly for X ²+²1 ≤ h ≤ X 1−²−²1 for some ²2 > 0. Here C = (1 + B)/2.
2.2
Preparation for the more precise Pair Correlation Conjecture
Montgomery [13] proposed the Strong Pair Correlation Conjecture as follow:
(2.17)
F (X, T ) ∼
1
T log T as T → ∞
2π
holds uniformly when T ≤ X ≤ T A , for any fixed A > 1. Recall
(2.18)
F (X, T ) =
X
0
X i(γ−γ ) w(γ − γ 0 ),
0≤γ≤T
0≤γ 0 ≤T
and w(u) = 4/(4 + u2 ). In Goldston and Montgomery [9], they proved that, (2.17)
holding uniformly when T ≤ X ≤ T A , for any fixed A > 1, is equivalent to
Z X
¡
¢2
1
1
ψ(x + δx) − ψ(x) − δx dx ∼ δX 2 log as X → ∞
2
δ
1
holding uniformly when X −1 ≤ δ ≤ X −² , for any fixed ² > 0.
From section 2.1, we discover that, assuming Riemann Hypothesis and some cancellation among the error terms of Twin Prime Conjecture, for every 0 < ² < 1/2,
Z X
¡
¢2
1
1
ψ(x + δx) − ψ(x) − δx dx = δX 2 log + CδX 2 + o(δX 2 )
2
δ
1
holds uniformly for X −1+² ≤ δ ≤ X −² . Following the argument of Goldston and
Montgomery [9], we can obtain
12
Theorem II.5. Assume Riemann Hypothesis. For 0 < B1 ≤ B2 < 1, if
³ T
´
1
D
F (X, T ) =
T log T +
T +O
2π
2π
(log T )8
(2.19)
holds uniformly for X B1 (log X)−11 ≤ T ≤ X B2 (log X)11 , then
Z
X¡
(2.20)
1
³ δX 2 ´
1 2
1
2
ψ(x + δx) − ψ(x) − δx dx = δX log + CδX + O
2
δ
log 1/δ
¢2
holds uniformly for X −B2 ≤ δ ≤ X −B1 .
Conversely, for 1 < A1 ≤ A2 < ∞, if
Z
X¡
(2.21)
1
³ δX 2 ´
¢2
1
1
ψ(x + δx) − ψ(x) − δx dx = δX 2 log + CδX 2 + O
2
δ
(log 1/δ)5
holds uniformly for X −1/A1 (log X)−10 ≤ δ ≤ X −1/A2 (log X)10 , then
(2.22)
F (X, T ) =
³ T ´
D
1
T log T +
T +O
2π
2π
log T
holds uniformly for T A1 ≤ X ≤ T A2 . Here C and D are related by C =
D
2
−
C0
2
+ 1.
Before proving this, we need some lemmas. Basically they are from Goldston and
Montgomery [9] with some variations in order to make the error terms explicit.
R∞
Lemma II.6. If I(Y ) = −∞ e−2|y| f (Y + y)dy = 1 + O( Y18 ), and if f (y) ≥ 0 for all
R1
y as well as 0 f (Y + y)dy ¿ 1, then
Z
log 2
(2.23)
e2y f (Y + y)dy =
0
³ 1 ´
3
+O
.
2
Y2
Proof: Let Kc (y) = max(0, c − |y|). We would choose c ≈
1
.
Y2
From Lemma 1 of
Goldston and Montgomery [9], one has
1
1
1
Kc (y) = e−2|y| − e−2|y−c| − e−2|y+c| +
2
4
4
Z
c
−c
(c − |z|)e−2|y−z| dz.
13
For 0 ≤ a ≤ 1, let La,c (y) = Kc (y − a), then one has
Z
∞
La,c (y)f (Y + y)dy =
−∞
1
1
1
I(Y + a) − I(Y + a + c) − I(Y + a − c)
2
4
4
Z c
+
(c − |z|)I(Y + a + z)dz
−c
= c2 + O
³ 1 ´
.
Y8
Since,
¢
¢
1¡
1¡
La,c (y) − La,c−η (y) ≤ χ[a−c,a+c] (y) ≤ La,c+η (y) − La,c (y)
η
η
and f ≥ 0,
Taking η =
Z
1
,
Y4
Z
³ 1 ´
χ[a−c,a+c] (y)f (Y + y)dy = 2c + O(η) + O
.
ηY 8
−∞
∞
we have
Z
∞
∞
χ[a−c,a+c] (y)f (Y + y)dy =
−∞
χ[a−c,a+c] (y)dy + O
−∞
³ 1 ´
.
Y4
Now, we approximate e2y over 0 ≤ y ≤ log 2 by a sum of N = [Y 2 ] step functions,
2 (n+1) log 2
,
] with value e(2n log 2/N ) for n =
called the sum S(y), each support on [ n log
N
N
R1
0, 1, ..., N − 1. Then, since f ≥ 0 and 0 f (Y + y)dy ¿ 1,
Z
log 2
Z
log 2
2y
S(y)f (Y + y)dy = O
e f (Y + y)dy −
0
0
Z
log 2
Z
log 2
e2y dy −
0
S(y)dy = O
0
³ 1 ´
,
Y2
³ 1 ´
.
Y2
So,
Z
log 2
Z
2y
³ 1 ´
Y2
Z−∞
³ 1 ´
³N ´
∞
+
O
=
S(y)dy + O
Y4
Y2
−∞
Z log 2
³ 1 ´
2y
=
e dy + O
Y2
0
e f (Y + y)dy =
0
which gives the lemma.
∞
S(y)f (Y + y)dy + O
14
Lemma II.7. Let f (t) be a continuous non-negative function defined for all t ≥ 0
RT
with f (t) ¿ (log (t + 2))2 . If J(T ) = 0 f (t)dt = T log T + DT + ²(T )T where
²(T ) = O( (log1T )λ ) for κ−1 (log κ1 )−(λ+2) ≤ T ≤ κ−1 (log κ1 )(λ+2) , with some λ > 0, then
Z ∞³
³ κ log log 1/κ ´
sin κu ´2
1
π
.
(2.24)
f (u)du = κ log + πC 0 κ + Oλ
u
2
κ
(log 1/κ)λ
0
Here C 0 and D are related by C 0 = 12 D − 12 (C0 + log 2) + 1.
Proof: Just like Lemma 2 of Goldston and Montgomery [9], we divide the range
of integration of (2.24) into four subintervals:
−1
I1 = {u : 0 ≤ u ≤ U1 = κ
³
1 ´−(λ+2)
log
}
κ
I2 = {u : U1 ≤ u ≤ U2 = cκ−1 }
³
1 ´(λ+2)
I3 = {u : U2 ≤ u ≤ U3 = κ−1 log
}
κ
I4 = {u : U3 ≤ u ≤ ∞}
Here c, (log κ1 )−(λ+2) ≤ c ≤ (log κ1 )(λ+2) , is a parameter to be chosen later.
Since f (t) ¿ (log (t + 2))2 , we see that
Z
Z U1
(2.25)
¿
κ2 (log (u + 2))2 du ¿ κ2 U1 (log U1 )2 ¿
I1
0
Z
Z
(2.26)
∞
¿
I4
u−2 (log u)2 du ¿ U3 −1 (log U3 )2 ¿
U3
κ
,
(log κ1 )λ
κ
.
(log κ1 )λ
By writing f (u) = log κ1 + log κu + (D + 1) + (f (u) − log u − (D + 1)), we can express
the integral on I2 as a sum of four integrals. Note that
Z U2 ³
Z ∞³
³
´
³κ´
sin κu ´2
sin κu ´2
κ
du =
du + O
+
O
u
u
c
(log κ1 )(λ+2)
U1
0
³
´
³
´
π
κ
κ
=
κ+O
,
1 (λ+2) + O
2
c
(log κ )
Z
U2 ³
U1
sin κu ´2
log κudu = κ
u
= κ
Z
³ sin v ´2
c
(log 1/κ)−(λ+2)
Z ∞³
´2
0
sin v
v
v
log vdv
log vdv + O
³ κ log log 1 ´
³ κ log c ´
κ
+
O
.
c
(log κ1 )(λ+2)
15
Put r(u) = J(u) − u log u − Du, then r0 (u) = f (u) − log u − (D + 1). By integrating
by parts, we have
Z
Z U2 ³
¢
sin κu ´2 ¡
sin κu ´2
f (u) − log u − (D + 1) du =
dr(u)
u
u
U1
U1
h sin κu
iU2 Z U2
³ sin κu ´2
2
= (
) r(u)
−
r(u)d
u
u
U1
U1
Z U2
³
hZ U2 κ sin κu cos κu
³ κ ´
¡ sin κu ¢2 i´
+
O
max
|²(u)|
= O
|
|du
+
du
U1 ≤u≤U2
u
u
(log κ1 )λ
U1
U1
U2 ³
with c > 1. Estimating the remaining integrals and using ²(u) ¿
Z
U2 ³
U1
1
,
(log u)λ
we have
¢
κ log log κ1
κ log c
sin κu ´2 ¡
f (u) − log u − (D + 1) du ¿λ
+
.
u
(log κ1 )λ
(log κ1 )λ
Hence,
Z
I2
(2.27)
Z ∞³
1
sin v ´2
π
π
log vdv + (D + 1)κ
= κ log + κ
2
κ
v
2
0
³ κ log 1
κ log log κ1 ´
κ log c
κ log c
κ
+ Oλ
+
+
+
.
c
c
(log κ1 )λ
(log κ1 )λ
Over I3 , we write f (u) = log κ1 + log κu + (D + 1) + (f (u) − log u − (D + 1)) and
break down the integral into four pieces again. Note that
Z
U3 ³
U2
Z
U3 ³
U2
sin κu ´2
du = κ
u
sin κu ´2
log κudu = κ
u
Z
(log 1/κ)(λ+2) ³
c
Z
(log 1/κ)(λ+2) ³
c
sin v ´2
κ
dv ¿ ,
v
c
sin v ´2
κ log c
log vdv ¿
,
v
c
and by a similar calculation as I2 ,
Z
U3 ³
U2
¢
κ log log κ1
sin κu ´2 ¡
f (u) − log u − (D + 1) du ¿λ
.
u
(log κ1 )λ
Thus,
Z
(2.28)
¿λ
I3
κ log κ1
κ log c κ log log κ1
+
+
.
c
c
(log κ1 )λ
16
Therefore, choosing c ≈ (log κ1 )(λ+1) , we have from (2.25), (2.26), (2.27) and (2.28)
that
Z
∞³
0
i
³ κ log log 1 ´
sin κu ´2
π
1 hπ
π
κ
f (u)du = κ log + (D +1)− (C0 +log 2−1) κ+Oλ
u
2
κ
2
2
(log κ1 )λ
because
R∞
0
( sinv v )2 log vdv = − π2 (C0 + log 2 − 1) (see [11] page 590). The lemma
follows from the relationship between C 0 and D.
Lemma II.8. If K is even, K 00 continuous,
R∞
−∞
|K| < ∞, K(x) → 0 as x → +∞,
K 0 → 0 as x → +∞, and if K 00 (x) ¿ x−3 as x → +∞, then
Z
(2.29)
∞
K̂(t) =
0
³ sin πtx ´2
K (x)
dx.
πt
00
Proof: This is Lemma 3 of Goldston and Montgomery [9]. One can justify it by
integration by parts twice.
In Goldston and Montgomery [9], they used the function
(2.30)
Kη (x) =
sin 2πx + sin 2π(1 + η)x
2πx(1 − 4η 2 x2 )
for η > 0. Then
(2.31)




1,
if |t| ≤ 1,



K̂η (t) =
cos2 ( π(|t|−1)
), if 1 ≤ |t| ≤ 1 + η,
2η





 0,
if |t| ≥ 1 + η.
Lemma II.9. For 0 < η <
1
4
and Kη (x) defined as in (2.30), then Kη (x) satisfies
the conditions in Lemma II.8. In particular,
³
1 ´
Kη 00 (x) ¿ min 1, 3 3 .
η x
Here the implicit constant is absolute.
17
Proof: One can easily see that Kη is even,
R∞
−∞
|Kη | < ∞, and Kη → 0 as
x → +∞. Breaking it down into partial fractions, we have
η
−η i
+
,
x 1 − 2ηx 1 + 2ηx
h1
η
−η i
Kη 0 =2π[cos 2πx + (1 + η) cos 2π(1 + η)x] +
+
x 1 − 2ηx 1 + 2ηx
h −1
i
2η 2
2η 2
+ [sin 2πx + sin 2π(1 + η)x] 2 +
+
,
x
(1 − 2ηx)2 (1 + 2ηx)2
h1
η
−η i
00
2
2
Kη =4π [− sin 2πx − (1 + η) sin 2π(1 + η)x] +
+
x 1 − 2ηx 1 + 2ηx
h −1
i
2η 2
2η 2
+ 4π[cos 2πx + (1 + η) cos 2π(1 + η)x] 2 +
+
x
(1 − 2ηx)2 (1 + 2ηx)2
h2
8η 3
−8η 3 i
+ [sin 2πx + sin 2π(1 + η)x] 3 +
+
.
x
(1 − 2ηx)3 (1 + 2ηx)3
Kη =[sin 2πx + sin 2π(1 + η)x]
h1
+
Clearly, Kη 0 → 0 as x → +∞. Moreover, Kη 00 is continuous everywhere except
possibly the points 0,
1
2η
1
and − 2η
. Now, using Taylor expansions for sin and cos, we
have, for −1 ≤ x ≤ 1,
h1
η
−η i
Kη 00 = 4π 2 [−2πx − 2π(1 + η)3 x + O(x3 )] +
+
x 1 − 2ηx 1 + 2ηx
h −1
i
2
2η
2η 2
+4π[2 + η + O(x2 )] 2 +
+
x
(1 − 2ηx)2 (1 + 2ηx)2
h2
8η 3
−8η 3 i
+
+[2π(2 + η)x + O(x3 )] 3 +
x
(1 − 2ηx)3 (1 + 2ηx)3
which shows that Kη 00 is well-defined and continuous at x = 0. As for
1
2η
1
2η
−1≤x≤
+ 1, we have
1
)+
2η
1
cos 2πx = cos (2π(x − ) +
2η
sin 2πx = sin (2π(x −
π
),
η
π
),
η
1
π
) + ),
2η
η
1
π
cos 2π(1 + η)x = − cos (2π(1 + η)(x − ) + ).
2η
η
sin 2π(1 + η)x = − sin (2π(1 + η)(x −
18
So,
h
π
π
1
Kη 00 = 4π 2 −(sin )(1 − (1 + η)2 ) − 2π(cos )(1 − (1 + η)3 )(x − )
η
η
2η
1
³
i
h
´ih
−2
1
π
π
+O (x − )2
− 2π(sin )(1 − (1 + η)2 )
1 + O(1) + 4π −η cos
2η
η
η
x − 2η
1
³
1
(2π)2
π
1 2
1 3 ´ih
3
2
(x − ) −
(cos )(1 − (1 + η) )(x − ) + O (x − )
1 2
2η
2
η
2η
2η
(x − 2η
)
i h
π
1
(2π)2
π
1
+O(1) + −2πη(cos )(x − ) −
(sin )(1 − (1 + η)2 )(x − )2
η
2η
2
η
2η
´ih
i
³
3
(2π)
π
1 3
1 4
−1
3
−
(cos )(1 − (1 + η) )(x − ) + O (x − )
1 3 + O(1) .
6
η
2η
2η
(x − 2η
)
The “pole terms” cancel out exactly. So, we see that Kη 00 is well-defined and continuous at x =
1
.
2η
Replacing η by −η, one gets the well-definedness and continuity at
1
x = − 2η
.
The above calculations show that Kη 00 ¿ 1 when 0 ≤ x ≤ 1 and
1
2η
+ 1. One can easily see that Kη 00 ¿ 1 when 1 ≤ x ≤
Kη 00 (x) ¿ 1 ¿
(2.32)
When
1
2η
1
η 3 x3
when 0 ≤ x ≤
1
2η
1
2η
−1 ≤ x ≤
− 1. Hence,
1
+ 1.
2η
+ 1 ≤ x ≤ η1 , after recombining the partial fractions, one has
Kη 00 (x) ¿
(2.33)
1
12η 2 x2 − 1
1
1
+
+ 3+
1 3.
2
2
2
2
2
2
x(4η x − 1) x (4η x − 1)
x
(x − 2η
)
1
))2 −1 = (2η +1)2 −1 ≥ η, ηx ≤ 1, and 12η 2 x2 −1 ≥ 4η 2 x2 .
But 4η 2 x2 −1 ≥ (2η(1+ 2η
Hence,
1
1
1
12η 2 x2 − 1
x2 η 2
1
≤
≤
and
¿
≤ 3 3.
2
2
3
3
2
2
2
2
2
2
x(4η x − 1)
ηx
η x
x (4η x − 1)
xη
η x
(2.34)
Meanwhile, (1 + 2η)(1 − η) = 1 + η − 2η 2 ≥ 1, so x ≥
that x −
(2.35)
1
2η
≥ ηx. Thus,
1
1
1 3 ¿ 3 3.
η x
(x − 2η )
1+2η
2η
≥
1
.
2η(1−η)
This implies
19
Combining (2.33), (2.34) and (2.35), we have
Kη 00 (x) ¿
(2.36)
Finally, when
1
η
1
η 3 x3
when
1
1
+1≤x≤ .
2η
η
≤ x, 4η 2 x2 − 1 ≥ 2η 2 x2 and x −
1
2η
≥ x2 . Putting these into (2.33)
(which is still valid), we get
Kη 00 (x) ¿
(2.37)
1
η 2 x2
1
1
+
+ 3 ¿ 3 3.
2
3
2
4
4
η x
xη x
x
η x
Consequently, the lemma follows from (2.32), (2.36) and (2.37).
Lemma II.10. If f is a non-negative function defined on [0, +∞), f (t) ¿ log (t + 2)2 ,
and if
Z
∞³
sin κt ´2
π
1
I(κ) =
f (t)dt = κ log + πC 0 κ + ²(κ)κ
t
2
κ
0
³
´
1
where ²(κ) = O (log 1/κ)
for T −1 (log T )−2 ≤ κ ≤ T −1 (log T )9 , then
5
Z T
³ T ´
J(T ) =
f (t)dt = T log T + DT + O
.
log T
0
Again, C 0 = 21 D − 12 (C0 + log 2) + 1.
Proof: Essentially, we follow Lemma 4 of Goldston and Montgomery [9]. Let
η=
1
.
(log T )2
Set X1 = (log T )−2 and X2 = 41 (log T )9 . Let Kη be the kernel defined as
in (2.30). Replace t by t/T in (2.29), multiply by f (t) − log t − (D + 1), and integrate
over 0 ≤ t < ∞, one finds that
Z ∞
Z
³ πx ´
¡
¢ ³t´
T 2 ∞ 00
f (t) − log t − (D + 1) K̂η
dt = 2
Kη (x)R
dx
T
π 0
T
0
where
Z
Z ∞³
sin κt ´2
sin κt ´2
R(κ) = I(κ) −
log tdt − (D + 1)
dt
t
t
0
0
Z
Z ∞³
1 ∞ ³ sin κt ´2
sin κt ´2
log κtdκt − κ log
dκt
= I(κ) − κ
κt
κ 0
κt
Z0 ∞ ³
sin κt ´2
dt
−(D + 1)
t
0
´
π
1 π ³
= I(κ) − κ log − κ (D + 1) − (C0 + log 2 − 1) ,
2
κ 2
∞³
20
R∞
because
0
( sinv v )2 dv =
Z
π
2
∞
I(κ) ¿
0
and
R∞
0
( sinv v )2 log vdv = − π2 (C0 + log 2 − 1). Since,
³
1 ´2
min(κ2 , t−2 )(log (t + 2))2 dt ¿ κ log (2 + )
κ
for all κ > 0, we see that, by Lemma II.9, Kη 00 (x) is bounded when 0 ≤ x ≤ X1 .
Thus,
Z
X1
00
Kη (x)R
³ πx ´
T
0
Z
X1
dx ¿
0
x³
T ´2
(log T )2 X1 2
1
log
dx ¿
¿
.
T
x
T
T (log T )2
On the other hand, by Lemma II.9,
Z
∞
X2
Z ∞
³ πx ´
1 x
(log T )2
1
2
Kη (x)R
dx ¿
(log
T
)
dx
¿
¿
.
3
3
3
T
η T X2
T log T
X2 η x T
00
Finally,
Z
X2
00
Kη (x)R
X1
³ πx ´
T
Z
¯ πx ¯ πx
¯
¯
dx ¿
|Kη (x)|¯²( )¯ dx
T T
X1
Z 1/η
Z X2
1
x
1
1
¿
dx
+
dx
5
5 3
2
X1 (log T ) T
1/η (log T ) η T x
1
¿
T log T
X2
00
by Lemma II.9. Therefore,
Z
∞
0
Z ∞
³t´
³ T ´
¡
¢ ³t´
f (t)K̂η
dt =
log t + (D + 1) K̂η
dt + O
T
T
log T
0
Z T
³Z (1+η)T
´
=
log t + (D + 1)dt + O
log t + (D + 1)dt
0
T
³ T ´
+O
log T
³ T ´
.
= T log T + DT + O(ηT log T ) + O
log T
Since f is non-negative, we see that
Z
∞
0
Z ∞
³ (1 + η)t ´
³t´
f (t)K̂η
dt ≤ J(T ) ≤
f (t)K̂η
dt,
T
T
0
which gives the desired result.
21
Lemma II.11. For 0 < δ < 1, let
(1 + δ)s − 1
a(s) =
.
s
(2.38)
If |c(γ)| ≤ 1 for all γ then
Z
(2.39)
+∞
¯X
|a(it)| ¯
2¯
−∞
γ
¯2
c(γ)
¯
¯ dt =
2
1 + (t − γ)
Z
+∞
−∞
³
¯ X a( 1 + iγ)c(γ) ¯2
¯
¯
2
¯
¯ dt
2
1 + (t − γ)
|γ|≤Z
³ (log Z)3 ´
2 3´
+ O δ (log ) + O
δ
Z
provided that Z ≥
1
.
δ
2
Here, the summations are over the imaginary parts of the
non-trivial zeros of the Riemann zeta function.
Proof: This is just Lemma 10 of Goldston and Montgomery [9].
We also need the explicit formula for ψ(x) (see Davenport [1] §17, as well as
Goldston and Montgomery [9]) to get:
ψ(x + δx) − ψ(x) − δx = −
X
³
a(ρ)xρ + O (log x)min(1,
|γ|≤Z
(2.40)
³
+ O (log x)min(1,
x ´
)
Zkxk
´
³ x(log xZ)2 ´
x
) +O
Zk(1 + δ)xk
Z
where a(s) is as in (2.38), and kθk = minn |θ −n| is the distance from θ to the nearest
integer. The error terms contribute a negligible amount if we take Z = X(log X)2 .
ρ = σ + iγ denotes zeros of Riemann zeta function.
Lemma II.12. For any t ≥ 0,
X
γ
1
¿ log (t + 2)
1 + (t − γ)2
where the sum is over the imaginary parts of the zeros of the Riemann zeta function.
Proof: This is a lemma in Davenport [1] on page 98.
22
Lemma II.13. Assume Riemann Hypothesis.
Z 2X ¯ X
Z 2X
¯2
¡
¢2
¯
ρ¯
a(ρ)x ¯ dx =
ψ((1 + δ)x) − ψ(x) − δx dx
¯
X
X
|γ|≤Z
³
´
2
+ O X 3/2 δ 1/2 log log X + X(log X)2
δ
where a(s) is defined by (2.38) and Z ≥ X(log X)2 .
Proof: From (2.40), we have for X ≤ x ≤ 2X,
−
X
¡
¢
a(ρ)xρ = ψ (1 + δ)x − ψ(x) − δx + O(log x).
|γ|≤Z
So
Z
2X
Z
¯X
¯2
¯
ρ¯
a(ρ)x ¯ dx =
¯
X
2X ¡
¢2
ψ((1 + δ)x) − ψ(x) − δx dx + error
X
|γ|≤Z
where
³Z
error = O
³Z
¿
³
´
2X
X
2X
³Z
2X
2
´
|ψ((1 + δ)x) − ψ(x) − δx| log xdx + O
(log x) dx
X
´ 12
´ 12 ³Z 2X
2
(log x)2 dx + X(log X)2
(ψ((1 + δ)x) − ψ(x) − δx) dx
X
2 ´ 12 1
¿ δX 2 (log )2 X 2 log X + X(log X)2
δ
3 1
2
¿ X 2 δ 2 log log X + X(log X)2
δ
X
by Cauchy’s inequality and (2.1).
2.3
Proof of main result
We are ready to prove Theorem II.5. We first assume (2.19) and derive (2.20).
Let
Z
T
J(T ) = J(X, T ) = 4
0
¯X
¯
¯
γ
¯2
X iγ
¯
¯ dt.
1 + (t − γ)2
Montgomery [13] (see his (26), but be aware of the changes in notation) showed that
¡
¢
J(X, T ) = 2πF (X, T ) + O (log T )3 .
23
Thus, (2.19) is equivalent to
³
(2.41)
J(X, T ) = T log T + DT + O
´
T
.
(log T )8
With a(s) defined by (2.38), we note that
2
|a(it)| = 4
where κ =
1
2
log (1 + δ). We have κ =
δ
2
³ sin κt ´2
t
+ O(δ 2 ) and log κ1 = log 1δ + log 2 + O(δ).
Then, by Lemma II.7 with λ = 8 and Lemma II.12, we deduce that
Z ∞
¯2
¯X
³
´
X iγ
π
1
κ
¯
¯
0
|a(it)|2 ¯
dt
=
κ
log
+
πC
κ
+
O
¯
1 + (t − γ)2
2
κ
(log 1/κ)7
0
γ
´
³
´
π δ³
1
δ
δ
(2.42)
=
log + log 2 + O(δ) + πC 0 + O(δ 2 ) + O
22
δ
2
(log 1/δ)7
³
´
1 ³ π log 2 πC 0 ´
δ
π
+
δ+O
.
= δ log +
4
δ
4
2
(log 1/δ)7
The values of T for which we have used (2.19) lie in the range
³
³
1 ´−10
1 ´10
−1
δ
log
≤ T ≤ 3δ
log
.
δ
δ
−1
(2.43)
The integrand is even, so the value is doubled if we integrate over negative values of
t as well. Then, by Lemma II.11,
Z
+∞
−∞
¯X
¯
¯
|γ|≤Z
´
³
´
a(ρ)X iγ ¯¯2
π
1 ³ π log 2
δ
0
+ πC δ + O
¯ dt = δ log +
1 + (t − γ)2
2
δ
2
(log 1/δ)7
provided that Z ≥ δ −1 (log 1δ )10 . Let S(t) denote the above sum over γ. Its Fourier
transform is
Z
+∞
Ŝ(u) =
S(t)e(−tu)dt = π
−∞
X
a(ρ)X iγ e(−γu)e−2π|u| .
|γ|≤Z
Hence, by Plancherel’s identity, the above integral is
Z
=π
2
+∞
−∞
¯X
¯2
¯
iγ ¯ −4π|u|
a(ρ)X ¯ e
du.
¯
|γ|≤Z
24
Let Y = log X, −2πu = y, we have
Z +∞ ¯ X
¯2
³
´
δ
1
¯
iγ(Y +y) ¯ −2|y|
(2.44)
a(ρ)e
dy = δ log + (log 2 + 2C 0 )δ + O
.
¯
¯e
δ
(log 1/δ)7
−∞
|γ|≤Z
Now, set
f (y) =
|
P
|γ|≤Z
a(ρ)eiγy |2
δ log 1δ + (log 2 + 2C 0 )δ
,
R1
then f (y) ≥ 0 if δ is small enough. From (2.44), we know that 0 f (Y + y)dy ¿ 1
R +∞
as well as −∞ e−2|y| f (Y + y)dy = 1 + O( Y18 ) for log 1δ ³ log X = Y . Therefore we
can use Lemma II.6. With the change of variable x = eY +y , we get
Z 2X ¯ X
¯2
³3
´³
1
1
¯
ρ¯
a(ρ)x
dx
=
+
O(
)
δ
log
+ (log 2 + 2C 0 )δ
¯
¯
2
2
(log
1/δ)
δ
X
|γ|≤Z
¢´ 2
¡
δ
+O
X
(log 1/δ)7
³ δX 2 ´
3 2
1 3
=
δX log + (log 2 + 2C 0 )δX 2 + O
.
2
δ 2
log 1/δ
By Lemma II.13 with Z = X(log X)2 ,
Z 2X
¡
¢2
ψ(x + δx) − ψ(x) − δx dx
X
3 2
1 3
=
δX log + (log 2 + 2C 0 )δX 2
2
δ 2
³ δX 2 ´
³
´
2
3/2 1/2
2
+O
+ O X δ log log X + X(log X) .
log 1/δ
δ
But as X −B2 ≤ δ ≤ X −B1 , 0 < B1 ≤ B2 < 1, the first error term is bigger than the
other one.
Z
Now, we replace X by X2−k and sum over 1 ≤ k ≤ K = [ 2 logloglog2 X ]. One obtains
X
X2−K
¡
´
¢2
1
1 − 2−2K ³ 2
δX log + (log 2 + 2C 0 )δX 2
ψ(x + δx) − ψ(x) − δx dx =
2
δ
³ δX 2 ´
+O
.
log 1/δ
To bound the contribution from the range 1 ≤ x ≤ X/2K , we use section 2.1 inequality (2.1):
Z X/2K
¡
1
¢2
ψ(x + δx) − ψ(x) − δx dx ¿ δ
2 ´2
δX 2
X2 ³
log
¿
.
(log X)4
δ
log (1/δ)
25
We finally get (2.20) as C = C 0 + log2 2 . The last thing to take care of is the range for
T and δ. Now for X −B2 ≤ δ ≤ X −B1 , putting this into (2.43), the whole argument
goes through as long as (2.19) is true for X B1 (log X)−10 ¿ T ¿ X B2 (log X)10 . But
we assume that this is true for X B1 (log X)−11 ≤ T ≤ X B2 (log X)11 to start with!
Therefore (2.19) implies (2.20) when X is sufficiently large.
We now deduce (2.22) from (2.21). Let X1 = X and X2 = X(log X)4 . Then
Z X̃
³ δ X̃ 2 ´
¡
¢2
1
1
(2.45)
ψ(x + δx) − ψ(x) − δx dx = δ X̃ 2 log + Cδ X̃ 2 + O
2
δ
(log 1/δ)5
1
for X1 ≤ X̃ ≤ X2 whenever
X −1/A1 (log X)−6 ≤ δ ≤ X −1/A2 (log X)10 .
(2.46)
By integration by parts and (2.45), we find that
Z X2
¡
¢2
1
1
ψ(x + δx) − ψ(x) − δx x−4 dx = X1 −2 δ log + CX1 −2 δ
2
δ
X1
³ δX −2
δX1 −2 log 1/δ ´
1
+O
+
.
(log 1/δ)5
(log X1 )8
From (2.1), we deduce that
Z ∞
¡
¢2
δX1 −2 (log 1/δ)2
ψ(x + δx) − ψ(x) − δx x−4 dx ¿
.
(log X1 )8
X2
We add these relations, and multiply through by X1 2 . By making a further appeal
to (2.45) with X̃ = X1 , we deduce that
Z ∞
³ x2 X 2 ´ ¡
¢2 −2
1
min
,
ψ(x
+
δx)
−
ψ(x)
−
δx
x dx
X1 2 x2
0
³
´
δ
1
.
= δ log + 2Cδ + O
δ
(log 1/δ)5
Write X for X1 , put Y = log X, x = eY +y , and appeal to the explicit formula (2.40)
with Z = X(log X)2 (the error term arises from the explicit formula can be treated
similarly as Lemma II.13 and it is small because of the range of δ). We have
Z +∞ ¯ X
¯2
³
´
1
δ
¯
iγ(Y +y) ¯ −2|y|
a(ρ)e
dy = δ log + 2Cδ + O
¯
¯e
δ
(log 1/δ)5
−∞
|γ|≤Z
26
which is almost (2.44) except for the error term. Now retracing our steps, we have
κ
(2.42) except that the error term is O( (log 1/κ)
5 ). Because of Lemma II.12, we can use
Lemma II.10 and obtain (2.41) except that the error term is O( logT T ). This works
provided that T −1 (log T )−2 ≤ κ ≤ T −1 (log T )9 . However, as T A1 ≤ X ≤ T A2 , from
(2.46), we have X −1/A1 (log X)−6 ≤ δ ≤ X −1/A2 (log X)10 which allows κ to be in the
range we want as δ ³ κ! So, the second half of the theorem is true.
One can modify the lemmas to get analogous results with smaller error terms.
Theorem II.14. Assume Riemann Hypothesis. For 0 < B1 ≤ B2 < 1, if, for some
² > 0,
(2.47)
F (X, T ) =
1
D
T log T +
T + O(T 1−² )
2π
2π
holds uniformly for X B1 −²1 ≤ T ≤ X B2 +²1 , then
Z
X¡
(2.48)
1
¢2
1
1
ψ(x + δx) − ψ(x) − δx dx = δX 2 log + CδX 2 + O(δ 1+²2 X 2 )
2
δ
holds uniformly for X −B2 ≤ δ ≤ X −B1 where ²1 , ²2 > 0 may depend on ².
Conversely, for 1 < A1 ≤ A2 < ∞, if, for some ² > 0,
Z
X¡
(2.49)
1
¢2
1
1
ψ(x + δx) − ψ(x) − δx dx = δX 2 log + CδX 2 + O(δ 1+² X 2 )
2
δ
holds uniformly for X −1/A1 −²1 ≤ δ ≤ X −1/A2 +²1 , then
(2.50)
F (X, T ) =
1
D
T log T +
T + O(T 1−²2 )
2π
2π
holds uniformly for T A1 ≤ X ≤ T A2 where ²1 , ²2 > 0 may depend on ². Here C and
D are related by C =
2.4
D
2
−
C0
2
+ 1.
Conclusion
From section 2.1 and 2.3, loosely speaking, the following are ”equivalent”:
27
(a) For every fixed ² > 0,
Z
X¡
1
¢2
X
ψ(x + h) − ψ(x) − h dx = hX log + BhX + o(hX)
h
holds uniformly for X ² ≤ h ≤ X 1−² .
(b) For every fixed ² > 0,
Z
X¡
1
¢2
1
1
ψ((1 + δ)x) − ψ(x) − δx dx = δX 2 log + CδX 2 + o(δX 2 )
2
δ
holds uniformly for X −1+² ≤ δ ≤ X −² .
(c) For every fixed ² > 0 and A ≥ 1 + ²,
F (X, T ) =
1
D
T log T +
T + o(T )
2π
2π
holds uniformly for T 1+² ≤ X ≤ T A where F is defined as (2.18).
Here B = −C0 − log 2π, C = 12 (B + 1) and C =
D
2
− C20 + 1. With a little algebra,
one has D = − log 2π − 1. Hence, we deduce a more precise Strong Pair Correlation
Conjecture
(2.51)
F (X, T ) =
T
T
T
log
−
+ o(T )
2π
2π 2π
which has the same main terms as the number of zeros of the Riemann zeta function
up to height T (it matches with the diagonal terms of F (X, T )).
Meanwhile, if one assumes the usual quantitative form of Twin Prime Conjecture
(see Chapter III Hypothesis HL with r = 2), one has (2.51) for T 1+² ≤ X ≤ T 2−² .
CHAPTER III
Beyond Pair Correlation
3.1
Introduction
Throughout this section, we shall use a strong quantitative form of Hardy-Littlewood
Prime k-tuple Conjecture:
Hypothesis HL Let D = {d1 , d2 , ..., dr } be a set of r distinct integers. Define
S({d1 , d2 , ..., dr }) :=
Y³
p
1 ´−r ³
νp (D) ´
1−
1−
p
p
where νp (D) is the number of distinct residue classes (mod p) among the numbers
di ’s. We write
r
N Y
X
Λ(n + di ) = S(D)N + E(N, D).
n=1 i=1
Then the conjecture states that, for any fixed r, and uniformly in |di | ≤ N ,
E(N, D) ¿ N 1/2+² .
We are interested in studying the moments
Z
(3.1)
N¡
µk (N, h) =
¢k
ψ(x + h) − ψ(x) − h dx
1
for any positive integer k.
28
29
Assuming Hypothesis HL, Montgomery and Soundararajan [15] were able to evaluate the moments µk (N, h) in N ² ≤ h ≤ N 1/k−² ,


k!
 ∼
N hk/2 (log Nh )k/2 , if k is even,
(k/2)!2k/2
µk (N, h)

 ¿k N hk/2−1/7k+² ,
if k is odd,
as N → ∞. And one expects that the above conclusions are valid uniformly in
N ² ≤ h ≤ N 1−² for each k. This suggests that µk (N, h)/N behaves like the k-th
moment of a normal random variable with mean 0 and variance h log Nh . Their result
is based on estimating the average value of a modified singular series. Let S(∅) = 1,
and define
U(D) :=
X
(−1)|I| S(I).
I⊂D
Set R0 (h) = 1 and define, for s > 0,
Rs (h) :=
X
U({d1 , ..., ds }).
d1 ,...,ds
1≤di ≤h
di distinct
Step 1 Reduction using Hypothesis HL to get
¶Z N
µ
k
X
k!hk−r
k
(log x − 1)k−r dxR2r−k (h) + Ok (hk/2 N 1−² )
(3.2) µk (N, h) =
r!2k−r k − r 1
r≥k/2
when N ² ≤ h ≤ N 1/k−² .
Step 2 Find an asymptotic formula for Rs (h) by first considering the sum without the distinct condition (treated in Montgomery and Vaughan [17]). Then, by a
combinatorial argument, one get, when s is even,
w
¢´s/2
s! ³ X µ(w)2 ¡ X ¯¯ a ¯¯2
H(
)
−
hφ(w)
+ Os (hs/2−1/7s+² ).
(3.3) Rs (h) = s s/2
( 2 )!2
φ(w)2 a=1
w
w|Qs
w>1
(a,w)=1
And, when s is odd,
Rs (h) ¿s hs/2−1/7s+² .
30
Here H(α) =
P
n≤h
e(nα), Qs =
Q
p≤hs+1
p. And they demonstrated that
w
´
X µ(w)2 ³ X
¯ a ¯2
¯H( )¯ − hφ(w) = −h log h + O(h(log h)1/2 ).
φ(w)2 a=1
w
(3.4)
w|Qs
w>1
(a,w)=1
Putting (3.4) into (3.3) and putting the appropriate information about Rs (h) into
(3.2), they got the desired results.
Our goal is to obtain a more precise asymptotic formula for µk (N, h) when k is
even. But first, we need a more precise asymptotic formula for (3.4).
3.2
An asymptotic formula
Lemma III.1. For k even,
S(k) := S({0, k}) = 2
Yµ
p>2
1
1−
(p − 1)2
¶Y
p|k
p>2
q
∞
p − 1 X µ(q)2 X ³ ak ´
=
e
.
p−2
φ(q)2 a=1
q
q=1
(a,q)=1
Note: S(k) = 0 when k is odd.
Proof: The second equality follows from definition. Using Lemma 1 of Montgomery and Soundararajan [15] with r = 2, d1 = 0, d2 = k, and letting y → ∞, one
has
∞
X
µ(q1 )µ(q2 )
S(k) = S({0, k}) =
φ(q1 )φ(q2 )
q ,q =1
1
2
X
1≤a1 ,a2 ≤q1
(a1 ,q1 )=1=(a2 ,q2 )
³a k ´
2
e
.
q2
a1
a
+ q 2 ∈Z
q1
2
But
a1
q1
+ aq22 is an integer only when q1 = q2 and a1 ≡ −a2 (mod q2 ). Hence the above
becomes
q
∞
X
µ(q)2 X
q=1
φ(q)2
a=1
(a,q)=1
³ ak ´
.
e
q
Lemma III.2.
h
X
1
1
(h − k)S(k) = h2 − h log h + Ah + O(h1/2+² )
2
2
k=1
where A = 21 (1 − C0 − log 2π) and C0 is Euler’s constant.
31
Proof: This is a theorem in Montgomery and Soundararajan [14].
Theorem III.3. For any positive integer s,
w
´
X µ(w)2 ³ X
¯ a ¯2
¯H( )¯ − hφ(w) = −h log h + B 0 h + O(h1/2+² )
Fs (h) =
φ(w)2 a=1
w
w|Qs
w>1
(a,w)=1
where B 0 = 2A+1 = 2−C0 −log 2π and recall H(α) =
P
n≤h e(nα), Qs =
Q
p≤hs+1
p.
Proof: We start with the left-hand side of Lemma III.2 and apply Lemma III.1
q
h
h
∞
X
X
X
µ(q)2 X ³ ak ´
e
(h − k)S(k) =
(h − k)
2
φ(q)
q
a=1
q=1
k=1
k=1
(a,q)=1
=
∞
X
q=1
q
µ(q)2 X
φ(q)2 a=1
³ ak ´
(h − k)e
q
k=1
h
X
(a,q)=1
=
∞
X
q=1
=
1
2
q
h
´
µ(q)2 X 1 ³¯¯ X ¡ ak ¢¯¯2
e
−
h
φ(q)2 a=1 2 k=1
q
∞
X
q=1
(a,q)=1
q
2
´
µ(q) X ³¯¯ ¡ a ¢¯¯2
−
hφ(q)
.
H
φ(q)2 a=1
q
(a,q)=1
Multiply both sides by 2, apply Lemma III.2 to left-hand side and truncate the
outermost sum of the right-hand side, we have
2
(3.5) h − h log h+2Ah + O(h
1/2+²
w
³¯ a ¯
´
X µ(w)2 X
2
¯
¯
) = h −h+
H( ) − hφ(w)
φ(w)2 a=1
w
w>1
2
(a,w)=1
w|Qs
µ X
w
¯¶
µ(w)2 ¯¯ X ¯¯ a ¯¯2
¯
+O
H( ) − hφ(w)¯ .
¯
2
φ(w)
w
s+1
a=1
w>h
(a,w)=1
Now we imitate Montgomery and Soundararajan [15] for the error term by noting
that
w
X
¯ ¡ a ¢¯2
¯ − hφ(w) =
¯H
w
a=1
(a,w)=1
=
X
w
X
d1 ,d2 ≤h
a=1
(a,w)=1
w
X
X
d1 ,d2 ≤h a=1
d1 6=d2 (a,w)=1
e
³ a(d − d ) ´
1
2
− hφ(w)
w
e
³ a(d − d ) ´
1
2
w
32
where the inner sum over a is the Ramanujan sum cw (d1 − d2 ) which is easily seen
to be µ( (w,dw
)φ((w, d1 − d2 )). Hence the above is
1 −d2 )
¿
X
φ((w, d1 − d2 )) ¿
d1 ,d2 ≤h
d1 6=d2
X
X
φ(l)
1 ¿ h2
d1 ,d2 ≤h
l|d1 −d2
l|w
l≤h
X φ(l)
l|w
l≤h
l
.
So the error term of (3.5) is
¿ h2
X µ(w)2 X φ(l)
X µ(l)2
2
¿
h
φ(w)2
l
lφ(l)
s+1
l≤h
l|w
l≤h
w>h
¿ h2
X
l≤h
1
hs+1 φ(l)
X
r≥hs+1 /l
µ(r)2
φ(r)2
¿ log h
as s ≥ 1. Putting this back to (3.5), we get the theorem.
3.3
A more precise formula for µk (N, h) when k is even
First we define a sequence of integer-coefficient polynomials as follow:
P0 (x) = 1; Pn (x) = xn − nPn−1 (x), n = 1, 2, ...
Lemma III.4. For any non-negative integer n,
Z
(log u)n du = uPn (log u) + C
for some constant C.
Proof: Induction on n. Clearly, the above is true for n = 0. Suppose that
R
(log u)n du = uPn (log u)+C for some n ≥ 0. Then by integration by parts, induction
hypothesis as well as the definition of Pk (x),
Z
Z
n+1
(log u)
n+1
du = u(log u)
−
(n + 1)(log u)n du
= u(log u)n+1 − u(n + 1)Pn (log u) + C
= uPn+1 (log u) + C
33
which completes our induction.
This lemma gives the reason why we would like to study the polynomials Pn (x)
R
as the integral (log x − 1)n dx occurs at (3.2).
Lemma III.5. For any non-negative integer n,
n
X
Pn (x) =
(−1)j
j=0
n!
xn−j .
(n − j)!
Proof: Induction on n. Clearly, it is true for n = 0. Suppose that it is true for
some n ≥ 0. From the definition of Pn (x) and induction hypothesis, we have
Pn+1 (x) = xn+1 − (n + 1)Pn (x)
n
X
n+1
= x
−
(−1)j (n + 1)
= xn+1 +
j=0
n
X
(−1)j+1
j=0
n!
xn−j
(n − j)!
(n + 1)!
x(n+1)−(j+1)
((n + 1) − (j + 1))!
which completes the induction by relabeling j + 1 as j.
The following is a crucial combinatorial identity!
Lemma III.6. For any non-negative integer N ,
N µ ¶
X
N
i=0
i
i
PN −i (x)y = PN (x + y) =
N µ ¶
X
N
i=0
i
xN −i Pi (y).
Proof: By the previous lemma, the left-hand side equals
N µ ¶³X
N −i
X
N
i=0
N
X
i
j=0
(−1)j
(N − i)! N −i−j ´ i
x
y
(N − i − j)!
N −j
X
(N − i)! N −i−j i
N!
x
y
i!(N − i)! (N − i − j)!
j=0
i=0
N −j µ
N
X N − j¶
X
N!
j
x(N −j)−i y i
=
(−1)
(N − j)! i=0
i
j=0
=
=
N
X
j=0
(−1)j
(−1)j
N!
(x + y)N −j = PN (x + y).
(N − j)!
34
One gets the last equality by simply exchanging x and y.
We are now ready to prove the main result.
Theorem III.7. Assume Hypothesis HL. For any positive even integer k and N ² ≤
h ≤ N 1/k−² ,
Z
N/h ¡
x ¢k/2
0
dx + Ok (hk/2 N 1−² )
E
E
µ
¶
k/2
³X
´
k ¡
N ¢k/2−i
k!
0
k/2
00
2
log
=
Nh
Pi (B ) + Ok (hk/2 N 1−² )
k/2
(k/2)!2
i
h
i=0
k!
µk (N, h) =
hk/2+1
(k/2)!2k/2
log
00
for some ²0 > 0 depending on ² and k. Here B 00 = 1 − C0 − log 2π and E = e−B =
2πeC0 −1 .
Proof: When N ² ≤ h ≤ N 1/k−² , we have from (3.2)
µ
¶Z N
k
X
k!hk−r
r
µk (N, h) =
(log x − 1)k−r dxR2r−k (h) + Ok (hk/2 N 1−² )
k−r
r!2
k−r 1
r≥k/2
·
¸
k!hk−r
(2r − k)!
r−k/2
r−k/2−1/7r+²
=
F2r−k (h)
+ Ok (h
)
2k−r (k − r)!(2r − k)! (r − k2 )!2r−k/2
r=k/2
k
X
Z
N
∗
(log x − 1)k−r dx + Ok (hk/2 N 1−² )
1
by (3.3) and the notation of Theorem III.3.
=
k
X
r=k/2
k!hk−r
F2r−k (h)r−k/2
2k/2 (k − r)!(r − k/2)!
k
³ X
+Ok N
r≥k/2
Z
N
(log x − 1)k−r dx
1
´
k!hk−r
r−k/2−1/7r+2²
+ Ok (hk/2 N 1−² )
h
2k−r (k − r)!(2r − k)!
k µ k ¶
¡
¢
k!hk/2 X
0
−1/2+² r−k/2
r−k/2
2
=
−
log
h
+
B
+
O(h
)
h
(k/2)!2k/2
r − k2
r=k/2
Z N
¢
¡
¢
¡
∗
(log x − 1)k−r dx + Ok N hk/2−1/7k+2² + Ok hk/2 N 1−²
1
35
by Theorem III.3. Hence, replacing r −
k/2
X
k!
k/2
h
µk (N, h) =
(k/2)!2k/2
i=0
k
2
by i, we have
µk¶
2
i
Z
N
0 i
(− log h + B )
1
N (log N ) ) + Ok (h N 1−²/8k )
Z N
¡
¢
k!
x
k/2
00 k/2
=
h
log
+
B
dx + Ok (hk/2 N 1−²/8k )
k/2
(k/2)!2
h
1
Z N/h
k!
hk/2+1
(log x + B 00 )k/2 dx + Ok (hk/2 N 1−²/8k )
=
(k/2)!2k/2
1/h
Z N/h
¡
k!
x ¢k/2
=
hk/2+1
log
dx + Ok (hk/2 N 1−²/8k )
k/2
(k/2)!2
E
E
+Ok (h
(k−1)/2+²
(log x − 1)k/2−i dx
k/2
k/2
which gives the first equality of the theorem. By a further substitution and Lemma
III.4, the main term of µk (N, h)
Z N/Eh
k!
k/2+1
h
E
(log x)k/2 dx
=
k/2
(k/2)!2
1
h
i
k!
N
N
k/2+1
=
h
E
Pk/2 (log
) + O(1)
(k/2)!2k/2
Eh
Eh
k!
N
k/2
=
N
h
P
(log
+ B 00 ) + Ok (hk/2 N 1−² )
k/2
(k/2)!2k/2
h
which gives the second equality by Lemma III.6.
Corollary III.8. Assume Hypothesis HL.
Z
X¡
1
¢2
X
0
ψ(x + h) − ψ(x) − h dx = hX log + BhX + O(hX 1−² )
h
for X ² ≤ h ≤ X 1/2−² . Here B = −C0 − log 2π.
Proof: One simply applies Theorem III.7 with k = 2, X = N , and because
P0 (x) = 1 and P1 (x) = x − 1.
Z
X¡
1
³
´
¢2
X
0
ψ(x + h) − ψ(x) − h dx = hX log + (B 00 − 1) + O(hX 1−² )
h
X
0
= hX log + BhX + O(hX 1−² )
h
36
as B 00 −1 = 1−C0 −log 2π −1 = B. Here, it only works for X ² ≤ h ≤ X 1/2−² because
one just assumes Hypothesis HL in contrast with considering further cancellations
among the error terms in Montgomery and Soundararajan [14].
3.4
Numerical evidence
In Montgomery and Soundararajan [14], they got some numerical data for µ̃k =
1
µ (1010 , 105 ).
1010 k
We try to compare them with values from the more precise asymp-
totic formula as well as the contribution from the first main term only. We find that
they are in good agreement!
k µ̃k
Result from new formula Result from first main term
0 1
1
1
2 9.0663 ∗ 105
9.0978 ∗ 105
1.1513 ∗ 106
4 2.4995 ∗ 1012 2.5131 ∗ 1012
3.9764 ∗ 1012
6 1.1573 ∗ 1019 1.1675 ∗ 1019
2.289 ∗ 1019
CHAPTER IV
“Equivalence” of Higher Even Moments
4.1
A brief introduction
In Chapter II, we deduce the “equivalence” of two asymptotic formulas concerning
the second moments
Z X
Z
¡
¢2
ψ(x + h) − ψ(x) − h dx and
1
X¡
¢2
ψ(x + δx) − ψ(x) − δx dx.
1
In Chapter III, we get a more precise asymptotic formula for the higher moments
Z X
¡
¢k
(4.1)
ψ(x + h) − ψ(x) − h dx
1
where k ≥ 2 is any positive even integer. So, one would be curious if it is possible to
establish an asymptotic formula for
Z X
¡
¢k
ψ(x + δx) − ψ(x) − δx dx
1
and vice versa just like what we did in Chapter II. The answer is yes but the validity
of the asymptotic formulas is more restrictive due to the fact that there are no higher
moment analogue for inequalities (2.1) and (2.2). But, in any case, the results may
provide new insight or evidence for the distribution of the zeros of the Riemann zeta
function. Moreover, our method replaces the brute-force calculations in Chapter II
with a more systematic approach. Throughout this chapter, we will assume Riemann
Hypothesis.
37
38
4.2
Some preparations
First of all, from Riemann Hypothesis (see [21]), ψ(x) = x + O(x1/2 (log x)2 ), one
can get the following: For any integer k ≥ 2,
Z X
¡
¢k
(4.2)
ψ(x + δx) − ψ(x) − δx dx ¿ X k/2+1 (log X)2k
1
for 0 < δ ≤ 1, and
Z
(4.3)
X¡
¢k
ψ(x + h) − ψ(x) − h dx ¿ X k/2+1 (log X)2k
1
for 0 < h ≤ X. These provide substitutes for inequalities (2.1) and (2.2).
We also need some lemmas.
Lemma IV.1. For any differentiable function f (u) and any 0 ≤ η ≤ 1,
Z (1+η)T
¡
¢
f (u)du = ηT f (T ) + O η 2 T 2 max |f 0 (t)| .
T ≤t≤(1+η)T
T
Proof: By mean-value theorem, we have
Z (1+η)T
f (u)du = ηT f (T + ξT ) where 0 ≤ ξ ≤ η
T
¡
¢
= ηT f (T ) + ξT f 0 (T + ξ 0 T ) where 0 ≤ ξ 0 ≤ ξ,
and the result follows.
Lemma IV.2. For any positive integer k, we have
xk − y k = (x − y)P (x, y) + (x − y)k
where P (x, y) is some homogeneous polynomial of degree k − 1.
Proof: By Factor Theorem, z − 1 divides z k − 1 − (z − 1)k . So,
(z − 1)P (z) = z k − 1 − (z − 1)k
for some integer polynomial P (z) of degree k − 1. Set z =
by y k , we get the desired result.
x
y
and multiply both sides
39
Lemma IV.3. For any positive integer k, and any non-negative real numbers α,
a1 , a2 , ..., ak , we have
(a1 + a2 + ... + ak )α ¿α,k aα1 + aα2 + ... + aαk .
Here, ¿α,k means that the implicit constant may depend on α and k but not on any
ai ’s.
Proof: Without loss of generality, suppose that a1 is the largest among the ai ’s.
Then
(a1 + a2 + ... + ak )α ≤ (ka1 )α ≤ k α (aα1 + aα2 + ... + aαk ).
4.3
Main result for k-th moments
Throughout this section and the next section, k is some even integer ≥ 4.
Theorem IV.4. Assume Riemann Hypothesis. If, for some small ², ²1 > 0 (²1 may
depend on ² and k, ²1 ≤ ²),
Z X
Z
¡
¢k
k/2
(4.4)
ψ(x + h) − ψ(x) − h dx = Dk h
1
X/h ¡
log
E
x ¢k/2
dx + Ok (hk/2 X 1−²1 )
E
holds uniformly for X ² ≤ h ≤ X 1−² , then
(4.5)
Z X
¡
1
³
Dk
1 ´k/2
k/2+1 k/2
ψ(x + δx) − ψ(x) − δx dx = k
X
δ
log
+ Ok (δ k/2 X k/2+1−²2 )
Eδ
+1
2
¢k
holds uniformly for X −1/2+²+2²1 ≤ δ ≤ X −² /2 with some ²2 > 0. Here, Dk =
k!
,
(k/2)!2k/2
E = 2πeC0 −1 , and C0 is Euler’s constant.
Proof: Again, our method is that of Saffari and Vaughan [20] employed in Goldston and Montgomery [9]. Let f (x, h) = ψ(x + h) − ψ(x) − h. Let X −1/2+²+²1 ≤ ∆ ≤
X −² . We want to calculate
Z V Z
(4.6)
V /2
∆¡
0
¢k
ψ(x + δx) − ψ(x) − δx dδdx.
40
By substituting h = δx, (4.6) becomes
Z
∆V
Z
V
∆V /2
h/∆
Z
Z
∆V
f (x, h)k
dxdh +
x
Z
V
=
∆V /2
Z
Z
0
V
V /2
f (x, h)k
dxdh
x
V /2
V²
Z
V
+
V²
h/∆
∆V /2
Z
V
+
∆V /2
Z
= I1 + I2 + I3 .
0
V /2
By integration by parts, we have from (4.4) that
Z
V
f (x, h)k
dx
x
U
h1 Z x
iV Z V ³Z x
´1
k
k
=
f (u, h) du +
f (u, h) du 2 dx
x 1
x
U
U
1
Z U/h
h 1 Z V /h ¡
¡
x ¢k/2
1
x ¢k/2 i
= Dk hk/2+1
log
dx −
log
dx
V E
E
U E
E
Z V
Z
1 x/h ¡
u ¢k/2
k/2+1
+Dk h
log
dudx + Ok (U −²1 hk/2 )
2
E
U x
E
= T1 + T2 + Ok (U −²1 hk/2 )
as long as V ² ≤ h ≤ U 1−² with U ≤ V ≤ 2U .
Z
k/2
V /h
T2 = D k h
Z
y¡
log
U/h
E
u ¢k/2
−1
dud( )
E
y
Z
h h Z V /h ¡
h U/h ¡
u ¢k/2
u ¢k/2
k/2
= Dk h
−
du +
du
log
log
V E
E
U E
E
Z V /h
(log y/E)k/2 i
dy .
+
y
U/h
Therefore,
Z
V
U
f (x, h)k
Dk k/2 h¡
V ¢k/2+1 ¡
U ¢k/2+1 i
dx = k
h
log
− log
+ Ok (U −²1 hk/2 )
x
Eh
Eh
+
1
2
as long as V ² ≤ h ≤ U 1−² with U ≤ V ≤ 2U . For I1 and I2 to work, we need
(4.7)
V² ≤
³ V ´1−²
³ h ´1−²
∆V
∆V
≤
, and V ² ≤ h ≤
for
≤ h ≤ ∆V.
2
2
∆
2
Since X −1/2+²+²1 ≤ ∆ ≤ X −² , for X 1/2 ≤ V ≤ X, one can check that (4.7) are
41
satisfied. Therefore,
Dk h
k
+1
2
I1 =
I2
Z
∆V
h
k/2
¡
∆V /2
V ¢k/2+1
log
dh −
Eh
Z
∆V
h
∆V /2
k/2
¡
1 ¢k/2+1 i
log
dh
E∆
+Ok (∆k/2+1 V k/2+1−²1 ),
Z
Z ∆V /2
¡
Dk h ∆V /2 k/2 ¡
V ¢k/2+1
V /2 ¢k/2+1 i
= k
h
log
dh −
dh
hk/2 log
Eh
Eh
+1 0
0
2
+Ok (V (k/2+1)² (log V )k/2+1 ) + Ok (∆k/2+1 V k/2+1−²1 ),
I3 ¿ V ² V k² (log V )k ,
Combining these, (4.6) equals
Dk
k
+1
2
·Z
∆V
h
k/2
0
¡
1 ¢k/2+1
− log
E∆
Z
¡
V ¢k/2+1
log
dh −
Eh
Z
∆V /2
0
¡
V /2 ¢k/2+1
hk/2 log
dh
Eh
¸
∆V
h
k/2
¡
¢
dh + O V (k+1)² (log V )k + Ok (∆k/2+1 V k/2+1−²1 )
∆V /2
for X 1/2 ≤ V ≤ X. Now, replacing V by X2−l in the above, summing over 0 ≤ l ≤
X
M = [ 2log
],
log 2
Z
Z
X
X/2M
(4.8)
Dk
=k
+1
2
∆¡
¢k
ψ(x + δx) − ψ(x) − δx dδdx
0
·Z
∆X
k/2
h
0
¡
X ¢k/2+1
log
dh −
Eh
k
2
1 ¡
1 ¢k/2+1
log
(∆X)k/2+1
E∆
+1
¸
+ Ok (X (k+1)² (log X)k ) + Ok (∆k/2+1 X k/2+1−²1 )
Z ∆X
¡
Dk
X ¢k/2
=k
hk/2 log
dh + Ok (∆k/2+1 X k/2+1−²1 )
Eh
+
1
0
2
by integration by parts, and as ∆ ≥ X −1/2+²+²1 ,
X (k+1)² (log X)k ¿ X (k/2+1)/2+² ¿ ∆k/2+1 X k/2+1−²1 .
Using (4.2),
Z
X/2M
Z
∆¡
(4.9)
1
0
¢k
ψ(x + δx) − ψ(x) − δx dδdx ¿k ∆X (k/2+1)/2 (log X)2k .
42
But, since ∆ ≥ X −1/2+²+²1 ,
(4.10)
∆k/2+1 X k/2+1−²1 À ∆(X −1/2+²+²1 )k/2 X k/2+1−²1 À ∆X k/4+1 À ∆X (k/2+1)/2 (log X)2k .
Combining (4.8), (4.9) and (4.10), we have
Z ∆Z X
¡
¢k
ψ(x + δx) − ψ(x) − δx dxdδ
0
1
(4.11)
Z ∆X
¡
Dk
X ¢k/2
=k
hk/2 log
dh + Ok (∆k/2+1 X k/2+1−²1 )
Eh
+
1
0
2
for X −1/2+²+²1 ≤ ∆ ≤ X −² .
We now deduce (4.5) from (4.11). Set η = X −2²1 /3 . By Lemma IV.1, one has for
X −1/2+²+2²1 ≤ ∆ ≤ X −² /2,
Z (1+η)∆ Z X
∆
(4.12)
Z
¡
¢k
ψ(x + δx) − ψ(x) − δx dxdδ
1
(1+η)∆X
¡
X ¢k/2
hk/2 log
dh + Ok (∆k/2+1 X k/2+1−²1 )
Eh
∆X
³
³
´
1 ´k/2
1
Dk
(∆X)k/2+1 log
η + Ok η 2 (∆X)k/2+1 (log )k/2
=k
E∆
∆
+1
2
Dk
+1
2
=k
+ Ok (∆k/2+1 X k/2+1−²1 ).
Let g(x, δx) = f (x, ∆x) for ∆ ≤ δ ≤ (1 + η)∆. Then one can easily check that
¡
¢
f (x, δx) − g(x, δx) = f (1 + ∆)x, (δ − ∆)x . So,
Z (1+η)∆ Z X
Z η∆ Z (1+∆)X
1+∆
¡
¢k
f (x, δx)k dxdδ
f (x, δx) − g(x, δx) dxdδ =
∆
1
0
1+∆
(4.13)
1
¿k (ηX∆)k/2+1 (log
)k/2
η∆
by (4.11), the choice of η and the range of ∆. Thus, by Lemma IV.3, (4.12) and
(4.13),
(4.14)
Z (1+η)∆ Z
X
Z Z
k
|f (x, δx)| +
|f (x, δx) − g(x, δx)|k
¿k ηX k/2+1 ∆k/2+1 (log
1 k/2
) + ∆k/2+1 X k/2+1−²1 .
∆
g(x, δx) dxdδ ¿k
∆
Z Z
k
1
43
By Lemma IV.2 and Holder’s inequality,
Z (1+η)∆ Z X
f (x, δx)k − g(x, δx)k dxdδ
∆
1
Z (1+η)∆ Z X
Z (1+η)∆ Z X
=
P (f, g)(f − g) +
(f − g)k
∆
1
∆
1
³Z Z
´(k−1)/k ³Z Z
´1/k Z Z
k
k/(k−1)
¿
|P (f, g)|
|f − g|
+
|f − g|k
(k−1)/k
= J1
1/k
J2
+ J2
where P (x, y) is a homogeneous polynomial of degree k − 1.
J2 ¿k η k/2+1 X k/2+1 ∆k/2+1 (log
1 k/2
)
η∆
by (4.13). And
Z Z
¡ X
¢k/(k−1)
J1 ¿k
|f |i |g|j
i+j=k−1
¿k
X Z Z ¡
|f | + |g|
¢k
by binomial theorem
i+j=k−1
Z Z
¿k
f k + g k by Lemma IV.3
1 k/2
) + ∆k/2+1 X k/2+1−²1 by (4.12) and (4.14)
η∆
1 k/2
ηX k/2+1 ∆k/2+1 (log
) as η = X −2²1 /3 .
η∆
¿k ηX k/2+1 ∆k/2+1 (log
¿k
Consequently, by Lemma IV.3,
Z Z
1 k/2
(4.15)
f k − g k ¿k X k/2+1 ∆k/2+1 η 3/2 (log
) .
η∆
Therefore, by (4.15) and (4.12),
Z Z
Z X
¡
¢k
ψ(x + ∆x) − ψ(x) − ∆x dx =
gk
η∆
Z Z1
³
1 k/2 ´
k
k/2+1 k/2+1 3/2
=
f +O X
∆
η (log
)
η∆
³
Dk
1 ´k/2
= k
η + Ok (∆k/2+1 X k/2+1−²1 )
(X∆)k/2+1 log
E∆
+1
2
³
1 k/2 ´
k/2+1 k/2+1 3/2
.
)
+Ok X
∆
η (log
η∆
44
Dividing through by η∆, we have
Z X
¡
¢k
ψ(x + ∆x) − ψ(x) − ∆x dx
1
³
³ ∆k/2 X k/2+1−²1 ´
Dk
1 ´k/2
k/2+1 k/2
=k
X
∆
log
+ Ok
E∆
η
+1
2
³
´
1 k/2
+ Ok X k/2+1 ∆k/2 η 1/2 (log
)
.
η∆
Finally, recall η = X −2²1 /3 , one has the error terms ¿ X k/2+1−²1 /4 ∆k/2 . So, the
theorem is true with ²2 =
4.4
²1
!
4
Its converse
Surprisingly, there is not much loss in the converse direction!
Theorem IV.5. Assume Riemann Hypothesis. If, for some small ², ²1 > 0 (²1 may
depend on ² and k, and ²1 ≤ ²),
(4.16)
Z X
³
¡
¢k
1 ´k/2
Dk
X k/2+1 δ k/2 log
+ Ok (δ k/2 X k/2+1−²1 )
ψ(x + δx) − ψ(x) − δx dx = k
Eδ
+1
1
2
holds uniformly for X −1+² ≤ δ ≤ X −² , then
Z
X¡
(4.17)
¢k
Z
ψ(x+h)−ψ(x)−h dx = Dk h
k/2+1
1
X/h ¡
E
log
x ¢k/2
dx+Ok (hk/2 X 1−²2 )
E
holds uniformly for X 2²+2²1 ≤ h ≤ X 1−(k/2+1)(²+²1 ) with some ²2 > 0. Here, Dk =
k!
,
(k/2)!2k/2
E = 2πeC0 −1 and C0 is Euler’s constant.
Proof: Let f (x, h) = ψ(x + h) − ψ(x) − h. Let X 2²+²1 ≤ H ≤ 2X 1−(k/2+1)(²+²1 ) .
Then there is some µ with k2 (² + ²1 ) ≤ µ ≤
1−2²
1+2/k
such that X 1−µ(1+2/k)+²1 ≤ H ≤
2X 1−µ−² . The lower bound for µ ensures 1 − µ(1 + k2 ) + ²1 < 1 − µ − ². First, we
calculate
Z
V
Z
H¡
(4.18)
V /2
0
¢k
ψ(x + h) − ψ(x) − h dhdx.
45
By substituting δ = hx , (4.18) becomes
Z
2H/V
Z
Z
H/δ
H/V
k
Z
V
f (x, δx)k xdxdδ
f (x, δx) xdxdδ +
H/V
Z
V /2
2H/V
Z
0
Z
H/δ
=
H/V
Z
+
H/V
( V2 )−1+²
+
( V2 )−1+²
V /2
Z
V
V /2
V /2
0
Z
V
= I1 + I2 + I3 .
V /2
By integration by parts, we have from (4.16) that
Z
V
f (x, δx)k xdx
U
=
i
Dk k/2 ³
1 ´k/2 h k/2+2
k/2+2
δ
log
V
−U
+ Ok (δ k/2 V k/2+2−²1 )
k
Eδ
+2
2
as long as U −1+² ≤ δ ≤ V −² with U ≤ V ≤ 2U . In order for this to work for I1 and
I2 , we need
(4.19)
³ V ´−1+²
2
≤
³ V ´−1+²
³ H ´−²
2H
H
H
≤ V −² , and
≤δ≤
.
≤δ≤
for
V
2
δ
V
V
Since X 1−µ(1+2/k)+²1 ≤ H ≤ 2X 1−µ−² , one can check that (4.19) are satisfied for
X 1−µ ≤ V ≤ X. Thus,
I1 =
I2
Dk
k
+2
2
Z
2H/V h¡
H/V
H ¢k/2+2 ¡ V ¢k/2+2 i k/2 ³
1 ´k/2
−
δ
log
dδ
δ
2
Eδ
+Ok (H k/2+1 V 1−²1 ),
Z
Dk h k/2+2 ¡ V ¢k/2+2 i H/V k/2 ³
1 ´k/2
= k
V
−
δ
log
dδ
2
Eδ
+2
0
2
+Ok (V 1+(k/2+1)² (log V )k/2 ) + Ok (H k/2+1 V 1−²1 ),
I3 ¿ V −1+² V 2 V k² (log V )k = V 1+(k+1)² (log V )k .
Let Ek = Dk /( k2 + 2), then
Z
1³
1 ´k/2
(4.18) = Ek H
log
dδ + Ek V k/2+2
2
δ
Eδ
H/V
³
³ V ´k/2+2 Z 2H/V
1 ´k/2
k/2
log
−Ek
δ
dδ
2
Eδ
0
k/2+2
2H/V
+Ok (V 1+(k+1)² (log V )k ) + Ok (H k/2+1 V 1−²1 )
Z
H/V
δ
0
k/2
³
1 ´k/2
log
dδ
Eδ
46
when X 1−µ ≤ V ≤ X. Now, replacing V by X2−l in the above, summing over
log X
0 ≤ l ≤ M = [ µlog
],
2
(4.20)
Z
Z
X
X/2M
H¡
0
Z
¢k
ψ(x + h) − ψ(x) − h dhdx
Z H/X
³
1³
1 ´k/2
1 ´k/2
k/2+2
k/2
=Ek H
log
dδ + Ek X
δ
log
dδ
δ2
Eδ
Eδ
H/X
0
³ X ´k/2+2 Z 2M +1 H/X
³
1 ´k/2
− Ek M +1
dδ + Ok (H k/2+1 X 1−²1 )
δ k/2 log
2
Eδ
0
Z H/X
Z 1/E ³
³
1
1 ´k/2
1 ´k/2
k/2+2
k/2+2
k/2
log
dδ
+
E
X
δ
log
dδ
=Ek H
k
2
Eδ
Eδ
0
H/X δ
k/2+2
2M +1 H/X
+ Ok (H k/2+1 X 1−²1 ).
because, as X 2²+²1 ≤ H, X 1+(k+1)² (log X)k ¿ H k/2+1 X 1−²1 . Also, the terms involving 2M +1 are absorbed into the error term as µ ≥ ² + ²1 . Using (4.3),
Z
X/2M
Z
H¡
(4.21)
¢k
ψ(x + h) − ψ(x) − h dhdx ¿ H(X 1−µ )k/2+1 (log X)2k .
0
1
But, since H ≥ X 1−µ(1+2/k)+²1 ,
H k/2+1 X 1−²1 À HX k/2+1−µ(k/2+1)+²1 Àk H(X 1−µ )k/2+1 (log X)2k .
(4.22)
Combining (4.20), (4.21) and (4.22), we have
Z
(4.23)
H
Z
X¡
¢k
ψ(x + h) − ψ(x) − h dxdh
0
1
Z 1/E ³
Z H/X
³
1 ´k/2
1
1 ´k/2
k/2+2
k/2+2
k/2
log
=Ek H
dδ + Ek X
dδ
δ
log
2
Eδ
Eδ
H/X δ
0
+ Ok (H k/2+1 X 1−²1 )
for X 2²+²1 ≤ H ≤ 2X 1−(k/2+1)(²+²1 ) .
We now deduce (4.17) from (4.23). Set η = X −2²1 /3 . For X 2²+2²1 ≤ H ≤
47
X 1−(k/2+1)(²+²1 ) ,
Z
(1+η)H
Z
X¡
¢k
ψ(x + h) − ψ(x) − h dxdh
H
1
Z 1/E
1 ´k/2
1³
k/2+2
= Ek ((1 + η)H)
log
dδ
2
Eδ
(1+η)H/X δ
Z 1/E ³
1
1 ´k/2
k/2+2
−Ek H
log
dδ
2
Eδ
H/X δ
Z (1+η)H/X
³
1 ´k/2
k/2+2
+Ek X
δ k/2 log
dδ + Ok (H k/2+1 X 1−²1 )
Eδ
H/X
Z 1/E ³
Z (1+η)H/X ³
1
1
1 ´k/2
1 ´k/2
k/2+2
k/2+2
= −Ek H
log
dδ
+
D
ηH
log
dδ
k
2
δ2
Eδ
Eδ
H/X
H/X δ
Z (1+η)H/X
³
1 ´k/2
k/2+2
+Ek X
δ k/2 log
dδ
Eδ
H/X
³
´
X
+Ok η 2 H k/2+1 X(log )k/2 + Ok (H k/2+1 X 1−²1 )
H
Z 1/E ³
1
1 ´k/2
= Dk ηH k/2+2
log
dδ
2
Eδ
H/X δ
³
´
X
+Ok η 2 H k/2+1 X(log )k/2 + Ok (H k/2+1 X 1−²1 )
H
by Lemma IV.1. Therefore
Z
(1+η)H
H
Z
X¡
1
Z
¢k
ψ(x + h) − ψ(x) − h dxdh
X/H ¡
u ¢k/2
du
E
E
³
´
X
+ Ok η 2 H k/2+1 X(log )k/2 + Ok (H k/2+1 X 1−²1 ).
H
(4.24)
=Dk ηH
k/2+2
log
Let g(x, h) = f (x, H) for H ≤ h ≤ (1 + η)H. Again, one can check that f (x, h) −
g(x, h) = f (x + H, h − H). So,
Z
(4.25)
(1+η)H
H
Z
X¡
1
¢k
f (x, h) − g(x, h) dxdh =
Z
ηH
0
Z
X+H
f (x, h)k dxdh
1+H
³
X ´k/2
¿k η k/2+1 XH k/2+1 log
ηH
by (4.23) as well as the choice of η and the range of H. Thus, by Lemma IV.3, (4.24)
48
and (4.25),
Z (1+η)H Z
Z Z
X
k
g(x, h) dxdh ¿k
(4.26)
H
Z Z
k
|f (x, h) − g(x, h)|k
|f (x, h)| +
1
¿k ηXH k/2+1 (log
X k/2
) + H k/2+1 X 1−²1 .
ηH
By Lemma IV.2 and Holder’s inequality,
Z (1+η)H Z X
f (x, h)k − g(x, h)k dxdδ
H
1
Z (1+η)H Z X
Z (1+η)H Z X
=
P (f, g)(f − g) +
(f − g)k
H
1
H
1
(4.27)
³Z Z
´(k−1)/k ³Z Z
´1/k Z Z
k/(k−1)
k
¿
|P (f, g)|
|f − g|
+
|f − g|k
(k−1)/k
=K1
1/k
K2
+ K2
where P (x, y) is a homogeneous polynomial of degree k − 1. From (4.25),
(4.28)
K2 ¿k η
k/2+1
XH
k/2+1
³
X ´k/2
log
.
ηH
And similar to the proof in Theorem IV.4,
Z Z
(4.29)
f k + g k ¿k ηXH k/2+1 (log
K1 ¿k
X k/2
)
ηH
by (4.24) and (4.26). Consequently, by (4.27), (4.28), (4.29) and Lemma IV.3,
Z Z
(4.30)
³
X ´k/2
f k − g k ¿k η 3/2 XH k/2+1 log
.
ηH
Therefore, by (4.30) and (4.24),
Z
Z Z
¢k
ψ(x + h) − ψ(x) − h dx =
gk
ηH
Z Z 1
³
X k/2 ´
=
f k + Ok XH k/2+1 η 3/2 (log
)
ηH
Z X/H
¡
u ¢k/2
k/2+2
log
= Dk ηH
du + Ok (H k/2+1 X 1−²1 )
E
E
³
X k/2 ´
.
)
+Ok XH k/2+1 η 3/2 (log
ηH
X¡
Divide through by ηH, and recall η = X −2²1 /3 , we get the theorem with ²2 =
²1
.
4
49
4.5
Concluding remarks and some numerical evidence
One can improve (4.2) and (4.3) slightly by using (2.1) and (2.2) of Chapter II
respectively as follow: For k > 2, assuming Riemann Hypothesis,
Z
X¡
¢k
ψ(x + δx) − ψ(x) − δx dx
1
¿
Z
max |ψ(x + δx) − ψ(x) − δx|
k−2
1≤x≤X
¿ δX
X¡
¢2
ψ(x + δx) − ψ(x) − δx dx
X¡
¢2
ψ(x + h) − ψ(x) − h dx
1
k/2+1
2k−2
(log X)
for 0 < δ ≤ 1, and
Z
X¡
¢k
ψ(x + h) − ψ(x) − h dx
1
¿
Z
max |ψ(x + h) − ψ(x) − h|
k−2
1≤x≤X
¿ hX
k/2
1
2k−2
(log X)
for 0 < h ≤ X. These, in turn, can give us slight improvements on the ranges for δ
and h in Theorem IV.4 and Theorem IV.5.
In general, one would expect that (4.4) uniformly in X ² ≤ h ≤ X 1−² is “equivalent” to (4.5) uniformly in X −1+² ≤ δ ≤ X −² . To prove this, one needs the following
analogue of (2.1) and (2.2) in Chapter II: For some nk > 0 (depending on k),
Z
X¡
(4.31)
¢k
ψ(x + δx) − ψ(x) − δx dx ¿k δ k/2 X k/2+1 (log X)nk
1
for 0 < δ ≤ 1, and
Z
X¡
¢k
ψ(x + h) − ψ(x) − h dx ¿k hk/2 X(log X)nk
(4.32)
1
for 0 < h ≤ X.
However, (4.31) and (4.32) cannot be true in those ranges unless k = 2. For
50
instance, when k > 2 and δ ¿ X −1 ,
Z
=
X¡
¢k
ψ(x + δx) − ψ(x) − δx dx
1
Z
X Z n
k
(Λ(n) − δx) dx + O(
n≤X
Λ(n)6=0
=
X
n/(1+δ)
¡
Λ(n)k n 1 −
n≤X
=
X
(δx)k dx)
1
1 ¢
+ O(X(log X)k−1 δ 2 ) + O(δ k X k+1 )
1+δ
1 2
δX (log X)k−1 + O(δ 2 X(log X)k−1 )
2
À δ k/2 X k/2+1 (log X)n
for any n, by prime number theorem. Similarly, when k > 2 and h ¿ 1,
Z
X¡
¢k
ψ(x + h) − ψ(x) − h dx
1
= hX(log X)k−1 + O(h2 X(log X)k−1 ) À hk/2 X(log X)n
for any n. So, the best one can hope for is that, for any ² > 0, (4.31) holds uniformly
for X −1+² ≤ δ ≤ X −² and (4.32) holds uniformly for X ² ≤ h ≤ X 1−² . But then one
has to modify the proofs for Theorem IV.4 and Theorem IV.5 in order to prove
Theorem IV.6. Fix some small ² > 0 and assume inequalities (4.31) and (4.32)
in the appropriate ranges as described above. For any positive even integer k, if for
some ²1 > 0,
Z
X¡
¢k
ψ(x + h) − ψ(x) − h dx = Dk h
Z
k/2+1
X/h ¡
log
E
1
x ¢k/2
dx + Ok (hk/2 X 1−²1 )
E
holds uniformly for X ² ≤ h ≤ X 1−² , then
Z
X¡
1
³
Dk
1 ´k/2
k/2+1 k/2
ψ(x + δx) − ψ(x) − δx dx = k
+ Ok (δ k/2 X k/2+1−²2 )
X
δ
log
Eδ
+
1
2
¢k
holds uniformly for X −1+²+3²1 ≤ δ ≤ X −² /2 with some ²2 > 0.
51
Conversely, if for some ²1 > 0,
Z
X¡
1
³
Dk
1 ´k/2
k/2+1 k/2
ψ(x + δx) − ψ(x) − δx dx = k
X
δ
log
+ Ok (δ k/2 X k/2+1−²1 )
Eδ
+1
2
¢k
holds uniformly for X −1+² ≤ δ ≤ X −² , then
Z
X¡
Z
¢k
ψ(x + h) − ψ(x) − h dx = Dk h
X/h ¡
k/2+1
1
log
E
x ¢k/2
dx + Ok (hk/2 X 1−²2 )
E
holds uniformly for X ²+2²1 ≤ h ≤ X 1−²−²1 with some ²2 > 0.
Sketch of proof of Theorem IV.6:
Let f (x, h) = ψ(x + h) − ψ(x) − h. Let X −1+²+2²1 ≤ ∆ ≤ X −² . One just follows
the proof of Theorem IV.4. One can check that the conditions (4.7) are satisfied
when X 1−²1 ≤ V ≤ X. Hence, one has
Z ∆Z X
¡
¢k
ψ(x + δx) − ψ(x) − δx dxdδ
1
0
Z ∆X
¡
Dk
X ¢k/2
=k
hk/2 log
dh + Ok (∆k/2+1 X k/2+1−²1 )
Eh
+1 0
2
for X −1+²+2²1 ≤ ∆ ≤ X −² . The only differences are the range of summation (0 ≤
log X
l ≤ M = [ ²1log
]) and error estimations: For X 1−²1 ≤ V ≤ X,
2
V (k/2+1)² (log V )nk +k ¿ V (k/2+1)²+²1 ¿ ∆k/2+1 V k/2+1−²1
(4.33)
as V −1+²+²1 ≤ X −1+²+2²1 ≤ ∆.
Z
I3
V²
Z
V
=
0
V /2
Z
f (x, h)k
dxdh =
x
V²
Z
Z
V
V ²/2
Z
V
+
V ²/2
V /2
0
V /2
¿k V ² (V ² )k/2 (log V )nk + V ²/2 (V ²/2 )k (log V )k ¿k ∆k/2+1 V k/2+1−²1 .
Z
Z
Z
∆
0
X/2M ¡
1
¢k
ψ(x + δx) − ψ(x) − δx dxdδ
Z
∆
Z
X 1−²1
X −1+²/2+²1
Z
X 1−²1
+
≤
X −1+²/2+²1
k/2
¿k ∆∆
(X
1
0
1−²1 k/2+1
)
1
nk
(log X)
+X
−1+²/2+²1
X 1−²1 X k²/2 (log X)k
¿k ∆k/2+1 X k/2+1−²1 + X (k+1)²/2 (log X)k ¿k ∆k/2+1 X k/2+1−²1
52
by (4.31), (4.32) and (4.33). Then, one restricts attention to X −1+²+3²1 ≤ ∆ ≤ X −² /2
and set η = X −2²1 /3 . The theorem follows in the same way as Theoerm IV.4.
Conversely, let 2X ²+²1 ≤ H ≤ 2X 1−²−²1 . One can check that the conditions (4.19)
are satisfied when X 1−²1 ≤ V ≤ X. Following Theorem IV.5, one has
Z HZ X
¡
¢k
ψ(x + h) − ψ(x) − h dxdh
0
1
Z 1/E ³
Z H/X
³
1
1 ´k/2
1 ´k/2
k/2+2
k/2+2
k/2
=Ek H
log
dδ
+
E
X
δ
log
dδ
k
2
Eδ
Eδ
H/X δ
0
+ Ok (H k/2+1 X 1−.99²1 )
for 2X ²+²1 ≤ H ≤ 2X 1−²−²1 . The only differences are the range of summation (same
as above) and error estimations: For X 1−²1 ≤ V ≤ X,
V 1+(k/2+1)² (log V )nk +k ¿ V V (k/2+1)²+²1 ¿ H k/2+1 V 1−²1
(4.34)
as 2X ²+²1 ≤ H ≤ 2X 1−²−²1 .
Z
I3
Z
( V2 )−1+²
=
Z
V
k
V −1+²
Z
f (x, δx) xdxdδ ≤
0
V −1+²/2
Z
V
+
V −1+²/2
V /2
Z
V
V /2
0
V /2
¿k V ² (V −1+² )k/2 V k/2+1 (log V )nk + V −1+²/2 V 2 (V ²/2 )k (log V )k ¿k H k/2+1 V 1−²1 .
Z
Z
H
Z
0
X/2M ¡
¢k
ψ(x + h) − ψ(x) − h dxdh
1
H
Z
Z
X 1−²1
X ²/2
Z
X 1−²1
+
≤
X ²/2
¿k H
0
1
k/2+1
X
1−²1
1
nk
(log X)
+ X ²/2 X 1−²1 X k²/2 (log X)k
¿k H k/2+1 X 1−.99²1 + X 1+(k+1)²/2−²1 (log X)k ¿k H k/2+1 X 1−.99²1
by (4.32), (4.31) and (4.34). Then, one restricts attention to X ²+2²1 ≤ H ≤ X 1−²−²1
and set η = X −2²1 /3 . The theorem follows the same way as Theorem IV.5.
Meanwhile, using a C program (see Appendix), we get some numerical evidence
53
in support of the truth of the asymptotic formula (4.16):
Z
X¡
1
³
Dk
1 ´k/2
k/2+1 k/2
ψ(x + δx) − ψ(x) − δx dx = k
X
δ
log
Eδ
+1
2
¢k
+ Ok (δ k/2 X k/2+1−²1 )
For X = 108 and δ = 0.0001:
k Actual value for k-th moment Result from formula
2 4.0075 ∗ 1012
3.8976 ∗ 1012
4 6.5161 ∗ 1017
6.0766 ∗ 1017
6 1.9592 ∗ 1023
1.7763 ∗ 1023
For X = 1010 and δ = 0.00001:
k Actual value for k-th moment Result from formula
2 5.0527 ∗ 1015
5.0485 ∗ 1015
4 1.0210 ∗ 1022
1.0195 ∗ 1022
6 3.8645 ∗ 1028
3.8602 ∗ 1028
CHAPTER V
“Equivalence” of Odd Moments
5.1
Introduction
In Chapter III, we mentioned the result of Montgomery and Soundararajan [15]
concerning odd moments under Riemann Hypothesis and Hypothesis HL: For any
positive odd integer k,
Z
X¡
(5.1)
¢k
ψ(x + h) − ψ(x) − h dx ¿k Xhk/2−1/7k+²
1
when X ² ≤ h ≤ X 1/k−² . One expects (5.1) to be true even for X ² ≤ h ≤ X 1−² . Then
the question arises if one can prove a similar thing for
Z
X¡
¢k
ψ(x + δx) − ψ(x) − δx dx
1
when k is odd, even in the range X −1+² ≤ δ ≤ X −1+1/k−² . Or can one prove an
“equivalence” statement between the two? In this Chapter, we should try to establish
such results assuming that we have good knowledge about the even moments already.
Apart from Saffari and Vaughan’s method, our key ingredient is the method of Ghosh
[4] & [5]. We would get some new results in the process as well.
54
55
5.2
Results on λ-th moments
As mentioned in the introduction, we assume the following weaker asymptotic
formulas for even moments: For any ² > 0 and any positive even integer k,
Z X
¡
¢k
X k/2
Γ(k + 1)
k/2
ψ(x + h) − ψ(x) − h dx = k
Xh
(log
)
h
Γ( 2 + 1)2k/2
1
(5.2)
¡
¢
+ Ok Xhk/2 (log X)k/2−1
for X ² ≤ h ≤ X 1−² . First, note that this is the first main term of the formula in
Theorem III.7. Second, if one traces the argument in [17] and [15] for the implicit con¡
¢
¡ k
¢
stant in the error term, one has Ok Xhk/2 (log X)k/2−1 = O AB Xhk/2 (log X)k/2−1
where A, B are some absolute constants greater than 1. (Of course, the implicit
constant depends on ² but we treat ² as fixed at the beginning). This is needed for
the argument of Ghosh and leads to a poor error term so that the lower order terms
of Theorem III.7 will be absorbed into the error term as the argument proceeds.
Lemma V.1.
√
1
πΓ(2z) = 22z−1 Γ(z)Γ(z + ).
2
Proof: This is Legendre’s double formula for gamma function.
Theorem V.2. Assume (5.2) for every even positive integer k. For any ² > 0 and
any λ > 0,
Z X
|ψ(x+h)−ψ(x)−h|λ dx =
1
³ Xhλ/2 (log X )λ/2 ´
Γ(λ + 1)
X λ/2
λ/2
h
Xh
(log
)
+O
λ
h
log4 X
Γ( λ2 + 1)2λ/2
for X ² ≤ h ≤ X 1−² . Here logn x stands for n times iterated logarithm (log1 x = log x
and logi+1 x = log logi x).
Proof: Note that this includes (5.2) with poorer error term. The method of proof
is essentially that in Ghosh [4] & [5]. There are two cases depending on the size of
λ:
56
(i) 0 < λ ≤ 1.
Let F (x) =
ψ(x+h)−ψ(x)−h
2−1/2 h1/2 (log X/h)1/2
1
Gλ
1
=
Gλ
|F (x)|λ =
R∞
u)2
and Gλ = 0 (sin
du. For any µ ≥ 1,
1+λ
u
Z ∞
(sin |F (x)|u)2
du
u1+λ
0
Z µ
³ 1
´
(sin |F (x)|u)2
−λ
du
+
O
µ
.
u1+λ
λGλ
0
Hence,
Z
Z µZ X
³X ´
1
(sin |F (x)|u)2
dxdu + Oλ λ .
(5.3)
|F (x)| dx =
Gλ 0 1
u1+λ
µ
1
P
2N +2
(−1)j+1 (2x)2j
Note that (sin x)2 = 12 N
+ O( (2x)
). So the main term of (5.3) is
j=1
(2j)!
(2N +2)!
(5.4)
X
λ
Z µ
Z
N
´
1
1 X (−1)j+1 (2u)2j ³ X
2j
|F
(x)|
dx
du
2Gλ 0 u1+λ j=1
(2j)!
1
Z µ
Z X
³
´
4N
2N
2N +2
+O
u du
|F (x)|
dx .
(2N + 2)! 0
1
Using assumption (5.2), the error term in (5.4) is bounded by
N
(5.5)
(2µ)2N
AB (2µ)2N Xµ
Xµ +
.
(N + 2)!
(2N + 3)! log X
Using (5.2), the main term of (5.4) contributes
(5.6)
Z µ
j Z µ
N
N
´
³ X X
1 X (−1)j+1 Γ(2j + 1)
AB
X
2j−2
2j
u
du .
(2u) du + O
2Gλ 0 u1+λ j=1 (2j)! Γ(j + 1)
log X j=1 (2j)! 0
The above error term is bounded by
(5.7)
N
X
X
µ2j−1
X
N
BN
A
¿
AB eµ .
log X
(2j)!
log X
j=1
Using Lemma V.1 with z = j + 12 , the main term of (5.6) becomes
Z µ
N
1 X (−1)j+1 22j
X
1
√ Γ(j + )(2u)2j du
1+λ
2Gλ 0 u
(2j)!
2
π
j=1
Z µ
N
³Z ∞
´
X
1 X (−1)j+1
j−1/2 −x
2j
x
e dx du
= √
(4u)
2 πGλ 0 u1+λ j=1 (2j)!
0
√
Z ∞
Z µ
¡ (4 xu)2N +2 ¢´
√ 2
X
1 ³
−1/2 −x
= √
x
e
(sin 2 xu) + O
dudx.
1+λ
(2N + 2)!
πGλ 0
0 u
57
The contribution from the above error term is
42N +2 (N + 1)! 2N +2−λ
¿λ
µ
X.
(2N + 2)!
(5.8)
The contribution from the main term is
Z ∞
³Z ∞ (sin 2√xu)2
X
1 ´
−1/2 −x
√
=
x
e
du + O( λ ) dx
u1+λ
λµ
πGλ 0
0
³
´
λ
λ+1
2
X
= √ Γ(
)X + Oλ λ
2
µ
π
³
´
Γ(λ + 1)
X
=
X + Oλ λ
λ
µ
Γ( 2 + 1)
by the definition of Gλ and Lemma V.1. Combining this with (5.3), (5.4), (5.5),
(5.6), (5.7) and (5.8), we have
Z
X
(5.9)
|F (x)|λ dx =
1
Γ(λ + 1)
X + error,
Γ( λ2 + 1)
where
N
X
(2µ)2N +1
AB (2µ)2N +2 X
X
N
X+
+ AB eµ
.
error ¿λ λ +
µ
(N + 1)!
(2N + 3)! log X
log X
p
N
log3 X
λ
Now, we choose N = log
and
µ
=
log4 X, then one can check that AB ¿
4X
√
1
log X, eµ ¿λ (log X) 4 , and using Stirling’s formula,
(2µ)2N +2
(N + 1)!
¿
¿λ
e(2N +2) log 2µ
1
√
¿ √ e(2N +2) log 2µ−N log (N/e)
N
N (N/e)
N
1
1
√ ¿
.
log3 X
N
Consequently, (5.9) becomes
Z
X
1
|F (x)|λ dx =
³ X ´
Γ(λ + 1)
X
+
O
λ
log4 X
Γ( λ2 + 1)
which gives the theorem for 0 < λ ≤ 1 after multiplying through by ( √12 h1/2 (log Xh )1/2 )λ .
(ii) 1 < λ.
58
Let λ = 2m + 1 + θ where m is a non-negative integer and 0 < θ ≤ 2. Let
R∞
u)4
Dθ = 0 (sin
du. Since
2+θ
u
Z
|F (x)|2m ∞ (sin |F (x)|u)4
λ
|F (x)| =
du
Dθ
u2+θ
0
Z
|F (x)|2m Y (sin |F (x)|u)4
=
du + Oλ (|F (x)|2m Y −1−θ ).
2+θ
Dθ
u
0
Then, similar to the calculation in case (i), for some Y ≥ 1,
(5.10)
Z X
1
1
|F (x)| dx =
Dθ
Z
Y
λ
0
Z
1 ³
u2+θ
X
´
|F (x)|2m (sin |F (x)|u)4 dx du + Oλ (Y −1−θ X)
1
by (5.2). From (7) of Ghosh [5], we have
³ (4u)2N +2 ´
1X
(−4)j+1 j−1
(sin u) =
where bj =
(4 − 1),
bj u2j + O
8 j=2
(2N + 2)!
(2j)!
N
4
and N is an integer, exceeding 2, which will be chosen later. Using this Taylor series
and (5.2), the main term of (5.10) equals
Z Y
N
X
1 X
2j Γ(2(m + j) + 1)
(5.11)
b
u
du
j
8Dθ 0 u2+θ j=2
Γ((m + j) + 1)
Z Y
N
´
³ X X
´ Z X
2N +2 Z Y
Bj
2j−2−θ
2m (4|F (x)|)
u2N −θ dudx .
+O
A |bj |
u
du +
|F (x)|
log X j=2
(2N + 2)!
0
0
1
The error term of (5.11) is
(5.12)
N
m+N +1 ´
AB e4Y X
42N +2 Y 2N +1 ³ (2(m + N + 1))! AB
¿
+
+
X.
Y 1+θ log X
(2N + 2)!
(m + N + 2)!
log X
By Lemma V.1, the main term of (5.11)
Z Y
Z
N
1 X bj 2j 22(m+j) ³ ∞ m+j−1/2 −x ´
X
√
u
=
x
e dx du
Dθ 0 u2+θ j=2 8
π
0
√
Z
Z
√ 4
(8 xu)2N +2 ´
22m X ∞ m−1/2 −x Y 1 ³
x
e
(sin 2 xu) + O(
) dudx.
= √
2+θ
(2N + 2)!
πDθ 0
0 u
Contribution from the error is
(5.13)
¿λ
(N + m + 1)! 4N 2N +1
2 Y
X
(2N + 2)!
59
while the main term
√
Z
Z
´
22m X ∞ m−1/2 −x ³ ∞ (sin 2 xu)4
−1−θ
√
=
du + O(Y
) dx
x
e
u2+θ
πDθ 0
0
Z ∞
22m+1+θ
√
=
X
xm+θ/2 e−x dx + Oλ (XY −1−θ )
π
0
Γ(λ + 1)
X + Oλ (XY −1−θ )
=
Γ( λ2 + 1)
by Lemma V.1. Therefore, combining this with (5.10), (5.11), (5.12) and (5.13), we
have
Z
X
(5.14)
|F (x)|λ dx =
1
Γ(λ + 1)
X + error,
Γ( λ2 + 1)
where
³1
N
AB e4Y
(2(m + N + 1))!
+
+ 42N Y 2N +1
error ¿λ X
Y
Y log X
(2N + 2)!(m + N + 2)!
2N B N
2N +1
2N
4 A
Y
4 (N + m + 1)! 2N +1 ´
+
+
Y
.
(2N + 2)! log X
(2N + 2)!
(5.15)
By Stirling’s formula, (5.15)
³1
AB e4Y
44N
AB
Y 2N +1 ´
2N +1
¿λ X
+
+
Y
+
.
Y
Y log X
(N + 1)N −m
(N + 1)2N +2 log X
Now, pick N =
log3 X
log4 X
N
N
and Y =
p
N
(N + 1)N −m Àλ log log X, e4Y ¿ log log X and Y 2N +1
X
Oλ ( √
3
log3 X
√
log X, 44N ¿ 4 log log X,
√
¿ log log X. Thus, (5.15) is
log3 X, we have AB ¿
√
) and we get the theorem after multiplying (5.14) by ( √12 h1/2 (log Xh )1/2 )λ .
However, one should expect more to be true, namely,
Conjecture V.3. For every ² > 0 and λ > 0,
Z
X
1
Γ(λ + 1)
|ψ(x + h) − ψ(x) − h| dx = λ
hλ/2+1
Γ( 2 + 1)2λ/2
λ
for X ² ≤ h ≤ X 1−² . E = 2πeC0 −1 .
Z
X/h ¡
log
E
x ¢λ/2
dx + o(Xhλ/2 )
E
60
Similarly, using almost the same calculations, if one assumes that, for every ² > 0
and any positive even integer k,
Z X
³
k!
1 ´k/2
k
k/2+1 k/2
(ψ(x + δx) − ψ(x) − δx) dx = k
δ
log
k X
δ
2
(
+
1)!2
1
2
(5.16)
´
³ k
1
+ O AB X k/2+1 δ k/2 (log )k/2−1
δ
for X −1+² ≤ δ ≤ X −² and A, B are some absolute constants greater than 1. Then
Theorem V.4. Assuming (5.16) for every positive even integer k. For every ² > 0
and λ > 0,
Z
X
1
1 ´λ/2
|ψ(x + δx) − ψ(x) − δx| dx =
δ
log
λ X
δ
Γ( λ2 + 2)2 2
³ X λ/2+1 δ λ/2 (log 1 )λ/2 ´
δ
+ Oλ
log4 X
λ
Γ(λ + 1)
λ/2+1 λ/2
³
for X −1+² ≤ δ ≤ X −² .
Again, one can conjecture the following:
Conjecture V.5. For every ² > 0 and λ > 0,
Z
X
|ψ(x+δx)−ψ(x)−δx|λ dx =
1
³
Γ(λ + 1)
1 ´λ/2
λ/2+1 λ/2
X
δ
log
+o(X λ/2+1 δ λ/2 )
Eδ
Γ( λ2 + 2)2λ/2
for X −1+² ≤ δ ≤ X −² . E = 2πeC0 −1 .
5.3
“Equivalence” for odd moments
In this section, we assume that, for every ² > 0 and every positive even integer
k, (5.2) is true for X ² ≤ h ≤ X 1−² and (5.16) is true for X −1+² ≤ δ ≤ X −² . By
Cauchy’s inequality, we have, for every positive integer n,
Z
1
X
Z
1
X
|ψ(x + h) − ψ(x) − h|n dx ¿n Xhn/2 (log X)n/2 for X ² ≤ h ≤ X 1−² ,
|ψ(x + δx) − ψ(x) − δx|n dx ¿n X n/2+1 δ n/2 (log X)n/2 for X −1+² ≤ δ ≤ X −² .
61
These inequalities will be needed in the proof of the “equivalence” statements just
like what happens in Chapter II and IV. Now, suppose for any positive odd integer
n,
Assumption V.6. For every 0 < ² < 1/2,
Z X
(ψ(x + h) − ψ(x) − h)n dx = o(Xhn/2 (log X)n/2 )
1
when X ² ≤ h ≤ X 1−² ,
then as
Z
Z
X
X
n
|ψ(x + h) − ψ(x) − h| dx +
1
(ψ(x + h) − ψ(x) − h)n dx
1
Z
X
= 2
max{ψ(x + h) − ψ(x) − h, 0}n dx,
1
we have
(5.17)
Z
X
max{ψ(x + h) − ψ(x) − h, 0}n dx =
1
1 Γ(n + 1)
X n/2
n/2
Xh
(log
)
n
2 Γ( 2 + 1)2n/2
h
+ o(Xhn/2 (log X)n/2 )
by (5.2) and Theorem V.2. Now, for any ²1 > 0, one can imitate Saffari and
Vaughan’s argument as in Theorem IV.6 and get
Z ∆Z X
max{ψ(x + δx) − ψ(x) − δx, 0}n dxdδ
0
1
Z ∆X
³
Γ(n + 1)
X ´n/2
1 n/2
n/2
n/2+1 n/2+1
=
h
log
dh
+
o(∆
X
(log
) )
n
2Γ( 2 + 2)2n/2 0
h
∆
for X −1+²+²1 ≤ ∆ ≤ X −²−²1 . Let f (x, h) = max{ψ(x + h) − ψ(x) − h, 0}, g(x, δx) =
f (x, ∆x) for ∆ ≤ δ ≤ (1 + η)∆. Following the argument in [9], one has, for
X −1+²+2²1 ≤ ∆ ≤ X −²−2²1 ,
(5.18)
Z X
1 n/2
1 Γ(n + 1)
n/2+1 n/2
X
∆
(log
)
max{ψ(x + ∆x) − ψ(x) − ∆x, 0}n dx =
n
2 Γ( 2 + 2)2n/2
∆
1
³
´
1
+ o X n/2+1 ∆n/2 (log )n/2
∆
62
by letting η approach 0 sufficiently slowly. The only difference in the argument is
the use of
|f (x, δx) − g(x, δx)| ≤ |(ψ(x + δx) − ψ(x) − δx) − (ψ(x + ∆x) − ψ(x) − ∆x)|
¡
¢
= |ψ (x + ∆x) + (δ − ∆)x − (δ − ∆)x|
for ∆ ≤ δ ≤ (1 + η)∆ when one estimates the integral
RR
|f − g|n . The above is
justified by
|max{a, 0} − max{b, 0}| ≤ |a − b|
which can be easily verified by considering different signs for a and b. Similarly, one
has
(5.19)
Z X
1 Γ(n + 1)
1
min{ψ(x + ∆x) − ψ(x) − ∆x, 0}n dx = −
X n/2+1 ∆n/2 (log )n/2
n
n/2
2 Γ( 2 + 2)2
∆
1
³
´
1
+ o X n/2+1 ∆n/2 (log )n/2 .
∆
Consequently, adding (5.18) and (5.19), we prove the following:
Theorem V.7. Assume (5.2) and (5.16) in the ranges X ² ≤ h ≤ X 1−² and X −1+² ≤
δ ≤ X −² respectively for any ² > 0 and any even integer k > 0. Then, for any positive
odd integer n and ²1 > 0,
Z
X
(ψ(x + h) − ψ(x) − h)n dx = o(Xhn/2 (log X)n/2 )
1
for X ² ≤ h ≤ X 1−² implies that
Z
X
1
´
³
1
(ψ(x + δx) − ψ(x) − δx)n dx = o X n/2+1 δ n/2 (log )n/2
δ
for X −1+²+2²1 ≤ δ ≤ X −²−2²1 .
Conversely, one also has:
63
Theorem V.8. Assume (5.2) and (5.16) in the ranges X ² ≤ h ≤ X 1−² and X −1+² ≤
δ ≤ X −² respectively for any ² > 0 and any even integer k > 0. Then, for any positive
odd integer n and ²1 > 0,
Z
X
1
³
1 n/2 ´
n/2+1 n/2
(ψ(x + δx) − ψ(x) − δx) dx = o X
δ (log )
δ
n
for X −1+² ≤ δ ≤ X −² implies that
Z
X
(ψ(x + h) − ψ(x) − h)n dx = o(Xhn/2 (log X)n/2 )
1
for X ²+2²1 ≤ h ≤ X 1−²−2²1 .
5.4
Numerical evidence
Again, we use the C program at Appendix to support the truth of our results.
Firstly, regarding
Z
X
|ψ(x+δx)−ψ(x)−δx|λ dx =
1
³
Γ(λ + 1)
1 ´λ/2
λ/2+1 λ/2
X
δ
log
+o(X λ/2+1 δ λ/2 ).
λ
λ/2
Eδ
Γ( 2 + 2)2
For X = 108 and δ = 0.0001:
λ
Actual value for λ-th moment Result from formula
1.0 1.5009 ∗ 1010
1.4851 ∗ 1010
2.1 7.1441 ∗ 1012
6.9344 ∗ 1012
3.2 4.8737 ∗ 1015
4.6213 ∗ 1015
4.3 4.1913 ∗ 1018
3.8864 ∗ 1018
5.4 4.2519 ∗ 1021
3.8768 ∗ 1021
6.5 4.8884 ∗ 1024
4.4213 ∗ 1024
For X = 1010 and δ = 0.00001:
64
λ
Actual value for λ-th moment Result from formula
1.0 5.3464 ∗ 1012
5.3452 ∗ 1012
2.1 1.0218 ∗ 1016
1.0210 ∗ 1016
3.2 2.7871 ∗ 1019
2.7835 ∗ 1019
4.3 9.5892 ∗ 1022
9.5764 ∗ 1022
5.4 3.9120 ∗ 1026
3.9079 ∗ 1026
6.5 1.8248 ∗ 1030
1.8232 ∗ 1030
Secondly, for n odd,
Z
X¡
1
³
´
¢n
1
ψ(x + δx) − ψ(x) − δx dx = o X n/2+1 δ n/2 (log )n/2 .
δ
For X = 108 and δ = 0.0001:
n Actual value for n-th moment
Γ(n+1)
X n/2+1 δ n/2 (log 1δ )n/2
Γ( n
+2)2n/2
2
1 −4.9574 ∗ 107
1.6143 ∗ 1010
3 −2.0632 ∗ 1013
1.7842 ∗ 1015
5 −3.3174 ∗ 1018
4.6952 ∗ 1020
For X = 1010 and δ = 0.00001:
n Actual value for n-th moment
Γ(n+1)
X n/2+1 δ n/2 (log 1δ )n/2
Γ( n
+2)2n/2
2
1 7.2371 ∗ 108
5.7074 ∗ 1012
3 −1.3468 ∗ 1016
7.8851 ∗ 1018
5 −2.5587 ∗ 1023
2.5937 ∗ 1025
Note:
Γ(n+1)
Γ(n/2+2)2n/2
λ-th moment.
acts as a normalization constant coming from the main term of
CHAPTER VI
More precise Pair Correlation Conjectures
6.1
Introduction
In Chapter II, we found a more precise asymptotic formula for the Strong Pair
Correlation Conjecture to be F (x, T ) =
T
2π
T
log 2π
−
T
2π
+ o(T ) when T ≤ x. It would
be interesting to do a similar thing when 1 ≤ x ≤ T . From Montgomery [13], we
have
F (x, T ) =
³ T log T ´
T
T (log T )2
log x +
+
O(x
log
x)
+
O(T
)
+
O
2π
2πx2
x2
for 1 ≤ x ≤ T 1−² where ² is any positive real number . Our goal is to determine
the second main terms as well. Note that this will provide insight on how F (x, T )
behaves from x ≤ T to x ≥ T . Essentially, we try to improve the estimates in
Montgomery [13] and use its method to arrive at a more precise asymptotic formula
when 1 ≤ x ≤ T 1−² . When T 1−² ≤ x, we use the method of Goldston and Gonek
[7] for mean values of Dirichlet polynomials and tails of Dirichlet series assuming
Hardy-Littlewood Conjecture on twin primes (see Chapter III).
6.2
Some preparations
Lemma VI.1. For 1 < σ < 2,
ζ0
1
ζ0
(1 − σ + it) = − (σ − it) − log τ + log 2π + Oσ ( )
ζ
ζ
τ
65
66
where τ = |t| + 2 and the implicit constant in the error term may depend on σ.
Proof: Starting from the functional equation of ζ(s) (see Davenport [1]):
³s´
³ (1 − s) ´
π −s/2 Γ
ζ(s) = π −(1−s)/2 Γ
ζ(1 − s)
2
2
1
s
1
1−s
− s log π + log Γ( ) + log ζ(s) = − (1 − s) log π + log Γ(
) + log ζ(1 − s)
2
2
2
2
by taking logarithm on both sides. Then, differentiating both sides, we have
1
1 Γ0 s
ζ0
1
1 Γ0 1 − s
ζ0
− log π +
( ) + (s) = log π −
(
) − (1 − s).
2
2Γ 2
ζ
2
2Γ
2
ζ
Thus,
ζ0
ζ0
(1 − s) = − (s) + log π −
ζ
ζ
ζ0
= − (s) + log π −
ζ
1 Γ0 s
1 Γ0 1 − s
( )−
(
)
2Γ 2
2Γ
2
³1
1
s
1
1−s
1 ´
log ( ) − log (
)+O
+
2
2
2
2
|s| |1 − s|
Γ0
(s)
Γ
= log s + O(|s|−1 ) for −π + δ < args < π − δ,
³1
ζ0
1
1
1 ´
= − (s) + log 2π − log s − log (1 − s) + O
+
.
ζ
2
2
|s| |1 − s|
because
With s = σ − it and τ = |t| + 2, one has
ζ0
ζ0
1
(1 − σ + it) = − (σ − it) − log τ + log 2π + Oσ ( ).
ζ
ζ
τ
Lemma VI.2. Assume Riemann Hypothesis. If 1 < σ < 2 and x ≥ 1, then
X
xiγ
(σ − 12 )2 + (t − γ)2
γ
³X
¡ x ¢σ+it ´
¡ x ¢1−σ+it X
+
Λ(n)
= −x−1/2
Λ(n)
n
n
n>x
n≤x
³
ζ0
1 ´
+x1/2−σ+it log τ + (σ − it) − log 2π + O( )
ζ
τ
1/2
1/2
x
x
+
+
+ O(x−2 τ −1 )
σ − 1 + it σ − it
(2σ − 1)
where τ = |t| + 2, and the implicit constants depend on σ only.
67
Proof: Basically, we make use of (22) in Montgomery [13]
X
xiγ
(σ − 12 )2 + (t − γ)2
γ
³X
¡ x ¢1−σ+it X
¡ x ¢σ+it ´
= −x−1/2
Λ(n)
+
Λ(n)
n
n
n>x
n≤x
(2σ − 1)
ζ0
x1/2 (2σ − 1)
− (1 − σ + it)x1/2−σ+it +
ζ
(σ − 1 + it)(σ − it)
∞
X
(2σ − 1)x−2n
−1/2
−x
(σ − 1 − it − 2n)(σ + it + 2n)
n=1
which holds for all x ≥ 1 because of continuity on both sides. Now for 1 < σ < 2,
applying Lemma VI.1, using partial fractions, and bounding the last sum, we get the
lemma.
Lemma VI.3.
³ x2 ´
1 2
1 2
Λ(n) n = x log x − x + OB
2
4
(log x)B
n≤x
X
2
for any B > 0 and the implicit constant may depend on B.
Proof: Let θ(x) =
P
p≤x
log p. Then, by integration by parts and prime number
theorem, for any A > 1, the left hand side
Z
x
=
u log udθ(u) + O
1
h
ix
Z
³ X
X
(log n)2 nk
´
2≤k≤log x n≤x1/k
x
¢
¡ 3
θ(u)u log u −
θ(u)(log u + 1)du + O x 2 (log x)3
1
1
Z x
³
´
x2
= x2 log x −
u log u + udu + OA
(log x)A−1
1
³
´
1
1
1
x2
= x2 log x − x2 log x + x2 − x2 + OA
2
4
2
(log x)A−1
=
which gives the lemma with B = A − 1.
Lemma VI.4.
X Λ(n)2
n>x
n3
=
³
´
1 log x 1 1
1
+
+
O
B
2 x2
4 x2
x2 (log x)B
68
for any B > 0 and the implicit constant may depend on B.
Proof: Similar to Lemma VI.3, one has for any A > 1, the left hand side
Z ∞
³X X (log n)2 ´
log u
=
dθ(u)
+
O
u3
n3k
x
2≤k n>x1/k
Z ∞
h
³ (log x)2 ´
log u i∞
1 − 3 log u
= θ(u) 3
−
θ(u)
du
+
O
u x
u4
x5/2
x
Z ∞
³
´
1 − 3 log u
1
log x
du
+
O
= − 2 −
A
x
u3
x2 (log x)A−1
x
³
´
log x
1
3 log x
3
1
= − 2 − 2+
+ 2 + OA 2
x
2x
2 x2
4x
x (log x)A−1
which gives the lemma with B = A − 1.
Lemma VI.5.
Z
T
0
∞
¯2
¯ ζ0 3
X
Λ(n)2
¯
¯
+ O(1).
¯ ( − it)¯ dt = T
ζ 2
n3
n=1
Proof: This follows directly from the following quantitative form (see [16] of Parseval’s identity for Dirichlet series:
Z
T
(6.1)
0
with an =
6.3
¯X
¯2
X
¡
¢
¯
¯
an n−it ¯ dt =
|an |2 T + O(n)
¯
n
n
Λ(n)
.
n3/2
A formula for F (x, T )
First, let us recall the definition of F (x, T ):
F (x, T ) =
X
0
xi(γ−γ ) w(γ − γ 0 )
0<γ,γ 0 ≤T
where w(u) =
4
.
4+u2
Theorem VI.6. Assume Riemann Hypothesis. For 1 ≤ x,
Z T
1
F (x, T ) =
|P (x, t)|2 dt
2π 0
∞
¡X
¢i
¡
¢
T h
T 2
T
Λ(n)2
3
+
(log
)
−
2
log
+
+
2
+
O
(log
T
)
2πx2
2π
2π
n3
n=1
69
where P (x, t) =
1
x
P
Λ(n)
n≤x n−1/2+it
+x
P
Λ(n)
n>x n3/2+it
−
x1/2−it
1/2+it
−
x1/2−it
.
3/2−it
Proof: Our method is essentially that of Montgomery [13]. Use Lemma VI.2 with
σ = 32 , we have
2
X
γ
X Λ(n)
xit X Λ(n)
xiγ
1+it
=
−
−
x
1 + (t − γ)2
x n≤x n−1/2+it
n3/2+it
n>x
x1/2
x1/2
+
+ x−1+it log τ
1
3
+
it
−
it
2
2
³ ζ0 3
1 ´
+ x−1+it
( − it) − log 2π + O( ) + O(x−2 τ −1 )
ζ 2
τ
£
¤
=xit −P (x, t) + Q(x, t) + R(x, t) + S(x, t)
+
(6.2)
where
X Λ(n)
1 X Λ(n)
x1/2−it x1/2−it
P (x, t) =
+x
− 1
− 3
,
x n≤x n−1/2+it
n3/2+it
+ it
− it
2
2
n>x
Q(x, t) = x−1 log τ,
´
³ ζ0 3
( − it) − log 2π ,
R(x, t) = x−1
ζ 2
S(x, t) = O(x−1 τ −1 ).
Write (6.2) as LHS(x, t) = RHS(x, t). From page 188 of Montgomery [13],
Z
T
(6.3)
¡
¢
|LHS(x, t)|2 dt = 2πF (x, T ) + O (log T )3 .
0
Meanwhile,
Z T
Z
2
|RHS(x, t)| dt =
0
(6.4)
Z
2
Z
2
Z
2
Z
2
|P | + |Q| + |R| + |S| + 2Re QR̄
Z
Z
Z
³ Z
´
+ O Re P Q̄ + Re P R̄ + Re QS̄ + Re RS̄ .
For the following, we make use of the prime number theorem whenever appropriate.
Z
T
0
Z T
1 X
1/2
P Q̄ = 2
Λ(n)n
n−it log (t + 2)dt
x n≤x
0
Z T
Z
³
X Λ(n) T
log τ ´
−it
−1/2
+
n
log
(t
+
2)dt
+
O
x
dt .
3/2
n
τ
0
0
n>x
70
By integration by parts twice, one has
Z
T
(6.5)
−it
n
0
³ 1 ´
¤
log (T + 2) £
log (t + 2)dt =
sin (T log n) + i cos (T log n) + O
.
log n
log n
So,
Z
(6.6)
T
Re
P Q̄ ¿
0
Z
T
0
Z
∞
´³X
1 T ³X
Λ(n) −it ´
1/2 −it
P R̄ = 2
n
dt
Λ(n)n n
x 0 n≤x
n3/2
n=1
Z T ³X
∞
Λ(n) −it ´³X Λ(n) −it ´
+
n
n
dt
n3/2
n3/2
0
n>x
n=1
Z T
X Λ(n) Z T
log 2π X
1/2
−it
− 2
Λ(n)n
n dt − log 2π
n−it dt
3/2
x n≤x
n
0
0
n>x
Z
∞
X
1 X
Λ(n2 ) T
(n1 n2 )−it dt
= 2
Λ(n1 )n1 1/2
3/2
x n ≤x
n2
0
n2 =2
1
Z
∞
³ 1 ´
X Λ(n1 ) X
Λ(n2 ) T
−it
(n
n
)
dt
+
O
+
.
1 2
3/2
3/2
1/2
n
n
x
1
2
0
n >x
n =2
1
Therefore, since
RT
0
2
(n1 n2 )−it dt ¿ 1 as n2 ≥ 2,
Z
T
(6.7)
P R̄ ¿
0
Z
(6.8)
(log T )2
.
x1/2
T
0
1
x1/2
Z T
Z T
X Λ(n) Z T 1
1 X
1
1
1
1/2
P S̄ ¿ 2
Λ(n)n
dt +
dt + 1/2
dt
3/2
2
x n≤x
n
τ
x
τ
0 τ
0
0
n>x
¿
log T
.
x1/2
Also,
Z
Z
T
(6.9)
QS̄ ¿ x
T
−2
0
0
Z
Z
T
(6.10)
(log T )2
log τ
dt ¿
.
τ
x2
RS̄ ¿ x
T
−2
0
0
1
log T
dt ¿
.
τ
x2
Combining (6.6), (6.7), (6.8), (6.9) and (6.10), we have the error term for (6.4)
(6.11)
¿
(log T )2
.
x1/2
71
Now, we try to calculate the remaining terms,
Z T
Z T
Z
1 T +2
2
−2
2
|Q| = x
(log τ ) dt = 2
(log t)2 dt
x 2
0
0
£
1
= 2 (T + 2)(log (T + 2))2 − 2(T + 2) log (T + 2)
x
(6.12)
¤
+ 2(T + 2) − 2(log 2)2 + 4 log 2 − 4 by Lemma III.4,
³ (log T )2 ´
¤
T£
2
= 2 (log T ) − 2 log T + 2 + O
.
x
x2
Z
T
|R|
Z
¯2
1 T ¯¯ ζ 0 3
¯
= 2
¯ ( − it) − log 2π ¯ dt
x 0 ζ 2
Z
¯2
i
³ 1 Z T ζ0 3
´
1 h T ¯¯ ζ 0 3
¯
2
= 2
( − it)dt .
¯ ( − it)¯ dt + (log 2π) T + O 2
x
ζ 2
x 0 ζ 2
0
2
0
So, by Lemma VI.5,
Z
T
(6.13)
2
|R| =
∞
hX
Λ(n)2
0
n3
n=1
Z
T
(6.14)
0
³1´
+ (log 2π)
+O 2
x2
x
2
1
|S| ¿ 2
x
Z
T
2
0
iT
1
1
dt
¿
.
τ2
x2
Finally,
Z
T
QR̄ = x
−2
0
= x−2
Z
∞
hX
Λ(n)
n=1
∞
hX
n=1
n3/2
Λ(n)
n3/2
Z
Z
T
log τ n
−it
0
i
T
dt − log 2π
log τ dt
0
T
log (t + 2)n−it dt
0
i
− log 2π[(T + 2) log (T + 2) − (T + 2)] + O(1) .
Hence, by (6.5),
Z
T
(6.15)
QR̄ = −
0
³ log T ´
log 2π(T log T − T )
+
O
.
x2
x2
72
Since x ≥ 1, combining (6.3), (6.4), (6.11), (6.12), (6.13), (6.14) and (6.15),
Z
T
¤
T£
2
(log
T
)
−
2
log
T
+
2
x2
0
∞
i T
T hX Λ(n)2
2
+ 2
+
(log
2π)
+ 2 [−2 log 2π log T + 2 log 2π]
x n=1 n3
x
¡
¢
+O (log T )3
Z T
∞
i
X
T ¢2
T
Λ(n)2
T h¡
2
=
− 2 log
+
+2
|P (x, t)| dt + 2 log
x
2π
2π n=1 n3
0
¡
¢
+O (log T )3
2πF (x, T ) =
|P (x, t)|2 dt +
which gives the theorem after dividing through by 2π.
6.4
Improvement on asymptotic formula when 1 ≤ x ≤
T
log T
Theorem VI.7. Assume Riemann Hypothesis. For any B > 0,
1
1hT
T
T
T i
T log x + 2
(log )2 − 2 log
2π
x 2π
2π
2π
2π
³ T
´
+ O(x log x) + OB
(log x)B
F (x, T ) =
for 1 ≤ x ≤
T
.
log T
Proof: In view of Theorem VI.6, it suffices to show
Z
T
|P (x, t)|2 dt = T log x + O(x log x) + OB
0
³
´
T
.
(log x)B
First, observe that
P (x, t) = x−1/2−it
³X
¡ x ¢−1/2+it X
¡ x ¢3/2+it ´
Λ(n)
+
Λ(n)
+ O(x1/2 τ −2 ).
n
n
n>x
n≤x
73
Thus,
Z
T
0
Z
¯X
¡ x ¢−1/2+it X
¡ x ¢3/2+it ¯¯2
¯
Λ(n)
+
Λ(n)
¯
¯ dt
n
n
0
n>x
n≤x
³Z T X
X
¡ x ¢−1/2 1
¡ x ¢3/2 1 ´
+O
Λ(n)
+
Λ(n)
dt
n
τ 2 n>x
n
τ2
0 n≤x
³Z T
´
+O
xτ −4 dt
1
|P (x, t)| dt =
x
2
T
0
¡ x ¢−1
1X
(T + O(n))
=
Λ(n)2
x n≤x
n
¡ x ¢3
1X
+
Λ(n)2
(T + O(n)) + O(x)
x n>x
n
³ T
´
+ O(x log x)
= T log x + OB
(log x)B
by (6.1), Lemma VI.3, Lemma VI.4 and prime number theorem.
Remark: In Montgomery [13], one has the main term
T
2π
log x +
1 T
(log T )2
x2 2π
and
T
error term O(T ) + O(x log x) + O( T log
). So, we do get a more precise asymptotic
x2
formula for F (x, T ) when 1 ≤ x ≤
T
.
log T
Also, if one uses the form of prime number
theorem under Riemann Hypothesis, the last error term of Theorem VI.7 becomes
T
).
O( x1/2−²
6.5
Preparation for the case
T
(log T )M
≤x
Fix a small positive real number ² and let K be a large integer (e.g. K = [10/²]
should work). We are going to define a smooth weight ΨU (t) almost that in [7]. We
require ΨU (t) to have support in [−1/U, 1 + 1/U ], 0 ≤ ΨU (t) ≤ 1, ΨU (t) = 1 for
(j)
1/U ≤ t ≤ 1 − 1/U , and ΨU (t) ¿ U j for j = 1, 2, ..., K. Here U = (log T )M for
some M > 1.
Let ∆ = 1/(2K U ). We define a sequence of functions as follow (which is Vino-
74
gradov’s construction) :
χ0 (t) =


 1, if 0 ≤ t ≤ 1,

 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 VI.8. χi (t) exist and are continuous, and χi (t) ≤ ∆−j for 0 ≤ j ≤ i − 1
and 2 ≤ i ≤ K + 1.
Proof: Induction on i. First note that χ1 (t) is continuous because
Z ∆
¯
¯ 1 Z ∆
1
¯
¯
χ0 (t + δ + x)dx −
χ0 (t + x)dx¯
|χ1 (t + δ) − χ1 (t)| = ¯
2∆ −∆
2∆ −∆
Z ∆
¯ 1 Z ∆+δ
¯
1
¯
¯
= ¯
χ0 (t + x)dx −
χ0 (t + x)dx¯
2∆ −∆+δ
2∆ −∆
Z −∆+δ
¯ 1 Z ∆+δ
¯
1
¯
¯
= ¯
χ0 (t + x)dx −
χ0 (t + x)dx¯
2∆ ∆
2∆ −∆
δ
≤
.
∆
Similarly,
Z ∆+h
Z −∆+h
i
χ2 (t + h) − χ2 (t)
1
1h 1
χ1 (t + x)dx −
χ1 (t + x)dx
=
h
h 2∆ ∆
2∆ −∆
1
=
[χ1 (t + ∆ + ξ1 ) − χ1 (t − ∆ + ξ2 )]
2∆
for some 0 ≤ ξ1 , ξ2 ≤ h by mean-value theorem. So χ02 (t) exists and equals to
1
[χ (t
2∆ 1
+ ∆) − χ1 (t − ∆)] which is continuous and ≤
(j)
1
.
∆
(j)
Assume that χi (t) are
continuous and satisfy χi ¿ ∆−j for some 2 ≤ i ≤ K and all 0 ≤ j ≤ i − 1. Now,
R ∆ (j)
(j)
1
for 0 ≤ j ≤ i − 1, χi+1 (t) = 2∆
χ (t + x)dx ≤ ∆−j by induction hypothesis. For
−∆ i
75
j = i,
(i−1)
(i)
χi+1 (t)
(i−1)
χ
(t + h) − χi+1 (t)
= lim i+1
h→0
h
Z ∆+h
Z −∆+h
i
1h 1
1
(i−1)
(i−1)
= lim
χi (t + x)dx −
χi (t + x)dx
h→0 h 2∆ ∆
2∆ −∆
1 (i−1)
(i−1)
[χ
(t + ∆) − χi (t − ∆)]
=
2∆ i
which is continuous and ≤ ∆−i by induction hypothesis.
2π∆y
Lemma VI.9. χ̂0 (y) = eπiy sinπyπy and χ̂i+1 (y) = χ̂i (y) sin2π∆y
for 0 ≤ i ≤ K. Here
R∞
fˆ(y) denotes the inverse Fourier transform of f (t), fˆ(y) = −∞ f (t)e(yt)dt.
Note: We use fˆ to stand for inverse Fourier transform in this and the next sections.
We do this so that the notation matches with [7] and [8].
R1
2πiy
Proof: χ̂0 (y) = 0 e(yt)dt = e 2πiy−1 = eπiy sinπyπy .
Z
∞
χ̂i+1 (y) =
χi+1 (t)e(yt)dt
Z ∆Z ∞
1
=
χi (t + x)e(yt)dtdx
2∆ −∆ −∞
Z ∆
1
χ̂i (y)e(−yx)dx
=
2∆ −∆
χ̂i (y) e(−y∆) − e(y∆)
sin 2π∆y
=
= χ̂i (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 VI.8. From Lemma VI.9, we know that Ψ̂U (y) =
2π∆y K+1
eπiy sinπyπy ( sin2π∆y
)
. It follows that
sin 2πy ³ sin 2π∆y ´K+1
,
ReΨ̂U (y) =
2πy
2π∆y
(6.16)
Ψ̂U (y) ¿ y −K for y À T ² ,
and Ψ̂U (T y) ¿ T −K² for y À τ −1 where τ = T 1−² .
76
These are similar to (18) and (19) in [7]. Also, by Lemma VI.8, it follows from the
discussion in [7] that
³
U K´
Ψ̂U (y), Ψ̂0U (y) ¿ min 1, (
)
2πy
which is (17) in [7]. Consequently, the results in [7] 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 ² ) and Theorem
3 as long as τ = T 1−² ≤ x.
6.6
Asymptotic formula of F (x, T ) when
T
(log T )M
Throughout this section, we assume τ = T 1−² ≤
≤x
T
(log T )M
≤ x, U = (log T )M
for M > 3, H ∗ = τ −2 x2/(1−²) , and ΨU (t) is defined as in the previous section. Let
s = σ + it and
A(s) :=
X Λ(n)
n≤x
ns
and A∗ (s) :=
X Λ(n)
n>x
ns
.
Then, from Theorem VI.6, with slight modifications, one has
F (x, T ) =
1
2π
Z
T
0
Z x
Z ∞
¯1³
´
³
´¯2
1
¯
¯
1/2−it
∗ 3
u
du + x A ( + it) −
u−3/2−it du ¯ dt
¯ A(− + it) −
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
(6.17)
where
F (x, T ) =
³ T (log T )2 ´
³ x1+6² ´
1
x2
I
(x,
T
)
+
I
(x,
T
)
+
O
+
O
1
2
2πx2
2π
U
T
Z x
¯2
³ t ´¯
1
¯
¯
u1/2−it du¯ dt,
I1 (x, T ) =
ΨU
¯A(− + it) −
T
2
1
−∞
Z
∞
77
and
Z
Z ∞
¯2
³ t ´¯
¯ ∗ 3
¯
−3/2−it
I2 (x, T ) =
ΨU
u
du¯ dt.
¯A ( + it) −
T
2
−∞
x
∞
This is essentially Lemma 1 of [8] with modification that V = − UT and T − UT , and
W =
2T
.
U
Riemann Hypothesis is assumed here so that the contribution from the
cross term is estimated via Theorem 3 of [7].
Now, we assume a quantitative form of the Hardy-Littlewood conjecture for twin
primes (see Chapter III Hypothesis HL with r = 2). By Corollary 1 of [7] (see also
the calculations at the end of [7] and [8]) and Lemma VI.3, one has,
I1 (x, T ) = Ψ̂U (0)T
X
Λ2 (n)n
n≤x
+4π
−4π
+O
³ T ´3 Z
2π
³T
∞
T /2πx h≤2πxv/T
´3 Z
2π
³ x3+6² ´
³ X
∞
T /2πτ x
³Z
´
dv
S(h)h2 ReΨ̂U (v) 3
v
2πxv/T
0
+ O(x5/2+7² )
´
dv
u2 du ReΨ̂U (v) 3
v
T
1 2
1
=
T x log x − T x2
2
4
Z 2πxv/T
´
³ T ´3 Z ∞ ³ X
dv
2
2
S(h)h −
u du ReΨ̂U (v) 3
+4π
2π
v
T /2πx h≤2πxv/T
0
Z
³ x3+6² ´
4π 3 T /2πx
ReΨ̂U (v)dv + O
− x
+ O(x5/2+7² )
3
T
T /2πτ x
1 2
1
=
T x log x − T x2
2
4
Z 2πxv/T
³ T ´3 Z ∞ ³ X
´
dv
2
+4π
S(h)h −
u2 du ReΨ̂U (v) 3
2π
v
T /2πx h≤2πxv/T
0
Z
³ KT x2 ´
³ x3+6² ´
2 3 T /x sin v
− x
dv + O
+
O
+ O(x5/2+7² )
M
3
v
(log
T
)
T
0
78
because, from (6.16),
Z
Z
T /2πx
ReΨ̂U (v)dv =
T /2πτ x
³T ´
sin 2πv ³ sin 2π∆v ´K+1
dv + O
2πv
2π∆v
τx
Z T /x
³
sin u
T ´
(1 + O(K∆2 u2 ))du + O
u
τx
0
Z T /x
³
³
2 2´
K∆ T
T ´
sin u
du + O
+
O
.
u
x2
τx
0
T /2πx
0
1
2π
1
=
2π
=
Similarly, by Corollary 2 of [7] and Lemma VI.4,
I2 (x, T ) = Ψ̂U (0)T
8π 2
+
T
8π 2
+
T
8π 2
−
T
X Λ2 (n)
n3
x<n
T /2πx ³
Z
0
Z
X S(h) ´
ReΨ̂U (v)vdv
h2
1≤h≤H ∗
T H ∗ /2πx ³
T /2πx
Z
X
2πxv/T <h≤H ∗
T H ∗ /2πx ³Z H ∗
−2
S(h) ´
ReΨ̂U (v)vdv
h2
´
u du ReΨ̂U (v)vdv
0
2πxv/T
+O(T −1 x−1+6² ) + O(x−3/2+6² ) + O(T 1−²/2 x−2 )
=
T log x 1 T
+
2x2
4 x2
Z
Z H∗
∗
X
8π 2 T H /2πx ³
S(h)
du ´
+
−
ReΨ̂U (v)vdv
T 0
h2
u2
2πxv/T
∗
2πxv/T <h≤H
³ x−1+6² ´
+O
+ O(x−3/2+6² ).
T
Therefore, by a change of variable y =
2πxv
T
and putting back to (6.17), we have
Z
³ T y ´ dy
T ∞ ³X
T
y3 ´
log x +
F (x, T ) =
S(h)h2 −
ReΨ̂U
2π
π 1 h≤y
3
2πx y 3
Z ∗
Z H∗
³ Ty ´
T H ³ X S(h)
du ´
+
−
ReΨ̂U
ydy
π 1
h2
u2
2πx
y
y<h≤H ∗
Z
Z H∗
³ Ty ´
T 1 ³ X S(h)
du ´
+
−
ReΨ̂U
ydy
π 0 h≤H ∗ h2
u2
2πx
y
Z T /x
³
´
³ x1+6² ´
sin v
KT
x
dv + O
+O
+ O(x1/2+7² )
−
M
−2
3π 0
v
(log T )
T
79
Z
T
T
T
4x T /x sin v
(6.18)
=
log x + I1 + I2 −
dv+
2π
π
π
3π 0
v
³
´
³ x1+6² ´
x2 ³ X S(h) ´³
T´
KT
1
−
cos
+
O
+
O
+ O(x1/2+7² ),
πT h≤H ∗ h2
x
(log T )M −2
T
where I1 and I2 are the first and second integral respectively. This is because
Z
Z H∗
³ Ty ´
T 1 ³ X S(h)
du ´
ReΨ̂U
ydy
−
π 0 h≤H ∗ h2
u2
2πx
y
Z
Z
³ KT ´
4πx2 ³ X S(h) ´ T /2πx
x T /x sin u
=
ReΨ̂U (v)vdv −
du + O
T h≤H ∗ h2
π 0
u
(log T )M
0
Z
¡
¢
x2 ³ X S(h) ´ T /x
2 2
=
sin
u
1
+
O(K∆
u
)
du
πT h≤H ∗ h2
0
Z
³ KT ´
x T /x sin u
−
du + O
π 0
u
(log T )M
Z
³ KT ´
x2 ³ X S(h) ´³
T ´ x T /x sin u
=
1
−
cos
−
du
+
O
πT h≤H ∗ h2
x
π 0
u
(log T )M
by a similar calculation as before and T ∆ ≤ x.
Before going any further, let us introduce some functions (see [8]) and study them
a bit. Set
X
y α+1
S(h)h −
Sα (y) :=
for α ≥ 0,
α+1
h≤y
α
and
Tα (y) :=
X S(h)
h>y
hα
for α > 1.
From [3],
(6.19)
1
S0 (y) = − log y + O((log y)2/3 ).
2
Suppose S0 (y) = − 21 log y + ²(y). By partial summation,
Z y
³ 1
yα
α ´
α
(6.20)
Sα (y) = −
+ ²(y)y − α
²(u)uα−1 du +
+
,
2α
2α α + 1
1
and
(6.21)
²(y)
1
1
− α −
+α
Tα (y) =
α−1
(α − 1)y
y
2αy α
Z
∞
y
²(u)
du.
uα+1
80
Lemma VI.10.
Z
y
B
y + O(y 1/2+² )
2
²(u)du =
1
where B = −C0 − log 2π as defined in Chapter II with C0 being Euler’s constant.
Proof:
Z
y
²(u)du =
1
Z y ³X
1
=
X
´
1
S(h) − u + log u du
2
h≤u
1
1
1
(y − h)S(h) − y 2 + y log y − y + 1
2
2
2
h≤y
1
= Ay − y + O(y 1/2+² )
2
1
by Lemma III.2 of Chapter III with A = (1 − C0 − log 2π)
2
B
=
y + O(y 1/2+² ).
2
Now, let us introduce
Z
y
f (y) :=
²(u) −
1
B
du.
2
By integration by parts and Lemma VI.10, one has
Z
y
1
B
²(u)udu = y 2 + yf (y) −
4
Z
∞
(6.22)
Z
y
f (u)du −
1
B
B
= y 2 + O(y 3/2+² ),
4
4
and
(6.23)
y
²(u)
B
f (y)
du = 2 − 3 + 3
3
u
4y
y
Z
∞
y
f (u)
B
du = 2 + O(y −5/2+² ).
4
u
4y
Lemma VI.11. For integer n ≥ 1,
Z
∞
1
³ Ty ´
1
x
Re
Ψ̂
dy =
U
n
y
2πx
T
Z
∞
1
³
sin Txy
1´
dy
+
O
K∆
log
.
y n+1
∆
81
Proof: By a change of variable v =
=
=
=
=
=
³ T ´n−1 Z
∞
Ty
x
and (6.16), the left hand side
1 sin v ³ sin ∆v ´K+1
dv
vn v
∆v
x
T /x
³ T ´n−1 Z 1/∆ sin v
³³ T ´n−1 Z ∞ 1
´
2 2
(1 + O(K∆ v ))dv + O
dv
n+1
n+1
x
x
T /x v
1/∆ v
Z 1/∆
³ T ´n−1 Z 1/∆ sin v
³³ T ´n−1
´
³³ T ´n−1 ´
1
2
dv + O
K∆
dv + O
∆n
n+1
n−1
x
x
x
T /x v
T /x v
³ T ´n−1 Z ∞ sin v
³
´
1
dv + O K∆ log
n+1
x
∆
T /x v
Z ∞
³
sin Txy
1´
x
dy
+
O
K∆
log
T 1 y n+1
∆
because T ∆ ≤ x. Note that the error term comes about when n = 2. If n is not 2,
then we can replace the error term by O(K∆).
Lemma VI.12. If F (y) ¿ y −3/2+² for y ≥ 1, then
Z
∞
1
Z ∞
³ Ty ´
sin T y
F (y)ReΨ̂U
dy =
F (y) T x dy + O(K∆).
2πx
y
1
x
Proof: By a change of variables v =
x
=
T
x
=
T
=
=
x
T
Z
Z
x sin v ³ sin ∆v ´K+1
F ( v)
dv
T
v
∆v
T /x
Z ∞
³ x
¢
|F ( Tx v)| ´
x sin v ¡
2 2
F ( v)
1 + O(K∆ v ) dv + O
dv
T
v
T ∆K+1 1/∆ v K+2
T /x
Z 1/∆
³ ¡T ¢
´
x sin v
1/2−² 3/2−²
F ( v)
dv + O K
∆
T
v
x
T /x
Z ∞
Z ∞
³
|F ( Tx v)| ´
x sin v
x
F ( v)
dv + O
dv + O(K∆)
T
v
T 1/∆
v
T /x
1/∆
F (y)
1
and (6.16), the left hand side
∞
x
T
Z ∞
=
Ty
x
sin Tx y
dy + O(K∆).
T
y
x
82
Now, we are ready to proceed. With the notation of Sα (y) and Tα (y),
Z
³ T y ´ dy
2πx y 3
1
Ry
Z ∞h
³ Ty ´
−1 ²(y) 2 1 ²(u)udu i
=
+
−
ReΨ̂U
dy
4y
y
y3
2πx
1
R
Z ∞ y
Z ∞
³ Ty ´
³ Ty ´
f (u)du
f (y)
1
Re
Ψ̂
dy
+
2
Re
Ψ̂
dy
−2
U
U
y2
2πx
y3
2πx
1
1
³ B 11 ´ x Z ∞ sin T y
x
+
+
dy + O(K∆)
4
2
12 T 1
y
R
Z ∞h
y
³ Ty ´
−1 ²(y) 2 1 ²(u)udu i
dy
=
+
−
Re
Ψ̂
U
4y
y
y3
2πx
1
Z ∞
Z ∞ Ry
f (u)du sin Tx y
f (y) sin Tx y
1
−2
dy
+
2
dy
T
T
3
y2
y
y
y
1
1
x
x
³ B 11 ´ x Z ∞ sin T y
x
+
+
dy + O(K∆)
4
2
12 T 1
y
∞
I1 =
S2 (y)ReΨ̂U
by (6.20), (6.22), Lemma VI.11 and Lemma VI.12. As for I2 , note that by (6.19)
and (6.21),
(6.24)
T2 (z) =
³ (log z)2/3 ´
1
+O
,
z
z2
and
X S(h) Z H ∗ du
1
1
−
= T2 (y) − T2 (H ∗ ) − − ∗
2
2
h
u
y H
y
y<h≤H ∗
³ (log H ∗ )2/3 ´
1
.
= T2 (y) − + O
y
(H ∗ )2
Therefore,
Z
³ (log H ∗ )2/3 ´´
³ Ty ´
1
+O
Re
Ψ̂
ydy
U
y
(H ∗ )2
2πx
1
Z H∗ ³
³ Ty ´
1´
=
ReΨ̂U
ydy
T2 (y) −
y
2πx
1
³ (log H ∗ )2/3 x2 Z T H ∗ /2πx
´
+O
|
Ψ̂
(v)|vdv
U
(H ∗ )2 T 2
T /2πx
Z ∞³
³ Ty ´
³1´
1´
=
T2 (y) −
ReΨ̂U
ydy + O ²
y
2πx
T
1
I2 =
H∗ ³
T2 (y) −
83
because of (6.24) and the formula of Ψ̂U (y) in the previous section that the integral
R∞
∗ )2/3
¿ x(logT H
¿ T1² by the definition of H ∗ (similar estimation for the error
H∗
H∗
term). Applying (6.21), (6.23) and Lemma VI.12,
Z
I2
Z ∞
³ Ty ´
³1´
−1 ²(y)
²(u) i
=
−
+ 2y
du
Re
Ψ̂
dy
+
O
U
4y
y
u3
2πx
T²
1
y
Z ∞h
³ Ty ´
−1 ²(y)
Bi
=
−
+
ReΨ̂U
dy
4y
y
2y
2πx
1
Z ∞ Z ∞
Z ∞
³ Ty ´
³ Ty ´
³1´
f (u)
f (y)
Re
Ψ̂
dy
+
6
y
duRe
Ψ̂
dy
+
O
−2
U
U
y2
2πx
u4
2πx
T²
1
y
1
Z ∞h
³ Ty ´
−1 ²(y)
Bi
=
−
+
ReΨ̂U
dy
4y
y
2y
2πx
1
Z ∞
Z ∞ Z ∞
f (u) sin Tx y
f (y) sin Tx y
dy
+
6
y
du T dy + O(K∆).
−2
T
y2
u4
y
y
1
y
1
x
x
∞h
Consequently, with miraculous cancellations, one has
I1 + I2
³ B 11 ´ x Z ∞ sin T y
sin Txy
x
dy +
+
dy
2
4
y
2
12
T
y
1
1
Z ∞ Ry
∞
f (u)du sin Tx y
f (y) sin Tx y
1
dy
+
2
dy
T
T
3
y2
y
y
y
1
1
x
x
Z ∞ Z ∞
f (u) sin Tx y
+6
y
du T dy + O(K∆)
u4
y
1
y
x
x
= −
2T
Z
−4
Z
∞
by Lemma VI.11 again. Putting this back to (6.18), we have
Theorem VI.13. Assume Riemann Hypothesis and Hypothesis HL for twin primes.
For any small positive real number ² and
T
(log T )M
≤ x,
Z
4x T /x sin v
x2 ³ X S(h) ´³
T´
T
log x −
dv +
1
−
cos
F (x, T ) =
2π
3π 0
v
πT h≤H ∗ h2
x
Z ∞
Z
³ B 11 ´ x Z ∞ sin T y
sin Txy
x
4T ∞ f (y) sin Tx y
x
−
dy
dy +
+
dy −
T
2π 1
y2
2
12 π 1
y4
π 1
y2
y
x
Z Ry
Z
Z ∞
2T ∞ 1 f (u)du sin Tx y
6T ∞
f (u) sin Tx y
+
dy +
y
du T dy
T
4
π 1
y3
π
u
y
y
1
y
x
x
³ x1+6² ´
³
´
1
KT
+ O
+ O(x 2 +7² ) + O
.
T
(log T )M −2
84
6.7
Conclusion
From the Theorem VI.13, one can easily see that, for any large M > 3,
(6.25)
F (x, T ) =
³
´
T
T
T
log x + O(x) + O
when
≤x≤T
M
−2
2π
(log T )
(log T )M
which improves Theorem VI.7. As for T ≤ x, we need the following lemmas for our
analysis.
Lemma VI.14.
Z
2n−1
i
¡
sin ax
a2n−1 h X (2n − k − 1)!
π¢
n
dx =
sin a + (k − 1) + (−1) ci(a)
x2n
(2n − 1)! k=1
a2n−k
2
1
R∞
Rx
where ci(x) = − x cost t dt = C0 + log x + 0 cos tt−1 dt and C0 is Euler’s constant.
∞
Proof: This is formula 3.761(3) on P.430 of [11].
Lemma VI.15. If F (y) ¿ y −3/2+² for y ≥ 1, then
Z
∞
1
sin Tx y
F (y) T dy =
y
x
Z
∞
F (y)dy + O
1
³¡ T ¢
´
1/2−²
x
when T ≤ x.
Proof: Since T ≤ x, we have the left hand side
Z
=
=
=
=
³
³¡ T ¢ ´´
³Z ∞ |F (y)| ´
2 2
F (y) 1 + O
y
dy + O
dy
T
x
y
1
x/T
x
Z x/T
´
³¡ T ¢
1/2−²
F (y)dy + O
x
Z1 ∞
³Z ∞
´
³¡ T ¢
´
1/2−²
F (y)dy + O
|F (y)|dy + O
x
1
x/T
Z ∞
´
³¡ T ¢
1/2−²
.
F (y)dy + O
x
1
x/T
Lemma VI.16.
Z ∞
³ 1 ´
X S(h)
7 B
f (u)
=
+
+
6
du
+
O
h2
4
2
u4
H∗
1
h≤H ∗
where B = −C0 − log 2π and C0 is Euler’s constant again.
85
Proof: First, from (6.19),
X S(h)
h2
h>H ∗
Z
∞
1
d(S0 (u) + u)
2
H∗ u
Z ∞
1
log u
1
¿
+
du ¿ ∗
∗
3
H
u
H
H∗
=
which accounts for the error term. It remains to see that
∞
X
S(h)
h=1
h2
Z
∞
=
1
Z
1
d(S0 (u) + u)
u2
∞
S0 (u) + u
du
u3
1
Z ∞
u − 12 log u + ²(u)
2
du
u3
1
Z ∞
1
²(u)
2− +2
du
4
u3
1
Z ∞
²(u) − B2
7 B
+ +2
du
4
2
u3
1
Z ∞
7 B
1
+ +2
df (u)
4
2
u3
1
Z ∞
7 B
f (u)
+ +6
du
4
2
u4
1
= 2
=
=
=
=
=
by integration by parts and the definitions of ²(u) and f (u).
Theorem VI.17. Assume Riemann Hypothesis and Hypothesis HL for twin primes.
For any small ² > 0,
(6.26)
F (x, T ) =
³ ¡T ¢
´
T
T
T
T
1/2−²
log
−
+ O² T
+
2π
2π 2π
x
(log T )M −2
when T ≤ x ≤ T 2−29² .
Proof: First observe that when x is in the required range, the error terms in
´
³
T
Theorem VI.13 is O² (log T )M −2 . Rewrite Theorem VI.13 as
F (x, T ) =
T
log x − T1 + T2 − T3 + T4 − T5 + T6 + T7 + error.
2π
86
Then, by Lemma VI.14, Lemma VI.15 and Lemma VI.16,
T1 =
T2 =
=
T3 =
=
T4 =
=
T5 =
T6 =
=
T7 =
=
Z
³T 3 ´
4x T /x
4T
1 + O(v 2 )dv =
+O 2 ,
3π 0
3π
x
Z ∞
³¡ T ¢ ´i
2 ³
x 7 B
f (u)
1 ´h 1 ¡ T ¢2
4
+ +6
du
+
O(
)
+
O
4
∗
πT 4
2
u
H
2 x
x
Z 1∞
³ ¡T ¢ ´
T ³7 B
f (u) ´
2
+ +6
du + O(T 1−2² ) + O T
,
4
2π 4
2
u
x
1
T h 1
T
T i
sin − ci( )
2π T /x
x
x
³ T ´
T
C0 T
T
T
+
+ O T( ) ,
− log −
2π
x
2π
2π
x
³ B 11 ´ x ³ T ´3 h ¡ x ¢
¡ x ¢2
T
T
π
3
+
2
sin +
sin ( + )
2
12 6π x
T
x
T
x
2
i
¡x¢
T
T
+
sin ( + π) + ci( )
T
x
x
³ B 11 ´ T
³ ¡ T ¢´
+
+O T
,
2 Z 12 2π
x
³ ¡T ¢
´
4T ∞ f (y)
1/2−²
dy
+
O
T
,
π 1
y2
x
Z Ry
³ ¡T ¢
´
2T ∞ 1 f (u)du
1/2−²
dy
+
O
T
π 1
y3
x
Z ∞
³
¡ T ¢1/2−² ´
f (y)
T
dy
+
O
T
,
π 1
y2
x
Z ∞
Z
³ ¡T ¢
´
f (u)
6T ∞
1/2−²
y
dudy
+
O
T
π 1
u4
x
y
Z ∞
Z ∞
³ ¡T ¢
´
3T
f (u)
3T
f (y)
1/2−²
−
du +
dy + O T
.
π 1
u4
π 1
y2
x
Combining these, we get
T
T h 8 7 B
B 11 i
log T +
− + + + C0 − 1 + +
2π
2π 3 4
2
2
12
³ ¡T ¢
´
³
´
T
1/2−²
+O² T
+ O²
x
(log T )M −2
³ ¡T ¢
´
³
´
T
T
T
1/2−²
log T +
[−1 − log 2π] + O² T
+ O²
=
2π
2π
x
(log T )M −2
F (x, T ) =
which gives the theorem.
Combining Theorem VI.7, (6.25), (6.26) and the more precise formula for F (X, T )
in Chapter II, we have the following more precise Strong Pair Correlation Conjecture:
87
Conjecture VI.18. For any small ² > 0 and any large M > 1,

h
i

1
T
T 2
T
T
T

log
x
+
(log
)
−
2
log

2

2π
x
2π
2π
2π
2π


³
´


T


+O(x log x) + O x1/2−²
,
if 1 ≤ x ≤ logT T ,



T
F (x, T ) =
log x + O(x),
if logT T ¿ x ≤ T,
2π





T
T
T
T 1/2−²

) + O( (logTT )M ), if T ≤ x ≤ T 1+² ,

 2π log 2π − 2π + O(T ( x )




 T log T − T + O(T 1−²1 ),
if T 1+² ¿ x.
2π
2π
2π
where ²1 > 0 may depend on ², and the implicit constants may depend on ² and M .
From this, one can deduce a more precise Weak Pair Correlation Conjecture:
Theorem VI.19. Assume Conjecture VI.18. For 0 < α,
Z
X
T
2π
T
log 2π
−
1−
−α
0<γ,γ 0 ≤T
|γ−γ 0 |≤ N (Tα)/T
where N (T ) =
α
1 = N (T ) + N (T )
³ sin πu ´2
πu
du + Oα
³ T ´
log T
T
.
2π
Corollary VI.20. Assume Conjecture VI.18. For 0 < α < β,
Z
X
where N (T ) =
1−
α
0<γ,γ 0 ≤T
α
N (T )/T
β
1 = N (T )
³ sin πu ´2
πu
du + Oβ
³ T ´
log T
≤γ−γ 0 ≤ N (Tβ)/T
T
2π
T
log 2π
−
T
.
2π
Proof of Corollary: Follow from the theorem straightforwardly.
Proof of Theorem: Let α > 0 be fixed, δ =
F (T (1−
1
(log T )2
and ²T =
log log T
.
log T
Let F (t) =
log 2π+1
)t
log T
χα,δ (x) =
, T ) and ² > 0. Define:


 1, if − α − δ ≤ x ≤ α + δ ,
2
2

 0,
and ψδ (x) =
otherwise.
Let rα,δ (x) = χα,δ ∗ ψδ (x) =
R∞
−∞


 1,
δ
if −

 0,
otherwise.
δ
2
≤ x ≤ 2δ ,
χα,δ (x − u)ψδ (u)du be their convolution which is
a trapezoidal function. Then r̂α,δ (t) = χ̂α,δ (t)ψ̂δ (t) =
sin πt(2α+δ) sin πtδ
πt
πtδ
where fˆ(t) =
88
r α,δ (x)
1
x
−α−δ −α
α
0
α+δ
Figure 6.1: graph of rα,δ
R∞
−∞
f (u)e−2πiut du is the Fourier transform of f . Hence, by the definition of F (x, T ),
(6.27)
X
¡ (γ − γ 0 )N (T ) ¢
rα,δ
w(γ − γ 0 ) =
T
0<γ,γ 0 ≤T
Z
Z
∞
∞
F (t)r̂α,δ (t)dt = 2
−∞
F (t)r̂α,δ (t)dt = 2I
0
because F (x, T ) = F ( x1 , T ) and r̂α,δ is even. Now we break down the integral I into
four pieces
Z
Z
1−²T
I=
+
0
Z
1
Z
1+²
+
∞
+
1−²T
1
= I1 + I2 + I3 + I4 .
1+²
By Conjecture VI.18, we have
Z
∞
I4 = N (T )
Z
1+²
1+²
³ T ´
³
´
T
+O
,
log T
(log T )M
1
Z 1
Z 1
³ T ´
,
= N (T )
r̂α,δ (t)dt − N (T )
(1 − t)r̂α,δ (t)dt + O
log T
1−²T
1−²T
Z 1−²T
Z 1−²T
= N (T )
r̂α,δ (t)dt − N (T )
(1 − t)r̂α,δ (t)dt
0
0
Z
hT
³ T ´
T 2
T
T i 1−²T e2(log 2πe)t
+
(log ) − 2 log
r̂
(t)dt
+
O
.
α,δ
α
2π
2π
2π
2π 0
T 2t
log T
I3 = N (T )
I2
I1
r̂α,δ (t)dt + O(T 1−²1 /2 ),
r̂α,δ (t)dt + O
Thus,
Z
Z
∞
1
I = N (T )
r̂α,δ (t)dt − N (T )
(1 − t)r̂α,δ (t)dt
0
0
hT
³
´i
³ T ´
T 2
T
T ih
α
1
+
(log ) − 2 log
+ Oα
+ Oα
2π
2π
2π
2π log (T /2πe)
(log T )3
log T
R
R
because, f ĝ = fˆg, and the transform pair
f (t) = e−2a|t| ,
fˆ(x) =
4a
,
4a2 + (2πx)2
89
Z
1−²T
0
e2(log 2πe)t
r̂α,δ (t)dt =
T 2t
=
=
=
=
Also,
Z
∞
e−2 log (T /2πe)t r̂α,δ (t)dt + O
³ log T ´
T2
³ log T ´
1 ∞ −2 log (T /2πe)|t|
e
r̂α,δ (t)dt + O
2 −∞
T2
Z ∞
³ log T ´
1
4 log (T /2πe)
r
(x)dx
+
O
α,δ
2 −∞ (2 log (T /2πe))2 + (2πx)2
T2
Z α
³ δ ´
1
1
dx
+
O
2 log (T /2πe) −α 1 + ( log (Tπx/2πe) )2
log T
³ 1 + α3 ´
α
+O
.
log (T /2πe)
(log T )3
0
Z
R∞
r̂ (t)dt = r(0) = 1. So, by
−∞ α,δ
R
f ĝ =
R
fˆg, the defintion of rα,δ (t), and the
transform pair
³ sin πu ´2
ˆ
,
f (u) =
πu
f (t) = max{1 − |t|, 0},
Z
Z
∞
2I = N (T )
1
r̂α,δ (t)dt − N (T )
−∞
(1 − |t|)r̂α,δ (t)dt
−1
hT
³ T ´
T
T
T i
2α
+
(log )2 − 2 log
+ Oα
2π
2π
2π
2π log (T /2π) − 1
log T
Z α³
´
h
T
T
T i
sin πu 2
du +
(log ) − 2
= N (T ) − N (T )
πu
2π
2π
2π
−α
Z α
´
³
´
³
T
1
+ Oα
∗
1du 1 +
T
log T
log
−α
Z α 2π ³
³ T ´
sin πu ´2
= N (T ) + N (T )
1−
du + Oα
.
πu
log T
−α
Putting this into (6.27), we have
X
¡ (γ − γ 0 )N (T ) ¢
w(γ − γ 0 )
T
0<γ,γ 0 ≤T
Z α
³ T ´
³ sin πu ´2
du + Oα
.
= N (T ) + N (T )
1−
πu
log T
−α
rα,δ
90
On the other hand,
X
¡ (γ − γ 0 )N (T ) ¢
rα,δ
w(γ − γ 0 )
T
0<γ,γ 0 ≤T
³
³
X
¡ α2 ¢´
=
1+O
+
O
(log T )2
0<γ,γ 0 ≤T
|γ−γ 0 |≤ N (Tα)/T
0<γ,γ 0 ≤T
α<|γ−γ 0 |
³
X
=
1+O
0<γ,γ 0 ≤T
|γ−γ 0 |≤ N (Tα)/T
¡
´
1
X
α2 ¢´
(log T )2
N (T )
<α+δ
T
´
¡
N (T ) ¢
(rα,2δ − rα−δ,2δ ) (γ − γ 0 )
w(γ − γ 0 )
T
0<γ,γ 0 ≤T
³
³ T ´
X
¡ α2 ¢´
1+O
+
O
.
α
2
(log
T
)
log
T
0
0<γ,γ ≤T
+O
=
³ X
|γ−γ 0 |≤ N (Tα)/T
Consequently,
1
r α,2δ
r
α−δ,2δ
r α,2δ
−α−2δ
−α
−α−δ
−α+δ
α−δ α
0
r α−δ,2δ
α+δ α+2δ
Figure 6.2: graph of rα,2δ − rα−δ,2δ
X
0<γ,γ 0 ≤T
|γ−γ 0 |≤ N (Tα)/T
Z
α
1 = N (T ) + N (T )
1−
−α
³ sin πu ´2
πu
³ T ´
du + Oα
log T
which gives the theorem. This reinforces the belief that the zeros of the Riemann zeta
function behave like the eigenvalues of a random matrix from the Gaussian Unitary
Ensemble (GUE).
APPENDIX
The following is the C program for computing
Z
X¡
Z
¢k
X
ψ(x + δx) − ψ(x) − δx dx and
1
|ψ(x + δx) − ψ(x) − δx|λ dx
1
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
short int *setarray(long int length) {
long int i=0;
short int *ip;
ip = new (short int)[(length+1)*sizeof(short int)];
if(ip==NULL)
exit;
else {
while(i<=length) {
ip[i]=1;
i=i+1;
}
return ip;
}
}
int nextindex(long int i, long int length, short int *ip){
91
92
long int j=i;
while(j<length && ip[j]==0)
j++;
return j;
}
sieve(double N, long int length, short int *ip, int *p){
long int i,j;
short int k;
double temp;
j=2;
while(j<=sqrt(N+length)){
if(p[j]!=0)
for(i=(long int)(floor((N+j-1)/j)*j-N); i<=length; i=i+j)
ip[i]=0;
j++;
}
for(k=2; k<=(log(N+length)/log(2)); k++){
j=(long int)(floor(exp(log(N-1)/k))+1);
temp=exp(log(N+length)/k);
while(j<=temp)
if(p[j]!=0) ip[(long int)(pow(j,k)-N)] = k;
j++;
}
}
93
}
double power(double d, int k){
int i;
double result=1.0;
for(i=1; i<=k; i++)
result = result * d;
return result;
}
integral(double a, double b, double c, double d, double *result){
int k;
double i = c-d*a;
double j = c-d*b;
if(b>a){
for(k=0; k<=5; k++)
result[k] = result[k] + (power(i,(k+2))-power(j,(k+2)))/ (d*(k+2));
/* This is used to compute non-integral moments
if(i>=0 && j>=0)
for(k=0; k<=5; k++)
result[k] = result[k] +
(exp((k*1.1+2)*log(i))-exp((k*1.1+2)*log(j)))/(d*(k*1.1+2));
else if(i>=0 && j<0)
for(k=0; k<=5; k++)
result[k] = result[k] +
94
(exp((k*1.1+2)*log(i))+exp((k*1.1+2)*log(-j)))/(d*(k*1.1+2)) ;
else
for(k=0; k<=5; k++)
result[k] = result[k] +
(exp((k*1.1+2)*log(-j))-exp((k*1.1+2)*log(-i)))/(d*(k*1.1+2) ); */
}
}
double check(long int i, short int *ip){
if(ip[i]==0) return 0;
else return (1/ip[i]);
}
compute(double N1, double N2, long int length1, long int length2,
short int *ip1, short int *ip2, double d, double *result){
long int i1 = 1;
long int i2 = 1;
long int i,temp1, temp2;
double sum = 0;
double a = N1;
for(i=1; i<=length1; i++)
sum = sum + log(N1+i)*check(i,ip1);
temp1 = nextindex(i1,length1,ip1);
temp2 = nextindex(i2,length2,ip2);
while(i1 <= length1){
95
if((N1+temp1) > (N2+temp2)/(1+d)){
if(a < (N2+temp2)/(1+d)){
integral(a, (N2+temp2)/(1+d), sum, d, result);
a = (N2+temp2)/(1+d);
sum = sum + log(N2+temp2)*check(temp2,ip2);
temp2 = temp2 + 1;
if(temp2 < length2){
i2 = temp2;
temp2 = nextindex(i2,length2,ip2);
}
}
else
sum = sum + log(N2+temp2)*check(temp2,ip2);
i2 = temp2 + 1;
temp2 = nextindex(i2,length2,ip2);
}
}
else if((N1+temp1) < (N2+temp2)/(1+d)){
if(a < (N1+temp1))
integral(a, (N1+temp1), sum, d, result);
a = N1 + temp1;
sum = sum - log(N1+temp1)*check(temp1,ip1);
i1 = temp1 + 1;
temp1 = nextindex(i1,length1,ip1);
}
96
else
sum =sum - log(N1+temp1)*check(temp1,ip1);
i1 = temp1 + 1;
temp1 = nextindex(i1,length1,ip1);
}
}
else{
integral(a, (N1+temp1), sum, d, result);
a = N1 + temp1;
sum = sum - log(N1+temp1)*check(temp1,ip1) +
log(N2+temp2)*check(temp2,ip2);
i1 = temp1 + 1;
temp1 = nextindex(i1,length1,ip1);
i2 = temp2 + 1;
if(temp2 < length2){
temp2 = nextindex(i2,length2,ip2);
}
}
}
}
main(){
double X = 100000000.0;
double delta = 0.0001;
short int *ip1, *ip2;
97
double N1 = floor(1/delta) + 1;
double N2 = floor(N1*(1+delta));
double N3 = floor(N2*(1+delta));
long int i,j;
double *result;
result = new double[6*sizeof(double)];
for(i=0; i<=5; i++) result[i]=0;
int *p = new int[(long int)(floor(sqrt(X*(1+delta)))+1)* sizeof(int)];
for(i=0; i<=floor(sqrt(X*(1+delta))); i++)
p[i] = 1;
for(i=2; i<=floor(sqrt(X*(1+delta)))/2; i++)
p[2*i]=0;
for(j=3; j<=sqrt(sqrt(X*(1+delta))); j=j+2){
if(p[j]==1)
for(i=2; i<=floor(sqrt(X*(1+delta)))/j; i++)
p[i*j]=0;
}
ip1 = setarray((long int)(N2-N1));
ip2 = setarray((long int)(N3-N2));
sieve(N1, (long int)(N2-N1), ip1, p);
sieve(N2, (long int)(N3-N2), ip2, p);
while(N1 < X){
compute(N1,N2,(long int)(N2-N1),(long int)(N3-N2),
ip1,ip2,delta,result);
N1 = N2;
98
free(ip1);
ip1 = ip2;
if(N3 < X) N2 = N3;
else N2 = X;
N3 = floor(N2*(1+delta));
ip2 = setarray((long int)(N3-N2));
sieve(N2, (long int)(N3-N2), ip2, p);
}
for(i=0; i<=5; i++)
printf(”%f\n”, result[i]);
free(ip1);
free(ip2);
free(p);
free(result);
}
BIBLIOGRAPHY
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100
BIBLIOGRAPHY
[1] H. Davenport. Multiplicative number theory. Springer-Verlag, second edition edition, 1980.
[2] P.J. Forrester and A.M. Odlyzko. Gaussian unitary ensemble eigenvalues and Riemann ζ
function zeros: a nonlinear equation for a new statistic. Phys. Rev. E., 54(3):R4493–R4495,
1996.
[3] J.B. Friedlander and D.A. Goldston. Some singular series averages and the distribution of
Goldbach numbers in short intervals. Illinois J. Math., 39:158–180, 1995.
[4] A. Ghosh. On Riemann zeta-function - sign changes of S(T). In Recent Progress in analytic
number theory, volume 1, pages 25–46. Academic Press, New York, 1981.
[5] A. Ghosh. On the Riemann zeta-function - Mean value theorems and the distribution of |S(T)|.
J. Number Theory, 17:93–102, 1983.
[6] D.A. Goldston. Linnik’s theorem on Goldbach numbers in short intervals. Glasgow Math. J.,
32:285–297, 1990.
[7] D.A. Goldston and S.M. Gonek. Mean value theorems for long Dirichlet polynomials and tails
of Dirichlet series. Acta Arith., 84(2):155–192, 1998.
[8] D.A. Goldston, S.M. Gonek, A.E. Ozluk, and C. Synder. On the pair correlation of zeros of
the Riemann zeta-function. Proc. London Math. Soc., 80(3):31–49, 2000.
[9] D.A. Goldston and H.L. Montgomery. On pair correlations of zeros and primes in short intervals. In Analytic Number Theory and Diophantine Problems (Stillwater, OK, July 1984),
volume 70 of Prog. Math., pages 183–203. Birkauser, Boston, 1987.
[10] D.A. Goldston and C.Y. Yildirim. Primes in short segments of arithmetic progressions. Can.
J. Math., 50(3):563–580, 1998.
[11] I.S. Gradshteyn and I.M. Ryzhik. Table of Integrals, Series and Products. Academic Press,
sixth edition, 2000.
[12] D.A. Hejhal. On the triple correlation of zeros of the zeta function. Internat. Math. Res.
Notices, pages 293–302, 1994.
[13] H.L. Montgomery. The pair correlation of zeros of the zeta function. In Analytic Number
Theory (St. Louis Univ., 1972), volume 24 of Proc. Sympos. Pure Math., pages 181–193. Amer.
Math. Soc., Providence, 1973.
[14] H.L. Montgomery and K. Soundararajan. Beyond pair correlation. Preprint.
[15] H.L. Montgomery and K. Soundararajan. Primes in short intervals. To appear.
[16] H.L. Montgomery and R.C. Vaughan. Hilbert’s inequality. J. London Math. Soc., 8(2):73–82,
1974.
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[17] H.L. Montgomery and R.C. Vaughan. On the distribution of reduced residues. Annals of
Math., 123:311–333, 1986.
[18] A.M. Odlyzko. On the distribution of spacings between zeros of the zeta function. Math.
Comp., 48:273–308, 1987.
[19] Z. Rudnick and P. Sarnak. Zeros of principal L-functions and random matrix theory. Duke
Math. J., 81:269–322, 1996.
[20] B. Saffari and R.C. Vaughan. On the fractional parts of x/n and related sequences II. Ann.
Inst. Fourier (Grenoble), 27(2):1–30, 1977.
[21] H. von Koch. Sur la distribution des nombres premiers. Acta Math., 24:159–182, 1901.
ABSTRACT
Pair correlation and distribution of prime numbers
by
Tszho Chan
Chair: Professor Hugh Montgomery
In early 1970s, Hugh Montgomery initialized the study of the pair correlation
function F (X, T ) on zeros of Riemann zeta function and made the famous Strong
Pair Correlation Conjecture on its asymptotic size when T ≤ X. Later, assuming
Riemann Hypothesis, Goldston and Montgomery proved that this and two other
asymptotic formulas for second moments of primes in short intervals are equivalent
to each other. Recently, Montgomery and Soundararajan proposed a more precise
asymptotic formula for one of the above second moments. They also worked on
higher moments under Hardy-Littlewood Prime k-tuple Conjecture and came to the
conclusion that primes in short intervals obey normal distribution.
The first goal of this thesis is to make use of the more precise asymptotic formula described above to get more precise formulas for the other second moment and
Strong Pair Correlation Conjecture. Our method is essentially that of Goldston &
Montgomery but with explicit error terms. The second goal is to improve the asymp-
1
totic formulas for higher even moments which tells us how they deviate from normal
distribution. The improved formulas match closely with actual data. The third goal
is to get “equivalence” results on higher moments of primes in short intervals. For
even moments, we can use the same method as that for second moments. However,
difficulties arise for odd moments. The best one can do is to assume good information
on even moments and use Ghosh’s method. Despite many unproved assumptions,
our results agree strongly with actual data.
Finally, we try to get second order terms for F (X, T ) in different ranges of X.
When 1 ≤ X ≤ T 1−² , we make more precise the original calculation by Montgomery.
When T 1−² ≤ X, we use Goldston & Gonek’s method for computing mean values of
Dirichlet polynomials and tails of Dirichlet series assuming Twin Prime Conjecture.
From these, one gains better understanding on the behavior of F (X, T ), especially
when X ≈ T . Consequently, one has more precise formulations for the Strong and
Weak Pair Correlation Conjectures which reinforce the belief that the zeros of the
Riemann zeta function behave like the eigenvalues of a random matrix from the
Gaussian Unitary Ensemble.