On the mean square of the Riemann zeta-function in short intervals

0.21 MB·English

by  Aleksandar Ivić

PUBLICATIONSDEL’INSTITUTMATHE´MATIQUE
Nouvelles´erie,tome??(??) (2009),??–??
On the mean square of
the Riemann zeta-function in short intervals
9 Aleksandar Ivi´c
0
0
2 Communicated by

n
a
J Abstract. Itisprovedthat, forTε6G=G(T)6 12√T,

5 2T 2 3 √T
1 ZT “I1(t+G,G)−I1(t,G)” dt = TGjX=0ajlogj“ G ”

] + Oε(T1+εG1/2+T1/2+εG2)
T N
N withsomeexplicitlycomputableconstantsaj (a3 >0)where,forfixedk∈ ,
h. Ik(t,G)= √1πZ−∞∞|ζ(12 +it+iu)|2ke−(u/G)2du.
at ThegeneralizationstothemeansquareofI1(t+U,G)−I1(t,G)over[T,T+H]
m and the estimation of the mean square of I2(t+U,G)−I2(t,G) are also
discussed.
[

3
v
2 1. Introduction
3
1 The mean values of the Riemann zeta-function ζ(s), defined as
0
. ∞
3 ζ(s)= n s (σ = es>1),
−
0 ℜ
n=1
8 X
0 (and otherwise by analytic continuation) occupy a central place in the theory of
: ζ(s). Ofparticularsignificanceisthe meansquareonthe“criticalline”σ = 1,and
v 2
i a vast literature exists on this subject (see e.g., the monographs [3], [4], and [18]).
X One usually defines the error term in the mean square formula for ζ(1 +it) as
| 2 |
r
a T T
(1.1) E(T):= ζ(1 +it)2dt T log +2γ 1 ,
| 2 | − 2π −
Z0 (cid:16) (cid:17)
2000 Mathematics Subject Classification. 11M06;11N37.
Key words and phrases. The Riemann zeta-function, the mean square in short intervals,
upperbounds.
1
2 SURNAME
where γ = Γ(1) = 0.5772156649... is Euler’s constant. More generally, one
′
−
hopes that for a fixed k the function (E(T) E (T) in this notation)
1
≡
T
(1.2) E (T):= ζ(1 +it)2kdt TP (logT) (k N)
k | 2 | − k2 ∈
Z0
representstheerrortermintheasymptoticformulaforthe2k-thmomentof ζ(1+
| 2
it),whereP (z)isasuitablepolynomialinzofdegreeℓ. Thisisknown,besidesthe
ℓ
|
case k =1, only in the case k =2 (see e.g., [4], [17]), andany further improvement
wouldbe of greatsignificance, inview of numerousapplications of powermoments
of ζ(1+it). Bymeansofrandommatrixtheoryplausiblevaluesofthecoefficients
| 2 |
ofthepolynomialP (z)thatoughttobestandingin(1.2)aregivenbyJ.B.Conrey
k2
et al. [2]. However, these values are still conjectural.
As forexplicit formulasfor E (T)andrelatedfunctions, we beginby mention-
k
ing the famous formula of F.V. Atkinson [1] for E(T). Let 0 < A < A be any
′
two fixed constants such that AT <N <AT, let d(n)= 1 be the number of
′ δn
|
divisors of n, and finally let N =N (T)=T/(2π)+N/2 (N2/4+NT/(2π))1/2.
′ ′ −P
Then
(1.3) E(T)= (T)+ (T)+O(log2T),
1 2
where X X
(1.4) (T)=21/2(T/(2π))1/4 ( 1)nd(n)n 3/4e(T,n)cos(f(T,n)),
−
1 −
n6N
X X
(1.5)
(T)= 2 d(n)n 1/2(log(T/(2πn)) 1cos(T log(T/(2πn)) T +π/4),
− −
2 − −
n6N′
X X
with
(1.6)
f(T,n) = 2Tarsinh πn/(2T) + 2πnT +π2n2 1π
− 4
= 1π+2√(cid:0)2pπnT + 1√(cid:1)2π3pn3/2 +β n5/2 +β n7/2 +... ,
−4 6 T1/2 5T3/2 7T5/2
where the β ’s are constants,
j
(1.7)
1
e(T,n) = (1+πn/(2T))−1/4 (2T/πn)1/2arsinh( πn/(2T)) −
= 1+O(n/T) n(16n<T), p o
and arsinhx=log(x+√1+x2).Atkinson’sformulacameintoprominenceseveral
decades after its appearance, and incited much research (see e.g., [3],[4] for some
of them). The presence of the function d(n) in (1.4) and the structure of the sum
RUNNING TITLE 3
Σ (T)pointouttheanalogybetweenE(T)and∆(x),theerrortermintheclassical
1
divisor problem, defined as
∆(x)= d(n) x(logx+2γ 1).
− −
n6x
X
This analogy was investigated by several authors, most notably by M. Jutila [13],
[14], and then later by the author [6]–[8] and [9]–[10]. Jutila [13] proved that
(1.8)
T+H
2
∆(x+U) ∆(x) dx
−
TZ (cid:16) (cid:17)
1 d2(n) T+H n 2
= x1/2 exp 2πiU 1 dx
4π2 n3/2 x −
nX62TU TZ (cid:12)(cid:12) (cid:18) r (cid:19) (cid:12)(cid:12)
(cid:12) (cid:12)
+ O (T1+ε+HU1/2Tε), (cid:12) (cid:12)
ε
for 1 6 U T1/2 H 6 T, and an analogous result holds also for the integral
≪ ≪
of E(x+U) E(x) (the constants in front of the sum and in the exponential will
−
be 1/√2π and √2π, respectively). Here and later ε(>0) denotes arbitrarilysmall
constants, not necessarily the same ones at each occurrence, while a b means
ε
≪
that the implied –constant depends on ε. From (1.8) one deduces (a b means
≪ ≍
that a b a)
≪ ≪
T+H 2 √T
(1.9) ∆(x+U) ∆(x) dx HUlog3
ZT (cid:16) − (cid:17) ≍ U !
for HU T1+ε and Tε U 6 1√T. In [14] Jutila proved that the integral in
≫ ≪ 2
(1.9) is
Tε(HU +T2/3U4/3) (1 H,U T).
ε
≪ ≪ ≪
This bound and (1.9) hold also for the integral of E(x+U) E(x). Furthermore
−
Jutila op. cit. conjectured that
2T 4
(1.10) E(t+U) E(t U) dt T1+εU2
ε
− − ≪
ZT (cid:16) (cid:17)
holds for 1 U T1/2, and the analogous formula should hold for ∆(t) as well.
≪ ≪
In fact, using the ideas of K.-M. Tsang [19] who investigated the fourth moment
of ∆(x), it can be shown that one expects the integral in (1.10) to be of the order
TU2log6(√T/U). As shown in [11], the truth of Jutila’s conjecture (1.10) implies
the hitherto unknown eighth moment bound
T
(1.11) ζ(1 +it)8dt T1+ε,
| 2 | ≪ε
Z0
which would have important consequences in many problems from multiplicative
number theory, such as bounds involving differences between consecutive primes.
4 SURNAME
Despite several results on E (T) (see e.g., [17]), no explicit formula is known
2
forthisfunction, whichwouldbe analogoustoAtkinson’sformula(1.3)-(1.7). This
is probably due to the complexity of the function in question, and it is even not
certain that such a formula exists. However,when one works not directly with the
moments of ζ(1 +it), but with smoothed versions thereof, the situation changes.
Let, for k N| fi2xed, |
∈
(1.12) Ik(t,G):= √1π ∞ |ζ(12 +it+iu)|2ke−(u/G)2du (1≪G≪t).
Z−∞
Y. Motohashi’s monograph [17] contains explicit formulas for I (t,G) and I (t,G)
1 2
in suitable ranges for G = G(t). The formula for I (t,G) involves quantities from
2
the spectral theory of the non-Euclidean Laplacian (see op. cit.). Let, as usual,
λ =κ2+1 0 bethediscretespectrumofthenon-EuclideanLaplacianacting
{ j j 4}∪{ }
on SL(2,Z)–automorphicforms,andα = ρ (1)2(coshπκ ) 1,whereρ (1)isthe
j j j − j
| |
firstFouriercoefficientoftheMaasswaveformcorrespondingtotheeigenvalueλ to
j
which the Hecke series Hj(s) is attached. Then, for T1/2log−DT 6G6T/logT,
and for an arbitrary constant D >0, Motohashi’s formula gives
(1.13)
G 1I (T,G)=O(log3D+9T) +
− 2
+ √π2T ∞ αjHj3(12)κj−1/2sin κjlog4κejT exp −41 GTκj 2 .
Xj=1 (cid:16) (cid:17) (cid:16) (cid:16) (cid:17) (cid:17)
For our purposes the range for which (1.13) holds is not large enough. We shall
use a more precise the expressionfor I (T,G) that can be derivedby following the
2
proof of Theorem 3, given in [12], and then taking σ 1 +0. Namely if
→ 2
κ κ 2 κ
j j j
(1.14) Y =Y (T;κ ):= 1+ + ,
0 0 j
T 4T 2T
r !
(cid:16) (cid:17)
then, for Tε G=G(T) T1 ε, it follows that
−
≪ ≪
(1.15)
I (T,G) F (T,G)+O(1)
2 0
∼
πG
+ √2T αjHj3(12)κj−1/2e−41G2log2(1+Y0)
κj6TXG−1logT
κ
sin κ log j +c κ3T 2+ +c κNT1 N .
× j 4eT 3 j − ··· N j −
(cid:16) (cid:17)
Here N(> 3) is a sufficiently large integer, and all the constants c in (1.15) may
j
be effectively evaluated. The meaning of the symbol is that besides the spectral
∼
sums in (1.15) a finite number of other sums appear, each of which is similar in
RUNNING TITLE 5
nature to the corresponding sum above, but of a lower order of magnitude. The
function F (t,G) is given explicitly e.g., by eq. (5.112) of [4]. We have
0
(1.16)
F (t,G)= e G ∞ B Γ′ + +B (Γ′)2Γ′′ (1 +it+iu)e (u/G)2du ,
0 ℜ √π 1 Γ ··· 11 Γ2 2 −
(cid:26) Z−∞(cid:20) (cid:21) (cid:27)
where B , ... ,B are suitable constants. The main contribution to F (t,G) is of
1 11 0
the form GP (logt), where P (z) is a polynomial of degree four in z, whose coeffi-
4 4
cients can be explicitly evaluated. This is easily obtained by using the asymptotic
formula
Γ(k)(s) k
(1.17) = bj,k(s)logjs+c 1,ks−1+...+c r,ks−r+Or(s−r−1)
Γ(s) − − | |
j=0
X
foranyfixedintegersk >1,r>0,whereeachb (s)( b forasuitable constant
j,k j,k
∼
b ) has an asymptotic expansion in non-positive powers of s. One obtains (1.17)
j,k
from Stirling’s classical formula for Γ(s).
2. Statement of results
In[11]theauthorimproved(1.9)anditsanalogueforE(T)toatrueasymptotic
formula. Namely it wasshownthat, for 1 U=U(T)61√T, wehave(c =8π 2)
≪ 2 3 −
(2.1)
2T 2 3 √T
∆(x+U) ∆(x) dx = TU c logj +
j
− U
ZT (cid:16) (cid:17) Xj=0 (cid:16) (cid:17)
+ O (T1/2+εU2)+O (T1+εU1/2),
ε ε
asimilarresultbeingtrueif∆(x+U) ∆(x) isreplacedbyE(x+U) E(x), with
− −
different constants c . It follows that, for Tε 6U =U(T)6T1/2 ε, (2.1) is a true
j −
asymptotic formula. Moreover, for T 6 x 6 2T and Tε 6 U = U(T) 6 T1/2 ε,
−
from (2.1) it follows that
(2.2)
√x √x
∆(x+U) ∆(x)=Ω √Ulog3/2 , E(x+h) E(x)=Ω √Ulog3/2 .
− U − U
n (cid:16) (cid:17)o n (cid:16) (cid:17)o
These omega results (f(x) = Ω(g(x)) means that lim f(x)/g(x) = 0) show
x
→∞ 6
that Jutila’s conjectures made in [13], namely that
(2.3) ∆(x+U) ∆(x) xε√U, E(x+U) E(x) xε√U
ε ε
− ≪ − ≪
for xε 6 U 6 x1/2 ε are (if true), close to being best possible. It should be also
−
mentionedthattheformula(2.1)canbefurthergeneralized,andtherepresentative
cases are the classicalcircle problem and the summatory function of coefficients of
holomorphic cusp forms, which were also treated in [11].
InthisworktheproblemofthemeansquareofthefunctionI (t,G)(see(1.12))
k
in short intervals is considered when k =1 or k =2. The former case is much less
difficult,andinfactanasymptoticformulainthemostimportantcaseofthemean
6 SURNAME
square of I (t+G,G) I (t,G) can be obtained. The result, which is similar to
1 1
−
(2.1), is
THEOREM 1. For Tε 6G=G(T)6 1√T we have
2
(2.4)
2T
2
I (t+G,G) I (t,G) dt=
1 1
−
Z
T (cid:0) (cid:1)
3 √T
= TG a logj +O (T1+εG1/2+T1/2+εG2)
j ε
G
Xj=0 (cid:16) (cid:17)
with some explicitly computable constants a (a >0).
j 3
Corollary 1. For Tε 6 G = G(T) 6 T1/2 ε, (2.4) is a true asymptotic
−
formula.
Corollary 2. For Tε 6G=G(T)6T1/2 ε we have
−
(2.5) I (T +G,G) I (T,G)=Ω(√Glog3/2T).
1 1
−
Namely if (2.5) did not hold, then replacing T by t, squaring and integrating
we would obtain that the left-hand side of (2.4) is o(TGlog3T) as T , which
→∞
contradicts the right-handside of (2.4). The formula givenby Theorem 1 makes it
then plausible to state the following conjecture, analogous to (2.3).
Conjecture. For Tε 6G=G(T)6T1/2 ε we have
−
I (T +G,G) I (T,G)=O (Tε√G).
1 1 ε
−
The generalizationof Theorem 1 to the mean square of I (t+U,G) I (t,G)
1 1
−
over [T, T +H] is will be discussed in Section 4, subject to the condition (4.1).
This is technically more involved than (2.5), so we have chosen not to formulate
our discussion as a theorem.
The mean square of I (t+U,G) I (t,G) is naturally more involvedthan the
2 2
−
meansquareofI (t+U,G) I (t,G). Atthepossiblestate ofknowledgeinvolving
1 1
−
estimates with κ and related exponential integrals, it does not seem possible to
j
obtain an asymptotic formula, but only an upper bound. This is
THEOREM 2. For Tε 6U 6GT ε T1/2 ε,U =U(T),G=G(T) we have
− −
≪
2T 2 U 2
(2.6) I (t+U,G) I (t,G) dt T2+ε .
2 2 ε
− ≪ G
ZT (cid:16) (cid:17) (cid:18) (cid:19)
RUNNING TITLE 7
3. The proof of Theorem 1
We shall prove first the bound
2T 2
(3.1) I (t+G) I (t,G) dt TGL3 (Tε 6G6 1√T),
1 − 1 ≪ 2
ZT (cid:16) (cid:17)
where henceforth L=logT for brevity. This shows that the sum on the left-hand
side of (2.5) is indeed of the order given by the right-hand side, a fact which will
be needed a little later. We truncate (1.12) at u = GL, then differentiate (1.1)
±
and integrate by parts to obtain
(3.2)
1 GL t+u
I (t,G) = log +2γ+E (t+u) e (u/G)2du+O(e L2/2)
1 ′ − −
√π 2π
Z−GL(cid:18) (cid:19)
1 GL t+u 2u
= log +2γ+ E(t+u) e−(u/G)2du+O(e−L2/2).
√π 2π G2
Z−GL(cid:18) (cid:19)
This gives, for T 6t62T,Tε G6 1√T,
≪ 2
(3.3)
GL
2 G2
I1(t+G,G)−I(t,G)= √πG2 ue−(u/G)2 E(t+u+G)−E(t+u) du+O T .
GZL (cid:16) (cid:17) (cid:16) (cid:17)
−
We square (3.3), noting that G2/T 1, and then integrate over [T, 2T]. We use
≪
the Cauchy-Schwarz inequality for integrals together with Jutila’s bound (1.9) (or
(2.1)) for E(x+U) E(x), and then (3.1) will follow.
−
To prove (2.4) we need a precise expression for I (t+G,G) I (t,G). One
1 1
−
waytoproceedistostartfrom(3.2)anduseAtkinson’sformula(1.3)–(1.7). Inthe
course of the proof, various expressions will be simplified by Taylor’s formula, and
one has to use the well known integral (see e.g., the Appendix of [3])
(3.4) ∞ exp Ax Bx2 dx= π exp A2 ( eB >0).
− B 4B ℜ
Z−∞ (cid:0) (cid:1) r (cid:18) (cid:19)
However, it seems more expedient to proceed in the following way. We start from
Y. Motohashi’s formula [17, eq. (4.1.16)], namely
(3.5)
Z (g) = ∞ e Γ′(1 +it) +2γ log(2π) g(t)dt+2π e g(1i)
1 ℜ Γ 2 − ℜ { 2 }
Z−∞(cid:20) (cid:26) (cid:27) (cid:21)
+ 4 ∞ d(n) ∞(y(y+1)) 1/2g (log(1+1/y))cos(2πny)dy,
− c
n=1 Z0
X
where, for k N fixed,
∈
(3.6) Z (g)= ∞ ζ(1 +it)2kg(t)dt, g (x)= ∞ g(t)cos(xt)dt.
k | 2 | c
Z−∞ Z−∞
8 SURNAME
Here the function g(r) is realfor r real,and there exist a large positive constantA
such that g(r) is regularand g(r) (r +1) A for mr 6A. The choice in our
−
≪ | | |ℑ |
case is
1
g(t):= exp ((T t)/G)2 , g (x)=Gexp 1(Gx)2 cos(Tx),
√π − − c −4
(cid:0) (cid:1) (cid:0) (cid:1)
and one verifies without difficulty that the above assumptions on g are satisfied.
In this case Z (g) becomes our I (T,G) (see (2.4)). To evaluate the integral on
1 1
the right-hand side of (3.5) we use (1.17) with k = 1. Thus when we form the
difference I (t+G,G) I (t,G) in this way, the integralon the right-hand side of
1 1
−
(3.5) produces the terms O(G2/T)+O(1) = O(1), since G2/T 1. The second
≪
integral on the right-hand side of (3.5) is evaluated by the saddle-point method
(see e.g., [3, Chapter 2]). A similar analysis was made in [8] by the author, and an
explicitformulaforI (T,G)isalsotobefoundinY.Motohashi[17,eq. (5.5.1)]. As
1
these results are either less accurate, or hold in a more restrictive range of G than
what we require for the proof of Theorem 1, a more detailed analysis is in order.
A convenient result to use is [3, Theorem 2.2 and Lemma 15.1] (due originally to
b
Atkinson [1]) for the evaluation of exponential integrals ϕ(x)exp(2πiF(x))dx,
a
where ϕ and F are suitable smooth, real-valued functions. In the latter only the
R
exponentialfactorexp( 1G2log(1+1/y))ismissing. Inthe notationof[1]and[3]
−4
we have that the saddle point x (root of F (x)=0) satisfies
0 ′
1 T 1 1/2 1
x =U = + ,
0
− 2 2πn 4 − 2
(cid:18) (cid:19)
and the presence of the above exponential factor makes it possible to truncate the
series in (3.5) at n = TG 2logT with a negligible error. Furthermore, in the
−
remaining range for n we have (in the notation of [3])
Φ0µ0F0−3/2 ≪(nT)−3/4,
which makes a total contribution of O(1), as does error term integral in Theorem
2.2 of [3]. The error terms with Φ(a), Φ(b) vanish for a 0+, b + . In this
→ → ∞
waywe obtaina formula,which naturallyhas a resemblanceto Atkinson’s formula
(compare it also to [8, eq. (19)]). This is
(3.7)
I (t+G,G) I (t,G)=O(1)+
1 1
−
+ √2G ( 1)nd(n)n 1/2 u(t+G,n)H(t+G,n) u(t,n)H(t,n) ,
−
− −
n6TXG−2L n o
where
t 1 1/2 1 −1/2
u(t,n):= + t T,16n6TG 2L ,
−
((cid:18)2πn 4(cid:19) − 2) (cid:16) ≍ (cid:17)
RUNNING TITLE 9
and (in the notation of (1.6))
2
πn
H(T,n):=exp G2 arsinh sinf(T,n) t T,16n6TG 2L .
−
− (cid:18) r2T (cid:19) ! (cid:16) ≍ (cid:17)
Now we square (3.7) and integrate over T 6 t 6 2T, using the Cauchy-Schwarz
inequality and (3.1) to obtain
2T 2
(3.8) I (t+G,G) I (t,G) dt = S+O(TG1/2L3/2),
1 1
−
ZT (cid:16) (cid:17)
where we set
(3.9)
2T d(n)
S := 2G2 ( 1)n u(t+G,n)H(t+G,n)
ZT (n6TXG−2L − n1/2h −
2
u(t,n)H(t,n) dt.
− )
i
Squaring out the sum over n in (3.9) it follows that
(3.10)
2T d2(n) 2
S = 2G2 u(t+G,n)H(t+G,n) u(t,n)H(t,n) dt+
n −
ZT n6TXG−2L h i
2T
d(m)d(n)
+ 2G2 ( 1)m+n u(t+G,m)H(t+G,m)
− √mn −
TZ m6=n6XTG−2L h
u(t,m)H(t,m) u(t+G,n)H(t+G,n) u(t,n)H(t,n) dt.
− −
ih i
The main term in Theorem 1 comes from the first sum in (3.10) (the diagonal
terms), while the sum over the non-diagonal terms m = n will contribute to the
6
errorterm. Toseethisnotefirstthatthefunctionsu(t,n)( (n/t)1/4 inourrange)
≍
and
2
πn
exp G2 arsinh
− (cid:18) r2T (cid:19) !
are monotonic functions of t when t [T, 2T], and moreover,since
∈
∂f(t,n) πn
= 2arsinh ,
∂t 2t
r
it follows that, for U,V =0 or G,
∂[f(t+U,m) f(t+V,n)] √m √n
(3.11) ± | ± | (m=n).
∂t ≍ √T 6
10 SURNAME
In the sum overm=n we can assume,by symmetry, that n<m62n or m>2n.
6
Hence by the first derivative test (see e.g., Lemma 2.1 of [3]) and (3.11) we have
that the sum in question is
d(m)d(n) 1
G2
≪ (mn)1/4 · √m √n
16n<mX6TG−2L −
d(m)
G2 d(n) +
≪ m n
n6TXG−2L n<Xm62n −
d(m) d(n)
+ G2
m3/4 n1/4
m6XTG−2L nX<2m
G2Tε 1+ d(m)L T1+ε.
ε
≪ ≪
(cid:16)n6TXG−2L m6XTG−2L (cid:17)
Note that, by the mean value theorem,
u(t+G,n) u(t,G)=O(Gn1/4T 5/4) (t T,16n6TG 2L).
− −
− ≍
Hence we obtain, by trivial estimation,
(3.12)
2T d2(n) 2
S = 2G2 u2(t,n) H(t+G,n) H(t,n) dt
n −
ZT n6TXG−2L (cid:16) (cid:17)
2T d2(n)
+ O G2 Gn1/2T 3/2dt +O (T1+ε)
− ε
ZT n6TXG−2L n · !
2T d2(n) 2
= 2G2 u2(t,n) H(t+G,n) H(t,n) dt+O (T1+ε),
ε
n −
ZT n6TXG−2L (cid:16) (cid:17)
since G2 T. Now note that
≪
1/2
t − n
u2(t,n)= +O (T 6t62T),
2πn T
(cid:18) (cid:19) (cid:16) (cid:17)
hence the error term above makes a contribution to S which is
G2 d2(n) TL4.
≪ ≪
n6TXG−2L
Similarly, replacing t+G by t in the exponential factor in H(t+G,n), we make a
total error which is T1+ε. Therefore (3.8) and (3.12) give
ε
≪