1 Introduction and statement of the main results
In 1946, Selberg [18] proved a central limit theorem for the real part of the logarithm of the Riemann zeta function on the critical line, showing that the distribution of log|ζ(12+it)| is approximately Gaussian with mean 0 and variance 12loglogT, i.e.
(1.1)1Tmeas{t∈[T,2T]:log|ζ(12+it)|12loglogT≥V}∼∫V∞e-x22dx2π
for any fixed V∈ℝ, as T goes to infinity.
The author [9] studied the distribution of log|ζ(12+it)| with respect to the weighted measure |ζ(12+it)|2dt and, assuming the Riemann hypothesis (RH), proved that it is asymptotically Gaussian with mean loglogT and variance 12loglogT.
In this paper, we investigate the value distribution of log|ζ(12+it)| with respect to the measure
(1.2)|ζ(m)(12+it)|2kdt
for any fixed non-negative integers m,k.
The motivation is due to the study of the large values of the Riemann zeta function, i.e. the uniformity in V in the central limit theorem (1.1). In [19], Soundararajan speculates that an upper bound like
1Tmeas{t∈[T,2T]:log|ζ(12+it)|12loglogT≥V}≪1Vexp(-V22)
holds in a large range for V. In particular, we expect (see [17, Conjecture 2] and [11]) that for any fixed k,
(1.3)1Tmeas{t∈[T,2T]:log|ζ(12+it)|≥kloglogT}≪k1loglogT1(logT)k2.
Expressing the characteristic function of positive reals as a Mellin transform, the left-hand side of (1.3) can be written as
1(logT)k212π∫-∞+∞1T(logT)k2∫T2Teiu(log|ζ(12+it)|-kloglogT)|ζ(12+it)|2k𝑑tdu2k+iu
and is then related to the distribution of log|ζ(12+it)| with respect to the weighted measure (1.2), with m=0.
In the case k=1,2, we can prove a central limit theorem for log|ζ(12+it)| with respect to the measure |ζ(m)(12+it)|2kdt, assuming RH only.
Corollary 1.1.
Assume the Riemann hypothesis, let m be a non-negative integer and let k=1 or k=2. As t varies in T≤t≤2T, the distribution of log|ζ(12+it)| is asymptotically Gaussian with mean kloglogT and variance 12loglogT, with respect to the weighted measure |ζ(m)(12+it)|2kdt.
In the case k>2, since not even the moments of zeta are known, we cannot expect to prove unconditionally a central limit theorem. However,
assuming the asymptotic formula [13, Conjecture 7.1] suggested by the recipe [6] for the twisted and shifted (2k)-th moments of the Riemann zeta function
∫T2T(ab)-itζ(12+α1+it)⋯ζ(12+αk+it)ζ(12+β1-it)⋯ζ(12+βk-it)𝑑t,
one can deal with the general case too. More precisely, we use the strategy of [4, Theorem 1.2] in order to re-write the conjecture of Hughes and Young and we assume the following statement.
Conjecture 1.2 (Hughes–Young).
Let T be a large parameter, let α1,…,αk,β1,…,βk≪(logT)-1, let Φj be the set of subsets of {α1,…,αk} of cardinality j, for j=0,…,k, and similarly let Ψj be the set of subsets of {β1,…,βk} of cardinality j. If 𝒮∈Φj and 𝒯∈Ψj, then write 𝒮={αi1,…,αij} and 𝒯={βl1,…,βlj}, where i1<i2<…<ij and l1<l2<…<lj. Let (α𝒮;β𝒯) be the tuple obtained from (α1,…,αk;β1,…,βk) by replacing αlr with -βlr and replacing βir with -αir
for 1≤r≤j. Consider
A(s)=∑a≤Tθf(a)as and B(s)=∑b≤Tθg(b)bs,
where f(a)≪aε and g(b)≪bε for any ε>0. We then conjecture that there exists a δ>0, depending on k, such that if θ<δ, then
∫T2TA(12+it)B(12+it)¯ζ(12+it+α1)⋯ζ(12+it+αk)ζ(12-it+β1)⋯ζ(12-it+βk)𝑑t
equals
∑a,b≤Tθf(a)g(b)¯∫T2T∑0≤j≤k∑𝒮∈Φj𝒯∈ΨjZα𝒮;β𝒯,a,b(t)(t2π)-𝒮-𝒯dt+O(T1-η)
for some η>0, where we have written (t/2π)-𝒮-𝒯 for (t/2π)-∑x∈𝒮x-∑y∈𝒯y and
Zα;β,a,b(t):=∑am1⋯mk=bn1⋯nk1abm11/2+α1⋯mk1/2+αkn11/2+β1⋯nk1/2+βkV(m1⋯mkn1⋯nktk)
with
V(x):=12πi∫(1)G(s)s(2π)-ksx-s𝑑s,
where G(s) is an even entire function of rapid decay in any fixed strip |ℜ(s)|≤C satisfying G(0)=1.
With this assumption we can prove the main theorem.
Theorem 1.3.
Let k,m∈N and assume the Riemann hypothesis and Conjecture 1.2 for k. As t varies in T≤t≤2T, the distribution of log|ζ(12+it)| is asymptotically Gaussian with mean kloglogT and variance 12loglogT, with respect to the weighted measure |ζ(m)(12+it)|2kdt.
In particular, since Conjecture 1.2 is known in the cases k=1 (see [2, 5] and, e.g., [3] for the easy modifications needed to account for the shifts) and k=2 (see [13, 4]), we notice that Corollary 1.1 trivially follows from Theorem 1.3.
We notice that Theorem 1.3 shows that the m-th derivative has no effect in the weighted distribution of log|ζ(12+it)|. This is consistent with the conjecture (see [12, Conjecture 6.1] and [8])
∫T2T|ζ(12+it)|2k-2h|ζ′(12+it)|2h𝑑t∼c(h,k)T(logT)k2+2h,
which indicates that |ζ′(12+it)|2h amplifies the contribution coming from the large values in the same way as |ζ(12+it)|2h would do (up to a normalization of (logT)2h). More generally, as far as moments are concerned, the m-th derivative of zeta should behave like zeta itself (see, e.g., [7]), up to a normalization of logmT, in accordance with Theorem 1.3.
We also note that, while in Selberg’s classical case the mean is 0 because the contribution of the small values and that of the large values of zeta balance out, by tilting with |ζ(12+it)|2k, the mean of log|ζ(12+it)| moves to the right as k grows, and this reflects the fact that the measure |ζ(12+it)|2kdt gives more and more weight to the large values of the Riemann zeta function.
Moreover, we look at the shifted weighted measure |ζ(12+it+iα)|2kdt with a real number α such that |α|<1. As we will see, the distribution of log|ζ(12+it)| is quite sensitive to the parameter α. Indeed, in computing the integral
(1.4)∫T2Tlog|ζ(12+it)||ζ(12+it+iα)|2k𝑑t,
one expects the same magnitude as in the unshifted case if |α| is smaller than (logT)-1, which is the typical scale for the Riemann zeta function. On the other hand, if |α| is larger, then the two factors in the integral (1.4) start decorrelating, and thus the size of the integral decreases. This phenomenon is shown in the following result, in which we use the notation
μ~α={loglogT+O(1)if |α|logT≤1,-log|α|+O(1)if |α|logT>1,
for any α∈(-1,1).
Theorem 1.4.
Let k∈N and assume the Riemann hypothesis and Conjecture 1.2 for k. As t varies in T≤t≤2T, for any fixed and real α such that |α|<1, the distribution of log|ζ(12+it)| is asymptotically Gaussian with mean kμ~α and variance 12loglogT, with respect to the measure |ζ(12+it+iα)|2kdt.
This theorem shows that the shift has no effect if it is smaller than (logT)-1. On the contrary, for larger values of the shift the mean gets smaller; for instance, if
α=(logT)δlogT
with δ∈(0,1), then the mean is ∼(1-δ)kloglogT. In this shifted case too, if k≤2, then Theorem 1.4 holds assuming RH only.
Lastly, we show that in the random matrix theory setting, an analogous weighted central limit theorem can be proved unconditionally. We consider the characteristic polynomials
Z=Z(U,θ)=det(I-Ue-iθ)
of N×N unitary matrices U and we investigate their distribution of values with respect to the circular unitary ensemble (CUE). It has been conjectured that the limiting distribution of the non-trivial zeros of the Riemann zeta function, on the scale of their mean spacing, is the same as that of the eigenphases θn of matrices in the CUE in the limit as N→∞ (see, e.g., [14]). Then we consider a tilted version of the Haar measure and we have the following theorem.
Theorem 1.5.
As N→∞, the value distribution of log|Z| is asymptotically Gaussian with mean klogN and variance 12logN with respect to the measure |Z|2kdHaar .
As usual, the correspondence with Theorem 1.3 holds if we identify the mean density of the eigenangles θn, N/2π, with the mean density of the Riemann zeros at a height T up the critical line 12πlogT2π, i.e. if
N=logT2π.
2 Proof of Theorems 1.3 and 1.4
To prove both the theorems, we introduce a set of shifts α1,…,αk,β1,…,βk and we denote for the sake of brevity
ζα,β(t):=ζ(12+α1+it)⋯ζ(12+αk+it)ζ(12+β1-it)⋯ζ(12+βk-it).
The general strategy of the proof is similar to the one of [9, Theorem 1], but here we avoid the detailed combinatorial analysis we performed in [9], by working with Euler products instead of Dirichlet series, inspired by [1, Proposition 5.1]. The first step is then approximating the logarithm of the Riemann zeta function with a suitable Dirichlet polynomial. Let us set
P~(t):=∑p≤x1p1/2+it,
where x:=Tε, with ε:=(logloglogT)-1. Now we show that log|ζ(12+it)| has the same distribution as ℜP~(t) with respect to the measure ζα,β(t)dt. This is achieved in the following proposition by bounding the second moment of the difference.
Proposition 2.1.
For a non-negative integer k, assume Conjecture 1.2 and RH. Let T be a large parameter and let α1,…,αk, β1,…,βk∈C such that |αi|,|βj|<1, |ℜ(αi)|,|ℜ(βj)|≤(logT)-1 and |αi-βj|≪(logT)-1 for all 1≤i,j≤k. Then we have
∫T2T|log|ζ(12+it)|-ℜP~(t)|2ζα,β(t)𝑑t≪kT(logT)k2(logloglogT)5/2.
Proof.
From Tsang’s work [20, equation (5.15)], we know that
logζ(12+it)-P~(t)=S1+S2+S3+O(R)-L(t)
with
S1:=∑p≤x(p-1/2-4/logx-p-1/2)p-it,
S2:=∑pr≤xr≥2p-r(1/2+4/logx+it)r,
S3:=∑x<n≤x3Λ(n)lognn-1/2-4/logx-it,
R:=5logx(|∑n≤x3Λ(n)n1/2+4/logx+it|+logT),
L(t):=∑ρ∫1/21/2+4/logx(12+4logx-u)1u+it-ρ112+4logx-ρ𝑑u,
where the sum in the definition of L(t) is over all non-trivial zeros of ζ.
Then we have to bound the second moments of the terms on the right-hand side with respect to the weighted measure ζα,β(t)dt by using Conjecture 1.2 (note that we are allowed to apply the conjecture since the shifts are small up to a change of variable, being |αi-βj|≪(logT)-1).
Let us start with
∫T2T|S1|2ζα,β(t)𝑑t≪|∫T2T∑0≤j≤k∑𝒮∈Φj𝒯∈Ψj(t2π)-𝒮-𝒯∑p,q≤x(p-4/logx-1)(q-4/logx-1)Zα𝒮,β𝒯,p,q(t)dt|
≪∫T2T∑0≤j≤k∑𝒮∈Φj𝒯∈Ψj|∑a,b≤x𝟙x(a)𝟙x(b)(a-4logx-1)(b-4logx-1)Zα𝒮,β𝒯,a,b(t)|dt,
where 𝟙x(⋅) is the indicator function of primes up to x. If we denote by Ω(n) the function which counts the number of prime factors of n with multiplicity, then we have
∑n≤x𝟙x(n)f(n)=∑n:p|n⟹p≤x∂z[zΩ(n)f(n)]z=0.
Hence we get
(2.1)∫T2T|S1|2ζα,β(t)dt≪∫T2T∑0≤j≤k∑𝒮∈Φj𝒯∈Ψj|∂z∂w[12πi∫(1)G(s)s(t2π)ksIα,βj,𝒮,𝒯(z,w;s)ds]z=0w=0|dt,
where
(2.2)Iα,βj,𝒮,𝒯(z,w;s):=∑a,b:p|ab⟹p≤xzΩ(a)wΩ(b)gx(a)gx(b)Z~α𝒮,β𝒯,a,b(s)
with gx being the multiplicative function defined by gx(pα)=p-4α/logx-1 and
(2.3)Z~α,β,a,b(s):=∑am1⋯mk=bn1⋯nk1abm11/2+α1+s⋯mk1/2+αk+sn11/2+β1+s⋯nk1/2+βk+s.
We now analyze the first term Iα,β0(z,w;s) and then we will see how to apply the method to deal with all others.
Since the sum in the definition (2.2) is multiplicative, we have
Iα,β0(z,w;s)=∑am1⋯mk=bn1⋯nk:p|ab⟹p≤xzΩ(a)wΩ(b)gx(a)gx(b)abm112+α1+s⋯mk12+αk+sn112+β1+s⋯nk12+βk+s
=∏p≤x∑a+m1+…+mk=b+n1+…+nkzΩ(pa)wΩ(pb)gx(pa)gx(pb)pa2+b2+m1(12+α1+s)+…+mk(12+αk+s)+n1(12+β1+s)+…+nk(12+βk+s)
(2.4)⋅∏p>x∑m1+…+mk=n1+…+nk1pm1(12+α1+s)+…+mk(12+αk+s)+m1(12+β1+s)+…+nk(12+βk+s)
By putting in evidence the first terms in the Euler products, this equals
Aα,β(z,w;s)∏p≤x(1+zwgx(p)2p+zgx(p)p1+β1+s+…+wgx(p)p1+αk+s)
⋅∏p≤x(1+1p1+α1+β1+2s+…+1p1+αk+βk+2s)∏p>x(1+1p1+α1+β1+2s+…+1p1+αk+βk+2s)
(2.5)=Aα,β*(z,w;s)∏i,j=1kζ(1+αi+βj+2s)⋅exp(∑p≤x{zwgx(p)2p+zgx(p)p1+β1+s+…+wgx(p)p1+αk+s}),
where Aα,β(z,w;s) and Aα,β*(z,w;s) are arithmetical factors (Euler products) converging absolutely in a half-plane ℜ(s)>-δ for some δ>0, uniformly for |z|,|w|≤1, such that their derivatives with respect to z and w at 0 also converge in the same half-plane.
We now have extracted the polar part, and hence we are ready to shift the integral over s in (2.1) to the left of zero. To do so, it is convenient to prescribe the same conditions as in [4, remarks after Lemma 2.1], assuming that G(s) vanishes at s=-αi+βj2 for all i,j, so that the only pole we pick in the contour shift is at s=0. Moreover, we assume that the shifts are such that |αi+βj|≫(logT)-1 for every 1≤i,j≤k, so that
(2.6)∏i,j=1k|ζ(1+αi+βj)|≪k(logT)k2.
Hence by (2.4)–(2.6), we get
∂z∂w[12πi∫(1)G(s)s(t2π)ksIα,β0(z,w;s)ds]z=0w=0
≪k(logT)k2|∂z∂w[exp(∑p≤x{zwgx(p)2p+zgx(p)p1+β1+⋯+wgx(p)p1+αk})]z=0w=0|
≪k(logT)k2(∑p≤x|gx(p)|p)2
≪k(logT)k2(logloglogT)2,
where (see, e.g., [9, p. 5])
|ℜ(αi)|,|ℜ(β)|≤(logT)-1 and ∑p≤x|p-4/logx-1|p≪logloglogT.
This last bound can be obtained by using the Taylor approximation e-z=1+O(z) for z≪1, which yields
∑p≤x|p-4/logx-1|p=∑p≤x|p-4logp/logx-1|p≪1logx∑p≤xlogpp≪1
by Merten’s first theorem.
All the other terms Iα,βj,𝒮,𝒯(z,w) can be treated exactly in the same way as Iα,β0(z,w) by assuming that |αi±βj|≫(logT)-1 for every i,j, since they only differ from the first case by permutations and changes of signs of the shifts. Therefore, we get
∂z∂w[12πi∫(1)G(s)s(t2π)ksIα,βj,𝒮,𝒯(z,w;s)ds]z=0w=0≪k(logT)k2(logloglogT)2
provided that
(2.7)|αi±βj|≫(logT)-1 for every 1≤i,j≤k.
Plugging this into (2.1), we prove that
(2.8)∫T2T|S1|2ζα,β(t)𝑑t≪kT(logT)k2(logloglogT)2.
Moreover, since the left-hand side in (2.8) is holomorphic in terms of the shifts, although we have proved the above for αi,βj such that (2.7) holds, the maximum modulus principle can be applied to obtain the bound to the enlarged domain we need.
We will treat the other pieces similarly. As regards the second one, we have that
∫T2T|S2|2ζα,β(t)𝑑t≪∫T2T∑0≤j≤k∑𝒮∈Φj𝒯∈Ψj|∑p1r1,p2r2≤xr1,r2≥2p1-4r1logxr1p2-4r2logxr2Zα𝒮,β𝒯,p1r1,p2r2(t)|dt.
As before, we analyze the first term only since all others are completely analogous. The term for j=0 is
12πi∫(1)G(s)s(t2π)ks∑p1r1m1⋯mk=p2r2n1⋯nk:p1r1,p2r2≤x,r1,r2≥2p1-4r1logxp2-4r2logxr1r21p1r1p2r2m112+α1+s⋯nk12+βk+sds.
This time, because of the condition r1,r2≥2, when we estimate the sum via the first terms of its Euler product, we just get that the above is
≪k∏i,j=1k|ζ(1+αi+βj)|
Applying the same machinery as before, this yields
∫T2T|S2|2ζα,β(t)𝑑t≪kT(logT)k2.
We use the same approach in order to bound the second moment of S3 as well, which is
≪k∫T2T|12πi∫(1)G(s)s(t2π)ks∑am1⋯mk=bn1⋯nkx<a,b<x3Λ(a)Λ(b)logalogba-4/logxb-4/logxabm112+α1+s⋯nk12+βk+s|𝑑t
≪kT(∑x<p≤x31p)2∏i,j=1k|ζ(1+αi+βj)|
≪kT(logT)k2
since the sum ∑x<p≤x3p-1 is bounded.
We deal with R in the same way and we get that
∫T2T|R|2ζα,β(t)𝑑t≪kT(logT)k2(logloglogT)2,
where the extra factor with the triple log comes from the second term in the definition of R, while the first one can be treated analogously to S3.
Finally, we have to bound the second moment of ℜL(t). To do so, in view of equations [9, (2.8) and (2.9)], it suffices to study
1(logx)2∫T2T(log+1ηtlogx)2(|∑n≤x3Λ(n)n-4/logxn1/2+it|+logT)2ζα,β(t)𝑑t,
where ηt:=minρ|t-γ| and log+t:=max(logt,0), with the aim of proving that this is
≪kT(logT)k2(logloglogT)5/2.
By applying the Cauchy–Schwarz inequality, the above is
≤(∫T2T(1logx|∑n≤x3Λ(n)n-4/logxn1/2+it|+1ε)4|ζ(12+β1-it)|2⋯|ζ(12+βk-it)|2𝑑t)12
(2.9)⋅(∫T2T(log+1ηtlogx)4|ζ(12+α1+it)|2⋯|ζ(12+αk+it)|2𝑑t)12
and the first term can be treated as R above and it is
≪kT(logT)k2(logloglogT)4.
We now conclude the proof, bounding the second term in (2.9). If we denote
𝒯:=[T-1logx,2T+1logx],
then we have
∫T2T(log+1ηtlogx)4|ζ(12+α1+it)|2⋯|ζ(12+αk+it)|2𝑑t
≤∑γ∈𝒯∫-1/logx1/logx(log+1|w|logx)4∏j=1k|ζ(12+αj+i(w+γ))|2dw
=∑γ∈𝒯∫-11(log+1|t|)4∏j=1k|ζ(12+αj+i(γ+tlogx))|2dtlogx
(2.10)≪1logx∫-11(log|t|)4∏j=1k(∑γ∈𝒯|ζ(12+iγ+αj+itlogx)|2k)1/kdt
by the Hölder inequality. The remaining sum can be bounded under RH in view of Kirila’s [15, Theorem 1.2] and Milinovich’s [16] works, which generalize the well-known result due to Gonek about the sum over the non-trivial zeros of zeta [10, Corollary 1]. Indeed, since the shifts αj+itlogx in the sum over zeros in (2.10) have modulus ≤1 and real part ≤(logT)-1 in absolute value, then we have
∑γ∈𝒯|ζ(12+iγ+αj+itlogx)|2k≪kTlogT(logT)k2
for every j=1,…,k. Putting this into (2.10), we get
(2.11)∫T2T(log+1ηtlogx)4|ζ(12+α1+it)|2⋯|ζ(12+αk+it)|2𝑑t≪kTlogTlogx(logT)k2.
Plugging (2.11) into (2.9), we prove that
∫T2T|ℜL(t)|2ζα,β(t)𝑑t≪kT(logT)k2(logloglogT)4T(logT)k2logloglogT
≪kT(logT)k2(logloglogT)5/2,
concluding the proof of the proposition.
∎
The second step is getting rid of the small primes, showing that their contribution does not affect the distribution asymptotically. This simple fact will simplify the third and last step of the proof, as we will see in the following. Let us define
P(t):=∑p∈X1p1/2+it,
where X:=(logT,x] (we recall that x=Tε and ε=(logloglogT)-1).
Proposition 2.2.
For a non-negative integer k, assume Conjecture 1.2 and RH. Let T be a large parameter and let α1,…,αk, β1,…,βk∈C such that |αi|,|βj|<1, |ℜ(αi)|,|ℜ(βj)|≤(logT)-1 and |αi-βj|≪(logT)-1 for all 1≤i,j≤k. Then we have
∫T2T|ℜP~(t)-ℜP(t)|2ζα,β(t)𝑑t≪kT(logT)k2(logloglogT)2.
Proof.
This can be proved with the same method used in Proposition 2.1. We recall that 𝟙logT(⋅) denotes the indicator function of primes up to logT. We start by studying
(2.12)12πi∫(1)G(s)s(t2π)ks∑am1⋯mk=bn1⋯nk𝟙logT(a)𝟙logT(b)abm112+α1+s⋯nk12+βk+sds
and, as usual, we estimate the sum with the first terms of its Euler product, extract the polar part, shift the integral over s to the left, getting that (2.12) is
≪k∏i,j=1k|ζ(1+αi+βj)|⋅|∂z∂w[exp(∑p≤logT{zwp+zp1+β1+…+wp1+αk})]z=0w=0|.
By the same argument as in the proof of Proposition 2.1 the above is
≪k(logT)k2(∑p≤logT1p)2≪(logT)k2(logloglogT)2,
and this concludes the proof.
∎
Finally, we investigate the distribution of the polynomial ℜP(t), which, thanks to Propositions 2.1 and 2.2, has the same distribution as log|ζ(12+it)|. The most natural method to do so is studying the moments, and this is achieved in the following result.
Proposition 2.3.
For a non-negative integer k, assume Conjecture 1.2 and RH. Let T be a large parameter and let α1,…,αk, β1,…,βk∈C such that |αi|,|βj|<1, |ℜ(αi)|,|ℜ(βj)|≤(logT)-1 and |αi-βj|≪(logT)-1 for all 1≤i,j≤k. Set L:=∑p∈X1p∼loglogT and μ∈R such that μ≪loglogT. Then for every fixed integer n we have
∫T2T(ℜP(t)-kμ)nζα,β(t)𝑑t
=∫T2T∑0≤j≤k∑𝒮∈Φj𝒯∈Ψj(t2π)-𝒮-𝒯Mα𝒮,β𝒯(0)∂zn[ez24ℒ-kzμexp(z2∑p∈Xgp(𝒮,𝒯)p)]z=0dt+Ok,n(T(logT)k2-1+ε)
where
gp(𝒮,𝒯):=∑x1∉𝒮p-x1+∑x2∉𝒯p-x2+∑x3∈𝒮px3+∑x4∈𝒯px4
and
Mα,β(s):=∑m1⋯mk=n1⋯nk1m112+α1+s⋯mk12+αk+sn112+β1+s⋯nk12+βk+s,
so that Mα,β(0) is the first term of the moment of ζα,β, predicted by the recipe [6] (the sum which defines Mα,β(s) does not converge for s=0, so we appeal to [6, Theorem 2.4.1] for the analytic continuation).
Proof.
Expanding out the powers, since 2ℜ(z)=z+z¯, we get
(2.13)∫T2T(ℜP(t)-kμ)nζα,β(t)𝑑t=∑j+h=n(nh)(-kμ)j2-h∑r+s=h(hr)∫T2TP(t)rP(t)¯sζα,β(t)𝑑t.
By using Conjecture 1.2 and ignoring the error term which is negligible in this context, the inner integral in (2.13) is
(2.14)∫T2T∑0≤j≤k∑𝒮∈Φj𝒯∈Ψj(t2π)-𝒮-𝒯∑a,b𝟙X*r(a)𝟙X*s(b)Zα𝒮,β𝒯,a,b(t)dt,
where 𝟙X*r(a) denotes the indicator function of primes in the interval X, self-convoluted r times. If we define temporarily the multiplicative function g given by g(pn)=1/n!, then the inner sum over a,b in (2.14) equals
(2.15)∂zr∂ws[∑a,b:p|ab⟹p∈XzΩ(a)wΩ(b)g(a)g(b)Zα𝒮,β𝒯,a,b(t)]z=0w=0.
Then, plugging (2.14) and (2.15) into (2.13) and recollecting together the powers we expanded before, we get
(2.16)∫T2T12πi∫(1)G(s)s(t2π)ks∑0≤j≤k∑𝒮∈Φj𝒯∈Ψj(t2π)-𝒮-𝒯∂zn[e-kμzIα,βj,𝒮,𝒯(z;s)]z=0dsdt,
with
Iα,βj,𝒮,𝒯(z;s):=∑a,b:p|ab⟹p∈X(z2)Ω(a)+Ω(b)g(a)g(b)Z~α𝒮,β𝒯,a,b(s);
see (2.3) for the definition of Z~α,β,a,b(s). Now study the first term in (2.16), i.e. j=0, since all other terms can be understood from the first one with a slight modification. Then we look at
Iα,β0(z;s)=∑am1⋯mk=bn1⋯nkp|ab⟹p∈X(z/2)Ω(a)+Ω(b)g(a)g(b)abm112+α1+s⋯mk12+αk+sn112+β1+s⋯nk12+βk+s
=∏p∈X∑a+m1+…+mk=b+n1+…+nk(z/2)a+bg(pa)g(pb)pa2+b2+m1(12+α1+s)+…+nk(12+βk+s)∏p∉X∑m1+…+mk=n1+…+nk1pm1(12+α1+s)+…+nk(12+βk+s)
=exp(∑p∈X{(z/2)2p+z/2p1+α1+s+…+z/2p1+αk+s+z/2p1+β1+s+z/2p1+βk+s})FX,α,β(z;s)Mα,β(s),
where FX,α,β(z;s) is an arithmetical factor (Euler product) converging absolutely in a product of half-planes containing the origin, such that FX,α,β(z;0) is holomorphic at z=0, FX,α,β(0,0)=1 and all its derivatives at z=0 are small, i.e.
∂zc[FX,α,β(z,0)]z=0≪(logT)-1
for any positive integer c.
Now we want to shift the integral over s in (2.16)
to the left of zero, picking the contribution of the (only) pole at s=0.
To do so, we appeal to the meromorphic continuation of the function Mα,β(s); see [6, Theorem 2.4.1].
Thus we can shift the path of integration to the vertical line (say) ℜ(s)=-110, where the integral is trivially bounded by ≪T1-1/10+ε for any positive ε.
Moreover, the contribution from the pole at s=0 gives
T∂zn[e-kμzez24ℒexp(z2∑p∈Xp-α1+⋯+p-βkp)FX,α,β(z;0)]z=0Mα,β(0).
Thanks to the bounds for μ, ℒ and the derivatives of FX,α,β being FX,α,β(0;0)=1 and Mα,β(0)≪(logT)k2
(again this is due to an argument similar to the one in the proof of Proposition 2.1,
when we assume extra conditions on the shifts and then we appeal to the maximum modulus principle),
we get that the term for j=0 in (2.16) equals
T∂zn[e-kμzez24ℒexp(z2∑p∈Xp-α1+⋯+p-βkp)]z=0Mα,β(0)+Ok,n(T(logT)k2-1+ε).
Analogously, the general term turns out to be
∫T2T(t2π)-𝒮-𝒯∂zn[ez24ℒ-kμzexp(z2∑p∈Xgp(𝒮,𝒯)p)]z=0Mα𝒮,β𝒯(0)dt+Ok,n(T(logT)k2-1+ε),
and putting this into (2.14), i.e. summing over j,𝒮,𝒯, we get the thesis.
∎
Proof of Theorem 1.3.
This proof follows easily from the three propositions we have proved above. If we take μ=ℒ and all shifts small enough, i.e.
αi,βj≪(logT)-1 for any i,j,
then the exponent on the right-hand side of the thesis in Proposition 2.3 becomes
z24ℒ+z2∑p∈Xgp(𝒮,𝒯)p-zkℒ=z24ℒ+O(z∑p∈X|p-α1-1|+…+|p-βk-1|p)
=z24ℒ+Ok(z∑p∈X|p-α1-1|p)
=z24ℒ+Ok(z)
and does not depend on 𝒮 and 𝒯 asymptotically, in fact. Hence we can bring that factor outside and reconstruct the moment of ζα,β, as follows:
∫T2T(ℜP(t)-kℒ)nζα,β(t)𝑑t
=∂zn[ez24ℒ+Ok(z)]z=0∫T2T∑0≤j≤k∑𝒮∈Φj𝒯∈Ψj(t2π)-𝒮-𝒯Mα𝒮,β𝒯(0)dt+Ok,n(T(logT)k2-1+ε)
=(∂zn[ez24ℒ]z=0+Ok,n((loglogT)n-12))∫T2Tζα,β(t)dt+Ok,n(T(logT)k2-1+ε).
The thesis follows by analyzing
∂zn[ez24ℒ]z=0=[∑m1+2m2=nn!m1!m2!(2zℒ4)m1(12!ℒ2)m2]z=0.
If n is odd then the coefficient
∂zn[ez24ℒ]z=0
vanishes, while if n is even, then only the term for m1=0 and m2=n2 survives and gives
n!(n/2)!(ℒ4)n/2=(n-1)!!(ℒ2)n/2,
i.e.
1∫T2Tζα,β(t)𝑑t∫T2T(ℜP(t)-kℒ)nζα,β(t)𝑑t={(1+ok,n(1))(n-1)!!(ℒ2)n/2if n is even,ok,n((loglogT)n/2)if n is odd.
This then matched with the Gaussian coefficient proves that, in the limit T→∞, ℜP(t) has Gaussian distribution, with mean kloglogT and variance 12loglogT, and then so does log|ζ(12+it)|, in view of Propositions 2.1 and 2.2. Theorem 1.3 follows by taking the derivatives with respect to the shifts.
∎
Proof of Theorem 1.4.
To derive Theorem 1.4, in Proposition 2.3 we set
α1=…=αk=iα and β1=…=βk=-iα,
with α∈ℝ,|α|<1, and we take μ as
μα:=∑p∈Xcos(αlogp)p=μ~α+O(1)={loglogT+O(1)if |α|logT≤1,-log|α|+O(1)if |α|logT>1,
by partial summation. Then we get
∫T2T(ℜP(t)-kμα)n|ζ(12+iα+it)|2kdt=(1+ok,n(1))∂zn[ez24ℒ]z=0∫T2T|ζ(12+iα+it)|2kdt
since the quantity
z2∑p∈Xgp(𝒮,𝒯)p-zkμα
vanishes for all 𝒮,𝒯 for this choice of the shifts. The thesis follows as in the proof of Theorem 1.3.
∎
3 Proof of Theorem 1.5
In the usual notations we set in the introduction, let us define the moment generating function
(3.1)MN(s)=〈|Z|s〉=∑j=0∞〈(log|Z|)j〉j!sj,
where the mean has to be considered over the group U(N) with respect to the Haar measure, and the cumulants Qj=Qj(N) by
logMN(s)=∑j=1∞Qjj!sj.
In [14], among other things, Keating and Snaith studied the cumulants showing that
Qn={0if n=1,12logN+O(1)if n=2,O(1)if n≥3,
and deduced a central limit theorem proving that the limiting distribution of log|Z| is Gaussian with mean 0 and variance 12logN. Here, for any k∈ℕ, we study the distribution of the random variable log|Z| with respect to the tilted measure |Z|2kdHaar. Before starting with our analysis, we recall that the moments of |Z| are known also for a non-negative integer k (see [14, (6) and (16)]):
(3.2)MN(2k)=〈|Z|2k〉=exp(∑j=1∞(2k)jj!Qj)=∏j=1NΓ(j)Γ(j+2k)Γ(j+k)2∼Nk2G2(1+k)G(1+2k),
where k∈ℝ and G denotes the Barnes G-function. We set
ℳ2k:=∏j=1NΓ(j)Γ(j+2k)Γ(j+k)2.
Now we are ready to consider the first moment
〈|Z|2klog|Z|〉=ddx[〈|Z|2k+x〉]x=0=ddx[∏j=1NΓ(j)Γ(j+2k+x)Γ(j+k+x2)2]x=0
by (3.1) and (3.2). We compute the derivative by Leibniz’s rule, writing
∏j=1NΓ(j)Γ(j+2k+x)Γ(j+k+x2)2=exp(∑j=1N{logΓ(j)+logΓ(j+2k+x)-2logΓ(j+k+x2)}),
and we get
(3.3)〈|Z|2klog|Z|〉=∏j=1NΓ(j)Γ(j+2k)Γ(j+k)2∑j=1N{Γ′Γ(j+2k)-Γ′Γ(j+k)}.
Moreover, an application of Stirling’s formula yields
(3.4)Γ′Γ(j+2k)-Γ′Γ(j+k)=kj+Ok(1j2).
Hence, by (3.3) and (3.4), the weighted mean of the random variable log|Z| is
μ2k:=1ℳ2k〈|Z|2klog|Z|〉=klogN+Ok(1).
Then we study the weighted n-th moment of the random variable log|Z|:
〈|Z|2k(log|Z|-μ2k)n〉=∑h+j=n(nh)(-μ2k)j〈|Z|2k(log|Z|)h〉
=∑h+j=n(nh)djdxj[e-xμ2k]x=0dhdxh[∏j=1NΓ(j)Γ(j+2k+x)Γ(j+k+x2)2]x=0
(3.5)=dndxn[exp(-xμ2k+∑j=1Nlog(Γ(j)Γ(j+2k+x)Γ(j+k+x2)2))]x=0.
If we set
fj(x)=fN,k,j(x):=logΓ(j)+logΓ(j+2k+x)-2logΓ(j+k+x2),
then we can carry on the computation in (3.5) by computing the derivative, getting
dndxn[exp(-xμ2k+∑j=1Nlog(Γ(j)Γ(j+2k+x)Γ(j+k+x2)2))]x=0
=∑m1,…,mnn!m1!⋯mn!e∑j=1Nfj(0)∏i=1n(1i!didxi[-xμ2k+∑j=1Nfj(x)]x=0)mi
(3.6)=ℳ2k∑m1,…,mnn!m1!⋯mn!(-μ2k+∑j=1Nfj′(0))m1(12∑j=1Nfj′′(0))m2∏i=3n(1i!∑j=1Nfj(i)(0))mi,
where the sums in (3.6) are over the n-tuple (m1,…,mn) such that
m1+2m2+⋯+nmn=n.
Using Stirling’s approximation formula, one can easily estimate the derivatives of fj(x) and prove
∑j=1Nfj′(0)=∑j=1N{kj+Ok(j-2)}=klogN+Ok(1),
∑j=1Nfj′′(0)=∑j=1N{1j+2k-1/2j+k+O(j-2)}=12logN+Ok(1),
(3.7)∑j=1Nfj(i)(0)=∑j=1NO(j-2)=O(1) for all i≥3.
Putting together (3.5), (3.6) and (3.7), one has
〈|Z|2k(log|Z|-μ2k)n〉=ℳ2k∑m1+2m2+⋯+nmn=nn!m1!⋯mn!(14logN+Ok(1))m2(Ok(1))m1+m3+⋯+mn.
Then, if n is even, the asymptotic is given by m2=n2 and mi=0 for i≠2, giving
〈|Z|2k(log|Z|-μ2k)n〉∼k,nℳ2kn!(n/2)!(14logN)n/2=ℳ2k(n-1)!!(12logN)n/2,
while if n is odd, then the n-th moment is surely ok,n(ℳ2k(logN)n/2).
