PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 83(97) (2008), 71–86 DOI: 10.2298/PIM0897071Y

ON THE MEAN SQUARE OF THE RIEMANN ZETA FUNCTION AND THE DIVISOR PROBLEM

Yifan Yang

Communicated by Aleksandar Ivić

Abstract. Let Δ(T) and E(T) be the error terms in the classical Dirichlet divisor problem and in the asymptotic formula for the mean square of the Riemann zeta function in the critical strip, respectively. We show that Δ(T) and E(T) are asymptotic integral transforms of each other. We then use this integral representation of Δ(T) to give a new proof of a result of M. Jutila.

1. Introduction and statement of results

Let ζ(s) be the Riemann zeta function, and let d(n) denote the number of positive divisors of n. The error terms Δ(T) and E(T) in the classical Dirichlet divisor problem and in the asymptotic formula for the mean square of ζ(s) on the critical line Re s = 1/2 are defined by

Δ(T) = ∑_{k<T} d(k) + \frac{1}{2}d(T) − T log T − (2γ − 1)T − \frac{1}{4}

with the convention that d(T) = 0 if T is not an integer and

E(T) = ∫_{0}^{T} |ζ(1/2 + it)|^{2} dt − T log \frac{T}{2π} − (2γ − 1)T,

respectively. The properties of Δ(T) and E(T) have been the subject of numerous papers. (For example, [2], [3], [4], [5], [7], [8], [9], [10], [13], [14], [16], [17], [18] and [19]. For a general overview of the subject see the book [6] or the survey article [15].)

Two of the most frequently used tools in the study of Δ(T) and E(T) are the following two remarkable formulas due to Voronoi [20] and Atkinson [1], respectively.

2000 Mathematics Subject Classification: Primary 11M06, 11N37; Secondary 11L07.

Lemma 1.1 (Voronoi). We have

Δ(T) = \frac{T^{1/4}}{π\sqrt{2}} ∑_{k⩽K} \frac{d(k)}{k^{3/4}} cos\{4π(kT)^{1/2} − π/4\} + O(\frac{T^{1/2+ϵ}}{K^{1/2}} ) + O(T^{ϵ})

for any K > 0.

Lemma 1.2 (Atkinson). Let 0 < A < A′ be constants, and suppose that AT ⩽ K ⩽ A′T. Put

(1.1)  K′ = K′(T) = \frac{T}{2π} + \frac{K}{2} − \sqrt{\frac{KT}{2π} + \frac{K^{2}}{4}} .

Then E(T) = Σ_{1} + Σ_{2} + O((log T)^{2}), where

Σ_{1} = \frac{1}{\sqrt{2}} ∑_{k⩽K}(−1)^{k}d(k)(\frac{kT}{2π} + \frac{k^{2}}{4} )^{−1/4}\{sinh^{−1}(\frac{πk}{2T} )^{1/2}\}^{−1} cos 2πθ_{k}(T/(2π))

with

θ_{k}(T) = 2T sinh^{−1} \frac{\sqrt{k}}{2\sqrt{T}} + \sqrt{kT + \frac{k^{2}}{4}} − \frac{1}{8},

and

Σ_{2} = 2 ∑_{k⩽K′} \frac{d(k)}{\sqrt{k}}(log \frac{T}{2πk} )^{−1} sin ρ_{k}(T) with ρ_{k}(T) = T log \frac{T}{2πk} − T − \frac{π}{4} .

We note that the contribution to E(T) is mainly from the first sum Σ_{1}. For instance, on the Lindelöf Hypothesis, the second sum Σ_{2} can be shown to be bounded by T^{ϵ}. In applications we can usually employ averaging techniques to show that the contribution from Σ_{2} is less significant than that from Σ_{1}. Thus, to study the properties of E(T), one would generally focus on Σ_{1}.

Using the Taylor expansions

2πθ_{k}(T/(2π)) = 4π(\frac{kT}{2π} )^{1/2} − \frac{π}{4} + O(k^{3/2}T^{−1/2})

\frac{1}{\sqrt{2}}(\frac{kT}{2π} + \frac{k^{2}}{4} )^{−1/4} \{sinh^{−1}(\frac{πk}{2T} )^{1/2}\}^{−1} = \sqrt{2}(\frac{T}{2π} )^{1/4}k^{−3/4} + O(T^{−3/4}k^{1/4})

it can be seen that, aside from the alternating factor (−1)^{k}, the first o(T^{1/3}) terms in Σ_{1} are asymptotically equal to the corresponding terms in Voronoi’s formula for 2πΔ(T/(2π)). This analogy between Δ(T) and E(T) motivates the work of Jutila [8], [9] and [13]. Jutila introduced a new function

Δ^{∗}(T) = −Δ(T) + 2Δ(2T) − \frac{1}{2}Δ(4T).

This function can be interpreted as the error term in the approximation of the summatory function of a certain arithmetic function, and there is a formula analogous to Voronoi’s formula for Δ^{∗}(T), namely,

Δ^{∗}(T) = \frac{T^{1/4}}{π\sqrt{2}} ∑_{k⩽K}(−1)^{k} \frac{d(k)}{k^{3/4}} cos(4π\sqrt{kT} − π/4) + O(T^{1/2+ϵ}K^{−1/2} + T^{ϵ}).

Since this formula also contains the alternating factor (−1)^{k}, the magnitude of the function Δ^{∗}(T) is more comparable to that of E(T) than that of Δ(T). In fact, Jutila [9] showed that

∫_{T}^{T +H}(E(u) − 2πΔ^{∗}(u/(2π)))^{2} du ≪ HT^{1/3+ϵ} + T^{1+ϵ}

for 2 ⩽ H ⩽ T, while the corresponding integrals for E(T)^{2} and Δ(u)^{2} (and hence Δ^{∗}(u)^{2}) are known to be bounded by HT^{1/2+ϵ}+T^{1+ϵ}. Using this similarity between 2πΔ^{∗}(T/(2π)) and E(T), Jutila [8] further proved that the truth of the conjecture Δ(T) ≪ T^{1/4+ϵ} implies the bound E(T) ≪ T^{5/16+ϵ}, and later [13] improved this conditional bound to T^{3/10+ϵ}.

The main purpose of the present paper is to provide a different perspective on the connection between Δ(T) and E(T). We will show that these two functions are in fact asymptotic integral transforms of each other.

Theorem 1.1. Define two functions f(u) and g(u) by

f(u) = f_{T} (u) = \frac{log(u/T)}{\sqrt{T} (u/T − 1)} exp\{−2πi(u log \frac{u}{T} − u + T − \frac{1}{8})\}
g(u) = g_{T} (u) = \frac{T/u − 1}{\sqrt{u} log(T/u)} exp\{2πi(T log \frac{T}{u} − T + u − \frac{1}{8})\} .

Let ϵ > 0, 0 < A < 1 and B > 0 be constants, and put B′ = 1+\sqrt{B + B^{2}/4}+B/2. For T − AT ⩽ u ⩽ T + AT let E_{1}(2πu) denote the main sum in Atkinson’s formula

E_{1}(2πu) = ∑_{k⩽BT} \frac{d(k) cos \{2πθ_{k}(u)\}}{a_{k}(u)},

where

θ_{k}(u) = 2u sinh^{−1} \frac{\sqrt{k}}{2\sqrt{u}} + \sqrt{ku + \frac{k^{2}}{4}} − \frac{k}{2} − \frac{1}{8},
a_{k}(u) = (4ku + k^{2})^{1/4} sinh^{−1} \frac{\sqrt{k}}{2\sqrt{u}}.

Set T_{1} = T − AT, T_{2} = T + AT, T_{3} = T/B′, T_{4} = B′T. Then we have

Δ(T) = \frac{1}{2π} ∫_{T_{1}}^{T_{2}} E_{1}(2πu)f(u) du + O(T^{ϵ}),

E_{1}(2πT) = 2π ∫_{T_{3}}^{T_{4}} Δ(u)g(u) du + O(T^{ϵ}),

where the O-constants depend only on ϵ, A and B.

The underlying idea of our approach evolves from the fact that the function χ(1 − s) is in fact the Mellin transform of 2 cos(2πx), where

χ(s) = \frac{(2π)^{s}}{2Γ(s) cos(πs/2)} (= 2^{s}π^{s−1} sin \frac{πs}{2}Γ(1 − s))

is the function in the functional equation ζ(s) = χ(s)ζ(1 − s) for ζ(s). Thus, by the Mellin inversion formula, we have

\frac{1}{2πi} ∫_{σ−i∞}^{σ+i∞} χ(1 − s)x^{−s} ds = 2 cos(2πx)

whenever the integral converges absolutely. Moreover, for the truncated integral

\frac{1}{2πi} ∫_{σ}^{σ+2πiT} χ(1 − s)x^{−s} ds = ∫_{0}^{T} χ(1 − σ − 2πit)x^{−σ−2πit} dt,

using the asymptotic expansion

χ(1 − σ − 2πit) = t^{σ−1/2} exp{2πi(t log t − t − 1/8)} (1 + O ((1 + |t|)^{−1}))

and the stationary phase method (see Lemma 2.1 below), we see that

∫_{0}^{T} χ(1 − σ − 2πit)x^{−σ−2πit} dt = exp\{−2πix\} × \begin{cases}1 + A_{σ,T} (x), & if x ⩽ T, \\ A_{σ,T} (x), & if x > T, \end{cases}

where A_{σ,T} (x) is differentiable for x ≠ T and satisfies

lim_{x→T −} A_{σ,T} (x) = −\frac{1}{2} + O_{σ}(1/\sqrt{T} ),

lim_{x→T +} A_{σ,T} (x) = \frac{1}{2} + O_{σ}(1/\sqrt{T} ),

A_{σ,T} (x) ≪_{σ} (\frac{T}{x} )^{σ} min (1, \frac{1}{\sqrt{T} | log(T/x)|}) .

In particular, if we write ζ(s)^{2} as

ζ(s)^{2} = ∑_{k⩽2T} d(k)k^{−s} + B(s, T),

then we have

\begin{aligned}∫_{0}^{T} |ζ(1/2 + 2πit)|^{2} dt & = ∫_{0}^{T} ζ(1/2 + 2πit)^{2}χ(1/2 − 2πit) dt \\ & = ∑_{k⩽T} d(k) + ∑_{k⩽2T} d(k)A_{1/2,T} (k) + C_{1}(T), \end{aligned}

where C_{1}(T) may be thought of as a secondary error term. Thus, integrating by parts on the second sum yields

\begin{aligned}∫_{0}^{T} |ζ(1/2 + 2πit)|^{2} dt = T (log T + 2γ − 1) & + ∫_{0}^{T −} Δ(u)A′_{1/2,T} (u) du \\ & + ∫_{T +}^{2T} Δ(u)A′_{1/2,T} (u) du + C_{2}(T). \end{aligned}

This shows that E(2πT) is representable asymptotically as an integral transform of Δ(u). Conversely, we can express Δ(T) asymptotically in terms of the “inverse” integral transform of E(2πu), and hence to study the properties of Δ(T) we may employ this integral representation of Δ(T), instead of the usual Voronoi’s formula. As an illustration we will give a new proof of a result of Jutila [10].

Theorem 1.2 (Jutila). Suppose that HU ≫ T^{1+ϵ} and U ⩽ \sqrt{T} /2. We have

∫_{T}^{T +H}(Δ(u + U) − Δ(u))^{2} du ≪ HU log^{3} \frac{\sqrt{T}}{U} .

The problem of estimating integrals of (Δ(u + U) − Δ(u))^{2} over an interval is closely related to that of sign changes of Δ(u) (see [4]).

There are other possible applications of our main result. For instance, we may use our integral representation of Δ(T) to show that ∫_{T}^{T +H} Δ(u)^{4} du ≪ T^{ϵ}(HT + H^{1/5}T^{8/5}) holds for all H ⩽ T. However, this result is inferior to that obtainable by the method of Ivić [5]. It seems to us that in order to achieve a stronger result, properties that are specifically related to d(n), or equivalently, to the Riemann zeta function, must be utilized. Another natural question to ask is whether our result will yield a good bound for |E(2πT) − 2πΔ(T)|, or a result that connects a bound for Δ(T) with that for E(T). We are unable to give an affirmative answer at present either.

As usual, the notations f(x) ≪ g(x) and f(x) = O(g(x)) mean that there is a positive constant c such that |f(x)| ⩽ c|g(x)| for x in the range under consideration. When lim_{x→a} f(x)/g(x) = 0, we use the notation f(x) = o(g(x)). The letter ϵ will always denote a small, but fixed positive number, though the number may not be the same at each occurrence. For example, we may write T^{ϵ} log T ≪ T^{ϵ}.

Acknowledgments. The author wishes to thank Prof. A. Hildebrand of the University of Illinois and Prof. K.-M. Tsang of the University of Hong Kong for providing valuable comments and suggestions. The author would also like to thank Professor A. Ivić for his interest in the work.

2. Proof of Theorem 1.1

We first quote an analytic lemma regarding exponential integrals. The first part of the lemma is due to Atkinson, and the second part is due to Jutila [12].

Lemma 2.1. Let μ(x) be a positive differentiable function in the interval [a, b]. Suppose that f(z) and g(z) are functions satisfying the following conditions:

(1) the function f(x) is real and f′′(x) > 0 for x ∈ [a, b];

(2) f(z) and g(z) are analytic for all z in the domain

⋃_{x∈[a,b]} {z : |z − x| ⩽ μ(x)};

(3) there exist positive functions F (x) and G(x) such that for x ∈ [a, b] and |z − x| ⩽ μ(x) we have

F (x) ≫ 1, |g(z)| ≪ G(x),
|f′(z)| ≪ F (x)μ(x)^{−1}, f′′(x) ≫ F (x)μ(x)^{−2};

(4) μ′(x) ≪ 1.

Let H_{J}(x) denote the function H_{J}(x) = G(x) (|f′(x)| + |f′′(x)|^{1/2})^{−J−1} . Assume that f′(x) has a zero c in the interval [a, b]. We then have

∫_{a}^{b} g(x) exp\{2πif(x)\} dx = \frac{e^{πi/4}g(c)}{\sqrt{|f′′(c)|}} exp\{2πif(c)\}

+ O(∫_{a}^{b} G(x) exp{−CF (x)} dx) + O(G(c)μ(c)F (c)^{−3/2})
+ O(H_{0}(a)) + O(H_{0}(b)),

where C is a positive number determined by the O-constants in condition (3).

Furthermore, if U is a positive number and J is a positive integer such that JU < (b − a)/2, a + JU < c < b − JU and U ≫ μ(c)F (c)^{−1/2}, we have

U^{−J} ∫_{0}^{U} du_{1} ⋯ ∫_{0}^{U} du_{J} ∫_{a+u_{1}+···+u_{J}}^{b−u_{1}−···−u_{J}} g(x) exp{2πif(x)} dx

= \frac{e^{πi/4}g(c)}{\sqrt{|f′′(c)|}} exp\{2πif(c)\} + O( ∫_{a}^{b}(1 + (\frac{μ(x)}{U})^{J})G(x) exp\{−CF (x)\} dx)
+ O(G(c)μ(c)F (c)^{−3/2}) + O(H_{J}(a)) + O(H_{J}(b)).

In the case when f′(x) does not vanish in [a, b], the above estimates hold without the terms involved with c. Moreover, if the condition f′′(x) > 0 is replaced by f′′(x) < 0, then the factor e^{πi/4} in the main terms is replaced by e^{−πi/4}.

The next lemma constitutes the essential part of the proof of Theorem 1.1.

Lemma 2.2. (i) Let A < 1 be a positive constant, and T and K be positive numbers with K ⩽ AT. Let T_{1} denote T − \sqrt{KT}, and T_{2} denote T + \sqrt{KT}. Set

θ(u) = θ_{k}(u) = 2u sinh^{−1} \frac{\sqrt{k}}{2\sqrt{u}} + \sqrt{ku + k^{2}/4} − \frac{k}{2} − \frac{1}{8},

f(u) = f_{T} (u) = \frac{log(u/T)}{\sqrt{T} (u/T − 1)} exp\{−2πi(u log \frac{u}{T} − u + T − \frac{1}{8})\} ,

and a(u) = a_{k}(u) = (4ku + k^{2})^{1/4} sinh^{−1} \frac{\sqrt{k}}{2\sqrt{u}}. Then we have, for 0 < k ⩽ K,

∫_{T_{1}}^{T_{2}} \frac{cos\{2πθ(u)\}}{a(u)}f(u) du = \frac{\sqrt{2}T^{1/4}}{k^{3/4}} cos \{4π(kT)^{1/2} − π/4\} + O(δ(k, T)),

where

δ(k, T) = \frac{1}{T^{1/4}k^{3/4}} (1 + min (\sqrt{T} , \frac{T}{|\sqrt{KT} − \sqrt{kT} |})) .

If k > K, then the estimate holds without the leading term.

(ii) Conversely, let K be a positive number, and set

T_{3} = T − \sqrt{KT + K^{2}/4} + K/2, T_{4} = T + \sqrt{KT + K^{2}/4} + K/2, and

g(u) = g_{T} (u) = \frac{T/u − 1}{\sqrt{u} log(T/u)} exp\{2πi(T log \frac{T}{u} − T + u − \frac{1}{8})\} .

Then we have, for k ⩽ K,

\sqrt{2} ∫_{T_{3}}^{T_{4}} \frac{u^{1/4}}{k^{3/4}} cos \{4π(ku)^{1/2} − π/4\} g(u) du = \frac{cos \{2πθ(T)\}}{a(T)} + O(η(k, T)),

where

\begin{aligned}η(k, T) = \frac{1}{T^{1/4}k^{3/4}} (1 & + min(\sqrt{T} , \frac{T}{|T_{3} − T + \sqrt{kT + k^{2}/4} − k/2|}) \\ & + min(\sqrt{T} , \frac{T}{|T_{4} − T − \sqrt{kT + k^{2}/4} − k/2|})). \end{aligned}

When k > K, the estimate holds without the main term.

Proof. To prove the first part of the lemma, we first write the cosine function as a sum of two exponentials, and then evaluate two branches separately. Let I_{k} denote the integral

I_{k} = \frac{1}{2} ∫_{T_{1}}^{T_{2}} \frac{exp\{2πiθ(u)\}}{a(u)}f(u) du,

and set h(u) = θ(u) − (u log(u/T) − u + T − 1/8). Let u_{k} be the solution of the equation h′(u) = 0. Since

(2.1)  h′(u) = 2 sinh^{−1} \frac{\sqrt{k}}{2\sqrt{u}} − log \frac{u}{T} ,

we have

\sqrt{\frac{T}{u_{k}}} = \sqrt{1 + \frac{k}{4u_{k}}} − \frac{\sqrt{k}}{2\sqrt{u}_{k}} ,

and thus

(2.2)  u_{k} = T + \sqrt{kT} .

Note that when k ⩽ K, the stationary point u_{k} lies in the interval [T_{1}, T_{2}]. We now apply the first part of Lemma 2.1 with

f(u) = h(u), g(u) = \frac{log(u/T)}{a(u)\sqrt{T} (u/T − 1)},

μ(u) = T (1 − \sqrt{A})/2,  F (u) = T,  G(u) = 1/T^{1/4}k^{3/4},
a = T_{1},  b = T_{2},  c = u_{k}.

Since f′′(u) is of constant sign and |f′′(u)| ≫ 1/T, we have

|f′(b)| = |f′(b) − f′(c)| ≫ |b − c|/T = |\sqrt{KT} − \sqrt{kT} |/T,

and the same lower bound holds for |f′(a)|. Thus, Lemma 2.1 yields

78 YANG

(2.3)  I_{k} = \begin{cases}\begin{matrix}\frac{e^{−πi/4}}{2\sqrt{T |h′′(u_{k})|}} & \frac{log(u_{k}/T)}{u_{k}/T − 1} & \frac{exp\{2πih(u_{k})\}}{a(u_{k})} \\ + O(δ(k, T)) & , \end{matrix} & if 0 < k ⩽ K, \\ O(δ(k, T)), & if k > K, \end{cases}

since

(2.4)  h′′(u_{k}) = \frac{2}{\sqrt{1 + k/(4u_{k})}}(−\frac{\sqrt{k}}{4u_{k}^{3/2}} ) − \frac{1}{u_{k}} < 0.

We now show that the main term in (2.3) is actually equal to

\frac{T^{1/4}}{\sqrt{2}k^{3/4}} exp\{4πi(kT)^{1/2} − πi/4\}.

By (2.1) and the definition of u_{k}, we have

(2.5)  sinh^{−1} \frac{\sqrt{k}}{2\sqrt{u_{k}}} = log \sqrt{u_{k}/T} .

It follows that, by (2.2),

\begin{aligned}h(u_{k}) & = \sqrt{ku_{k} + k^{2}/4} + (u_{k} − T) − k/2 \\ & = \sqrt{kT + k^{3/2}T^{1/2} + k^{2}/4} + \sqrt{kT} − k/2 = 2\sqrt{kT} . \end{aligned}

Moreover, from (2.2) we have

\frac{u_{k}}{T} − 1 = \frac{1}{T} \sqrt{kT} = \sqrt{k/T} ,

(4ku_{k} + k^{2})^{1/4} = (4kT + 4k^{3/2}T^{1/2} + k^{2})^{1/4} = (2\sqrt{kT} + k)^{1/2},

and hence, by (2.5), a(u_{k}) = \frac{1}{2} log(u_{k}/T)(2\sqrt{kT} + k)^{1/2}. By (2.4), we have

h′′(u_{k}) = −\frac{1}{u_{k}} (\begin{matrix}√ & \overline{k} \\ √ & \overline{4u_{k} + k} \end{matrix} + 1)

= −\frac{1}{T + \sqrt{kT}} (\begin{matrix}√ & \overline{k} \\ √ & \overline{4T + 4(kT)^{1/2} + k} \end{matrix} + 1)

= −\frac{1}{T + \sqrt{kT}} (\frac{2\sqrt{k} + 2\sqrt{T}}{\sqrt{k} + 2\sqrt{T}} ) = −\frac{2}{\sqrt{kT} + 2T} .

Inserting these expressions into (2.3), we obtain, for k ⩽ K,

\begin{aligned}I_{k} & = \frac{(\sqrt{kT} + 2T )^{1/2}}{2^{3/2}\sqrt{T}} \begin{matrix}√ & \overline{T} \\ √ & \overline{k} \end{matrix} \frac{2}{(2\sqrt{kT} + k)^{1/2}} exp\{4πi(kT)^{1/2} − πi/4\} + O(δ(k, T)) \\ & = \frac{T^{1/4}}{\sqrt{2}k^{3/4}} exp\{4πi(kT)^{1/2} − πi/4\} + O(δ(k, T)). \end{aligned}

For the other integral

\frac{1}{2} ∫_{T_{1}}^{T_{2}} \frac{exp\{−2πiθ(u)\}}{a(u)}f(u) du,

we can show that the function −θ(u) − (u log(u/T) − u + T − 1/8) has a root at u = T − \sqrt{kT} , and Lemma 2.1 yields

\frac{1}{2} ∫_{T_{1}}^{T_{2}} \frac{exp\{−2πiθ(u)\}}{a(u)}f(u) du = \frac{T^{1/4}}{\sqrt{2}k^{3/4}} exp\{−4πi(kT)^{1/2} + πi/4\} + O(δ(k, T))

for 0 < k ⩽ K and

∫_{T_{1}}^{T_{2}} \frac{exp\{−2πiθ(u)\}}{a(u)}f(u) du ≪ δ(k, T)

for k > K. The first part of the lemma follows by combining these estimates with (2.3).

The proof of the second part is analogous, and the calculation is essentially the same as that in the proof of Theorem 7.2 of [6] and that in the proof of Theorem 1 of [11]. For completeness we sketch the proof as follows. We consider the integral

J_{k} = \frac{1}{\sqrt{2}} ∫_{T_{1}}^{T_{2}} \frac{u^{1/4}}{k^{3/4}} exp \{4πi(ku)^{1/2} − πi/4\} g(u) du.

Let h(u) denote h(u) = 2(ku)^{1/2} + T log(T/u) − T + u − 1/4. We have

h′(u) = \frac{(ku)^{1/2}}{u} − \frac{T}{u} + 1
(2.6)  h′′(u) = −\frac{\sqrt{k}}{2u^{3/2}} + \frac{T}{u^{2}} .

Thus, if u_{k} is the real root of the equation h′(u) = 0, then we have

(2.7)  \begin{matrix}√ & \overline{k} \\ √ & \overline{T} \end{matrix} = \sqrt{\frac{T}{u_{k}}} − \sqrt{\frac{u_{k}}{T}} .

It follows that

(2.8)  log \sqrt{\frac{T}{u_{k}}} = sinh^{−1} \frac{\sqrt{k}}{2\sqrt{T}} ,

(2.9)  u_{k} = T − \sqrt{kT + k^{2}/4} + \frac{k}{2} ,

and h′′(u_{k}) > 0. By Lemma 2.1, if k ⩽ K, then we have

J_{k} = \frac{1}{\sqrt{2u_{k}}} \frac{e^{πi/4}}{\sqrt{|h′′(u_{k})|}} \frac{u_{k}^{1/4}}{k^{3/4}} \frac{T/u_{k} − 1}{log(T/u_{k})} exp\{2πih(u_{k})\} + O(η(k, T)).

In light of (2.7), (2.8) and (2.9) we see that

\frac{T/u_{k} − 1}{u_{k}^{1/4}k^{3/4} log(T/u_{k})} = \frac{1}{2u_{k}^{3/4}k^{1/4} sinh^{−1}(\sqrt{k}/(2\sqrt{T}))}

and

h(u_{k}) = 2(ku_{k})^{1/2} + (T log(T/u_{k}) − T + u_{k}) − 1/4

= (ku_{k})^{1/2} + 2T sinh^{−1} \frac{\sqrt{k}}{2\sqrt{T}}

= \sqrt{k}(−\sqrt{k}/2 + \sqrt{T + k/4}) + 2T sinh^{−1} \frac{\sqrt{k}}{2\sqrt{T}} − 1/4

= 2T sinh^{−1} \frac{\sqrt{k}}{2\sqrt{T}} + \sqrt{kT + k^{2}/4} − \frac{k}{2} − 1/4 = θ(T) − 1/8

Furthermore, by (2.6) and (2.7), we have

h′′(u_{k}) = −\frac{\sqrt{k}}{2u_{k}^{3/2}} + \frac{T}{u_{k}^{2}} = (−\frac{1}{2}\sqrt{\frac{k}{T}} + \sqrt{\frac{T}{u_{k}}} )\frac{T^{1/2}}{u_{k}^{3/2}}

= \frac{1}{2}(\sqrt{\frac{T}{u_{k}}} + \sqrt{\frac{u_{k}}{T}} )\frac{T^{1/2}}{u_{k}^{3/2}} = \frac{T^{1/2}}{u_{k}^{3/2}} \sqrt{\frac{1}{4}(\sqrt{\frac{T}{u_{k}}} − \sqrt{\frac{u_{k}}{T}} )^{2} + 1}

= u_{k}^{−3/2}\sqrt{T + k/4}.

Hence, for k ⩽ K, the integral J_{k} can be estimated as

J_{k} = \frac{1}{\sqrt{2u_{k}}} \frac{e^{πi/4}}{\sqrt{|h′′(u_{k})|}} \frac{u_{k}^{1/4}}{k^{3/4}} \frac{T/u_{k} − 1}{log(T/u_{k})} exp\{2πih(u_{k})\} + O(η(k, T))

= \frac{u_{k}^{3/4}}{\sqrt{2}(T + k/4)^{1/4}} \frac{exp\{2πiθ(T)\}}{2u_{k}^{3/4}k^{1/4} sinh^{−1}(\sqrt{k}/(2\sqrt{T} ))} + O(η(k, T))

= \frac{exp\{2πiθ(T)\}}{2a(T)} + O(η(k, T)).

For the case where k ⩾ K, the same lemma implies that J_{k} ≪ η(k, T). Similarly, we can show that

\frac{1}{\sqrt{2}} ∫_{T_{1}}^{T_{2}} \frac{u^{1/4}}{k^{3/4}} exp\{−4πi(ku)^{1/2} + πi/4\}g(u) du

= \begin{cases}\frac{exp\{−2πiθ(T)\}}{2a(T)} + O(η(k, T)), & if k ⩽ K, \\ O(η(k, T)), & if k > K. \end{cases}

combining this with estimates for J_{k} the second part of the lemma follows, and the proof of the lemma is complete. □

Proof of Theorem 1.1. The proof of Theorem 1.1 is a straightforward application of Lemma 2.2. By Lemma 2.2, we have

\begin{matrix}\frac{1}{2π} ∫_{T_{1}}^{T_{2}} E_{1}(2πu)f(u) du \\ = \frac{T^{1/4}}{π\sqrt{2}} ∑_{k⩽min(A^{2}T,BT)} \frac{d(k)}{k^{3/4}} cos\{4π(kT)^{1/2} − π/4\} + O( ∑_{k⩽BT} δ(k, T)) \end{matrix}

In light of Voronoi’s formula (Lemma 1.1) the main term in the last expression is Δ(T) + O(T^{ϵ}), while the O-term is bounded by

\begin{matrix}≪ \frac{1}{T^{1/4}} ∑_{k⩽BT} \frac{d(k)}{k^{3/4}} + \frac{1}{T^{1/4}} ∑_{k⩽A^{2}T/2} \frac{d(k)}{k^{3/4}} + T^{1/4} ∑_{|k−A^{2}T |⩽AT^{1/2}} \frac{d(k)}{k^{3/4}} \\ + T^{3/4} ∑_{AT^{1/2}⩽|k−A^{2}T |⩽A^{2}T/2} \frac{d(k)}{k^{3/4}|\sqrt{KT} − \sqrt{kT} |} + T^{1/4} ∑_{k⩾3A^{2}T/2} \frac{d(k)}{k^{5/4}}. \end{matrix}

Using the bound d(k) ≪ k^{ϵ} for any fixed ϵ > 0 we see that the last expression is bounded by T^{ϵ}, and thus

\frac{1}{2π} ∫_{T_{1}}^{T_{2}} E_{1}(2πu)f(u) du = Δ(T) + O(T^{ϵ}).

This proves the first part of the theorem.

The proof of the other part of the theorem is very similar, and the details are omitted. However, we note that we need the following form of Voronoi’s formula

Δ(T) = \frac{T^{1/4}}{π\sqrt{2}} ∑_{k⩽K} \frac{d(k)}{k^{3/4}} cos\{4π(kT)^{1/2} − π/4\}

− \frac{3}{32\sqrt{2}π^{2}} T^{−1/4} ∑_{k⩽K} \frac{d(k)}{k^{5/4}} sin\{4π(kT)^{1/2} − π/4\} + O(T^{−3/4})

in order to show that the error term is of order T^{ϵ}. □

3. A new proof of Theorem 1.2

In this section we will give a new proof of Theorem 1.2 using the integral representation of Δ(T) obtained in the previous section. We shall provide details only when our arguments differ from the usual methods, and sketch the proof when the arguments are identical or similar to that in literature. We first prove a lemma that generalizes the Halász–Montgomery inequality.

Lemma 3.1. Let the inner product of two complex-valued functions ξ(u) and ϕ(u) be defined by (ξ, ϕ) = ∫ ξ\overline{ϕ} du, and let ∥ξ∥ denote ∥ξ∥ = (∫ |ξ|^{2} du)^{1/2} . Suppose that ξ_{λ}(u) = ξ(u, λ) and ϕ_{λ,r}(u) = ϕ_{r}(u, λ), r = 1, 2, …, R, are integrable with respect to λ for 0 ⩽ λ ⩽ L. We have

∑_{r⩽R}|\frac{1}{L} ∫_{0}^{L}(ξ_{λ}, ϕ_{λ,r}) dλ|^{2} ⩽ \frac{1}{L} ∫_{0}^{L} ∥ξ_{λ}∥^{2} dλ × max_{r⩽R} ∑_{s⩽R}|\frac{1}{L} ∫_{0}^{L}(ϕ_{λ,r}, ϕ_{λ,s}) dλ|

82 YANG

Proof. For any complex scalars c_{r} we have

∑_{r⩽R} \frac{c_{r}}{L} ∫_{0}^{L}(ξ_{λ}, ϕ_{λ,r}) dλ = \frac{1}{L} ∫_{0}^{L} (ξ_{λ}, ∑ \overline{c}_{r}ϕ_{λ,r}) dλ

⩽ \frac{1}{L} ∫_{0}^{L} ∥ξ_{λ}∥ ∥∑ \overline{c}_{r}ϕ_{λ,r}∥ dλ

⩽ (\frac{1}{L} ∫_{0}^{L} ∥ξ_{λ}∥^{2}dλ)^{1/2}(\frac{1}{L} ∫_{0}^{L} ∥∑ \overline{c}_{r}ϕ_{λ,r}∥^{2} dλ)^{1/2}

Expanding ∥ ∑ \overline{c}_{r}ϕ_{λ,r}∥^{2} and noting that |\overline{c}_{r}c_{s}| ⩽ (|c_{r}|^{2} + |c_{s}|^{2})/2 we obtain

∫_{0}^{L} ∥∑ \overline{c}_{r}ϕ_{λ,r}∥^{2} dλ ⩽ ∑_{r,s⩽R} |\overline{c}_{r}c_{s}||∫_{0}^{L}(ϕ_{λ,r}, ϕ_{λ,s}) dλ|

⩽ \frac{1}{2} ∑_{r,s⩽R}(|c_{r}|^{2} + |c_{s}|^{2})|∫_{0}^{L}(ϕ_{λ,r}, ϕ_{λ,s}) dλ|

⩽ max_{r} ∑_{s}|∫_{0}^{L}(ϕ_{λ,r}, ϕ_{λ,s}) dλ| × ∑_{r⩽R} |c_{r}|^{2}.

Choosing

c_{r} = \overline{\frac{1}{L} ∫_{0}^{L}(ξ_{λ}, ϕ_{λ,r}) dλ},

the claimed inequality follows immediately. □

Proof of Theorem 1.2. Let

S(u, K_{1}, K_{2}) = u^{1/4} ∑_{K_{1}<k⩽K_{2}} \frac{d(k)}{k^{3/4}} cos\{4π(ku)^{1/2} − π/4\}

denote the partial sum in Voronoi’s formula, and set

S_{U}(u, K_{1}, K_{2}) = S(u + U, K_{1}, K_{2}) − S(u, K_{1}, K_{2}).

In view of Voronoi’s formula, to prove the theorem it suffices to consider the integral

∫_{T}^{T +H} S_{U}(u, 0, T)^{2} du.

Assume that U ⩽ \sqrt{T} /2. Let m be the integer such that 2^{m} < T^{1/3}U^{−2/3} ⩽ 2^{m+1}, and let M denote 2^{m}. We have trivially

cos\{4π\sqrt{k(u + U)} − π/4\} − cos\{4π\sqrt{ku} − π/4\} ≪ \frac{\sqrt{k}U}{\sqrt{T}} ,

and thus

∫_{T}^{T +H} S_{U}(u, 0, M)^{2} du ≪ H(T^{1/4} ∑_{k⩽M} \frac{d(k)}{k^{3/4}} \frac{\sqrt{k}U}{\sqrt{T}} )^{2} ≪ HU log^{2}(\sqrt{T} /U),

which is contained in the claimed bound.

We next consider the case when M^{3} < k ⩽ T. Using standard arguments (see, for example, [6, p.363]) we see that

∫_{T}^{T +H} S_{U}(u, M^{3}, T)^{2} du ≪ HT^{1/2} ∑_{k⩾M^{3}} \frac{d(k)^{2}}{k^{3/2}} + T^{1+ϵ} ≪ HU log^{3} \frac{\sqrt{T}}{U} + T^{1+ϵ}.

Thus it remains to deal with the cases when M < k ⩽ M^{3}. We write S_{U}(u, M, M^{3}) as ∑_{K} S_{U}(u, K, 2K), where K runs over integers of the form 2^{m}M. When K_{1} ⩽ K_{2}/4, we have

∫_{T}^{T +H} S_{U}(u, K_{1}, 2K_{1})S_{U}(u, K_{2}, 2K_{2}) du ≪ T^{1+ϵ}K_{1}^{1/4}K_{2}^{−1/4}.

When K_{1} = K_{2}/2, we use the inequality 2|ab| ⩽ |a|^{2} + |b|^{2}, and obtain

∫_{T}^{T +H} S_{U}(u, M, M^{3})^{2} du ≪ ∑_{K} ∫_{T}^{T +H} S_{U}(u, K, 2K)^{2} du + T^{1+ϵ}.

Thus, the proof of the result will be complete if we can show that

(3.1)  ∫_{T}^{T +H} S_{U}(u, K, 2K)^{2} du ≪ \frac{HU^{2}\sqrt{K}}{\sqrt{T}} log^{3} K

for K ⩽ \sqrt{T} /U.

Let T_{h} ∈ [T + h, T + h + 1] denote a point with

|S_{U}(T_{h}, K, 2K)| = max_{T +h⩽u⩽T +h+1} |S_{U}(u, K, 2K)|.

We then have

∫_{T}^{T +H} S_{U}(u, K, 2K)^{2} du ≪ ∑_{h=0}^{H} |S_{U}(T_{h}, K, 2K)|^{2}.

For h ⩽ H we denote by T′_{h} and T′′_{h} the points T_{h} − 4\sqrt{KT} and T_{h} + 4\sqrt{KT}, respectively. Set L = \sqrt{KT}. Applying the second part of Lemma 2.1 and following the calculation in Lemma 2.2 we obtain

\begin{matrix}S_{U}(T_{h}, K, 2K) = \frac{1}{L} ∫_{0}^{L} ∫_{T′_{h}+λ}^{T′′_{h}−λ} Σ(u, K) (f(u, T_{h} + U) − f(u, T_{h})) du dλ \\ + O(\frac{T^{1/4}}{K^{9/4}} ∑_{K<k⩽2K} d(k)) & , \end{matrix}

where

Σ(u, K) = ∑_{K<k⩽2K} \frac{d(k) cos\{2πθ_{k}(u)\}}{\sqrt{2}a_{k}(u)},

f(u, v) = \frac{log(u/v)}{\sqrt{v}(u/v − 1)} exp\{−2πi(u log(u/v) − u + v − 1/8)\}

with

θ_{k}(u) = 2u sinh^{−1} \frac{\sqrt{k}}{2\sqrt{u}} + \sqrt{ku + k^{2}/4} − \frac{k}{2} − \frac{1}{8} ,

a_{k}(u) = (4ku + k^{2})^{1/4} sinh^{−1} \frac{\sqrt{k}}{2\sqrt{u}} .

Setting f_{U}(u, v) = f(u, v + U) − f(u, v), we obtain

\begin{matrix}∑_{h=0}^{H} S_{U}(T_{h}, K, 2K)^{2} ≪ ∑_{h=0}^{H}|\frac{1}{L} ∫_{0}^{L} ∫_{T′_{h}+λ}^{T′′_{h}−λ} Σ(u, K)f_{U}(u, T_{h}) du dλ|^{2} \\ +HT^{1/2}K^{−5/2} log^{2} K \end{matrix}

Since K ⩾ T^{1/3}U^{−2/3}, the term HT^{1/2}K^{−5/2} log^{2} K is bounded by HU log^{2}(\sqrt{T} /U). We now apply Lemma 3.1 with

ξ_{λ}(u) = Σ(u, K),

ϕ_{λ,h}(u) = \begin{cases}\overline{f_{U}(u, T_{h})}, & when T′_{h} + λ ⩽ u ⩽ T′′_{h} − λ, \\ 0, & else, \end{cases}

and the inner product (ξ, ϕ) given by

(ξ, ϕ) = ∫_{T −5L}^{T +H+5L} ξ(u)\overline{ϕ(u)} du.

It follows from Lemma 3.1 that

(3.2) ∑_{h=0}^{H} S(T_{h}, K, 2K)^{2} ≪ \frac{1}{L} ∫_{0}^{L} ∥ξ_{λ}∥^{2} dλ × max_{r⩽R} ∑_{s⩽R} |\frac{1}{L} ∫_{0}^{L}(ϕ_{λ,r}, ϕ_{λ,s}) dλ|

Using standard arguments ([6, p. 363]) again we see that

(3.3)  ∥ξ_{λ}∥^{2} ≪ HT^{1/2}K^{−1/2} log^{3} K + T^{1+ϵ}.

Moreover, we have

f_{U}(u, T_{h}) ≪ U max_{T′_{h}⩽u⩽T′′_{h}+L+U} \frac{∂}{∂u}f(u, T_{h}) ≪ \frac{U\sqrt{K}}{T}

for all T 1_{h} ⩽ u ⩽ T′′_{h} + L, and thus

\frac{1}{L} ∫_{0}^{L}(ϕ_{λ,h_{1}}, ϕ_{λ,h_{2}}) dλ ≪ \sqrt{T K}(\frac{U\sqrt{K}}{T} )^{2} = \frac{U^{2}K^{3/2}}{T^{3/2}}

for all h_{1}, h_{2} ⩽ H. On the other hand, integrating by parts, we see that the main contribution to the integral L^{−1} ∫_{0}^{L}(ϕ_{λ,h_{1}}, ϕ_{λ,h_{2}}) dλ can be written as a sum of quantities of the form

\begin{matrix}c_{1}\{g(T_{h_{1}} + U, T_{h_{2}} + U, t + c_{2}L) − g(T_{h_{1}}, T_{h_{2}} + U, t + c_{2}L) \\ −g(T_{h_{1}} + U, T_{h_{2}}, t + c_{2}L) + g(T_{h_{1}}, T_{h_{2}}, t + c_{2}L) & \} \end{matrix}

where t is

T′′_{h_{1},h_{2}} = min(T′′_{h_{1}}, T′′_{h_{2}}) or T′_{h_{1},h_{2}} = max(T′_{h_{1}}, T′_{h_{2}}),

c_{2} is 0 or 1 for t = T′_{h_{1},h_{2}}, 0 or −1 for t = T′′_{h_{1},h_{2}}, and c_{1} is 1 or −1, depending on t and c_{2}, and

g(u, v, t) = −\frac{log(t/u) log(t/v)}{4π^{2}\sqrt{uv} log^{2}(u/v)(t/u − 1)(t/v − 1)}.

It follows that, for |h_{1} − h_{2}| ⩾ \sqrt{T/K},

\frac{1}{L} ∫_{0}^{L}(ϕ_{λ,h_{1}}, ϕ_{λ,h_{2}}) dλ ≪ \frac{1}{LT log^{2}(T_{h_{1}}/T_{h_{2}})} (\frac{U\sqrt{K}}{\sqrt{T}} )^{2} ≪ \frac{U^{2}\sqrt{K}}{\sqrt{T} |h_{1} − h_{2}|^{2}} ,

and thus

∑_{h_{2}⩽H} \frac{1}{L} ∫_{0}^{L}(ϕ_{λ,h_{1}}, ϕ_{λ,h_{2}}) dλ ≪ \frac{U^{2}K}{T}

for all h_{1} ⩽ H. Inserting this estimate and (3.3) into (3.2), we hence obtain (3.1). This completes the proof of Theorem 1.2. □

References

[1] F. V. Atkinson, The mean-value of the Riemann zeta function, Acta Math. 81 (1949), 353– 376.

[2] D. R. Heath-Brown, The mean value theorem for the Riemann zeta-function, Mathematika 25 (1978), 177–184.

[3] D. R. Heath-Brown, The distribution and moments of the error term in the Dirichlet divisor problem, Acta Arith. 60 (1992), 389–415.

[4] D. R. Heath-Brown and K.-M. Tsang, Sign changes of E(T), δ(x), and P (x), J. Number Theory 49 (1994), 73–83.

[5] A. Ivić, Large values of the error term in the divisor problem, Invent. Math. 71 (1983), 513–520.

[6] A. Ivić, The Riemann zeta-function. The theory of the Riemann zeta-function with applications, Wiley, New York, 1985.

[7] A. Ivić, On some problems involving the mean square of ζ( \frac{1}{2} + it), Bull. Cl. Sci. Math. Nat. Sci. Math. 23 (1998), 71–76.

[8] M. Jutila, Riemann’s zeta function and the divisor problem, Ark. Mat. 21 (1983), 75–96.

[9] M. Jutila, On a formula of Atkinson, In Topics in classical number theory, Vol. I, II (Budapest, 1981), pp. 807–823; North-Holland, Amsterdam, 1984.

[10] M. Jutila, On the divisor problem for short intervals, Ann. Univ. Turku. Ser. A I 186 (1984), 23–30; Studies in honour of Arto Kustaa Salomaa on the occasion of his fiftieth birthday.

[11] M. Jutila, Transformation formulae for Dirichlet polynomials, J. Number Theory 18 (1984), 135–156.

[12] M. Jutila, Lectures on a method in the theory of exponential sums, Published for the Tata Institute of Fundamental Research, Bombay, 1987.

[13] M. Jutila, Riemann’s zeta-function and the divisor problem. II, Ark. Mat. 31 (1993), 61–70.

[14] Y.-K. Lau and K.-M. Tsang, Mean square of the remainder term in the Dirichlet divisor problem, J. Théor. Nombres Bordx 7 (1995), 75–92; Les Dix-huitièmes Journées Arithmétiques (Bordeaux, 1993).

[15] K. Matsumoto, Recent developments in the mean square theory of the Riemann zeta and other zeta-functions, In Number theory, pages 241–286. Birkhäuser, Basel, 2000.

[16] E. Preissmann, Sur la moyenne quadratique du terme de reste du problème du cercle, C. R. Acad. Sci. Paris Sér. I Math. 306 (1998), 151–154.

[17] K.-C. Tong, On divisor problems. III, Acta Math. Sinica 6 (1956), 515–541.

[18] K.-M. Tsang, Higher-power moments of δ(x), E(t) and P (x), Proc. London Math. Soc. (3) 65 (1992), 65–84.

[19] K.-M. Tsang, Mean square of the remainder term in the Dirichlet divisor problem, II, Acta Arith. 71 (1995), 279–299.

[20] G. F. Voronoi, Sur une fonction transcendante et ses applications à la sommation de quelques séries, Ann. Sci. École Norm. Sup. (3) 21 (1904), 207–268.

Department of Applied Mathematics (Received 24 11 2007) National Chiao Tung University Hsinchu Taiwan

yfyang@math.nctu.edu.tw