PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 95 (109) (2014), 173–188 DOI: 10.2298/PIM1409173B

AN UNEXPECTED PROPERTY OF ODD ORDER DERIVATIVES OF HARDY’S FUNCTION

Philippe Blanc

Communicated by Aleksandar Ivić

Abstract. Assuming the Riemann hypothesis, we show that the odd order derivatives of Hardy’s function have, under some condition, an unexpected behavior for large values of t.

1. Introduction and main result

Let ζ be the Riemann zeta function, and Z the Hardy function defined by

Z(t) = e^{iθ(t)} ζ ( \frac{1}{2} + it)

where

θ(t) = arg(π^{−i \frac{t}{2}} Γ( \frac{1}{4} + i\frac{t}{2}))

and the argument is defined by continuous variation of t starting with the value 0 at t = 0. It can be shown [6] that

θ(t) = \frac{t}{2} log \frac{t}{2π} − \frac{t}{2} − \frac{π}{8} + O( \frac{1}{t}) .

The real zeros of Z coincide with the zeros of ζ located on the line of real part \frac{1}{2}. If the Riemann hypothesis is true, then the number of zeros of Z in the interval ]0, t]is given by [6]

(1.1)  N(t) = \frac{1}{π} θ(t) + 1 + S(t)

where S(t) = \frac{1}{π} argζ(\frac{1}{2} + it) if t is not a zero of Z and argζ(\frac{1}{2} + it) is defined by continuous variation along the straight lines joining 2, 2 + it and \frac{1}{2} + it starting with the initial value argζ(2) = 0. If t is a zero of Z we set S(t) = lim_{ϵ→0_{+}} S(t+ϵ). Let us choose T, say T = 500, and let us plot the graphs of functions

2010 Mathematics Subject Classification: 11M26.

Key words and phrases: Riemann zeta function, distribution of zeros, Hardy’s function.

Dedicated to Professor Jean Descloux on the occasion of his 80th birthday.

f_{2k−1} (t) := (−1)^{k+1} \frac{Z^{(2k−1)} (t)}{θ′(T)^{2k−1}}

for k = 1, …, 5 on the interval [T − 10, T + 10]. Observe that the term 1/θ′(T)^{2k−1} is just a scaling factor. These graphs show that the functions (−1)^{k+1} Z^{(2k−1)} have

Figure 1. Graphs of f_{2k−1} for k = 1, …, 5

generally the same signs, at least for small values of k. This can be explained heuristically by a formula due to Lavrik [8], which asserts that for t sufficiently large and 1 ⩽ k ⩽ \frac{1}{4} logt, we have, uniformly in k,

Z^{(2k−1)} (t) = 2(−1)^{k} ∑_{1⩽n⩽\sqrt{t/2π}} \frac{1}{\sqrt{n}}(θ′(t) − logn)^{2k−1} sin(θ(t) − t logn)

+ O(t^{− \frac{1}{4}} (\frac{3}{2} logt)^{2k}).

Let us denote by M = M(T) the largest integer r, possibly infinite, such that

sign((−1)^{k+1} Z^{(2k−1)} (T)) = sign(Z′(T)) for k = 1, 2, …, r.

For some T the values of M are surprisingly large. Using Mathematica we compute Z with high precision and get for example M(100) = 26, M(1000.4) = 138 and M(9999.5) = 402.

Now let T be large enough such that Z(T) > 0 and let γ_{k}, where k ≠ 0, be the zeros of Z ordered in increasing order, taking their multiplicities into account, and numbered so that ⋯⩽ γ_{−2} ⩽ γ_{−1} < T < γ_{1} ⩽ γ_{2} ⩽ · · · . Further, let 4(loglogT)^{−1} ⩽ a ⩽ \sqrt{T} such that T + a and T − a are not zeros of Z and, finally, let m, n ⩾ 1 such that γ_{−m−1} < T − a < γ_{−m} and γ_{n} < T + a < γ_{n+1} . Note that the existence of m, n ⩾ 1 is an immediate consequence of a result of Goldston and Gonek [4]. We assume that (−1)^{m}Z′(T − a) ⩾ 0 and (−1)^{n}Z′(T + a) ⩽ 0 and we denote by K = K(T, a) the largest integer r, possibly infinite, such that

(1.2) (−1)^{m+k+1} Z^{(2k−1)} (T − a) ⩾ 0 and (−1)^{n+k+1} Z^{(2k−1)} (T + a) ⩽ 0

for k = 1, 2, …, r. For some T and a, the values of K are also surprisingly large: K(109.3, 9.4) = 21, K(1070.1, 8.5) = 108 and K(10025.5, 9.8) = 408. The goal of this paper is to give a conditional upper bound for K.

It should be observed that if we replace Z(t) by cost and choose T = 0 and a = lπ + \frac{π}{4} where l ∈ ℕ^{∗}, then m = n = l and conditions (1.2) hold for every k and hence K = ∞.

We now define a quantity which appears in our main result. By Lavrik’s formula [8], for t sufficiently large and 0 ⩽ k ⩽ \frac{1}{4} logt, we have, uniformly in k,

Z^{(2k)} (t) = 2(−1)^{k} ∑_{1⩽n⩽\sqrt{t/2π}} \frac{1}{\sqrt{n}}(θ′(t) − logn)^{2k} cos(θ(t) − t logn)

+ O(t^{− \frac{1}{4}} ( \frac{3}{2} logt)^{2k+1} )

and using θ′(t) = \frac{1}{2} log \frac{t}{2π} + O(1/t^{2}) we get

(1.3) |Z^{(2k)} (t)| = θ′(t)^{2k}( 2 ∑_{1⩽n⩽\sqrt{t/2π}} \frac{1}{\sqrt{n}}( 1 − \frac{logn}{θ′(t)})^{2k} cos(θ(t) − t logn))

+ O(3^{2k}θ′(t)^{2k}t^{− \frac{1}{4}} logt).

As Ivić says in [5], it is difficult to get good uniform bounds for Z^{(2k)} (t) from (1.3). Nevertheless, when Z(T) is large, relation (1.3) suggests that

(1.4)  ∫_{T −a}^{T +a} (Z^{(2k)} (t))^{2}dt = 2a(c_{2k}θ′(T)^{2k}Z(T))^{2}

where c_{2k} is small. For the aforementioned computations, we used the results of Kotnik [7] and chose T in a neighborhood of 10^{2}, 10^{3} and 10^{4} and a approximately equal to 10 such that Z(T) and K(T, a) are large. We obtain c_{42} = 0.56 … and c_{216} = 0.34 … which show that c_{2k} can be small even for some k > \frac{1}{4} logT despite the fact that, for fixed T, the sequence c_{2k} is unbounded. Numerical experiments indicate that Lavrik’s formula is probably true for larger values of k with a better error term.

Theorem 1.1. For T large enough, let 4(loglogT)^{−1} ⩽ a ⩽ \sqrt{T}, K be the number defined in the introduction, ΔS = S(T + a) − S(T − a) and further let K^{∗} = \frac{a}{π} logT + ΔS. If the Riemann hypothesis is true and if ΔS ⩾ 1, then

(1.5)  K ⩽ max( \frac{a}{2π} A_{K,T} \frac{logT}{ΔS}( 1 + O( \frac{logloglogT}{loglogT}) ) , K^{∗} logK^{∗})

where

A_{K,T} = logc_{2K} + ΔS log2 + \frac{1}{\sqrt{2}} \frac{logT}{loglogT}.

For T large enough such that Z(T) is large in the sense of [7], numerical experiments show that the bound (1.5) is probably true without the term logc_{2K} . If this is the case and if we neglect the big O in (1.5) and choose T ⩽ 10^{50} and a ⩽ 1 such that ΔS = 1, we get K ⩽ 327. Note that for the three values of K(T, a) already given, we have ΔS < 0. This suggests that the behaviour of K is different according to ΔS ⩽ 0 or ΔS ⩾ 1. This is unexpected.

This work stems from an observation of Ivić [5] about the values of the derivatives of Z in a neighborood of points where |Z| attains a large value. Some of the material used in our proof has been published by the author in [1].

The notations used in this paper are standard: ⌊x⌋ stand for the usual floor function and {x} := x − ⌊x⌋. Bernoulli and Chebyshev polynomials of degree n are denoted by B_{n}(x) and T_{n}(x); they are defined by

∫_{x}^{x+1} B_{n}(t) dt = x^{n} and T_{n}(cosθ) = cosnθ.

The organization of this paper is as follows: In Section 2 we prove the key identity, a property of the derivatives of Bernoulli polynomials and preparatory lemmas. Section 3 is devoted to the proof of our main result.

2. Preliminary results

We first prove an identity which will be used later to establish a relation between the value of a function f ∈ C^{2r} [−a, a] at 0, the zeros of f and the values of its derivatives of odd order on the boundaries of the interval.

Lemma 2.1. Let −a < x_{−m} < ⋯< x_{−1} < x_{0} < x_{1} < ⋯< x_{n} < a and for

l = 1, 2, … let Ψ_{2l−1} be the function defined on [−a, a] by

Ψ_{2l−1} (x) = \frac{(4a)^{2l−1}}{(2l)!} ∑_{k=−m}^{n} μ_{k}( B_{2l}( \frac{1}{2} + \frac{x + x_{k}}{4a} ) + B_{2l}( \{ \frac{x − x_{k}}{4a} \}) )

where ∑ _{k=−m}^{n} μ_{k} = 0. Then for f ∈ C^{2r} [−a, a] where r ⩾ 1, we have the identity

(2.1) ∑_{k=−m}^{n} μ_{k}f(x_{k}) = ∑_{k=1}^{r} f^{(2k−1)} (a)Ψ_{2k−1} (a) − ∑_{k=1}^{r} f^{(2k−1)} (−a)Ψ_{2k−1} (−a)

− ∫_{−a}^{a} f^{(2r)} (x)Ψ_{2r−1} (x) dx.

Proof. By definition the function Ψ_{2r−1} is C^{2r−2} , piecewise polynomial and the relation B′_{l}(x) = lB_{l−1} (x) for l = 1, 2, … leads to

Ψ_{2r−1}^{(j)} (x) = \frac{(4a)^{2r−j−1}}{(2r − j)!} ∑_{k=−m}^{n} μ_{k}( B_{2r−j} ( \frac{1}{2} + \frac{x + x_{k}}{4a} ) + B_{2r−j} ( \{ \frac{x − x_{k}}{4a} \}) )

for j = 1, …, 2r − 1 and x ≠ x_{−m} , …, x_{n} if j = 2r − 1. This implies that

(2.2)  Ψ_{2r−1}^{(2r−2j)} = Ψ_{2j−1} for j = 1, 2, …, r − 1

and that Ψ_{2r−1}^{(2r−2j+1)} (±a) (2.3)

= \frac{(4a)^{2j−2}}{(2j − 1)!} ∑_{k=−m}^{n} μ_{k}( B_{2j−1} ( \frac{1}{2} + \frac{±a + x_{k}}{4a} ) + B_{2j−1} ( \{ \frac{±a − x_{k}}{4a} \}) )

= \frac{(4a)^{2j−2}}{(2j − 1)!} ∑_{k=−m}^{n} μ_{k}( B_{2j−1} ( \frac{1}{2} + \frac{±a + x_{k}}{4a} ) + B_{2j−1} ( \frac{1}{2} − \frac{±a + x_{k}}{4a} ) )

= 0 for j = 1, 2, …, r.

Further for x ≠ x_{−m} , …, x_{n} we have

(2.4)  Ψ_{2r−1}^{(2r−1)} (x) = ∑_{k=−m}^{n} μ_{k}( B_{1}( \frac{1}{2} + \frac{x + x_{k}}{4a} ) + B_{1}( \{ \frac{x − x_{k}}{4a} \}) )

= ∑_{k=−m}^{n} μ_{k}( \frac{x + x_{k}}{4a} + \{ \frac{x − x_{k}}{4a} \} − \frac{1}{2})

and as ∑ _{k=−m}^{n} μ_{k} = 0 the function Ψ_{2r−1}^{(2r−1)} is piecewise constant. Explicitly, for x ∈ ]x_{j} , x_{j+1} [, we get

Ψ_{2r−1}^{(2r−1)} (x) = ∑_{k=−m}^{j} μ_{k}( \frac{x}{2a} − \frac{1}{2}) + ∑_{k=j+1}^{n} μ_{k}( \frac{x}{2a} + 1 − \frac{1}{2}) = ∑_{k=j+1}^{n} μ_{k} = − ∑_{k=−m}^{j} μ_{k}

which leads to

∫_{x_{j}}^{x_{j+1}} f′(x)Ψ_{2r−1}^{(2r−1)} (x) dx = −( ∑_{k=−m}^{j} μ_{k}) (f(x_{j+1} )−f(x_{j})) for j = −m, …, n−1.

Summing these equalities and using that Ψ_{2r−1}^{(2r−1)} = 0 on the intervals [−a, x_{−m} [ and ]x_{n}, a], which follows from (2.4), we have

∑_{k=−m}^{n} μ_{k}f(x_{k}) = ∫_{−a}^{a} f′(x) Ψ_{2r−1}^{(2r−1)} (x) dx

and we complete the proof by integrating 2r − 1 times the right-hand side by parts taking into account relations (2.2) and (2.3). □

For further use we recall some elementary facts concerning the divided differences.

Lemma 2.2. Let I = ]− T, T [ , f ∈ C^{m+n} (I) and let g be the function defined for pairwise distinct numbers t_{−m} , …, t_{n} ∈ I by

g(t_{−m} , …, t_{n}) = ∑_{k=−m}^{n} \frac{f(t_{k})}{∏_{\begin{matrix}−m⩽j⩽n \\ j≠k \end{matrix}} (t_{k} − t_{j})} .

Then a) The function g has a continuous extension g^{∗} defined for t_{−m} , …, t_{n} ∈ I

and there exists η = η(t_{−m} , …, t_{n}) ∈ I such that

g^{∗}(t_{−m} , …, t_{n}) = \frac{f^{(m+n)} (η)}{(m + n)!} .

b) Let y_{0}, y_{1}, …, y_{l} be the distinct values of t_{−m} , …, t_{n} considered as fixed and let r_{k} be the number of index j such that t_{j} = y_{k}. Then there exist α_{k,i}

depending on y_{0}, y_{1}, …, y_{l} such that

g^{∗}(t_{−m} , …, t_{n}) = ∑_{k=0}^{l} ∑_{i=0}^{r_{k} −1} α_{k,i} f^{(i)} (y_{k}).

Proof. Assertion a) is a consequence of the representation formula

(2.5) g(t_{−m} , …, t_{n}) =

∫_{0}^{1} dτ_{1} ∫_{0}^{τ_{1}} dτ_{2} ⋯ ∫_{0}^{τ_{m+n−1}} f^{(m+n)} ( t_{−m} + ∑_{k=1}^{m+n} τ_{k}(t_{−m+k} − t_{−m+k−1} )) dτ_{m+n} .

A proof of b) is given in [9]. □

In the next lemma we indicate the choice of coefficients μ_{k} for which the identity of Lemma 2.1 is of practical use for large values of a. The main reason of this choice will appear in the proof of (2.11) in Lemma 2.9.

Lemma 2.3. Let Ψ_{2l−1} be defined for pairwise distinct x_{−m} , …, x_{n} ∈]−a, a[ by

Ψ_{2l−1} (x_{−m} , …, x_{n}, x) = \frac{(4a)^{2l−1}}{(2l)!} ∑_{k=−m}^{n} μ_{k}( B_{2l}( \frac{1}{2} + \frac{x + x_{k}}{4a} ) + B_{2l}( \{ \frac{x − x_{k}}{4a} \}) )

where

μ_{k} = \frac{α_{k}}{α_{0}} and \frac{1}{α_{k}} = ∏_{\begin{matrix}−m⩽j⩽n \\ j≠k \end{matrix}} ( sin( π \frac{x_{k}}{2a}) − sin( π \frac{x_{j}}{2a}) ) for k = −m, …, n.

Then

a) For l ⩾ 1 the functions Ψ_{2l−1} (·, …, ·, ±a) have continuous extensions Ψ_{2l−1}^{∗} (·, …, ·, ±a) defined for x_{−m} , …, x_{n} ∈]− a, a[ .

b) If 2l ⩾ m + n + 2 the function Ψ_{2l−1} has a continuous extension Ψ_{2l−1}^{∗} defined for x_{−m} , …, x_{n} ∈]− a, a[ and x ∈ [−a, a].

c) If 2r ⩾ m + n + 2 and f ∈ C^{2r} [−a, a] is defined on [−a, a] and vanishes at x_{k} with k ≠ 0 where −a < x_{−m} ⩽ … ⩽ x_{−1} < x_{0} < x_{1} ⩽ … ⩽ x_{n} < a and the x_{k} are numbered taking into account their multiplicity, then we have the identity

(2.6)  f(x_{0}) = ∑_{k=1}^{r} f^{(2k−1)} (a)Ψ_{2k−1}^{∗} (a) − ∑_{k=1}^{r} f^{(2k−1)} (−a)Ψ_{2k−1}^{∗} (−a)

− ∫_{−a}^{a} f^{(2r)} (x)Ψ_{2r−1}^{∗} (x) dx

where for short Ψ_{2k−1}^{∗} (±a) and Ψ_{2r−1}^{∗} (x) stand for Ψ_{2k−1}^{∗} (·, …, ·, ±a) and

Ψ_{2r−1}^{∗} (·, …, ·, x).

Proof. Introducing the function h defined by

h(t, x) = \frac{(4a)^{2l−1}}{(2l)!} ( B_{2l}( \frac{1}{2} + \frac{x}{4a} + \frac{1}{2π} Arcsint) + B_{2l}( \{ \frac{x}{4a} − \frac{1}{2π} Arcsint\}) )

we have

Ψ_{2l−1} (x_{−m} , …, x_{n}, ±a) = \frac{1}{α_{0}} ∑_{k=−m}^{n} α_{k}h( sin( π \frac{x_{k}}{2a}) , ±a)

for pairwise distinct x_{−m} , …, x_{n} ∈]−a, a[ and assertion a) holds since the functions h(·, ±a) belong to C^{∞} ]− 1, 1[ .

By definition the function h belongs to C^{2l−2} ( ]−1, 1[ × [ −a, a]) and the assertion b) is a consequence of the representation formula (2.5) since we have 2l − 2 ⩾ m + n. For pairwise distinct x_{−m} , …, x_{n} ∈]−a, a[ the left-hand side of identity (2.1) reads

\frac{1}{α_{0}} ∑_{k=−m}^{n} α_{k}f( \frac{2a}{π} Arcsin( sin( π \frac{x_{k}}{2a}) ) )

and thanks to Lemma 2.2 this expression, and hence the identity (2.1), extend to x_{−m} , …, x_{n} ∈] − a, a[ . One completes the proof of c) by observing, thanks to Lemma 2.2, that the left-hand side reduces to f(x_{0}) when the x_{k} are zeros of multiplicity r_{k} of f. □

The results stated in Lemma 2.4 play a central role in the proof of properties of functions Ψ_{2l−1}^{∗} (·, …, ·, ±a).

Lemma 2.4. For all m, k ∈ ℕ^{∗} we have the inequality

(−1)^{m+1} \frac{d^{k}}{dx^{k}} B_{2m} ( \frac{1}{2} + \frac{1}{π} Arcsin \sqrt{x}) > 0 for x ∈ [0, 1[ .

The proof of Lemma 2.4 requires two technical results given in Lemmas 2.5 and 2.6.

Lemma 2.5. For all k ∈ ℕ we have the Taylor expansion

(Arcsinx)^{2k} = ∑_{l=0}^{∞} \frac{(2k)!}{(2l)!}2^{2l−2k} b_{k, l} x^{2l} for x ∈ [−1, 1]

where b_{k, l} are integers defined recursively by

b_{0, 0} = 1 and b_{k, 0} = b_{0, l} = 0 for k, l ⩾ 1
b_{k+1, l+1} = b_{k, l} + l^{2}b_{k+1, l} for k, l ⩾ 0 .

Proof. We note first that the functions f_{2k}(x) := (Arcsinx)^{2k} satisfy

(1 − x^{2})f′′_{2k+2} (x) − x f′_{2k+2} (x) − (2k + 2)(2k + 1)f_{2k}(x) = 0 for x ∈ ]−1, 1[ .

From the definition of f_{2k} and the above equality it follows that numbers c_{k, l} defined by f_{2k}(x) = ∑ _{l=0}^{∞} c_{k, l} x^{2l} for x ∈ [−1, 1] are uniquely determined by the recurrence relations

c_{0, 0} = 1 and c_{k, 0} = c_{0, l} = 0 for k, l ⩾ 1

(2l + 2)(2l + 1)c_{k+1, l+1} − 4l^{2}c_{k+1, l} − (2k + 2)(2k + 1)c_{k, l} = 0 for k, l ⩾ 0.

A simple check shows that c_{k, l} = \frac{(2k)!}{(2l)!}2^{2l−2k} b_{k, l}. □

Lemma 2.6. Let b_{k, l} be the numbers defined in Lemma 2.5. Then

(2.7)  lim_{l→∞} \frac{b_{k, l}}{((l − 1)!)^{2}} = \frac{π^{2k−2}}{(2k − 1)!} for all k ⩾ 1.

Proof. From the definition of numbers b_{k, l} we infer that b_{1, l} = ((l − 1)!)^{2} for l ⩾ 1. Thus relation (2.7) is trivially true for k = 1. We then assume k ⩾ 2. As b_{j, 1} = 0 for j ⩾ 2 the numbers d_{j, l} defined for j, l ⩾ 1 by d_{j, l} = \frac{b_{j,l}}{((l−1)!)^{2}} satisfy the recurrence relations

d_{j, 1} = 0 and d_{1, l} = 1 for j ⩾ 2 and l ⩾ 1,

d_{j+1, l+1} = \frac{1}{l^{2}} d_{j, l} + d_{j+1, l} for j, l ⩾ 1.

Using the fact that d_{j−1, l} = 0 for l = 1, …, j − 2 we get first for j ⩾ 2 the equality

d_{j, n_{j}} = ∑_{n_{j−1} =j−1}^{n_{j} −1} \frac{1}{n_{j−1}^{2}} d_{j−1, n_{j−1}}

which we iterate to obtain

d_{k, l} = ∑_{n_{k−1} =k−1}^{l−1} \frac{1}{n_{k−1}^{2}} ∑_{n_{k−2} =k−2}^{n_{k−1} −1} \frac{1}{n_{k−2}^{2}} ⋯ ∑_{n_{2} =2}^{n_{3} −1} \frac{1}{n_{2}^{2}} ∑_{n_{1} =1}^{n_{2} −1} \frac{1}{n_{1}^{2}} .

This leads to

lim_{l→∞} d_{k, l} = ∑_{n_{k−1} >n_{k−2} >···>n_{2} >n_{1} >0}  ∏_{j=1}^{k−1} \frac{1}{n_{j}^{2}}

and we recognize in the right-hand side the number ζ({2}_{(k−1)} ) whose value, given in [2], is equal to the right-hand side of (2.7). □

DERIVATIVES OF HARDY’S FUNCTION 181

Proof of Lemma 2.4. It suffices to prove that the numbers e_{m, l} defined by

(2.8)  (−1)^{m+1} B_{2m} ( \frac{1}{2} + \frac{1}{π} Arcsinx) = ∑_{l=0}^{∞} e_{m, l} x^{2l}

satisfy e_{m, l} > 0 for all m, l ∈ ℕ^{∗}. Using Taylor’s formula and the evenness of function B_{2m} (\frac{1}{2} + \frac{t}{π}), we have

B_{2m} ( \frac{1}{2} + \frac{t}{π} ) = ∑_{k=0}^{m} \frac{1}{(2k)!} B_{2m}^{(2k)} ( \frac{1}{2}) ( \frac{t}{π} )^{2k} = ∑_{k=0}^{m} \binom{2m}{2k} B_{2m−2k} ( \frac{1}{2}) ( \frac{t}{π} )^{2k}

and the Taylor expansion of (Arcsinx)^{2k} given in Lemma 2.5 leads to

B_{2m} ( \frac{1}{2} + \frac{1}{π} Arcsinx) = ∑_{k=0}^{m} (\binom{2m}{2k} B_{2m−2k} ( \frac{1}{2}) π^{−2k} ∑_{l=0}^{∞} \frac{(2k)!}{(2l)!}2^{2l−2k} b_{k, l} x^{2l})

= \frac{(2m)!}{(2π)^{2m}} ∑_{k=0}^{m} ( \frac{(2π)^{2m−2k}}{(2m − 2k)!} B_{2m−2k} ( \frac{1}{2}) ∑_{l=0}^{∞} \frac{2^{2l}}{(2l)!}b_{k, l} x^{2l}) .

We then change the order of summation to get

(2.9)  (−1)^{m+1} B_{2m} ( \frac{1}{2} + \frac{1}{π} Arcsinx) = \frac{(2m)!}{(2π)^{2m}} ∑_{l=0}^{∞} \frac{2^{2l}}{(2l)!}f_{m, l} x^{2l}

where

f_{m, l} = (−1)^{m+1} ∑_{k=0}^{m} \frac{(2π)^{2m−2k}}{(2m − 2k)!} B_{2m−2k} ( \frac{1}{2}) b_{k, l} .

We prove by recurrence over m that f_{m, l} > 0 for m, l ⩾ 1. To this end we set g_{m,l} = \frac{f_{m,l}}{((l−1)!)^{2}} for m, l ⩾ 1 and since b_{0, l} = 0 for l ⩾ 1 we have

g_{m+1, l+1} = \frac{(−1)^{m+2}}{(l!)^{2}} ∑_{k=1}^{m+1} \frac{(2π)^{2m+2−2k}}{(2m + 2 − 2k)!} B_{2m+2−2k} ( \frac{1}{2}) b_{k, l+1}

= \frac{(−1)^{m+2}}{(l!)^{2}} ∑_{k=1}^{m+1} \frac{(2π)^{2m+2−2k}}{(2m + 2 − 2k)!} B_{2m+2−2k} ( \frac{1}{2}) ( b_{k−1, l} + l^{2} b_{k, l} )

= \frac{(−1)^{m+2}}{(l!)^{2}} ∑_{k=1}^{m+1} \frac{(2π)^{2m+2−2k}}{(2m + 2 − 2k)!} B_{2m+2−2k} ( \frac{1}{2}) b_{k−1, l} + g_{m+1, l}

= −\frac{(−1)^{m+1}}{(l!)^{2}} ∑_{k=0}^{m} \frac{(2π)^{2m−2k}}{(2m − 2k)!} B_{2m−2k} ( \frac{1}{2}) b_{k, l} + g_{m+1, l}

= −\frac{1}{l^{2}} g_{m, l} + g_{m+1, l}

and this implies that

g_{m+1, l+1} + \frac{1}{l^{2}} g_{m, l} = g_{m+1, l} for l ⩾ 1.

We have g_{1, l} = f_{1, l} = 1 for all l ⩾ 1. Let us suppose that g_{m, l} > 0 for all l ⩾ 1. Then g_{m+1, l+1} < g_{m+1, l} and it follows that g_{m+1, l} > lim_{l→∞} g_{m+1, l} . Thanks to Lemma 2.6 we have

(2.10) lim_{l→∞} g_{m+1, l} = (−1)^{m+2} ∑_{k=1}^{m+1} \frac{(2π)^{2m+2−2k}}{(2m + 2 − 2k)!} B_{2m+2−2k} ( \frac{1}{2}) \frac{π^{2k−2}}{(2k − 1)!}

= (−1)^{m+2} π^{2m} ∑_{k=1}^{m+1} \frac{2^{2m+2−2k}}{(2m + 2 − 2k)!(2k − 1)!} B_{2m+2−2k} ( \frac{1}{2})

and using B_{j} (\frac{1}{2}) = 0 for all odd j and the formula

B_{n}(x + y) = ∑_{j=0}^{n} \binom{n}{j} B_{j} (x)y^{n−j}

we check that the sum which appears in (2.10) is equal to

∑_{j=0}^{2m+1} \frac{2^{j}}{j!(2m + 1 − j)!} B_{j} ( \frac{1}{2}) = \frac{2^{2m+1}}{(2m + 1)!} ∑_{j=0}^{2m+1} \binom{2m + 1}{j} B_{j} ( \frac{1}{2}) ( \frac{1}{2})^{2m+1−j}

= \frac{2^{2m+1}}{(2m + 1)!} B_{2m+1} (1) = 0.

Hence g_{m, l} > 0 for m, l ⩾ 1 and this implies, thanks to (2.9), that the numbers e_{m, l} defined by (2.8) are positive for m, l ⩾ 1. □

We are now in position to prove main properties of functions Ψ_{2l−1}^{∗} (·, …, ·, ±a).

Lemma 2.7. Let Ψ_{2l−1}^{∗} (·, …, ·, ±a) be the functions defined in Lemma 2.3. Then

a) (−1)^{n+l+1} Ψ_{2l−1}^{∗} (x_{−m} , …, x_{n}, a) > 0 for x_{−m} , …, x_{n} ∈]− a, a[ .

b) (−1)^{m+l+1} Ψ_{2l−1}^{∗} (x_{−m} , …, x_{n}, −a) > 0 for x_{−m} , …, x_{n} ∈]− a, a[ .

Proof. For pairwise distinct x_{−m} , …, x_{n} ∈]− a, a[ we have

Ψ_{2l−1}^{∗} (x_{−m} , …, x_{n}, ±a) = 2\frac{(4a)^{2l−1}}{(2l)!} ∑_{k=−m}^{n} μ_{k} B_{2l}( \frac{1}{2} + \frac{±a + x_{k}}{4a} )

since the function B_{2m} (\frac{1}{2} + t) is even and then

(−1)^{n+l+1} Ψ_{2l−1}^{∗} (x_{−m} , …, x_{n}, ±a)

= 2\frac{(4a)^{2l−1}}{(2l)!} ( \frac{(−1)^{n}}{α_{0}} ) ∑_{k=−m}^{n} α_{k}(−1)^{l+1} B_{2l}( \frac{1}{2} + \frac{±a + x_{k}}{4a} ) .

The first two terms of the right-hand side are positive and the third term reads ∑ _{k=−m}^{n} α_{k}h_{±} (sin(π \frac{x_{k}}{2a} )) where

h_{±} (t) = (−1)^{l+1} B_{2l}( \frac{3}{4} ± \frac{1}{2π} Arcsint) for t ∈ [−1, 1].

The identities

\frac{3}{4} ± \frac{1}{2π} Arcsint = \frac{1}{2} + \frac{1}{π} Arcsin \sqrt{\frac{1 ± t}{2}} for t ∈ [−1, 1]

together with Lemma 2.4 show that h_{+}^{(m+n)} and (−1)^{m+n} h_{−}^{(m+n)} are positive on ]−1, 1[ and the conclusion holds by Lemma 2.2. □

The last point is to bound the integral which appears in the right-hand side of the identity (2.6). This is the content of Lemma 2.9, whose proof needs the following result.

Lemma 2.8. Let b_{r,s} the numbers defined for integers r ⩾ 4 and s ⩾ 0 by

b_{r,s} = ( \frac{r}{r + s} )^{2r log r−1} \binom{2r + s − 1}{s} .

Then ∑ _{s=0}^{∞} b_{r,s}^{2} = 1 + o(1) as r → ∞.

Proof. We have b_{r,s} = O_{r}(s^{−2r log r+2r} ) = O_{r}(s^{−1} ) and hence ∑ _{s=0}^{∞} b_{4,s}^{2} is convergent. We now prove that b_{r,s} ⩽ b_{4,s} for r ⩾ 4. We have logb_{r,s} = g(r, s) where the function g is defined for (x, y) ∈ [4, ∞[ × [0, ∞[ by

g(x, y) = (2x logx − 1) log( \frac{x}{x + y} ) + logΓ(2x + y) − logΓ(y + 1) − logΓ(2x).

Straightforward computations lead to

\frac{∂g}{∂x}(x, y) = (2 logx + 2) log( \frac{x}{x + y} ) + (2x logx − 1)\frac{y}{x(x + y)} + 2Ψ(2x + y) − 2Ψ(2x)

and

\frac{∂^{2}g}{∂y∂x}(x, y) = −\frac{1 + 2x + 2y + 2y logx}{(x + y)^{2}} + 2Ψ′(2x + y)

where Ψ is the derivative of logΓ. We have \frac{∂g}{∂x} (x, 0) = 0 and moreover since Ψ′(z) = ∑ _{k=0}^{∞} \frac{1}{(z+k)^{2}} , we get Ψ′(z) ⩽ \frac{1}{z} + \frac{1}{z^{2}} for z > 0 and therefore

\frac{∂^{2}g}{∂y∂x}(x, y) ⩽ −\frac{1 + 2x + 2y + 2y logx}{(x + y)^{2}} + \frac{2}{2x + y} + \frac{2}{(2x + y)^{2}}

= − ( \frac{4x^{3} + 2(1 + 3y)x^{2} + (2x − 1)y^{2}}{(x + y)^{2}(2x + y)^{2}} + \frac{2y logx}{(x + y)^{2}} ) ⩽ 0.

Hence \frac{∂g}{∂x} (x, y) ⩽ \frac{∂g}{∂x} (x, 0) = 0 and this implies that g(x, y) ⩽ g(4, y) and hence b_{r,s} ⩽ b_{4,s} for r ⩾ 4.

Let ϵ > 0 and s_{0} be such that ∑ _{s=s_{0}}^{∞} b_{4,s}^{2} ⩽ \frac{ϵ}{2}. Since b_{r,s} → 0 as r → ∞ for s ⩾ 1, there exists r_{0} such that ∑ _{s=1}^{s_{0} −1} b_{r,s}^{2} ⩽ \frac{ϵ}{2} for r ⩾ r_{0}. Hence

1 ⩽ ∑_{s=0}^{∞} b_{r,s}^{2} ⩽ b_{r,0}^{2} + ∑_{s=1}^{s_{0} −1} b_{r,s}^{2} + ∑_{s=s_{0}}^{∞} b_{4,s}^{2} ⩽ 1 + ϵ for r ⩾ r_{0}.

The proof is complete. □

Lemma 2.9. For positive integers m, n and l which satisfy m + n ⩾ 4 and l ⩾ (m + n) log(m + n) and for x_{−m} , …, x_{n} ∈]−a, a[, let Ψ_{2l−1}^{∗} be the function defined in Lemma 2.3. Then

∥Ψ_{2l−1}^{∗} ∥_{2} = \frac{2^{m+n−1}}{|α_{0}|\sqrt{a}} ( \frac{2a}{(m + n)π} ) ^{2l} (1 + o(1)) as m + n → ∞

where

∥Ψ_{2l−1}^{∗} ∥_{2}^{2} = ∫_{−a}^{a} (Ψ_{2l−1}^{∗} (x_{−m} , …, x_{n}, x))^{2}dx.

Proof. The use of the Fourier series expansion

B_{2l}(x) = (−1)^{l+1} 2((2l)!) ∑_{j=1}^{∞} \frac{1}{(2jπ)^{2l}} cos(2jπx) for x ∈ [0, 1]

and the identity cosα + cosβ = 2 cos(\frac{α+β}{2} ) cos(\frac{α−β}{2} ) lead, for pairwise distinct x_{−m} , …, x_{n} ∈]−a, a[, to the expression

Ψ_{2l−1}^{∗} (x_{−m} , …, x_{n}, x)

= (−1)^{l+1} 2 \frac{(2a)^{2l−1}}{α_{0}π^{2l}} ∑_{j=1}^{∞} \frac{1}{j^{2l}} ( ∑_{k=−m}^{n} α_{k} cos( jπ( \frac{1}{2} + \frac{x_{k}}{2a}) ) ) cos( jπ( \frac{1}{2} + \frac{x}{2a}) ) .

Using the identity cos(jπ(\frac{1}{2} + y)) = (−1)^{j} T_{j} (sin(πy)) and introducing the numbers a_{j,k} = (−1)^{j} T_{j} (sin(π \frac{x_{k}}{2a} )) we have ∑ _{k=−m}^{n} α_{k} a_{j,k} = 0 for j = 1, …, m + n − 1, this is crucial, and, therefore

(2.11) Ψ_{2l−1}^{∗} (x_{−m} , …, x_{n}, x)

= (−1)^{l+1} 2 \frac{(2a)^{2l−1}}{α_{0}π^{2l}} ∑_{j=m+n}^{∞} \frac{1}{j^{2l}} ( ∑_{k=−m}^{n} α_{k}a_{j,k} ) cos( jπ( \frac{1}{2} + \frac{x}{2a}) ) .

Using Lemma 2.2, squaring (2.11) and integrating on the interval [−a, a], we get

(2.12)  ∥Ψ_{2l−1}^{∗} ∥_{2}^{2} = ( 2 \frac{(2a)^{2l−1}}{α_{0}π^{2l}} ) ^{2} ∑_{j=m+n}^{∞} \frac{1}{j^{4l}} ( \frac{T_{j}^{(m+n)} (τ_{j})}{(m + n)!}) ^{2}a

for some τ_{j} ∈]−1, 1[. It is well known [10] that

max_{−1⩽x⩽1} |T_{j}^{(m+n)} (x)| = T_{j}^{(m+n)} (1) = 2^{m+n−1} (m + n − 1)! j\binom{m + n + j − 1}{j − m − n}

for j = m + n, m + n + 1, … and then

∥Ψ_{2l−1}^{∗} ∥_{2}^{2} ⩽ ( \frac{2^{m+n−1}}{α_{0} \sqrt{a}} ( \frac{2a}{(m + n)π} )^{2l}) ^{2} ∑_{j=m+n}^{∞} ( \frac{m + n}{j} )^{4l−2} \binom{m + n + j − 1}{j − m − n} ^{2}.

We set j = m + n + s and since l ⩾ (m + n) log(m + n) we have

∥Ψ_{2l−1}^{∗} ∥_{2}^{2} ⩽ ( \frac{2^{m+n−1}}{α_{0} \sqrt{a}} ( \frac{2a}{(m+n)π} )^{2l})^{2} ∑_{s=0}^{∞} (( \frac{m + n}{m + n + s} )^{2l−1} \binom{2(m+n)+s−1}{s}) ^{2}

⩽ ( \frac{2^{m+n−1}}{α_{0} \sqrt{a}} ( \frac{2a}{(m + n)π} )^{2l}) ^{2} ∑_{s=0}^{∞} b_{m+n,s}^{2}

= ( \frac{2^{m+n−1}}{α_{0} \sqrt{a}} ( \frac{2a}{(m + n)π} )^{2l}) ^{2}(1 + o(1))

as m + n → ∞, thanks to Lemma 2.8. To complete the proof we compute a lower bound for ∥Ψ_{2l−1}^{∗} ∥_{2}^{2} using the first term in the series which appears in the right-hand side of (2.12). □

3. Proof of Theorem

In this section we assume that the Riemann hypothesis is true. Our result is a consequence of identity (2.6) which for Z(T) gives

(3.1) Z(T) = ∑_{k=1}^{K} Z^{(2k−1)} (T + a)Ψ_{2k−1}^{∗} (a) − ∑_{k=1}^{K} Z^{(2k−1)} (T − a)Ψ_{2k−1}^{∗} (−a)

− ∫_{−a}^{a} Z^{(2K)} (T + x)Ψ_{2K−1}^{∗} (x) dx

where 2K ⩾ m + n + 2 and for short Ψ_{2k−1}^{∗} (±a) and Ψ_{2m−1}^{∗} (x) stand respectively for Ψ_{2k−1}^{∗} (x_{−m} , …, x_{n}, ±a) and Ψ_{2m−1}^{∗} (x_{−m} , …, x_{n}, x) and x_{k} = γ_{k} − T. The main step in the proof is to bound the integral which appears in the right-hand side of (3.1). In our proof we use the bound

(3.2)  | ∫_{T}^{T +h} S(u) du| ⩽ \frac{π}{16} \frac{logT}{(loglogT)^{2}} + O( \frac{logT logloglogT}{(loglogT)^{3}})

for 0 < h ⩽ \sqrt{T}, due to Carneiro et al. [3].

Lemma 3.1. Let T be sufficiently large, 4(loglogT)^{−1} ⩽ a ⩽ \sqrt{T}, m and n be the integers defined in the introduction and let α_{0} be the coefficient defined in Lemma 2.3 where x_{0} = 0 and x_{k} = γ_{k} − T. Then

(3.3)  − log|α_{0}| ⩽ −\frac{2a}{π}θ′(T) log2 + \frac{1}{\sqrt{2}} \frac{logT}{loglogT}( 1 + O( \frac{logloglogT}{loglogT}) ) .

186 BLANC

Further, if S(T + a) − S(T − a) ⩾ 1, then

(3.4)  m + n = \frac{2a}{π}θ′(T) + ΔS( 1 + O( \frac{a^{3}}{T^{2}}) )

where ΔS = S(T + a) − S(T − a).

Proof. By definition

| \frac{1}{α_{0}} | = ∏_{\begin{matrix}−m⩽j⩽n \\ j≠0 \end{matrix}} | sin( π \frac{x_{j}}{2a}) |

and using (1.1) and Stieltjes integral we have

− log|α_{0}| = ∫_{T −a}^{T +a} log | sin( π \frac{t − T}{2a} ) | d( \frac{1}{π}θ(t) + 1 + S(t))

and an integration by parts leads to

− log|α_{0}| = \frac{1}{π} ∫_{T −a}^{T +a} θ′(t) log | sin( π \frac{t − T}{2a} ) | dt

− \frac{π}{2a} ∫_{T −a}^{T +a} cot( π \frac{t − T}{2a} ) (S(t) − S(T)) dt.

Now for t ∈ [T − a, T + a] we have

θ′(t) = θ′(T) + θ′′(T)(t − T) + \frac{1}{2}θ′′′(τ)(t − T)^{2} for some τ ∈ [T − a, T + a]

and using θ′′′(t) = O(1/t^{2}) together with

∫_{T −a}^{T +a} log | sin( π \frac{t − T}{2a} ) | dt = \frac{4a}{π} ∫_{0}^{π/2} logsinτ dτ = −2a log2

we get

(3.5)  − log|α_{0} | = −\frac{2a}{π}θ′(T) log2

− \frac{π}{2a} ∫_{T −a}^{T +a} cot( π \frac{t − T}{2a} ) (S(t) − S(T)) dt + O( \frac{a^{3}}{T^{2}}) .

Further, for η ∈]0, a] we have

∫_{T −a}^{T +a} cot( π \frac{t − T}{2a} ) (S(t) − S(T)) dt

= ∫_{T −a}^{T −η} cot( π \frac{t − T}{2a} ) S(t) dt + ∫_{T −η}^{T} cot( π \frac{t − T}{2a} ) (S(t) − S(T)) dt

+ ∫_{T}^{T +η} cot( π \frac{t − T}{2a} ) (S(t) − S(T)) dt + ∫_{T +η}^{T +a} cot( π \frac{t − T}{2a} ) S(t) dt

and we bound from below the third term using

S(t) − S(T) ⩾ −\frac{1}{π}(θ(t) − θ(T)) ⩾ −\frac{1}{π}θ′(T + a)(t − T) ⩾ −\frac{1}{2π} logT (t − T)

which, together with the inequality cotx < \frac{1}{x} for x ∈]0, \frac{π}{2} ], implies

∫_{T}^{T +η} cot( π \frac{t − T}{2a} ) (S(t) − S(T)) dt ⩾ −\frac{ηa}{π^{2}} logT.

Applied to the fourth term, the second mean-value theorem gives

∫_{T +η}^{T +a} cot( π \frac{t − T}{2a} ) S(t) dt = cot( π \frac{η}{2a}) ∫_{T +η}^{τ} S(t) dt
⩾ −\frac{2a}{πη} ( \frac{π}{16} \frac{logT}{(loglogT)^{2}} + O( \frac{logT logloglogT}{(loglogT)^{3}}) )

where τ ∈ [T + η, T + a] and (3.2) has been used. Proceeding in the same way to bound from below the first and second term we finally get

− \frac{π}{2a} ∫_{T −a}^{T +a} cot( π \frac{t − T}{2a} ) (S(t) − S(T)) dt

⩽ \frac{η}{π} logT + \frac{1}{η}( \frac{π}{8} \frac{logT}{(loglogT)^{2}} + O( \frac{logT logloglogT}{(loglogT)^{3}}) ) .

We choose η = π(2\sqrt{2}loglogT )^{−1} and we complete the proof of (3.3) using (3.5). Since m + n = N(T + a) − N(T − a), the Taylor formula leads to

m + n = \frac{1}{π}(θ(T + a) − θ(T − a)) + ΔS = \frac{2a}{π}θ′(T) + ΔS( 1 + O( \frac{a^{3}}{T^{2}}) ) .

which proves (3.4). □

Proof of Theorem 1.1. Either K ⩽ K^{∗} logK^{∗} and there is nothing to prove or K > K^{∗} logK^{∗} and this implies that K ⩾ (m + n) log(m + n). Assuming T large enough we have also m + n ⩾ 4. By definition of K and thanks to Lemma 2.7, the first sum in the right-hand side of identity (3.1) is nonpositive and the second one is nonnegative. Using Cauchy–Schwarz inequality, (1.4) and Lemma 2.9, we get

Z(T) ⩽ ∥Z^{(2K)} ||_{2} ∥Ψ_{2K−1}^{∗} ∥_{2} ⩽ C_{2K} Z(T)

where

C_{2K} = c_{2K} 2^{m+n− \frac{1}{2}} |α_{0}|^{−1} ( \frac{2aθ′(T)}{(m + n)π} ) ^{2K} (1 + o(1))

⩽ c_{2K} 2^{m+n} |α_{0}|^{−1} ( 1 + π \frac{ΔS}{2aθ′(T)}( 1 + O( \frac{a^{3}}{T^{2}}) ) ) ^{−2K}

for T sufficiently large. Since C_{2K} ⩾ 1 we have logC_{2K} ⩾ 0 and this implies that

logc_{2K} + (m + n) log2 − log|α_{0} | − 2K log( 1 + π \frac{ΔS}{2aθ′(T)}( 1 + O( \frac{a^{3}}{T^{2}}) ) ) ⩾ 0.

Finally, as ΔS = O( logT (loglogT)^{−1} ) and thanks to Lemma 3.1, we get

2Kπ \frac{ΔS}{2aθ′(T)}( 1 + O( \frac{1}{loglogT}) )

⩽ ( logc_{2K} + ΔS log2 + \frac{1}{\sqrt{2}} \frac{logT}{loglogT}) ( 1 + O( \frac{logloglogT}{loglogT}) )

and the proof is complete. □

Acknowledgements. I am grateful to Professor Jean Descloux for his encouragements, to my colleagues Jean-François Hêche, Fred Lang, Eric Thiémard and Jacques Zuber for their help and to my wife, Ariane, a very creative artist.

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Département des technologies industrielles (Received 19 12 2013) Haute École d’Ingénierie et de Gestion CH-1400 Yverdon-les-Bains Switzerland

 philippe.blanc@heig-vd.ch