The First Moment of $L(\frac{1}{2},\chi)$ for Real Quadratic Function Fields
J.C. Andrade, J. MacMillan

TL;DR
This paper refines the asymptotic understanding of the first moment of quadratic Dirichlet L-functions over function fields, identifying new main terms and bounding the error with advanced techniques.
Contribution
It extends Florea's methods to include additional main terms for the first moment of quadratic L-functions over function fields, addressing technical challenges.
Findings
Identifies new main terms of size $(2g+2)q^{(2g+2)/3}$, $q^{g/6+[rac{g}{2}]}$, and $q^{g/6+[rac{g-1}{2}]}$.
Bounds the error term by $q^{g/2(1+\epsilon)}$.
Improves asymptotic formulas for the first moment of quadratic L-functions in function fields.
Abstract
In this paper we use techniques first introduced by Florea to improve the asymptotic formula for the first moment of the quadratic Dirichlet L-functions over the rational function field, running over all monic, square-free polynomials of even degree at the central point. With some extra technical difficulties that doesn't appear in Florea's paper, we prove that there are extra main terms of size and , whilst bounding the error term by .
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Taxonomy
TopicsAnalytic Number Theory Research · Coding theory and cryptography · Algebraic Geometry and Number Theory
The First Moment of for Real Quadratic Function Fields
J.C. Andrade
J. MacMillan
††2010 Mathematics Subject Classification: Primary 11M38; Secondary 11M06, 11G20
Date: August 09, 2019
ABSTRACT: In this paper we use techniques first introduced by Florea to improve the asymptotic formula for the first moment of the quadratic Dirichlet L-functions over the rational function field, running over all monic, square-free polynomials of even degree at the central point. With some extra technical difficulties that doesn’t appear in Florea’s paper, we prove that there are extra main terms of size and , whilst bounding the error term by .
1 Introduction
An important and well-studied problem in analytic number theory is to understand the asymptotic behaviour of moments of families of L-functions. Considering the family of Dirichlet L-functions, , with a real primitive Dirichlet character modulo defined by the Jacobi symbol , a problem is to understand the asymptotic behaviour of
[TABLE]
summing over fundamental discriminants , as . In this context, Jutila, [15], proved, when , that
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where and
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Goldfeld and Hoffstein, [12], improved the error term to . Young, [24], showed that the error term is bounded by when considering the smoothed first moment. Jutila, [15], computed the second moment and Soundararajan, [22], computed the second and third moments, when averaging over real, primitive, even characters with conductors . It is conjectured that
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where the sum is over fundamental discriminants. Keating and Snaith, [18], conjectured a precise value for and Conrey et al, [8], conjectured the integral moments and formulas for the principal lower order terms.
In the function field setting, the analogue problem is to understand the asymptotic behaviour of
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as , where denotes the space of monic, square-free polynomials of degree over , which corresponds to the imaginary quadratic function field, and denotes the quadratic Dirichlet L-function for the rational function field. Since we are letting , there are two limits to consider. The first is to fix and let and the second is to fix and let . Katz and Sarnak, [16, 17] used equidistribution results to relate the limit of (1.3) to a random matrix theory integral, which was then computed by Keating and Snaith, [18]. Therefore we will concentrate on the other limit, namely when is fixed and we let . In this context, Andrade and Keating, [3], computed the first moment of (1.3), when , which is considered to be the function field analogue of Jutila’s result (1.2). In particular they proved the following result.
Theorem 1.1** (Andrade and Keating).**
Let be the fixed cardinality of the ground field and assume for simplicity that . Then
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where
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and denotes the zeta function associated with .
Andrade and Keating, [4], also conjectured asymptotic formulas for higher and integral moments for (1.3) which is considered to be the function field analogue of Keating and Snaith’s result, [18] and Conrey et al, [8]. Rubinstein and Wu, [20], provided numerical evidence for the conjecture given by Andrade and Keating, [4]. They numerically computed the moments for and , where , for various values of and compared them to the conjectured formulas.
Using a similar method to Young’s, [24], in the number field case, Florea, [9], improved the asymptotic formula obtained by Andrade and Keating (1.4) by obtaining a secondary main term of size and bounding the error term by .
Theorem 1.2** (Florea).**
Let be a prime with . Then
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where is a polynomial of degree 1 that can be explicitly calculated.
Using a similar method, Florea, [10, 11], computed the second, third and fourth moments of (1.3) at and showed that these asymptotic formulas agree with the formulas conjectured by Andrade and Keating in [4].
In [1], Andrade obtained an asymptotic formula for the first moment of at . In particular, he proved that
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Using the techniques presented by Florea, Andrade and Jung, [2], improved the asymptotic formula, (1.6), by obtaining a secondary main term of size and bounding the error term by , for any . In particular, they proved that
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where is a constant that can be explicitly calculated.
In a recent paper, Bae and Jung, [7], improved the asymptotic formula for the second derivative of (1.3) at , that was obtained by Andrade and Rajagopal, [5], using the techniques presented by Florea. In particular, compared to the asymptotic formula obtained by Andrade and Rajagopal, Bae and Jung were able to obtain a secondary main term of size whilst also bounding the error term by
Another problem in function fields is to understand the aymptotic behaviour of
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as , where denotes the space of monic, square-free polynomials of degree , which corresponds to the real quadratic function field. In particular we concentrate on when is fixed and letting . In this context, Jung, [13], obtained an asymptotic formula for the first moment of (1.8) at .
Theorem 1.3** (Jung).**
Assume that is odd and greater than 3. Then we have
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In this paper, we will use Florea’s method to improve the asymptotic formula obtained by Jung (1.9). In particular we will obtain secondary main terms of size and , whilst also bounding the error term by . The main result of this paper is the following Theorem.
Theorem 1.4**.**
Let be a prime with . Then
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where is a polynomial of degree 1 and and are constants that can explicitly be computed (see formulas (5.46), (5.47) and (5.48)).
The calculations in this paper will follow the techniques presented in Florea, [9]. However we encounter extra difficulties and extra terms are present when computing the first moment of for real quadratic function fields, compared to the calculations of the first moment of for imaginary function fields.
2 Preliminaries and Background
We first introduce the notation which will be used throughout the article and then provide some background information on Dirichlet L-functions in function fields. We denote to be the set of all monic polynomials in and we denote and to be the set of all monic polynomials of degree and degree at most in respectively. Let denote the space of monic, square-free monic polynomials over of degree . For a polynomial , we denote its degree by and its norm by . The letter denotes a monic, irreducible polynomial over .
2.1 Preliminaries on Dirichlet characters and Dirichlet L-functions for function fields
Most of the facts in this section are proved in [19]. For , the zeta function of , denoted by is defined by the infinite series
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There are monic polynomials of degree , therefore we have
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We will make use of the change of variables , so that we write and thus .
Assume that is odd with . For a monic irreducible polynomial, the quadratic residue symbol is defined by
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for coprime to . If , then . We can also define the Jacobi symbol for arbitrary monic . Let be coprime to and , then the Jacobi symbol is defined by
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Theorem 2.1** (Quadratic Reciprocity).**
Let be relatively prime and and . Then
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If we assume that , then the quadratic reciprocity gives
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For we have
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Definition 2.2**.**
Let be square-free. We define the quadratic character using the quadratic residue symbol for by
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Therefore, if , we have
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Definition 2.3**.**
The L-function corresponding to the quadratic character by
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which converges for . For the change of variables , we have
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Since is a monic, square-free polynomial, we have, from Proposition 4.3 in [19], that is a polynomial in of . From [21], has a trivial zero if and only if is even. This enables us to define the completed L-function, , by
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where if is even and otherwise. Then is a polynomial in of degree and satisfies the functional equation
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For a monic, square-free polynomial of degree or , the affine equation defines a projective and connected hyperelliptic curve of genus over . The zeta function associated to was first introduced by Artin, [6], and is defined by
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where is the number of points on with coordinates in a field extension of of degree . Weil, [23], showed that
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where is a polynomial of degree . In his thesis, Artin, proved that . Also, Weil, [23], proved the Riemann Hypothesis for function fields, thus all the zeros of lie on the circle .
2.2 Functional Equation and Preliminary Lemmas
For , the approximate functional equation was initially proved in Jung, [13], but has been corrected to match that of [20].
Lemma 2.4**.**
Let be a quadratic character, where . Then
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Proof.
See [13], Lemma 2.1. ∎
Using Lemma 2.4, it follows that
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We now state two Lemmas that will be used in the calculations later.
Lemma 2.5**.**
Let . Then we have
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where means that any prime factor of are among the prime factors of .
Proof.
See [9], Lemma 2.2 ∎
We now state a version of Poisson summation formula over function fields. For , let , where is the coefficient of in the expansion of (for more information, see [9], section 3). For a general character modulo , the generalised Gauss sum is defined as
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The following Poisson summation formula holds.
Lemma 2.6**.**
Let and let be a positive integer.
If is odd, then
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If is even, then
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Proof.
See [9], Proposition 3.1. ∎
Remark 2.7**.**
is nonzero if and only if is a square, in which case , where is Euler’s phi function for polynomials in .
Lemma 2.8** (The function field analogue of Perron’s formula).**
If the power series
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converges absolutely for , then
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and
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3 Setup of the Problem
Using Lemma 2.5 and the approximate functional equation (2.2), we write
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where
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and
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From section 4 in [9], we have
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we see that the terms in and corresponding to are bounded by . Therefore we can rewrite the terms as
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and
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For and , write
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where and denotes the sum over of odd and even degree respectively. If is odd, then using Lemma 2.6, we have
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and
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where
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If is even, then we rewrite as
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Using the remark from the previous section, we have
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and
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Similarly, for we have
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and
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where
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and
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Define to be the sum over square and to be the sum over non-square . Note that in Equation (3.22), when is even, is odd and so cannot be a square. Also note that in Equation (3.11), when is odd, is even, thus there is a contribution to the main term when is odd, which is not present when working in the imaginary function field case.
4 Main Term
In this section we evaluate the main terms, and . The main result in this section is the following result.
Proposition 4.1**.**
For any , we have that
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and
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where and
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Remark 4.2**.**
Let
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Remark 4.3**.**
is analytic in . We may further write
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which furnishes an analytic continuation of to the region .
Proof of Proposition 4.1.
From (3.13), (3.14), (3.15) and (3.16) and using the facts that (see [9], Section 5),
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we have
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and
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Using the function field analogue of Perron’s formula (Lemma 2.8), we have
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and
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where and
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By multiplicativity, we may write
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Inserting (4.14) into (4.9), (4.10), (4.11) and (4.12) the Proposition follows. ∎
5 Contribution from V-square
Let
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where
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and
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In this section we will evaluate the term . The next Proposition is the main result in this section.
Proposition 5.1**.**
Using the same notation as before, we have that
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where
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and
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with and
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Furthermore is a linear polynomial and and are constants that can be explicitly calculated, (see (5.46), (5.47) and (5.48)).
Before we prove Proposition 5.1, we need the following notation and subsequent results. For , let
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where
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Then we have the following results.
Lemma 5.2**.**
For , we have
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where
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Moreover converges absolutely for and .
Proof.
See [9], Lemma 6.2. ∎
Lemma 5.3**.**
We have
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where
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Moreover converges absolutely for and .
Proof.
See [9], Lemma 6.3 ∎
Outline of the Proof of Proposition 5.1: From the Poisson summation formula the sum over square polynomials will occur when the degree of is even and when the degree of is odd. In the next two subsections, we will find two integrals for each corresponding to simple poles at and . In the third subsection we will manipulate the integrals corresponding to the pole at , similar to that done in section 6, [9], which will yield the main terms. In the final subsection, we will evaluate the integrals corresponding to the pole at , which will yield the secondary main terms.
5.1 Degree even
In this subsection, we prove the following result.
Lemma 5.4**.**
We have
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where and are the integrals stated at the end of the subsection.
Proof.
From (3.21) and using the function field analogue of Perron’s formula, we obtain
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Also, using the fact that (see [9], Proof of Lemma 6.1),
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we have
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where
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Similarly we have
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and
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where
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and
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Using the function field analogue of Perron’s formula, we have
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and
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The second integrals in (5.8), (5.9), (5.10) and (5.11) are zero since the integrands have no poles inside the circle . Therefore we have
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and
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Using equation (5.5) in Lemma 5.2 we obtain
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and
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Using equation (5.6) in Lemma 5.3, we obtain
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and
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Shrinking the contour to , we do not encounter any poles. Enlarging the contour to , we encounter two simple poles, one at and one at . Evaluating the residues at and and writing
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whilst using Lemma 6.3 where we have for each , , then we have
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[TABLE]
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and
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∎
5.2 Degree odd
In this subsection, we prove the following result.
Lemma 5.5**.**
We have
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where and are the integrals stated at the end of the subsection.
Proof.
From (3.11) and using the function field analogue of Perron’s formula, we have
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Also, using the fact that (see, [9], Proof of Lemma 6.1),
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we have
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where
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Similarly we have
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and
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where
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and
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Using the function field analogue of Perron’s formula, we have
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[TABLE]
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and
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The second integrals in (5.14), (5.15), (5.16) and (5.17) are zero since the integrands have no poles inside the circle . Therefore we have
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and
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Using equation (5.5) in Lemma 5.2 we have
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and
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Using equation (5.6) in Lemma 5.3, we have
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and
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Shrinking the contour to , we do not encounter any poles. Enlarging the contour to , we encounter two simple poles, one at and one at . Evaluating the residues at and and writing
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whilst using Lemma 6.3 where we have for each , , then we have
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[TABLE]
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and
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∎
5.3 Contribution from Terms
In this subsection, we will focus on evaluating the terms which corresponds to the pole at , these will give the main terms in Proposition 5.1. Let
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then, the main result in this subsection is the following.
Lemma 5.6**.**
Using the same notation as before, we have
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where, in particular, the terms and are the integrals stated in Proposition 5.1.
Proof.
Rewrite and as
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and
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Then, for , let
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where
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and
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Using the change of variables , the contour of integration becomes the circle around the origin and note that (from Lemma 5.2) is absolutely convergent for . We have
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and
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Using the fact (see [9], section 6) that
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we get that
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and
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We see that that (5.20) and (5.21) are precisely the terms and given in the statement of Lemma 5.6. Similarly, using the substitution we have
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[TABLE]
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and
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Using a variant of (5.19) we have
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[TABLE]
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and
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Rewrite as
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Then, we let
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where
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and
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Combining and , we have
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Using the fact that (see [13], Proof of Main Theorem)
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we have
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Let
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where
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and
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Similarly combining and and using (5.27), we have
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where
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and
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We see that that and are precisely the terms and given in the statement of Lemma 5.6. From (4.3), we have that , thus, inside the circle , has a pole of order at . Using the Residue Theorem we have that
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Similarly, inside the circle , the integrals and have a simple pole at and a pole at of order and respectively. Thus we have
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and
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For the remaining integrals, we rewrite and as
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where
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and
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Using the substitution we have
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[TABLE]
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[TABLE]
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and
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Using a variant of (5.19), we have
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[TABLE]
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and
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Inside the circle , the integrals all have poles at and of varying orders, thus using the Residue Theorem, we have that
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[TABLE]
[TABLE]
[TABLE]
[TABLE]
[TABLE]
To complete the proof, we want to show that
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equals zero. Using the fact that (see [14], section 1)
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we see that the terms corresponding to residue at and equal zero. Finally, we use induction on (see appendix) to show that the terms corresponding to the residue at equals zero. Thus (5.44) equals zero. ∎
5.4 Contribution from Terms
We will now focus on evaluating the terms which corresponds to the pole at , these will give the secondary main terms in Proposition 5.1. Let
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then, the main result in this subsection is the following.
Lemma 5.7**.**
Using the same notation as before, we have that
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where is a polynomial of degree 1 given by (5.46) and and are constants given by (5.47) and (5.48) respectively.
Proof.
For each and we write
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Enlarging the contour to we encounter a double pole at of and a simple pole at of . From Lemma 5.3, is absolutely convergent when . Then, we have
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and
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where the second terms are bounded by . Then
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where each is a linear polynomial whose coefficients can be computed explicitly. Let
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where
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and
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Also we write
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where each are constants that can be explicitly computed. Let
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then we have
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and
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Moreover,
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and
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∎
Proposition 5.1 is immediate from Lemma 5.4, Lemma 5.5, Lemma 5.6, Lemma 5.7 and (5.1).
6 Error From Non-Square
Let
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where
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and
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Then, in this section we will bound the term . The next Proposition is the main result in this section.
Proposition 6.1**.**
Using the notation described previously, we have, for any
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To prove Proposition 6.1, we will need the following results (see [9], section 7). We have
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with . For a non-square and positive integer , let
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Then, if , then we have
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6.1 Bounding
For each and , we have
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Write
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and
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where , and denote the sum over non-square of degree and respectively. Similarly and denote the sum over with and respectively. Then, by (6.5), we can write
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and
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with . Using (6.6), we can bound and trivially bounding the sum over , we get that , , and , thus . Using the same calculations, we can bound and by , hence .
6.2 Bounding
For , let
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where and denotes the sum over non-square with and respectively. Then, using (6.5), we have
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and
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with . Using (6.6) to bound and trivially bounding the sum over , we get that and . Thus . Similar calculations can be used to bound , and by . Therefore , which proves Proposition 6.1.
7 Proof of Theorem 1.4
We combine the results from the previous sections to prove Theorem 1.4.
Proof of Theorem 1.4..
Using (3.1), we have
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Using equations stated in previous sections, we can rewrite (7.1) as
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Using Proposition 4.1, Proposition 5.1 and Proposition 6.1, we have
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By Remark 4.3, has an analytic continuation for and , therefore between the circles and , the integrands corresponding to the terms and have a double pole at . Similarly the integrands corresponding to the terms and have a simple pole at . Computing the residue at , we get that
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[TABLE]
[TABLE]
and
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where with the change of variables . Putting everything together and using equation (5.27) the Theorem follows. ∎
Appendix A Completing the Proof of Lemma 5.6
In the appendix we prove the claim that
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equals zero. For the terms corresponding to the residues at and we have shown that (A.1) equals zero, thus it remains to show that for the terms corresponding to the residue at , (A.1) equals zero. We prove this using induction on .To do this we consider two cases, the first when is even and second when is odd.
A.1 even
Let for , then we will show that equals zero for all For the base case, , we have that (A.1) is equal to
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which when cancelling the terms equals zero, hence the base case is true. Assume that (A.1), for . Then we have that
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It remains to show that (A.1) is equal to zero for . For , we have that (A.1) is equal to
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Rearranging (A.1), we have that (A.1) is equal to
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Using the inductive hypothesis, we have that (A.1) equals zero, therefore it remains to show that (A.6) equals zero. Rearranging (A.6) we get that it is equal to
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Thus (A.1) for and so, by induction, (A.1) for all even.
A.2 odd
Now let , then we want to show, using induction on that (A.1) equals zero for all For the base case, , we have that (A.1) is equal to
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which, when cancelling the terms equals zero, hence the base case is true. Assume that (A.1) for . Then we have that
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It remains to show that (A.1) is equal to zero for . For , we have that (A.1) is equal to
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Rearranging (A.2), we have that (A.1) is equal to
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Using the inductive hypothesis, we have (A.9) equals zero. Thus it remains to show that (A.10) equals zero. Rearranging (A.10), we have that it equals
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Thus (A.1) for and so by induction (A.1) for all odd. This completes the proof of Lemma 5.6.
Acknowledgement: The first author is grateful to the Leverhulme Trust (RPG-2017-320) for the support through the research project grant ”Moments of L-functions in Function Fields and Random Matrix Theory”. The second author is also grateful to the Leverhulme Trust (RPG-2017-320) for the support given during this research through a PhD studentship.
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