The fourth moment of quadratic Dirichlet L -functions
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Mathematische Zeitschrift
The fourth moment of quadratic Dirichlet L-functions Quanli Shen1 Received: 8 October 2019 / Accepted: 27 July 2020 © Springer-Verlag GmbH Germany, part of Springer Nature 2020
Abstract We study the fourth moment of quadratic Dirichlet L-functions at s = 21 . We show an asymptotic formula under the generalized Riemann hypothesis, and obtain a precise lower bound unconditionally. The proofs of these results follow closely arguments of Soundararajan and Young (J Eur Math Soc 12(5):1097–1116, 2010) and Soundararajan (Ann Math (2) 152(2):447–488, 2000). Keywords Moments of L-functions · Quadratic Dirichlet L-functions Mathematics Subject Classification 11M06 · 11M50
1 Introduction Let χd = d· be a real primitive Dirichlet character modulo d given by the Kronecker symbol, where d is a fundamental discriminant. The k-th moment of quadratic Dirichlet L-functions is
L( 21 , χd )k ,
(1.1)
0 1, L c (s, χ) is given by the Euler product of L(s, χ) with omitting all prime factors of c. The last equation follows by the definition of h(x, y, z) in (3.5). Moving the lines of the integral to Re(u) = Re(v) = log1 X , the double integral above is bounded by ∗ 2 4
(log X )τ (c) |g(u)g(v)| |L( 21 + u, χ8l )|4 |du| |dv|. (3.8) ( log1 X ) ( log1 X )
(l,2)=1 l≤ 5X2 2c
Here we use the inequalities 2ab ≤ and |L c ( 21 + u, χ8l )| ≤ τ (c)|L( 21 + u, χ8l )|. By Theorem 2.4, we see that for |Im(u)| ≤ cX2 , a2
∗
+ b2
|L( 21 + u, χ8l )|4
(l,2)=1 l≤ 5X2
X log11 X . c2
(3.9)
2c
Also, by Lemma 2.3, we get that ∗
|L( 21
+ u, χ8l )|
4
(l,2)=1 l≤ 5X2
X c2
1+ε (1 + |Im(u)|)1+ε .
2c
Substituting both (3.9) and (3.10) in (3.8), we can bound (3.8) by
123
τ 4 (c) X log13 X . c2
(3.10)
The fourth moment of quadratic Dirichlet L-functions
Together with (3.7), this yields τ 4 (c) τ 5 (c) 13 1
X log X
X Y −1 log44 X . c2 c2
S2 X log13 X
(c,2)=1
a>Y a|c
c>Y
This completes the proof of the conditional part of the lemma. The unconditional part follows similarly by substituting (3.10) in (3.8).
Now we consider S1 . Using the Poisson summation formula (see Lemma 2.1) for the sum over d in S1 , we obtain that S1 = 2X
×
μ(a) (−1)k a2
a≤Y (a,2)=1 ∞ −∞
k∈Z
(n 1 ,2a)=1 (n 2 ,2a)=1
h(x X , n 1 , n 2 )(cos + sin)
2πkx X 2n 1 n 2 a 2
τ (n 1 )τ (n 2 ) G k (n 1 n 2 ) √ n1n2 n1n2 d x.
(3.11)
Let S1 (k = 0) denote the sum above over k = 0, which are called diagonal terms. Let S1 (k = 0) denote the sum over k = 0. Write S1 (k = 0) = S1 (k = ) + S1 (k = ), where S1 (k = ) denotes the terms with square k, and S1 (k = ) denotes the remaining terms. We call S1 (k = ) off-diagonal terms. We will discuss S1 (k = 0), S1 (k = ), and S1 (k = ) in Sects. 4–6, respectively.
4 Evaluation of S1 (k = 0) In this section, we shall extract one main term of S1 from S1 (k = 0). The argument here is similar to [19, Section 3.2]. It follows from the definition of G k (n) in (2.4) that G 0 (n) = φ(n) if n = , and G 0 (n) = 0 otherwise. By this fact and (3.11), we
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