The symmetric-square L-function on the critical line

Journal of Number Theory 140 (2014) 196–214

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Journal of Number Theory www.elsevier.com/locate/jnt

The symmetric-square L-function on the critical line ✩ Qingfeng Sun School of Mathematics and Statistics, Shandong University, Weihai, Weihai, Shandong 264209, China

a r t i c l e

i n f o

Article history: Received 10 April 2011 Accepted 2 January 2014 Available online 3 March 2014 Communicated by Ph. Michel MSC: 11F67

a b s t r a c t We prove an asymptotic formula for the first moment of the symmetric-square L-functions L(sym2 f, s) on the critical line for cusp forms f of weight k, level N and real primitive character χ modulo N , as N → ∞. It follows that for any t ∈ R, there exists f such that L(sym2 f, 1/2 + it) = 0. © 2014 Elsevier Inc. All rights reserved.

Keywords: Symmetric-square L-function First moment

1. Introduction Let N  1 be an odd square-free integer and χ a real primitive character modulo N satisfying the consistency condition χ(−1) = (−1)k . Let Sk (N, χ) denote the finite dimensional Hilbert space of cusp forms of weight k, level N and nebentypus χ with respect to the Petersson inner product  f, g = f (z)g(z)y k−2 dx dy. Γ0 (N )\H

✩ This work is partially supported by the National Natural Science Foundation of China (Grant No. 11101239), the Program for Changjiang Scholars and Innovative Research Team in University (IRT1264) and the Independent Innovation Foundation of Shandong University (2012ZRYQ005).

http://dx.doi.org/10.1016/j.jnt.2014.01.002 0022-314X/© 2014 Elsevier Inc. All rights reserved.

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Since χ is primitive, we can choose an orthogonal basis Bk (N, χ) which consists of primitive forms, that is, normalized eigenforms for all Hecke operators. Any f ∈ Bk (N, χ) has the Fourier expansion f (z) =



λf (n)n(k−1)/2 e(nz),

n1

where the Hecke eigenvalues λf (n) satisfy 



λf (m)λf (n) =

χ(d)λf

d|(m,n)

 mn , d2

m, n  1,

and λf (n) = χ(n)λf (n) for (n, N ) = 1. For s > 1, the symmetric-square L-function associated to f is given by (see Shimura [8])  λf (n2 )   L sym2 f, s = ζ (N ) (2s) , ns

(1.1)

n1

where ζ (N ) (2s) = ζ(2s)



p|N (1

− p−2s ). Define

−s −3s/2

L∞ (s) = 2

π



s+1−ν Γ (s + k − 1)Γ 2

 (1.2)

with ν = 0 or 1 according as χ(−1) = 1 or −1. Then the complete function     Λ sym2 f, s = N s/2 L∞ (s)L sym2 f, s

(1.3)

can be continued to a meromorphic function on the whole complex plane with possible simple poles at s = 0, 1, and satisfies the functional equation     Λ sym2 f, s = Λ sym2 f, 1 − s .

(1.4)

In this paper, we are concerned with the non-vanishing behavior of L(sym2 f, s) on the critical line. Denote ωf =

(4π)k−1 f, f . Γ (k − 1)

Our main result is the following theorem.

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Theorem 1.1. Assume that k  3. Given an integer  1, ( , N ) = 1, we have 

ωf−1 λf

f ∈Bk (N,χ)

   2 1 2

L sym f, 2

√

 log p L∞ ( 12 ) 1 φ(N ) N log +2 + 2γ + =√

p−1 L∞ ( 12 )

N p|N   1 3 + Ok, − N − 4 + + 2+ N − 4 −

(1.5)

where φ(n) is the Euler totient function. Moreover, for any real number t = 0, we have 

ωf−1 λf

f ∈Bk (N,χ)

=

1

1 2 +it

   2 1 2

L sym f, + it 2

ζ(1 + 2it)

+ N −it



1 − p−1−2it



p|N

  L∞ ( 12 − it) 1 ζ(1 − 2it) 1 − p−1+2it 1 1 −it L∞ ( 2 + it) 2 p|N

 − 1 +   3 +  1 3 + Ok, 1 + |t| 4 − N − 4 + + 1 + |t| 4 2+ N − 4 − .

(1.6)

Theorem 1.1 generalizes some results in Blomer [2] where twisted moments of L(sym2 f, s) at s = 1/2 were studied. It is worth noting that Blomer [2] established for the first time an asymptotic formula for the second moment of L(sym2 f, s) at s = 1/2. For χ trivial, a sharp bound for the second moment of L(sym2 f, s) on the critical line was obtained by Iwaniec and Michel [5]. The proof of Theorem 1.1 starts from the approximate functional equation and Petersson trace formula. Then we obtain the so-called diagonal term D and non-diagonal term 2πi−k R from Petersson trace formula. The estimation of D is similar to that in [2]. To bound R, we will use some analytic properties of Hurwitz zeta function. Using some ideas of Khan [6], we derive from Theorem 1.1 the following non-vanishing result. Corollary 1.2. Assume that k  3. Given t ∈ R, there exists a constant N0 (t) such that for any N  N0 (t) there exists f ∈ Sk (N, χ) such that L(sym2 f, 1/2 + it) = 0. Proof. For t = 0, we take = 1. Then by (1.5), for large N , the first moment of L(sym2 f, 1/2) is non-zero, and thus not all values of L(sym2 f, 1/2) can equal zero.

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For t = 0 and large N , the main term in (1.6) is 1

1 2 +it

ζ(1 + 2it)

=√

    L∞ ( 12 − it) 1 −1+2it 1 − p−1−2it + N −it ζ(1 − 2it) 1 − p 1 1 L∞ ( 2 + it) 2 −it

p|N

p|N

    it   it 1 N− 2 −it −1−2it 2 L · 2

+ it ζ(1 + 2it) · N 1 − p . ∞ 2

L∞ ( 12 + it) p|N

Now we choose such that 

−it



 it 2

· N L∞

    1 −1−2it + it ζ(1 + 2it) 1−p = 0. 2 p|N

 Denote a = −it and b = N it/2 L∞ (1/2 + it)ζ(1 + 2it) p|N (1 − p−1−2it ). Note that (ab) = (a)(b) − (a)(b). If (b) = 0, we take = 1, then (ab) = 0; if (b) = 0, we take such that (a) = 0, i.e., t log = rπ,

for all r ∈ Z

which clearly holds for any , ( , N ) = 1 if eπ/t ∈ / Z and also holds for , ( , N eπ/t ) = 1 if eπ/t ∈ Z. Therefore, for t = 0, we can choose , ( , N ) = 1 such that the main term in (1.6) is nonzero. Thus not all values of L(sym2 f, 1/2 + it) can equal zero. 2 2. Approximate functional equation We need the following approximate functional equation for L(sym2 f, 1/2 + it). Lemma 2.1. For any t ∈ R, we have      1  λf (n2 )  n  1 λf (n2 ) n 2 −it L∞ ( 2 − it) √ +N , L sym f, + it = V W √ 1 1 1 2 2 +it 2 −it L ( + it) N N n n ∞ 2 n1 n1 where

V (y) =

1 2πi



y −u ζ (N ) (1 + 2it + 2u)

L∞ ( 12 + it + u) G(u) du, u L∞ ( 12 + it)

(2.1)

y −u ζ (N ) (1 − 2it + 2u)

L∞ ( 12 − it + u) G(−u) du u L∞ ( 12 − it)

(2.2)

(1)

1 W (y) = 2πi

 (1)

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with

 2 −4 −8it 2 u u + 1 eu + 2 2 4t + 1 4t + 1     2 −4 1 1 u− − it u+ + it eu . = 2 4t + 1 2 2 

G(u) =

(2.3)

Proof. Following Iwaniec and Kowalski [4], we consider the integral    1 1 G(u) 2 du. I= Λ sym f, + it + u 2πi 2 u (1)

Note that for G(u) in (2.3), Λ(sym2 f, 1/2 + it + u)G(u)u−1 has only a simple pole at u = 0. Moving the line of integration in I to u = −1, by the residue theorem, we have      1 1 1 G(u) du + Λ sym2 f, + it . (2.4) I= Λ sym2 f, + it + u 2πi 2 u 2 (−1)

Here we have used the fact that G(0) = 1. By the functional equation in (1.4), we have    G(u) 1 1 2 du Λ sym f, + it + u 2πi 2 u (−1)

=



1 2πi

(−1)

1 =− 2πi



  G(u) 1 du Λ sym2 f, − it − u 2 u   G(−u) 1 2 du. Λ sym f, − it + u 2 u

(2.5)

(1)

By (1.1), (1.3), (2.4) and (2.5), we have   1 L sym2 f, + it 2    1 + it + u) u L∞ ( G(u) 1 1 2 2 du N2 = L sym f, + it + u 1 2πi 2 u L∞ ( 2 + it) (1)

+

L∞ ( 12 N −it L∞ ( 12

1 = 2πi

 N (1)

+ N −it

u 2

− it) 1 + it) 2πi

 (1)

  L∞ ( 12 − it + u) G(−u) 1 2 du N L sym f, − it + u 1 2 u L∞ ( 2 − it) u 2

  λf (n2 )  G(u) + it + u) (N ) ζ (1 + 2it + 2u) du 1 u L∞ ( 12 + it) n 2 +it+u n1

L∞ ( 12

L∞ ( 12 − it) 1 L∞ ( 12 + it) 2πi

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  λf (n2 )  G(−u) L∞ ( 12 − it + u) (N ) ζ (1 − 2it + 2u) du N 1 u L∞ ( 12 − it) n 2 −it+u n1



u 2

× (1)

1  λf (n2 )  n   λf (n2 )  n  −it L∞ ( 2 − it) +N , = V √ W √ 1 1 L∞ ( 12 + it) n1 n 2 −it N N n 2 +it n1

where V (y) and W (y) are defined in (2.1) and (2.2), respectively. 2 3. Proof of Theorem 1.1 We need Petersson trace formula which states that 

ωf−1 λf (m)λf (n)

f ∈Bk (N,χ)



= δmn + 2πi−k

c≡0 (mod N )

 √  Sχ (m, n; c) 4π mn Jk−1 , c c

where δmn is the Krönecker symbol, Jk−1 (x) is the J-Bessel function, and 

Sχ (m, n; c) =

dd≡1 (mod c)

 χ(d)e

 md + nd . c

Since ( , N ) = 1, we have λf ( 2 ) = λf ( 2 ). By Lemma 2.1 and (3.1), we have  f ∈Bk (N,χ)

    1 ωf−1 λf 2 L sym2 f, + it 2

     λf (n2 ) n √ ωf−1 λf 2 V 1 N n 2 +it n1 f ∈Bk (N,χ) 

 1  λf (n2 ) n −it L∞ ( 2 − it) +N W √ 1 L∞ ( 12 + it) n1 n 2 −it N   

  1  4π n n Sχ ( 2 , n2 ; c) −k δ,n + 2πi = Jk−1 V √ 1 c c N n 2 +it n1 c≡0 (mod N )   1  1 n −it L∞ ( 2 − it) W √ +N 1 L∞ ( 12 + it) n1 n 2 −it N  

 Sχ ( 2 , n2 ; c) 4π n −k × δ,n + 2πi Jk−1 c c 

=

=

1 1

2 +it

 V

√ N



c≡0 (mod N )

+

L∞ ( 12 N −it L∞ ( 12

  − it) 1

W √ 1 + it) 2 −it N

(3.1)

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−k

+ 2πi

+





1

n1

n 2 +it

1

V

L∞ ( 12 2πi−k N −it L∞ ( 12 

×

c≡0 (mod N )

n √ N



 c≡0 (mod N )

  Sχ ( 2 , n2 ; c) 4π n Jk−1 c c

  − it)  1 n √ W 1 + it) n1 n 2 −it N

  4π n Sχ ( 2 , n2 ; c) Jk−1 c c

= D + 2πi−k R, where D = D1 + N −it

L∞ ( 12 − it) D2 L∞ ( 12 + it)

(3.2)

with 

1

D1 =

V 1 +it

2



√ , N

D2 =

 

√ , W 1 N

2 −it 1

(3.3)

and R = R1 + N −it

L∞ ( 12 − it) R2 L∞ ( 12 + it)

(3.4)

with R1 = R2 =



  Sχ ( 2 , n2 ; c) 4π n Jk−1 , c c

(3.5)

     4π n n Sχ ( 2 , n2 ; c) √ J . W k−1 1 c c N c≡0 (mod N ) n 2 −it n1

(3.6)



1

n1

n 2 +it

1



 V

n √ N

 c≡0 (mod N )

1

Then Theorem 1.1 follows from the following estimates: (i) For t = 0, √

 log p L∞ ( 12 ) 1 φ(N ) N log +2 + 2γ + D=√

p−1 L∞ ( 12 )

N p|N   1 + Ok, − N − 4 + .

(3.7)

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(ii) For t ∈ R, t = 0, D=

1

1 2 +it

+

ζ(1 + 2it)



1 − p−1−2it



p|N

L∞ ( 12 N −it L∞ ( 12

+ Ok,



  − it) 1 ζ(1 − 2it) 1 − p−1+2it 1 −it + it) 2 p|N

− 1 +  1 1 + |t| 4 − N − 4 + .

(3.8)

(iii) For t ∈ R,   3 +  3 R = Ok, 1 + |t| 4 2+ N − 4 − .

(3.9)

We will establish (3.7) and (3.8) in Section 4 and prove (3.9) in Sections 5–7. 4. Estimation of D First, we estimate D1 . By (2.1) and (3.3), we have D1 =

1

1 1

2 +it 2πi

 

√ N

−u ζ (N ) (1 + 2it + 2u)

(1)

L∞ ( 12 + it + u) G(u) du. u L∞ ( 12 + it)

(4.1)

We need the following result. Lemma 4.1. Let u = δ + iy with δ > −1/2 fixed. For any t ∈ R, we have   3δ   3δ +k− 34 3π |y| L∞ ( 12 + it + u) e4 . k,δ 1 + |t| 2 1 + |y| 2 1 L∞ ( 2 + it) Proof. By (1.2), for u = δ + iy, we have L∞ ( 12 + it + u) L∞ ( 12 + it) = 2−δ−iy π − 2 (δ+iy) 3

Γ (δ + k −

1 2

δ+ 32 −ν + i t+y 2 2 ) . 3 t 2 −ν + i ) 2 2

+ i(t + y))Γ (

Γ (k −

1 2

+ it)Γ (

(4.2)

By Stirling’s formula (see (5.113) in Iwaniec and Kowalski [4]), for |t + y| > 1, we have   1 Γ δ + k − + i(t + y) 2    

√  δ+k−1 − π |t+y| |t + y| i(t+y) 1 2 = 2π i(t + y) 1+O e e |t + y| |t + y|δ+k−1 e− 2 |t+y| . π

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For |t + y|  1, we have        Γ δ + k − 1 + i(t + y)   Γ δ + k − 1 k,δ 1.   2 2 Thus,    δ+k−1 − π |t+y| 1 Γ δ + k − + i(t + y) k,δ 1 + |t + y| e 2 . 2

(4.3)

On the other hand, for |t| > 1, applying Stirling’s formula for Γ (k − 1/2 + it), we have   π 1 Γ k − + it k |t|k−1 e− 2 |t| . 2 For |t|  1, using the relations Γ (z)Γ (1 − z) =

π , sin(πz)

zΓ (z) = Γ (z + 1),

we have −1    sin(π(k − 1 3 − k − it Γ k − + it = Γ 2 2 π

1 2

+ it))

Γ ( 32 − it) eiπ(k− 2 ) (e−πt + eπt ) 3 5 1 2πi ( 2 − k − it)( 2 − k − it) · · · ( 2 − it) 1

=

k 1. It follows that



1 Γ k − + it 2



 k−1 − π |t|

k 1 + |t| e 2 .

(4.4)

By (4.3) and (4.4), we have Γ (δ + k − 12 + i(t + y)) (1 + |t + y|)δ+k−1 e− 2 |t+y| k,δ π 1 Γ (k − 2 + it) (1 + |t|)k−1 e− 2 |t|  δ  δ+k−1 π |y| k,δ 1 + |t| 1 + |y| e2 . π

Similarly, δ+ 32 −ν 2 3 2 −ν Γ( 2

Γ(

+ i t+y 2 ) + i 2t )

By (4.2), Lemma 4.1 follows. 2

 δ  δ+1−ν π k,δ 1 + |t| 2 1 + |y| 2 4 2 e 4 |y|  δ  δ+1 π k,δ 1 + |t| 2 1 + |y| 2 4 e 4 |y| .

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Now we estimate D1 . We distinguish two cases according as t = 0 or t = 0. Case 1. If t = 0, we move the line of integration in (4.1) to u = −1/2 + , picking up a double pole at u = 0, 1 1 1 D1 = √ resu=0 + √

2πi





(− 12 +)

√ N

−u ζ (N ) (1 + 2u)

L∞ ( 12 + u) G(u) du. u L∞ ( 12 )

First we compute the residues in (4.5). Since ζ (N ) (s) = ζ(s)

resu=0



p|N (1

(4.5)

− p−s ), we have

L∞ ( 12 + u) d G(u) = lim · uζ(1 + 2u) u→0 du L∞ ( 12 ) −u

  

. 1 − p−1−2u · √ · N p|N

(4.6)

Note that for t = 0, G(u) = −4u2 + 1, G(0) = 1, G (0) = 0, lim uζ(1 + 2u) =

u→0

1 , 2

lim

u→0

 d  uζ(1 + 2u) = γ, du

where γ is Euler’s constant, and 

   φ(N ) , 1 − p−1−2u = 1 − p−1 = u→0 N p|N p|N 



   d  d −1−2u −1−2u exp lim log 1 − p 1−p = lim u→0 du u→0 du lim

p|N

p|N

   2p−1−2u log p = lim 1 − p−1−2u u→0 1 − p−1−2u p|N

=

p|N

φ(N )  2 log p N

p|N

p−1

.

By (4.6), we have resu=0 =

√

 log p L ( 1 ) φ(N ) N log +2 + 2γ + ∞ 12 . 2N

p−1 L∞ ( 2 ) p|N

Next we bound the integral in (4.5). Denote u = −1/2 +  + iy. Then       (N )   12 ω(N )  −2−2iy  ζ (1 + 2u) = ζ(2 + 2iy) 1−p ,   1 + |y| · 2 p|N

(4.7)

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where ω(n) denotes the number of distinct prime divisors of n. By Robin [7], we have ω(n)  1.3841

log n log log n

for n  3.

It follows that   (N )  1 ζ (1 + 2u)  N  1 + |y| 2

(4.8)

for any  > 0. By Lemma 4.1 and (4.8), we have 

 (− 12 +)

√ N

−u ζ (N ) (1 + 2u)

L∞ ( 12 + u) G(u) 1 1 du k, 2 − N − 4 + . 1 u L∞ ( 2 )

(4.9)

By (4.5), (4.7) and (4.9), we have √

 log p L ( 1 ) N 1 φ(N ) log +2 + 2γ + ∞ 12 D1 = √

p−1 L∞ ( 2 )

2N p|N   1 + Ok, − N − 4 + .

(4.10)

Case 2. If t = 0, we move the line of integration in (4.1) to u = −1/2 + , picking up simple poles at u = 0 and u = −it, D1 =

1 1

2 +it ×

resu=0 +

1 1

2 +it

1

1 resu=−it + 1 +it 2πi

2

 (− 12 +)



√ N

−u ζ (N ) (1 + 2it + 2u)

L∞ ( 12 + it + u) G(u) du, u L∞ ( 12 + it)

(4.11)

where 

resu=0

−u

L∞ ( 12 + it + u)

(N ) √ = lim ζ (1 + 2it + 2u) G(u) u→0 L∞ ( 12 + it) N   1 − p−1−2it , = ζ(1 + 2it)

(4.12)

p|N

and resu=−it

−u

 L∞ ( 12 + it + u) G(u)

(N ) = lim (u + it)ζ (1 + 2it + 2u) √ u→−it u L∞ ( 12 + it) N =

φ(N ) it L∞ ( 12 ) G(−it). 2N 1+it/2 −it L∞ ( 12 + it)

(4.13)

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It remains to bound the integral in (4.11). Denote u = −1/2 +  + iy. Then       (N )    −2i(t+y)  ζ (1 + 2it + 2u) = ζ 2i(t + y) 1−p   p|N

 

1  N 1 + |t + y| 2

(4.14)

for any  > 0. By Lemma 4.1 and (4.14), we have 

 (− 12 +)

√ N

−u ζ (N ) (1 + 2it + 2u)

L∞ ( 12 + it + u) G(u) du u L∞ ( 12 + it)

 − 1 + 1 1 k, 1 + |t| 4 2 − N − 4 + .

(4.15)

By (4.11), (4.12), (4.13) and (4.15), we have D1 =



1

ζ(1 + 2it)

1

2 +it



1−p

−1−2it



p|N



φ(N ) it L∞ ( 12 ) + G(−it) 2N 1+it/2 −it L∞ ( 12 + it)

 − 1 +  1 + Ok, 1 + |t| 4 − N − 4 + .

(4.16)

D2 can be estimated similarly. By (2.2) and (3.3), we have D2 =

 

1

1 1

2 −it 2πi

√ N

−u ζ (N ) (1 − 2it + 2u)

(1)

L∞ ( 12 − it + u) G(−u) du. u L∞ ( 12 − it)

(4.17)

If t = 0, then D2 = D1 . We have √

 log p L ( 1 ) 1 φ(N ) N log +2 + 2γ + ∞ 12 D2 = √

p−1 L∞ ( 2 )

2N p|N   1 + Ok, − N − 4 + .

(4.18)

If t = 0, we move the line of integration in (4.17) to u = −1/2 + , picking up simple poles at u = 0 and u = it, D2 =

1



2 −it 1

+ Ok,

ζ(1 − 2it)



 1 − p−1+2it +

p|N



− 1 +  1 1 + |t| 4 − N − 4 + .



φ(N ) −it L∞ ( 12 ) G(−it) 2N 1−it/2 it L∞ ( 12 − it)

(4.19)

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Therefore, for t = 0, by (3.2), (4.10) and (4.18), we have √

 log p   L ( 1 ) 1 1 φ(N ) N + Ok, − N − 4 + . log +2 + 2γ + ∞ 12 D= √

p−1 L∞ ( 2 )

N p|N For t = 0, by (3.2), (4.16) and (4.19), we have D=

1

1 2 +it

ζ(1 + 2it)

1 − p−1−2it



p|N

L∞ ( 12 N −it L∞ ( 12

+



  − it) 1 ζ(1 − 2it) 1 − p−1+2it 1 −it + it) 2 p|N

  + Ok, 1 + |t|

− 14 +

 1

− N − 4 + .

This proves (3.7) and (3.8). 5. Estimation of R Since the estimations of R1 and R2 are the same, we only estimate R1 in detail. By (2.1),  V

n √ N

 =

 

1 2πi

n √ N

−u ζ (N ) (1 + 2it + 2u)

(1)

L∞ ( 12 + it + u) G(u) du. u L∞ ( 12 + it)

(5.1)

By the Mellin inversion formula for the J-Bessel functions (see Gradshteyn and Ryzhik [3, 6.422-6]), we have  Jk−1

4π n c



  Γ ( k−1+s ) 2π n −s 2 ds, c Γ ( k+1−s ) 2



1 = 4πi

(α)

1 − k < α < 1.

(5.2)

Take α = −1/2 in (5.2). By (3.5), (5.1) and (5.2), we have 1 R1 = 2(2πi)2





u

(1+) (− 12 )

) N 2 ζ (N ) (1 + 2it + 2u) L∞ ( 12 + it + u) Γ ( k−1+s 2 1 k+1−s s (2π ) u L∞ ( 2 + it) Γ ( 2 )

× L(u, s)G(u) du ds,

(5.3)

where L(u, s) =

 n1

1 nu+s+1/2+it

 c≡0 (mod N )

Sχ ( 2 , n2 ; c) . c1−s

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Note that 

L(u, s) =

c≡0 (mod N )



=



c1−s



  Sχ 2 , a2 ; c

1 s+u+1/2+it (m + a/c) m0



3

c≡0 (mod N )

cu+ 2 +it

1 nu+s+1/2+it

   1 Sχ 2 , a2 ; c c−s−u− 2 −it

a (mod c)

1

 n1 n≡a (mod c)

a (mod c)

1

c≡0 (mod N )

=



1 c1−s

a (mod c)

   2 2  a 1 , Sχ , a ; c ζ s + u + + it, 2 c

(5.4)

where for w > 1, ζ(w, λ) =



(m + λ)−w ,

m0

is the Hurwitz zeta function. It is known that ζ(w, λ) is meromorphic on C with a simple pole at w = 1 of residue 1, and satisfies the functional equation (see Apostol [1, Theorem 12.6]) −1

ζ(w, λ) = i

Γ (1 − w)(2π)

w−1

   

w w F (1 − w, λ) − e − F (1 − w, −λ) , e 4 4

(5.5)

where for w > 1, F (w, λ) =

 e(mλ) . mw

(5.6)

m1

Thus for u = 1 + , L(u, s) is meromorphic on C with a simple pole at s = 1/2 − it − u. Assume that k  3. Moving the line of integration in s in (5.3) to s = −1 − 2, passing a simple pole at s = 12 − it − u, we have 1 R1 = 4πi

 (1)

 ×

2



c≡0 (mod N )

+

k− 1 −it−u

u

) N2 ζ (N ) (1 + 2it + 2u) L∞ ( 12 + it + u) Γ ( 2 2 1 1 1 −it−u k+ +it+u u L∞ ( 2 + it) Γ ( 2 (2π ) 2 )

1 2(2πi)2





(1) (−1−2)

× L(u, s)G(u) du ds = R11 + R21 ,



1 3 cu+ 2 +it

  2 2  Sχ , a ; c G(u) du

a (mod c) u

) N 2 ζ (N ) (1 + 2it + 2u) L∞ ( 12 + it + u) Γ ( k−1+s 2 (2π )s u L∞ ( 12 + it) Γ ( k+1−s ) 2

Q. Sun / Journal of Number Theory 140 (2014) 196–214

210

where R11

1 = 4πi

 (1)

k− 1 −it−u

u

) N2 ζ (N ) (1 + 2it + 2u) L∞ ( 12 + it + u) Γ ( 2 2 1 1 1 −it−u k+ +it+u u L∞ ( 2 + it) Γ ( 2 (2π ) 2 ) 2

× K(u)G(u) du,   1 R21 = 2(2πi)2

(5.7) u 2

N ζ s (2π )

(N )

(1 + 2it + 2u) u

(1) (−1−2)

L∞ ( 12

+ it + u) L∞ ( 12 + it)

× L(u, s)G(u) du ds,

Γ ( k−1+s ) 2 k+1−s Γ( 2 ) (5.8)

with K(u) =

 c≡0 (mod N )



1 u+ 32 +it

c

  Sχ 2 , a2 ; c .

(5.9)

a (mod c)

In Section 6, we will show that   3 + 5 R11 k, 1 + |t| 4 1+ N − 4 − . In Section 7, we will show that   3 + 3 R21 k, 1 + |t| 4 2+ N − 4 − . Therefore,   3 + 3 R1 k, 1 + |t| 4 2+ N − 4 − .

(5.10)

  3 + 3 R2 k, 1 + |t| 4 2+ N − 4 − .

(5.11)

Similarly, R1 satisfies

Then (3.9) follows from (3.4), (5.10) and (5.11). 6. Estimation of R11 Consider the unique factorization c = 2β N qq12 q22 , where β ∈ N ∪ {0}, q, q1 , q2 are odd, μ2 (q) = 1, q1 |q ∞ , (q2 , q) = 1. We need the following result which is Corollary 4 in Blomer [2]. Lemma 6.1. For m ∈ Z, we have  a (mod c)

      am c √ cq , 4 2 − m2 . Sχ 2 , a2 ; c e c q

Q. Sun / Journal of Number Theory 140 (2014) 196–214

211

Now we estimate R11 in (5.7). By Lemma 6.1, we have  a (mod c)

    c √ Sχ 2 , a2 ; c cq , 4 2 . q

Since ( , N ) = 1, we have 

Sχ



 √

, a ; c cq 2



2

a (mod c)

c , 4 2 qN



√

 cq ·

1 c · = N − 2 c. qN

Thus for u > 1/2, K(u) in (5.9) is bounded by 

  K(u) N − 12

c≡0 (mod N )

1 1 cu+ 2

N −u−1 .

(6.1)

Now we move the line of integration in (5.7) to u = 1/2 +  and denote u = 1/2 +  + iy. Then by (6.1), we have

R11



1+

N

− 54 − 2

+∞     ζ (N ) (2 + 2 + 2i(t + y))  L∞ (1 +  + i(t + y))        1/2 +  + iy L∞ ( 12 + it)

−∞

 k−1−−i(t+y)    Γ(  )  1 2  ×  k+1++i(t+y) G +  + iy  dy, 2 Γ( ) 2

(6.2)

where by (2.3),        G 1 +  + iy   1 + |y| 2 e−y2 .   2

(6.3)

We also have      (N )      −2−2−2i(t+y)    ζ 2 + 2 + 2i(t + y) = ζ 2 + 2 + 2i(t + y) 1−p  p|N

1,

(6.4)

and by Lemma 4.1,    L∞ (1 +  + i(t + y))   3     k, 1 + |t| 4 + 1 + |y| k+ e 3π 4 |y| . 1   L ( + it) ∞ 2

(6.5)

Q. Sun / Journal of Number Theory 140 (2014) 196–214

212

Moreover, by Stirling’s formula, as in Lemma 4.1,  k−1−−i(t+y)  k− π Γ( )  (1 + |t + y|) 2 −1 e− 4 (|t+y|) 2  k+ π  k+1++i(t+y)  k, (1 + |t + y|) 2 e− 4 (|t+y|) Γ( ) 2  −1− k, 1 + |t + y| .

(6.6)

By (6.2)–(6.6), we conclude that

R11

k,

+∞   k+1+ 3π |y| −y2 e4 e dy 1 + |y|



 3 + 5 1 + |t| 4 1+ N − 4 −



 3 + 5 1 + |t| 4 1+ N − 4 − .

−∞

k, 7. Estimation of R21

In this section, we estimate R21 in (5.8). Recall that R21 =



1 2(2πi)2



u

(1) (−1−2)

) N 2 ζ (N ) (1 + 2it + 2u) L∞ ( 12 + it + u) Γ ( k−1+s 2 1 k+1−s s (2π ) u L∞ ( 2 + it) Γ ( 2 )

× L(u, s)G(u) du ds,

(7.1)

where by (5.4), 

L(u, s) =

3

c≡0 (mod N )

Denote w = s + u + L(u, s) =



1



1 2

cu+ 2 +it

a (mod c)

    a 1 . Sχ 2 , a2 ; c ζ s + u + + it, 2 c

+ it. Then by (5.5) and (5.6), we have 1 u+ 32 +it



  Sχ 2 , a2 ; c i−1 Γ (1 − w)(2π)w−1

c a (mod c)         a w a w F 1 − w, −e − F 1 − w, − × e 4 c 4 c        2 2  ma Γ (1 − w) 1 1 w e Sχ , a ; c e = 3 i(2π)1−w 4 m1−w c cu+ 2 +it m1 c≡0 (mod N ) a (mod c)       2 2  ma w  1 . Sχ , a ; c e − −e − 4 m1−w c c≡0 (mod N )

m1

a (mod c)

Q. Sun / Journal of Number Theory 140 (2014) 196–214

213

By Lemma 6.1, we have  a (mod c)

   2 2  ma c. Sχ , a ; c e ± c

Moving the line of integration in (7.1) to u = 1/2 + . Then (1 − w) = 1 +  > 1. Thus,   L(u, s) 

 c≡0 (mod N )

1  N −1− . c1+

(7.2)

Denote s = −1 − 2 + iv. By Stirling’s formula, as in Lemma 4.1,  k  k−1+s   Γ( 2 )   = Γ(2 − 1 −  + Γ(k + 1 +  −  Γ ( k+1−s )  2 2  −2− k, 1 + |v| .

iv  2 ) iv  2)

(1 + |v|) 2 − 2 − e− 4 |v| k

k,

3

π

(1 + |v|) 2 + 2 + e− 4 |v| k

1

π

(7.3)

Recall that we have moved the line of integration in (7.1) to u = 1/2 + . By (6.3)–(6.5) and (7.1)–(7.3), we have

R21

k,

2+

N

− 34 −

+∞     ζ (N ) (2 + 2 + 2i(t + y))  L∞ (1 +  + i(t + y))        1/2 +  + iy L ( 1 + it) ∞ 2

−∞

 +∞        −2− 1 × G +  + iy  dv dy 1 + |v| 2 −∞

k,



 3 + 3 1 + |t| 4 2+ N − 4 −



 3 + 3 1 + |t| 4 2+ N − 4 − .

+∞   k+1+ 3π |y| −y2 1 + |y| e4 e dy

−∞

k, References

[1] T.M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, New York, Heidelberg, Berlin, 1976. [2] V. Blomer, On the central value of symmetric-square L-functions, Math. Z. 260 (2008) 755–777. [3] I.S. Gradshteyn, I.M. Ryzhik, Table of Integrals, Series, and Products, seventh ed., Academic Press, New York, 2007. [4] H. Iwaniec, E. Kowalski, Analytic Number Theory, Amer. Math. Soc. Colloq. Publ., vol. 53, Amer. Math. Soc., Providence, 2004. [5] H. Iwaniec, P. Michel, The second moment of the symmetric square L-function, Ann. Acad. Sci. Fenn. Math. 26 (2001) 465–482. [6] R. Khan, The first moment of the symmetric-square L-function, J. Number Theory 124 (2007) 259–266.

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[7] G. Robin, Estimation de la fonction de Tchebychef θ sur le k-ième nombre premier et grandes valeurs de la fonction ω(n) nombre de diviseurs premiers de n, Acta Arith. 42 (1983) 367–389. [8] G. Shimura, On the holomorphy of certain Dirichlet series, Proc. London Math. Soc. (3) 31 (1975) 79–98.