Note on Legendre Symbols Connecting with Certain Infinite Series

Notes on Number Theory and Discrete Mathematics Vol. 19, 2013, No. 2, 77–83

Note on Legendre symbols connecting with certain inﬁnite series

Ramesh Kumar Muthumalai

Department of Mathematics, D. G. Vaishnav College Chennai–600106, Tamil Nadu, India e-mail: [email protected]

Abstract: Some applications of Legendre symbols connecting with certain infinite series are studied. In addition, some identities for Dirichlet series through Legendre symbols and Hurwitz zeta function are given. Keywords: Legendre symbols, Infinite series, Dirichlet series. AMS Classification: 11A25, 40A05.

1 Introduction

In number theory, the Legendre symbol is a completely multiplicative function with values 1, -1, 0. Let p be any odd prime, if n 6≡ 0(modp) Legendre symbol (n|p) is deﬁned as follows[1, p.179]  +1 If nRp  (n|p) = . (1.1)  −1 If nRp

where nRp denote n is a quadratic residue mod p and nRp denote n is a quadratic nonresidue mod p. If n ≡ 0(modp), we define (n|p) = 0. Recently, some properties and applications of Mobius function and a new arithmetic function connecting with infinite series, infinite products and partition of an integer are studied in Ref[2]. The purpose of this work is to study some applications of Legendre symbols connecting with infinite series of rational functions and Dirichlet series.

77 2 Legendre symbols connecting with inﬁnite series

Lemma 2.1. For any odd prime p and |x| < 1,

∞ p−1 X 1 X F (x) = (k|p)xk = (m|p)xm. (2.1) p 1 − xp k=1 m=1 Proof. The Legendre symbol (k|p) is periodic. Then

∞ ∞ p−1 X X X (k|p)xk = (m|p)xkp+m. k=1 k=0 m=1 Rearranging right hand side of above equation, gives

∞ p−1 ∞ X X X (k|p)xk = (m|p)xm xkp. k=1 m=1 k=0 After simpliﬁcation, completes the Lemma.

Example 2.2. Let p = 3, 5 and 7 in (2.1). Then x(1 − x) F (x) = . 3 1 − x3

x(1 − x)(1 − x2) F (x) = . 5 1 − x5

x(1 − x)(1 + 2x + x2 + 2x3 + x4) F (x) = . 7 1 − x7 Theorem 2.3. For an odd prime p and |x| < 1,

∞ X k x = µ(k)(k|p)Fp(x ). (2.2) k=1 Proof. Using Lemma 2.1, that ∞ ∞ ∞ X k X X kn µ(k)(k|p)Fp(x ) = µ(k)(k|p) (n|p)x . (2.3) k=1 k=1 n=1 Rearranging above equation, gives ∞ ∞ X k X k X µ(k)(k|p)Fp(x ) = x µ(d)(d|p)(k/d|p). k=1 k=1 d|k Since (d|p)(k/d|p) = (k|p),

78 P∞ k Theorem 2.4. Let p be any odd prime and c be an arithmetic function. If P (x) = k=1 c(k)x , then ∞ X k P (x) = a(k)Fp(x ). (2.4) k=1 where X a(k) = µ(d)(d|p)c(k/d). (2.5) d|k Proof. Consider

∞ X P (x) = c(k)xk. (2.6) k=1 Using Theorem 2.3, that

∞ ∞ X X mk P (x) = c(k) µ(m)(m|p)Fp(x ). k=1 m=1 Rearranging above equation, gives

∞ X k X P (x) = Fp x µ(d)(d|p)c(k/d). k=1 d|k P Setting a(k) = d|k µ(d)(d|p)c(k/d) in the above equation, completes the theorem.

a(k) = (k|p)b(k). (2.9)

Substituting (2.9) in (2.4), completes the proof.

79 Example 2.6. Let c(k) = 1/ks in Theorem 2.4. Then

∞ X k Φ(x, s, 1) = a(k)Fp(x ). (2.10) k=1 where 1 X a(k) = µ(d)(d|p)ds ks d|k and Φ(x, s, b) is Lerch zeta function deﬁned by

∞ X xk+1 Φ(x, s, b) = . (k + b)s k=0 Example 2.7. The following identities are some of examples of Corollary 2.5.

1. Let f(k) = k in Corollary 2.5, then

∞ X k P (x) = k(k|p)x = xD [Fp(x)] . k=1

P k Also b(k) = d|k µ(d) d = ϕ(k), where ϕ is Euler totient function. Hence

∞ X k xD [Fp(x)] = (k|p)ϕ(k)Fp(x ). k=1 where D differential operator denotes differentiation with respect to x. For instance, if p = 3 1 − x2 1 − x x(1 − x2) x4(1 − x5) = ϕ(1) + ϕ(2) + ϕ(5) − .... 1 + x + x2 1 − x3 1 − x6 1 − x15

2. Let f(k) = 1/k, then

∞ X xk Z ∞ P (x) = (k|p) = F (xe−t)dt. k p k=1 0 Also 1 X ϕ−1(k) b(k) = µ(d)d = , k k d|k where ϕ−1 is inverse of Euler totient function with respect to convolution. Hence

∞ Z ∞ X ϕ−1(k) F (xe−t)dt = (k|p) F (xk). p k p 0 k=1 For instance, Z ∞ −t −t −1 2 2 xe (1 − xe ) −1 x(1 − x) ϕ (2) x (1 − x ) 3 −3t dt = ϕ (1) 3 + 6 + .... 0 1 − x e 1 − x 2 1 − x

80 3 Some identities for Dirichlet series through Legendre symbols

Theorem 3.1. Let Re(s) > 1, p be any odd prime and c be an arithmetic function. Then,

∞ p−1 ∞ X c(k) 1 X X a(k) = (m|p)ζ(s, m/p) . (3.1) ks ps ks k=1 m=1 k=1 P where a(k) = d|k µ(d)(d|p)c(k/d).

Proof. For |x| < 1, let

∞ X P (x) = c(k)xk. (3.2) k=1

Replace x by e−t, multiply by ts−1 for Re(s) > 1 and integrating on [0, ∞), then

∞ Z ∞ X Z ∞ ts−1P (e−t)dt = c(k) ts−1e−ktdt. 0 k=1 0

R ∞ s−1 −kt Γ(s) Since 0 t e dt = ks ,

∞ Z ∞ X c(k) ts−1P (e−t)dt = Γ(s) . (3.3) ks 0 k=1

Similarly, replace x by e−t in (2.4), multiply by ts−1 for Re(s) > 1 and integrating on [0, ∞), then Z ∞ ∞ Z ∞ s−1 −t X s−1 −kt t P (e )dt = a(k) t Fp(e )dt 0 k=1 0 Using (2.1). gives

∞ p−1 Z ∞ X Z ∞ ts−1 X ts−1P (e−t)dt = a(k) (m|p)e−mktdt 1 − e−pkt 0 k=1 0 m=1 Rearranging right hand side of above equation, yields

∞ p−1 Z ∞ X X Z ∞ ts−1 ts−1P (e−t)dt = a(k) (m|p) e−mktdt. 1 − e−pkt 0 k=1 m=1 0 Using solution of the integral [2, p. 349]

∞ p−1 Z ∞ Γ(s) X a(k) X ts−1P (e−t)dt = (m|p)ζ(s, m/p). (3.4) ps ks 0 k=1 m=1 Comparing (3.3) and (3.4), completes the theorem.

81 Corollary 3.2. For Re(s) > 1 and odd prime p,

Example 3.3. Let f(k) = 1 in corollary 3.2. Then

∞ p−1 X (k|p) 1 X = (m|p)ζ(s, m/p). (3.6) ks ps k=1 m=1 The following Dirichlet series are obtained by taking f(k) = k, kn, 1/k in (3.5) and using the identity (3.6), that

∞ p−1 X ϕ(k) P (m|p)ζ(s − 1, m/p) (k|p) = p m=1 . ks Pp−1 k=1 m=1(m|p)ζ(s, m/p)

∞ Pp−1 X Jn(k) (m|p)ζ(s − n, m/p) (k|p) = pn m=1 . ks Pp−1 k=1 m=1(m|p)ζ(s, m/p)

∞ p−1 X ϕ−1(k) 1 P (m|p)ζ(s + 1, m/p) (k|p) = m=1 . ks+1 p Pp−1 k=1 m=1(m|p)ζ(s, m/p)

α where Jn is Jordan totient function. Let f(k) = µ(k)/k in (3.5). Then using the identity −1 P α σα (k) = d|k d µ(d)µ(k/d)

∞ p−1 ∞ X (k|p) 1 X X σ−1(k) µ(k) = (m|p)ζ(s, m/p) (k|p) α . ks+α ps ks+α k=1 m=1 k=1

where σα is the divisor function. Using (3.6), gives the following identity

∞ ∞ Pp−1 0 X σα(k) X Λ(k) (m|p)ζ (s, m/p) (k|p)µ(k) = p−(2s+α) (k|p) = log p − m=1 . (3.7) ks+α ks Pp−1 k=1 k=1 m=1(m|p)ζ(s, m/p) where Λ is van Mangoldt function and dζ(s, m/p)/ds = ζ0(s, m/p).

Theorem 3.4. For any odd prime p, 1 (k|p) = F (k)(0). (3.8) k! p π Z 2π π Z 2π (k|p) = k Cp(r, θ) cos kθdθ = k Sp(r, θ) sin kθdθ. (3.9) r 0 r 0 P∞ k P∞ k Where Cp(r, θ) = k=1(k|p)r cos kθ and Sp(r, θ) = k=1(k|p)r sin kθ

82 Proof. Differentiating (2.1) with respect to k and setting x = 0, gives (3.9). For |r| < 1, 0 ≤ θ ≤ 2π and odd prime p, consider

iθ Fp(re ) = Cp(r, θ) + iSp(r, θ). (3.10)

Using (2.1), comparing real and imaginary parts, then

∞ X k Cp(r, θ) = (k|p)r cos kθ. (3.11) k=1 ∞ X k Sp(r, θ) = (k|p)r sin kθ. (3.12) k=1

Multiply (3.12) and (3.13) by cos kθ and sin kθ, integrating on (0, 2π), gives (3.10).

Remark 3.5. Squaring and adding (3.12) and (3.13) and then integrating on [0, 2π], then

∞ Z 2π 1 X C (r, θ)2 + S (r, θ)2 dθ = (k|p)2r2k. p p π 0 k=1

Since (k|p)2 = 1, ∞ X r2 r2p (k|p)2r2k = − . 1 − r2 1 − r2p k=1 Then using (3.11), we obtain for odd prime p and |r| < 1,

Z 2π 2 2p iθ 2 1 r r |Fp re | dθ = 2 − 2p . (3.13) 0 π 1 − r 1 − r

References

[1] Apostal, T. M. Introduction to Analytic Number Theory, Springer International Student Edi- tion, New York, 1989.

[2] Gradshteyn, I. S., I. M. Ryzhik, Tables of Integrals, Series and Products (6th ed.), Academic Press, USA, 2000.

[3] Muthumalai, R. K. Some properties of new arithmetic function and its applications to analytic number theory, Notes on Number Theory and Discrete Mathematics, Vol. 17, 2011, No. 3, 38–48.