Epstein Zeta Function

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The Epstein Zeta Function Johar M. Ashfaque The Epstein zeta function Z(s) is defined for <(s) > 1 by Z(s) =

∞ X m,n=−∞ (m,n)6=(0,0)

1 (am2 + bmn + cn2 )s

where a, b and c are real numbers with a > 0 and b2 − 4ac < 0. Z(s) can be continued analytically to the whole complex plane except for the simple pole at s = 1. The value of Z(k) for k = 2, 3, ... is determined in terms of the infinite series of the form ∞ X cotr (nπτ ) , (r = 1, 2, ..., k) n2k−1 n=1

where Ď„=

1

b+

√

b2 − 4ac . 2a

Introduction

Let a, b and c be real numbers with a > 0 and D = 4ac − b2 > 0 so that Q(u, v) = au2 + buv + cv 2 is a positive-definite binary quadratic form of discriminant D. The Epstein zeta function Z(s) is then defined by the double series ∞ X

Z(s) =

m,n=−∞ (m,n)6=(0,0)

1 Q(m, n)s

where s = Ďƒ + it with Ďƒ, t ∈ R and Ďƒ > 1. Since Q(u, v) ≼ Îť(u2 + v 2 ) with Îť=

p 1 a + c − (a − c)2 + b2 > 0 2

for all real numbers u and v, the double series converges absolutely for Ďƒ > 1 and uniformly in every half plane Ďƒ ≼ 1 + ( > 0). Thus Z(s) is analytic function of s for Ďƒ > 1. Furthermore, it can be continued analytically to the whole complex plane except for the simple pole at s = 1 and satisfies the functional equation √ s √ 1−s D D Γ(s)Z(s) = Γ(1 − s)Z(1 − s). 2Ď€ 2Ď€

1

2

The Functional Equation

Recall 2.1 τ= Setting

b+

√

b2 − 4ac 2a

√ √ b b+i D D x= , y= , τ = x + iy = 2a 2a 2a

we have τ +τ

=

ττ

=

b a c a

so that = am2 + bmn + cn2

Q(m, n)

= a(m + nτ )(m + nτ ) = a|m + nτ |2

and

∞ X

Z(s) =

m,n=−∞ (m,n)6=(0,0)

1 , σ > 1. as |m + nτ |2s

Separating the term with n = 0, we have Z(s) =

∞ ∞ ∞ 2 X 1 1 2 X X + , σ > 1. as m=1 m2s as n=1 m=−∞ |m + nτ |2s

We wish to evaluate the second term in and therefore apply the Poisson summation formula Z ∞ ∞ ∞ X X f (m) = f (u) cos 2mπu du m=−∞

−∞

m=−∞

to the function f (t) =

1 |t + τ |2s

to obtain ∞ X

1 |m + τ |2s m=−∞

= =

Z ∞ X

∞

m=−∞ −∞ Z ∞ ∞ X m=−∞ −∞ Z ∞ ∞ X

cos 2mπu du |u + τ |2s cos 2mπu du {(u + x)2 + y 2 }s

cos 2mπ(t − x) dt (t2 + y 2 )s m=−∞ −∞ Z ∞ ∞ X cos 2mπt = cos 2mπx dt 2 2 s −∞ (t + y ) m=−∞ =

since the integrals involving the sine function vanish which yields Z ∞ Z ∞ ∞ ∞ X 1 2 1 22 X cos 2mπyt = dt + cos 2mπx dt, σ > 1. 2s 2s−1 2 s 2s−1 2 s |m + τ | y (1 + t ) y 0 −∞ (1 + t ) m=−∞ m=1 2

Now, we wish to evaluate the two integrals by first making the substitution u=

t2 1 + t2

which gives 1 3 2tdt 1 = 1 − u, du = = 2u 2 (1 − u) 2 dt 2 2 2 1+t (1 + t )

and therefore Z

∞

0

dt 1 = 2 s (1 + t ) 2

1

Z

(1 − u)

s− 23

− 12

u

0

√ Γ(s − 21 ) π 1 1 1 du = B s − , = . 2 2 2 2Γ(s)

From an integral representation of the Bessel function Z ∞ ν 1 cos yt 1 2 1 Γ ν+ Kν (y) = √ y > 0, <(v) > − , 1 dt, ν+ 2 2 2 π y (1 + t ) 2 0 we have

∞

Z 0

cos 2mπyt dt = (1 + t2 )s

√

1

π(mπy)s− 2 1 Ks− 21 (2mπy), σ > . Γ(s) 2

Thus ∞ X

1 = |m + τ |2s m=−∞

√ Γ s − 21 π

√ ∞ X 1 4 π + 2s−1 (mπy)s− 2 cos(2mπx)Ks− 12 (2mπy), σ > 1 y Γ(s) m=1

y 2s−1 Γ(s)

such that for n ≥ 1 1 √ Γ s− 2 π

√ ∞ X 1 1 4 π = + (mnπy)s− 2 cos(2mnπx)Ks− 21 (2mnπy). 2s 2s−1 2s−1 2s−1 2s−1 |m + nτ | n y Γ(s) n y Γ(s) m=1 m=−∞ ∞ X

Hence

Z(s) = 2a−s ζ(2s)+2a−s y 1−2s

√ Γ s − 12 π ζ(2s−1)+

Γ(s)

1 ∞ ∞ 1 8a−s y 2 −s π s X 1−2s X n (mn)s− 2 cos(2mnπx)Ks− 12 (2mnπy). Γ(s) n=1 m=1

Collecting the terms with mn = k, we obtain

Γ s− Z(s) = 2a−s ζ(2s)+2a−s y 1−2s

1 2

Γ(s)

√

π ζ(2s−1)+

k=1

that is

Γ s− Z(s) = 2a−s ζ(2s) + 2a−s y 1−2s

where H(s) = 4

∞ X

1 ∞ 1 8a−s y 2 −s π s X X 1−2s n (k)s− 2 cos(2kπx)Ks− 12 (2kπy) Γ(s)

1 2

√

π

1

ζ(2s − 1) +

Γ(s)

n|k

1

2a−s y 2 −s π s H(s) Γ(s)

σ1−2s (k)(k)s− 2 cos(2kπx)Ks− 12 (2kπy)

k=1

and σν denotes the ν-th powers of the divisors of k σν (k) =

X

ν

d =

d|k

X k ν d|k

3

d

.

We can now express Z(s) in another form s s 1 1 ay y 1 Γ(s)Z(s) = 2 Γ(s)ζ(2s) + 2y 1−s π 2 −s Γ s − ζ(2s − 1) + 2y 2 H(s). π π 2 By the functional equation of the Riemann zeta function 1 ζ(2s − 1) = 2(2π)2s−2 sin s − πΓ(2 − 2s)ζ(2 − 2s) 2 and the basic properties of the gamma function √ Γ(2 − 2s) = we have

π Γ(1 − s) − 12 ) sin(s − 21 )π

22s−1 Γ(s

1 1 π 2 −s Γ s − ζ(2s − 1) = π s−1 Γ(1 − s)ζ(2 − 2s) 2

and

ay π

s

s 1−s 1 y y Γ(s)Z(s) = 2 Γ(s)ζ(2s) + 2 Γ(1 − s)ζ(2 − 2s) + 2y 2 H(s). π π

Recall 2.2 K−ν (y) = Kν (y) and

ν

ν

k − 2 σν (k) = k 2 σ−ν (k) leading to H(s) = H(1 − s). For

φ(s) =

ay π

s Γ(s)Z(s)

we have φ(s) = φ(1 − s). Since

√ ay =

we have

D 2

√ s √ 1−s D D Γ(s)Z(s) = Γ(1 − s)Z(1 − s) 2π 2π

which is the functional equation of Z(s).

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