- Email: [email protected]
On a functional equation
Acta Mathematica Scientia 2009,29B(2):225–231 http://actams.wipm.ac.cn
ON A FUNCTIONAL EQUATION∗ Ding Yi (
)
Mathematics Department, New Jersey City University, USA
Abstract
In this article, the author derives a functional equation 1 π η(s) = ( )s− 2 4
πs 2 Γ(1 − s) sin( ) η(1 − s) π 2
(1)
of the analytic function η(s) which is defined by η(s) = 1−s − 3−s − 5−s + 7−s + · · ·
(2)
for complex variable s with Re s > 1, and is defined by analytic continuation for other values of s. The author proves (1) by Ramanujan identity (see [1], [3]). Her method provides a new derivation of the functional equation of Riemann zeta function by using Poisson summation formula. Key words Functional equation, Zeta function, M¨ unts formula 2000 MR Subject Classification
1
39B
Introduction The Euler’s zeta function ζ(s) is defined for any complex number s with Re(s) > 1 as ζ(s) =
∞ 1 . ns n=1
This series defines a holomorphic function. Riemann extended this function to a holomorphic function ζ(s) defined for all complex numbers s with s = 1 through analytic continuation. The Riemann zeta function ζ(s) has certain “trivial” zeros for s = −2, s = −4, s = −6, · · ·. The Riemann hypothesis states: nontrivial zeros of the Riemann zeta function lie on the critical line 1/2 + it. The connection between this function and prime numbers was already realized by Eurler 1 ζ(s) = . 1 − p−s prime p Helge von Koch in 1901 proved that the Riemann hypothesis is equivalent to the following considerable strengthening of the prime number theorem x 1 π(x) = dx + O(x1/2 lnx) x → ∞. ln(x) 2 ∗ Received
February 5, 2007. Supported by Separated Budget Research from New Jersey City University
226
ACTA MATHEMATICA SCIENTIA
Vol.29 Ser.B
Here, π(x) denotes the prime counting function, and the O-notation is the Landau symbol. Hilbert and Polya conjectured that imaginary parts of zeros of the Riemann zeta function should be the eigenvalues of a Hermitian operator. Atle Selberg gave a resemblance to the explicit formula by proving a duality between the length spectrum of a Riemann surface and the eigenvalues of its Laplacian. The statistical distribution of the zeros on the critical line now is believed to be exactly the same as the pair correlation distribution for the eigenvalues of a random Hermitian matrix after the work of Hugh Montgomery and Freeman Dyson. The Riemann zeta function statisfies the following functional equation: s ζ(s) = 2s π s−1 Γ(1 − s) sin(π ) ζ(1 − s) 2 valid for all s in C − {0, 1}. In this article, we derive a functional equation for a function η(s) similar to Riemann zeta function. Function η(s) is defined by η(s) = 1−s − 3−s − 5−s + 7−s + · · · The functional equation for η(s) is the following: η(s) =
π s− 12 2
Γ(1 − s) sin
π
4
πs 2
η(1 − s)
(3)
This work uses Ramanujan’s identity and Hurwitz zeta function. This method provides a new derivation of functional equation of Riemman zeta function.
2
Generalized M¨ unts Formula
For a function F (x) such that F (x) and F (x) are continuous over any finite interval and of O(x−λ ), O(x−μ ) as x → ∞, with λ > 1 and μ > 1. M¨ unts showed the following formula:
∞
0
s−1
x
C dx = ζ(s) F (nx) − x n=1
∞
∞ 0
xs−1 F (x)dx.
Here, C is a constant, and s is a complex number in some region. In this section, we provide a generalized M¨ unts formula and give a proof by using Hurwitz zeta function. The generalized M¨ unts formula is stated as followsl. Theorem 1 For a given function F(x) such that F (x) and F (x) are continuous over any finite interval and of O(x−λ ), O(x−μ ) as x → ∞, with λ > 1 and μ > 1, then 0
∞
xs−1
∞
n=0
F ((n + a)x) −
1 x
∞
0
F (v)dv dx = ζ(s, a)
∞ 0
y s−1 F (y)dy,
(4)
where s = σ + it, 0 < σ < 1, 0 < a ≤ 1. The Hurwitz zeta-function ζ(s, a) is an analytic function and is defined when σ > 1 by ζ(s, a) =
∞
1 . (n + a)s n=0
(5)
No.2
227
Ding: ON A FUNCTIONAL EQUATION
Proof First, for λ > σ > 1, we can interchange the integration and summation to have
∞
0
∞
xs−1
F (nx + ax)dx =
n=0
∞ n=0 ∞
∞
0
xs−1 F (nx + ax)dx
∞ 1 y s−1 F (y)dy s (n + a) 0 n=0 ∞ y s−1 F (y)dy, = ζ(s, a) =
0
because of ∞ n=0
1 | (n + a)s
∞ s−1 y F (y)|dy ≤
∞
0
n=0
1 (n + a)σ
∞
0
y σ−1 |F (y)|dy < ∞.
Now, we have the following estimate
∞
F (nx + ax) −
n=1
=x =x
∞
0
F (ux + ax)du
F (ux + ax)(u − [u])du
0 1 x
0
∞
F (ux + ax)(u − [u])du + x
= xI1 + x
∞ 1 x
∞ 1 x
O((ux + ax)−μ )du
O((ux)−μ )du
= xI1 + O(1). By looking at I1 in the above estimate, when x ≤ a1 , it implies ux+ax ≤ 2 and |F (ux+ax)| ≤ A, −μ when x > a1 , it implies ux + ax > 1 and |F (ux + ax)| ≤ K|ux + ax | ≤ A. This implies |xI1 | ≤ A. This inequality gives us the following estimate 1 ∞ F (v + ax)dv + O(1) = F (v)dv + O(1) x ax 0 1 ax 1 ∞ 1 ∞ F (v)dv − F (v)dv + O(1) = F (v)dv + O(1) = x 0 x 0 x 0 c = + O(1). x
∞
1 F (nx + ax) = x n=0
∞
Hence, when σ > 1, the following integral is calculated as ∞ ζ(s, a) y s−1 F (y)dy
0
∞
= 0
=
0
1
xs−1 s−1
x
∞
F (nx + ax)dx
n=0 ∞
c c dx + + F (nx + ax)dx − x s − 1 n=0
1
∞
s−1
x
∞ n=0
F (nx + ax)dx.
(6)
228
ACTA MATHEMATICA SCIENTIA
Vol.29 Ser.B
Now, the right-hand side of the above equality is regular for σ > 0, s = 1. Further more, if σ < 1, we have ∞ c = −c xs−2 dx. s−1 0 So, if 0 < σ < 1, the right-hand side of (6) is ∞ ∞
c dx xs−1 F (nx + ax) − x 0 n=0
∞ and is regular. Because the integral ζ(s, a) 0 y s−1 F (y)dy is also regular when 0 < σ < 1, thus we have (4).
3
Riemann’s Functional Equation
In this section, we want to prove functional equation for Riemann zeta function ζ(s) by using Poisson summation formula and M¨ unts formula. We have Theorem 2 For the Riemann zeta function ζ(s), s = σ + it, we have the following equality: πs ζ(s) = 2s π s−1 Γ(1 − s) sin ζ(1 − s), 0 < σ < 1. (7) 2 Proof If we take a = 1 in (4), we have the M¨ unts formula (see [1], section 2.11), that is, for 0 < σ < 1: ∞ ∞ ∞
1 ∞ s−1 s−1 ζ(s) y F (y)dy = x F (nx) − F (v)dv dx. (8) x 0 0 0 n=1 From the Poisson summation formula (see [1], section 2.8), we have the following equality: ∞ ∞ √ 1 1 β Fc (0) + Fc (nβ) = α F (0) + F (nα) , 2 2 n=1 n=1 ∞ where αβ = 2π and Fc (x) = π2 0 F (t) cos xtdt. Now, rewrite the above equality as follows: ∞ ∞ ∞ 1 1 β β F (nα) − √ F (α)dα = Fc (nβ) − F (0). α α 2 2π 0 n=1 n=1
The above equality will become the following after rearrangement: ∞ ∞ 1 ∞ β 1 ∞ F (nα) − F (α)dα = √ Fc (nβ) − Fc (v)dv . α 0 β 0 2π n=1 n=1 Now, using (9) we can write the formula (8) as follows: ∞ ∞ ∞ 1 ∞ y s−1 F (y)dy = αs−1 F (nα) − F (v)dv dα ζ(s) α 0 0 0 n=1 ∞ ∞ β 1 ∞ = αs−1 √ Fc (nβ) − Fc (v)dv dα β 0 2π n=1 0 ∞ ∞ 1 ∞ s− 12 −s = (2π) β Fc (nβ) − Fc (v)dv dβ. β 0 0 n=1
(9)
(10)
No.2
229
Ding: ON A FUNCTIONAL EQUATION
If Fc (β) has the same properties as F (α), then we have the M¨ unts formula for Fc (β) (see Remark 2), and this means
∞
0
β
−s
∞
1 Fc (nβ) − β n=1
∞ 0
Fc (v)dv dβ = ζ(1 − s)
Then, combining (10) and (11), we have ∞ 1 y s−1 F (y)dy = (2π)s− 2 ζ(1 − s) ζ(s) 0
∞
0
∞
0
z −s Fc (z)dz.
z −s Fc (z)dz.
(11)
(12)
But
∞
0
z
−s
Fc (z)dz = =
2 π
∞
0
2 π
∞
0
z
−s
∞
dz 0
F (α)dα
F (α) cos zαdα
∞
0
z −s cos zαdz.
It is well known that ∞ Γ(1 − s) 1−s πs z −s cos zαdz = cos π = Γ(1 − s)αs−1 sin , 1−s α 2 2 0 so, we have
∞
0
z
−s
Fc (z)dz =
2 πs Γ(1 − s) sin π 2
0
∞
αs−1 F (α)dα
(13)
0 < σ < 1.
(14)
From (12) and (13), we get ζ(s) = ζ(1 − s)Γ(1 − s)2s π s−1 sin
πs , 2
This is Riemann’s functional equation (see [2], section 2.1).
4
Functional Equation for η(s)
In this section, we will prove a functional equation for η. Let us state the functional equation for η in the following: Theorem 3 For the analytic function η(s) defined by η(s) = 1−s − 3−s − 5−s + 7−s + · · ·
(15)
for complex variable s with Res > 1, and defined for other values of s by analytic continuation. We have the following functional equation: πs π s− 12 2 Γ(1 − s) sin η(1 − s). (16) η(s) = 4 π 2 Proof Putting F (x) = f (8x) in (4) gives us the following equality:
∞ 0
s−1
x
∞
1 f ((8m + 8a)x) − x m=0
0
∞
f (8v)dv dx = ζ(s, a)
∞ 0
y s−1 f (8y)dy.
(17)
230
ACTA MATHEMATICA SCIENTIA
Vol.29 Ser.B
Taking a = 18 , a = 38 , a = 58 , and a = 78 in the above equality, respectively, we get ∞ ∞
xs−1 [f ((8m + 1)x) − f ((8m + 3)x) − f ((8m + 5)x) + f ((8m + 7)x)] dx 0
m=0
3 5 7 ∞ s−1 1 y f (8y)dy = ζ(s, ) − ζ(s, ) − ζ(s, ) + ζ(s, ) 8 8 8 8 0 ∞ = η(s)8s y s−1 f (y)dy.
(18)
0
From the Ramanujan equality (see [1], section 2.8), we have the following equality for αβ = ∞
π 4:
[f ((8m + 1)α) − f ((8m + 3)α) − f ((8m + 5)α) + f ((8m + 7)α)]
m=0
=
∞ 4 β [fc ((8m + 1)β) − fc ((8m + 3)β) − fc ((8m + 5)β) + fc ((8m + 7)β)]. π m=0
The above equality gives the following calculation: ∞ ∞ ∞ 4 s−1 s−1 β η(s) y f (y)dy = α [fc ((8m + 1)β) − fc ((8m + 3)β) π m=0 0 0 −fc ((8m + 5)β) + fc ((8m + 7)β)]dα ∞ ∞ π s−1 4 π 1 β ( ) [fc ((8m + 1)β) − fc ((8m + 3)β) = 4β π 4 β 2 m=0 0 −fc ((8m + 5)β) + fc ((8m + 7)β)]dβ ∞ ∞ 1 π ( )s− 2 β −s [fc ((8m + 1)β) − fc ((8m + 3)β) = 4 0 m=0 −fc ((8m + 5)β) + fc ((8m + 7)β)]dβ ∞ π s− 1 s 2 8 η(1 − s) β −s fc (β)dβ, =( ) 4 0 by (18) again with changing s into 1 − s. Now, using (13) again, we should have ∞ π s− 1 2 πs ∞ s−1 s−1 2 η(s) y f (y)dy = ( ) α f (α)dα, η(1 − s)Γ(1 − s) sin 4 π 2 0 0 and finally η(s) =
π s−1 2 4
π
Γ(1 − s) sin
πs η(1 − s). 2
This is the functional equation for η(s).
5
Remarks
5.1
Remark 1 In this remark, we show that η(s) = 0 when σ > 1. Indeed, we have ∞ an η(s) = , ns n=1
(19)
No.2
Ding: ON A FUNCTIONAL EQUATION
where an is given as the following: an =
⎧ ⎪ ⎪ ⎨1
n ≡ 1, 7(mod 8),
⎪ ⎪ ⎩
otherwise
231
−1 n ≡ 3, 5(mod 8), 0
We see, by computing directly, that am an = amn , and we have the following: ∞ ∞ ∞ ∞ am an μ(n) amn ak = μ(n) = μ(n) = a1 = 1. ms n=1 ns (mn)s ks m=1 m,n=1 k=1
n/k
So, we know that η(s) = 0 when σ > 1 and ∞ 1 an μ(n) = . η(s) n=1 ns
5.2
Remark 2 We are going to prove the following statement: If F (α) = 0 for α > A and F (α), F (α), F (α) are continuous over (0, ∞), then Fc (β) = O(β −2 ), Fc (β) = O(β −2 )
as
β → +∞.
Proof A straight forward calculation gives the following: ∞ A 2 2 F (α) cos αβdα = F (α) cos αβdα, Fc (β) = π 0 π 0 A A 2 2 βFc (β) = F (α) cos αβdαβ = − sin αβF (α)dα, π 0 π 0 A A 2 2 2 2 F (0). β Fc (β) = − sin αβF (α)dαβ = − cos αβF (α)dα + π 0 π 0 π The following estimate is true by using the above calculation: A 2 2 2 |F (0)| ≤ M. |F (α)|dα + |β Fc (β)| ≤ π 0 π Thus, we have Fc (β) = O(β −2 ). A Similarly, we can work on Fc (β) = − π2 0 αF (α) sin αβdα to get Fc (β) = O(β −2 ). We have proved the statement. Note From the above statement, we know the generalized M¨ unts formula is also true for Fc (β) with λ = μ = 2 if F (α) is the function in Remark 2. References [1] Titchmarsh E C. Introduction to the Theory of Fourier Integrals. Oxford, 1948 [2] Titchmarsh E C. The Theory of the Riemann Zeta-Function. Oxford, 1951 [3] Ding Yi. Distribution form of Ramanujan Identity (preprint)
ON A FUNCTIONAL EQUATION∗ Ding Yi (
)
Mathematics Department, New Jersey City University, USA
Abstract
In this article, the author derives a functional equation 1 π η(s) = ( )s− 2 4
πs 2 Γ(1 − s) sin( ) η(1 − s) π 2
(1)
of the analytic function η(s) which is defined by η(s) = 1−s − 3−s − 5−s + 7−s + · · ·
(2)
for complex variable s with Re s > 1, and is defined by analytic continuation for other values of s. The author proves (1) by Ramanujan identity (see [1], [3]). Her method provides a new derivation of the functional equation of Riemann zeta function by using Poisson summation formula. Key words Functional equation, Zeta function, M¨ unts formula 2000 MR Subject Classification
1
39B
Introduction The Euler’s zeta function ζ(s) is defined for any complex number s with Re(s) > 1 as ζ(s) =
∞ 1 . ns n=1
This series defines a holomorphic function. Riemann extended this function to a holomorphic function ζ(s) defined for all complex numbers s with s = 1 through analytic continuation. The Riemann zeta function ζ(s) has certain “trivial” zeros for s = −2, s = −4, s = −6, · · ·. The Riemann hypothesis states: nontrivial zeros of the Riemann zeta function lie on the critical line 1/2 + it. The connection between this function and prime numbers was already realized by Eurler 1 ζ(s) = . 1 − p−s prime p Helge von Koch in 1901 proved that the Riemann hypothesis is equivalent to the following considerable strengthening of the prime number theorem x 1 π(x) = dx + O(x1/2 lnx) x → ∞. ln(x) 2 ∗ Received
February 5, 2007. Supported by Separated Budget Research from New Jersey City University
226
ACTA MATHEMATICA SCIENTIA
Vol.29 Ser.B
Here, π(x) denotes the prime counting function, and the O-notation is the Landau symbol. Hilbert and Polya conjectured that imaginary parts of zeros of the Riemann zeta function should be the eigenvalues of a Hermitian operator. Atle Selberg gave a resemblance to the explicit formula by proving a duality between the length spectrum of a Riemann surface and the eigenvalues of its Laplacian. The statistical distribution of the zeros on the critical line now is believed to be exactly the same as the pair correlation distribution for the eigenvalues of a random Hermitian matrix after the work of Hugh Montgomery and Freeman Dyson. The Riemann zeta function statisfies the following functional equation: s ζ(s) = 2s π s−1 Γ(1 − s) sin(π ) ζ(1 − s) 2 valid for all s in C − {0, 1}. In this article, we derive a functional equation for a function η(s) similar to Riemann zeta function. Function η(s) is defined by η(s) = 1−s − 3−s − 5−s + 7−s + · · · The functional equation for η(s) is the following: η(s) =
π s− 12 2
Γ(1 − s) sin
π
4
πs 2
η(1 − s)
(3)
This work uses Ramanujan’s identity and Hurwitz zeta function. This method provides a new derivation of functional equation of Riemman zeta function.
2
Generalized M¨ unts Formula
For a function F (x) such that F (x) and F (x) are continuous over any finite interval and of O(x−λ ), O(x−μ ) as x → ∞, with λ > 1 and μ > 1. M¨ unts showed the following formula:
∞
0
s−1
x
C dx = ζ(s) F (nx) − x n=1
∞
∞ 0
xs−1 F (x)dx.
Here, C is a constant, and s is a complex number in some region. In this section, we provide a generalized M¨ unts formula and give a proof by using Hurwitz zeta function. The generalized M¨ unts formula is stated as followsl. Theorem 1 For a given function F(x) such that F (x) and F (x) are continuous over any finite interval and of O(x−λ ), O(x−μ ) as x → ∞, with λ > 1 and μ > 1, then 0
∞
xs−1
∞
n=0
F ((n + a)x) −
1 x
∞
0
F (v)dv dx = ζ(s, a)
∞ 0
y s−1 F (y)dy,
(4)
where s = σ + it, 0 < σ < 1, 0 < a ≤ 1. The Hurwitz zeta-function ζ(s, a) is an analytic function and is defined when σ > 1 by ζ(s, a) =
∞
1 . (n + a)s n=0
(5)
No.2
227
Ding: ON A FUNCTIONAL EQUATION
Proof First, for λ > σ > 1, we can interchange the integration and summation to have
∞
0
∞
xs−1
F (nx + ax)dx =
n=0
∞ n=0 ∞
∞
0
xs−1 F (nx + ax)dx
∞ 1 y s−1 F (y)dy s (n + a) 0 n=0 ∞ y s−1 F (y)dy, = ζ(s, a) =
0
because of ∞ n=0
1 | (n + a)s
∞ s−1 y F (y)|dy ≤
∞
0
n=0
1 (n + a)σ
∞
0
y σ−1 |F (y)|dy < ∞.
Now, we have the following estimate
∞
F (nx + ax) −
n=1
=x =x
∞
0
F (ux + ax)du
F (ux + ax)(u − [u])du
0 1 x
0
∞
F (ux + ax)(u − [u])du + x
= xI1 + x
∞ 1 x
∞ 1 x
O((ux + ax)−μ )du
O((ux)−μ )du
= xI1 + O(1). By looking at I1 in the above estimate, when x ≤ a1 , it implies ux+ax ≤ 2 and |F (ux+ax)| ≤ A, −μ when x > a1 , it implies ux + ax > 1 and |F (ux + ax)| ≤ K|ux + ax | ≤ A. This implies |xI1 | ≤ A. This inequality gives us the following estimate 1 ∞ F (v + ax)dv + O(1) = F (v)dv + O(1) x ax 0 1 ax 1 ∞ 1 ∞ F (v)dv − F (v)dv + O(1) = F (v)dv + O(1) = x 0 x 0 x 0 c = + O(1). x
∞
1 F (nx + ax) = x n=0
∞
Hence, when σ > 1, the following integral is calculated as ∞ ζ(s, a) y s−1 F (y)dy
0
∞
= 0
=
0
1
xs−1 s−1
x
∞
F (nx + ax)dx
n=0 ∞
c c dx + + F (nx + ax)dx − x s − 1 n=0
1
∞
s−1
x
∞ n=0
F (nx + ax)dx.
(6)
228
ACTA MATHEMATICA SCIENTIA
Vol.29 Ser.B
Now, the right-hand side of the above equality is regular for σ > 0, s = 1. Further more, if σ < 1, we have ∞ c = −c xs−2 dx. s−1 0 So, if 0 < σ < 1, the right-hand side of (6) is ∞ ∞
c dx xs−1 F (nx + ax) − x 0 n=0
∞ and is regular. Because the integral ζ(s, a) 0 y s−1 F (y)dy is also regular when 0 < σ < 1, thus we have (4).
3
Riemann’s Functional Equation
In this section, we want to prove functional equation for Riemann zeta function ζ(s) by using Poisson summation formula and M¨ unts formula. We have Theorem 2 For the Riemann zeta function ζ(s), s = σ + it, we have the following equality: πs ζ(s) = 2s π s−1 Γ(1 − s) sin ζ(1 − s), 0 < σ < 1. (7) 2 Proof If we take a = 1 in (4), we have the M¨ unts formula (see [1], section 2.11), that is, for 0 < σ < 1: ∞ ∞ ∞
1 ∞ s−1 s−1 ζ(s) y F (y)dy = x F (nx) − F (v)dv dx. (8) x 0 0 0 n=1 From the Poisson summation formula (see [1], section 2.8), we have the following equality: ∞ ∞ √ 1 1 β Fc (0) + Fc (nβ) = α F (0) + F (nα) , 2 2 n=1 n=1 ∞ where αβ = 2π and Fc (x) = π2 0 F (t) cos xtdt. Now, rewrite the above equality as follows: ∞ ∞ ∞ 1 1 β β F (nα) − √ F (α)dα = Fc (nβ) − F (0). α α 2 2π 0 n=1 n=1
The above equality will become the following after rearrangement: ∞ ∞ 1 ∞ β 1 ∞ F (nα) − F (α)dα = √ Fc (nβ) − Fc (v)dv . α 0 β 0 2π n=1 n=1 Now, using (9) we can write the formula (8) as follows: ∞ ∞ ∞ 1 ∞ y s−1 F (y)dy = αs−1 F (nα) − F (v)dv dα ζ(s) α 0 0 0 n=1 ∞ ∞ β 1 ∞ = αs−1 √ Fc (nβ) − Fc (v)dv dα β 0 2π n=1 0 ∞ ∞ 1 ∞ s− 12 −s = (2π) β Fc (nβ) − Fc (v)dv dβ. β 0 0 n=1
(9)
(10)
No.2
229
Ding: ON A FUNCTIONAL EQUATION
If Fc (β) has the same properties as F (α), then we have the M¨ unts formula for Fc (β) (see Remark 2), and this means
∞
0
β
−s
∞
1 Fc (nβ) − β n=1
∞ 0
Fc (v)dv dβ = ζ(1 − s)
Then, combining (10) and (11), we have ∞ 1 y s−1 F (y)dy = (2π)s− 2 ζ(1 − s) ζ(s) 0
∞
0
∞
0
z −s Fc (z)dz.
z −s Fc (z)dz.
(11)
(12)
But
∞
0
z
−s
Fc (z)dz = =
2 π
∞
0
2 π
∞
0
z
−s
∞
dz 0
F (α)dα
F (α) cos zαdα
∞
0
z −s cos zαdz.
It is well known that ∞ Γ(1 − s) 1−s πs z −s cos zαdz = cos π = Γ(1 − s)αs−1 sin , 1−s α 2 2 0 so, we have
∞
0
z
−s
Fc (z)dz =
2 πs Γ(1 − s) sin π 2
0
∞
αs−1 F (α)dα
(13)
0 < σ < 1.
(14)
From (12) and (13), we get ζ(s) = ζ(1 − s)Γ(1 − s)2s π s−1 sin
πs , 2
This is Riemann’s functional equation (see [2], section 2.1).
4
Functional Equation for η(s)
In this section, we will prove a functional equation for η. Let us state the functional equation for η in the following: Theorem 3 For the analytic function η(s) defined by η(s) = 1−s − 3−s − 5−s + 7−s + · · ·
(15)
for complex variable s with Res > 1, and defined for other values of s by analytic continuation. We have the following functional equation: πs π s− 12 2 Γ(1 − s) sin η(1 − s). (16) η(s) = 4 π 2 Proof Putting F (x) = f (8x) in (4) gives us the following equality:
∞ 0
s−1
x
∞
1 f ((8m + 8a)x) − x m=0
0
∞
f (8v)dv dx = ζ(s, a)
∞ 0
y s−1 f (8y)dy.
(17)
230
ACTA MATHEMATICA SCIENTIA
Vol.29 Ser.B
Taking a = 18 , a = 38 , a = 58 , and a = 78 in the above equality, respectively, we get ∞ ∞
xs−1 [f ((8m + 1)x) − f ((8m + 3)x) − f ((8m + 5)x) + f ((8m + 7)x)] dx 0
m=0
3 5 7 ∞ s−1 1 y f (8y)dy = ζ(s, ) − ζ(s, ) − ζ(s, ) + ζ(s, ) 8 8 8 8 0 ∞ = η(s)8s y s−1 f (y)dy.
(18)
0
From the Ramanujan equality (see [1], section 2.8), we have the following equality for αβ = ∞
π 4:
[f ((8m + 1)α) − f ((8m + 3)α) − f ((8m + 5)α) + f ((8m + 7)α)]
m=0
=
∞ 4 β [fc ((8m + 1)β) − fc ((8m + 3)β) − fc ((8m + 5)β) + fc ((8m + 7)β)]. π m=0
The above equality gives the following calculation: ∞ ∞ ∞ 4 s−1 s−1 β η(s) y f (y)dy = α [fc ((8m + 1)β) − fc ((8m + 3)β) π m=0 0 0 −fc ((8m + 5)β) + fc ((8m + 7)β)]dα ∞ ∞ π s−1 4 π 1 β ( ) [fc ((8m + 1)β) − fc ((8m + 3)β) = 4β π 4 β 2 m=0 0 −fc ((8m + 5)β) + fc ((8m + 7)β)]dβ ∞ ∞ 1 π ( )s− 2 β −s [fc ((8m + 1)β) − fc ((8m + 3)β) = 4 0 m=0 −fc ((8m + 5)β) + fc ((8m + 7)β)]dβ ∞ π s− 1 s 2 8 η(1 − s) β −s fc (β)dβ, =( ) 4 0 by (18) again with changing s into 1 − s. Now, using (13) again, we should have ∞ π s− 1 2 πs ∞ s−1 s−1 2 η(s) y f (y)dy = ( ) α f (α)dα, η(1 − s)Γ(1 − s) sin 4 π 2 0 0 and finally η(s) =
π s−1 2 4
π
Γ(1 − s) sin
πs η(1 − s). 2
This is the functional equation for η(s).
5
Remarks
5.1
Remark 1 In this remark, we show that η(s) = 0 when σ > 1. Indeed, we have ∞ an η(s) = , ns n=1
(19)
No.2
Ding: ON A FUNCTIONAL EQUATION
where an is given as the following: an =
⎧ ⎪ ⎪ ⎨1
n ≡ 1, 7(mod 8),
⎪ ⎪ ⎩
otherwise
231
−1 n ≡ 3, 5(mod 8), 0
We see, by computing directly, that am an = amn , and we have the following: ∞ ∞ ∞ ∞ am an μ(n) amn ak = μ(n) = μ(n) = a1 = 1. ms n=1 ns (mn)s ks m=1 m,n=1 k=1
n/k
So, we know that η(s) = 0 when σ > 1 and ∞ 1 an μ(n) = . η(s) n=1 ns
5.2
Remark 2 We are going to prove the following statement: If F (α) = 0 for α > A and F (α), F (α), F (α) are continuous over (0, ∞), then Fc (β) = O(β −2 ), Fc (β) = O(β −2 )
as
β → +∞.
Proof A straight forward calculation gives the following: ∞ A 2 2 F (α) cos αβdα = F (α) cos αβdα, Fc (β) = π 0 π 0 A A 2 2 βFc (β) = F (α) cos αβdαβ = − sin αβF (α)dα, π 0 π 0 A A 2 2 2 2 F (0). β Fc (β) = − sin αβF (α)dαβ = − cos αβF (α)dα + π 0 π 0 π The following estimate is true by using the above calculation: A 2 2 2 |F (0)| ≤ M. |F (α)|dα + |β Fc (β)| ≤ π 0 π Thus, we have Fc (β) = O(β −2 ). A Similarly, we can work on Fc (β) = − π2 0 αF (α) sin αβdα to get Fc (β) = O(β −2 ). We have proved the statement. Note From the above statement, we know the generalized M¨ unts formula is also true for Fc (β) with λ = μ = 2 if F (α) is the function in Remark 2. References [1] Titchmarsh E C. Introduction to the Theory of Fourier Integrals. Oxford, 1948 [2] Titchmarsh E C. The Theory of the Riemann Zeta-Function. Oxford, 1951 [3] Ding Yi. Distribution form of Ramanujan Identity (preprint)