The Truth Of The Riemann Hypothesis

The truth of the Riemann hypothesis Werner Raab Abstract We shall show that the meromorphic function v(s) =

π sin(πs)(1/2 − s)ζ(3/2 − s)

is holomorphic within the complex strip: 0 < <s < 1, since it is the Mellin transform Z ∞ v(s) = ts−1 w(t) dt 0

of the function

1 w(t) = 2πi

Z 1/2+i∞ 1/2−i∞

t−s v(s) ds

with the properties w(t) = O(1)

when t → 0

and w(t) = O(1/t) when t → ∞.

Keyword: Riemann hypothesis Mathematics Subject Classification: 11M26

1

A Mellin transformation

We consider the reciprocal u(s) =

1 (s − 1)ζ(s)

of the regularized zeta-function (s − 1)ζ(s) with the complex variable s. 1

(1)

The famous Riemann hypothesis may be expressed by the claim that the meromorphic function π v(s) = u(3/2 − s) (2) sin(πs) is holomorphic within the complex strip: 0 < <s < 1. Theorem 1 Within the complex strip: 0 < <s < 1/2, the function v(s) is equal to the Mellin transform v(s) =

Z ∞ 0

ts−1 w(t) dt

of the series

(3) s

∞ 2 X t µ(ν) w(t) = √ arctan ν t ν=1 ν with the values µ(ν) of the M¨obius arithmetic function.

(4)

Proof. The M¨obius numbers µ(ν) can be defined by the Dirichlet series ∞ X 1 µ(ν) = . ζ(s) ν=1 ν s

This converges in the half plane: <s > 1 and vanishes at s = 1. If 0 < <s < 1/2, then it follows from the Mellin transformation s Z ∞ Z √t/ν Z ∞ t dx dt = ts−3/2 dt ts−3/2 arctan ν 1 + x2 0 0 0 √ Z = ν

√ Z 1 Z ∞ ts−1 dx t dt = ν dt dx ν + x2 t 0 0 ν + x2 t 0 0 Z ∞ s−1 Z 1 t dx ν s−1/2 s−1/2 =ν dt = π 1+t sin(πs)(1 − 2s) 0 0 x2s

that

∞

s−1

Z 1

∞ X π µ(ν) v(s) = sin(πs)(1/2 − s) ν=1 ν 3/2−s

∞ X µ(ν) Z ∞ s−3/2 t arctan =2 ν=1

ν

0

s

Z ∞ ∞ X µ(ν) t dt = 2 ts−3/2 arctan ν 0 ν=1 ν

s

t dt. ν

In addition, we remark that µ

∞ 2 X µ(ν) π w(t) = √ − arctan 2 t ν=1 ν

r

ν t

2

¶

∞ 2 X µ(ν) = −√ arctan t ν=1 ν

r

ν . t

(5)

2

Power series expansions

Theorem 2 If 0 ≤ t < 1, then w(t) is represented by the power series w(t) =

∞ X

u(3/2 + k)(−t)k .

(6)

k=0

Proof. According to the definition (4) we have ∞ ∞ 2 X µ(ν) X (−1)k w(t) = √ t ν=1 ν k=0 1 + 2k

=

∞ ∞ X (−t)k X µ(ν) k=0

=

1/2 + k ν=1 ν 3/2+k

µ ¶1/2+k

t ν

∞ X

(−t)k . k=0 (1/2 + k)ζ(3/2 + k)

Another way to derive this result is the Mellin inversion ∞ X 1 Z 1/2+i∞ −s t−s u(3/2 − s) w(t) = . t v(s) ds = π Ress=−k 2πi 1/2−i∞ sin(πs) k=0

Theorem 3 If 0 ≤ t < ∞, then w(t) is represented by the series µ

∞ 1 X t w(t) = ∆m u(3/2) 1 + t m=0 1+t

¶m

(7)

with the differences m

∆ u(3/2) =

m X k=0

Ã

!

(−1)k m . k (1/2 + k)ζ(3/2 + k)

Proof. The identities Ã

∞ 1 X m t = 1 + t m=k k

!µ

k

t 1+t

¶m

hold for t > −1/2. According to Theorem 2 we obtain the power series (1 + t)w(t) = =

∞ X

u(3/2 + k)(−1)

k=0 ∞ X m=0

µ

∞ X

k

m=k

t 1+t

à ! ¶m X m m k=0

3

k

Ã

!µ

m k

t 1+t

¶m

(−1)k u(3/2 + k).

3

Finite differences

In order to estimate the fuction w(t) when the positive real variable t tends to infinity, we glance at the general theory of the finite differences m

∆ u(s) =

m X

Ã

!

m (−1)k u(s + k) = ∆m+1 u(s) + ∆m u(s + 1). k

k=0

(8)

The sequence of the partial sums n−1 X

n−1 X

∆m u(s + 1) =

m=0

³

´

∆m u(s) − ∆m+1 u(s) = u(s) − ∆n u(s)

m=0

converges to the limit ∞ X

∆m u(s + 1) = u(s) − lim ∆n u(s), n→∞

m=0

if limn→∞ ∆n u(s) exists. Theorem 4 If s is a positive real number, then lim ∆n u(s) = 0.

(9)

n→∞

Proof. For each fixed positive value of the real variable s the series f (s, t) =

∞ X u(s + k) k t k=0

k!

represents an entire function of the variable t. The Cauchy product −t

f (s, t)e

∞ ∞ X u(s + k) k X (−1)m m = t t

k!

k=0

=

à ! ∞ m X (−t)m X m m=0

m!

k=0

k

m!

m=0

(−1)k u(s + k) =

∞ X ∆m u(s) m=0

m!

(−t)m

may be considered as a generating function of the differences ∆m u(s). The series Z ∞ ∞ ∞ X u(s + k) Z ∞ k −2t u(s + k) X = t e dt = f (s, t)e−2t dt k+1 2 k! 0 0 k=0 k=0 =

∞ X ∆m u(s) m=0

m!

m

(−1)

Z ∞ 0

tm e−t dt =

∞ X

(−1)m ∆m u(s)

m=0 m m

converges apparently. Therefore the terms (−1) ∆ u(s) tend to zero. 4

4

Conclusion

Theorem 5 The function µ

∞ X

t (1 + t)w(t) = ∆ u(3/2) 1+t m=0 m

¶m

of the positive real variable t is bounded, i. e. w(t) = O(1/t)

(10)

when t tends to infinity. Proof. From Theorem 4 it is plain that the limit ∞ X m=0

∆m u(3/2) = u(1/2) − n→∞ lim ∆n u(1/2) = u(1/2)

(11)

exists. According to Abel’s theorem of continuity for power series, we have u(1/2) =

∞ X m=0

µ m

∆ u(3/2) lim

t→∞

t 1+t

¶m

∞ X

µ

t = lim ∆ u(3/2) t→∞ 1+t m=0 m

¶m

.

Now we have arrived at our goal: the Mellin integral Z ∞ π v(s) = = ts−1 w(t) dt sin(πs)(1/2 − s)ζ(3/2 − s) 0 represents a holomorphic function not only within the strip: 0 < <s < 1/2 (Theorem 1), but moreover within the entire strip: 0 < <s < 1, as Riemann conjectured. Finally, we note an interesting consequence of our derivation: the formulas (4) and (5) on the one hand, and (6) and (11) on the other hand, show that ∞ 2 X µ(ν) √ u(3/2) = w(0) = lim arctan t→0 t ν=1 ν

and

s

∞ X t µ(ν) √ =2 ν ν=1 ν ν

r ∞ ∞ √ X X µ(ν) µ(ν) ν √ . arctan = −2 u(1/2) = lim w(t)t = −2 lim t t→∞ t→∞ t ν ν=1 ν=1 ν

The idea of our proof is due in substance to Marcel Riesz [2]. For details concerning the fundamentals of the Riemann zeta-function we refer to Landau’s classical Handbuch [1]. A.M.D.G. 5

References [1] E. Landau, Handbuch der Lehre von der Verteilung der Primzahlen. Third (corrected) edition, two volumes in one, Chelsea Publishing, New York, 1974. (First edition, in two volumes, 1909.) [2] M. Riesz, Sur l’hypoth`ese de Riemann. Acta Math. 40 (1916), 185-190. Collected Papers, Springer-Verlag, Berlin, Heidelberg, New York, 1988, 165-170.

Werner Raab Dr. phil., Professor, retired member of the Mathematical Institute of the University of Bonn Residence: Anton-Klieber-Str. 14, 6410 Telfs, Austria E-mail: [email protected]

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