The Dirichlet series that generates the Möbius

function is the inverse of the Riemann zeta

function in the right half of the critical strip
arXiv:1009.1092v13 [math.GM] 18 Jan 2015

Roupam Ghosh
April 3, 2019

Abstract:
In this paper I introduce
P∞ a criterion for the Riemann hypothesis, and then using
that I prove k=1 µ(k)/k s converges for ℜ(s) > 21 . I use a step function
ν(x) = 2{x/2} − {x} for the Dirichlet eta function ({x} is the fractional part of
x), which was at the core of my investigations, and hence derive the stated result
subsequently.

1
 In 1859, Bernhard Riemann showed the existence of a deep relationship between
two very different mathematical entities., viz. the zeros of an analytic function and
prime numbers.

The Riemann Hypothesis is usually stated as, the non-trivial zeros of the Rie-
mann zeta function lie on the line ℜ(s) = 21 . Although, this is the standard
formulation, one of the exciting features of this problem is, it can be formulated
in many different and unrelated ways.

The approach I take in this paper is influenced by Beurling’s 1955 paper: A closure
problem related to the Riemann zeta function and Báez-Duarte’s 2001 paper: New
versions of the Nyman-Beurling criterion for the Riemann Hypothesis, although
it takes a new approach. In this paper I would be studying a simple step function
ν relating it to the Dirichlet eta function η. I will show how the step function ν
convolves with the Möbius function µ(n) and gives a constant, which I think is a
new result significant at attacking RH.

TheoremP∞1: For1
the Dirichlet eta function is defined as, for all ℜ(s) > 0
1
η(s) = k=1 (2k−1)s − (2k) s We get an equivalent expression in the form for all

ℜ(s) > 0 where ν(x) = 2 {x/2} − {x}, and the expression being given by
Z ∞
η(s) = s ν(x)x−s−1 dx (1)
1

Proof: A simplification of the integral shall prove this case. We have for all x ∈
[1, ∞] ν(x) = 0 or 1. It is not hard to see that ν(x) = 0 whenever x ∈ [2k, 2k + 1)
and 1 whenever x ∈ [2k − 1, 2k) for all positive integers k. Hence, we can write
the integral as
Z ∞ X∞ Z 2k
−s−1
s ν(x)x dx = sx−s−1 dx
1 k=1 2k−1

Giving us the sum

∞
X 1 1
s
−
k=1
(2k − 1) (2k)s
which is nothing but η(s). Since we already know that this sum converges for
ℜ(s) > 0, we get our result.

2
Alternatively, we can write equation (1) as
Z 1  
η(s) 1
= ν xs−1 dx (2)
s 0 x

Discussion: ν(x) is a simple function that oscillates between 0 and 1 at every

integer. As mentioned in the above theorem, ν(x) = 0 whenever x ∈ [2k, 2k + 1)
and 1 whenever x ∈ [2k − 1, 2k) for all positive integers k. Note that ν(x/k) = 0
if x < k. The next theorem relates ν and Möbius µ in a very interesting way. I
think this might be a new and interesting result, where relating µ with the simple
oscillating step function gives a constant value of −1. I think this result is signifi-
cant for the proof of RH.



 0 if x ∈ [0, 1)
Theorem 2: ∞ x
P
k=1 µ(k)ν k
= 1 if x ∈ [1, 2)
−1 if x ∈ [2, ∞)


Proof:
We have for 0 < θ ≤ 1
θ
θs η(s)
 
θ
Z
ν xs−1 dx =
0 x s
Since ν(θ/x) = 0 where θ < x, or θ = 1 otherwise, hence
Z 1  
θ
ν xs−1 dx = 0
θ x

We get the following important equation,

Z 1  
θ θs η(s)
ν xs−1 dx = (3)
0 x s

If we set fµ (x) = ∞
P
k=1 µ(k)ν(x/k) we get

1   ∞
1 η(s) X µ(k)
Z
s−1
fµ x dx = (4)
0 x s k=1 k s

(Justification for the exchange of summation and integral: Let fn (x) = nk=1 µ(k)ν(x/k).
P
∞ R∞
Now since for ℜ(s) > 1, we have nk=1 0 |µ(k)ν(x/k)x−s−1 |dx ≤ nk=1 |µ(k)| ν(x)x−σ−1 dx ≤
P R P
k σ 0
ζ(σ)η(σ) R ∞ P ∞ −s−1
ζ(2σ)σ
< ∞. By Fubini-Tonelli’s theorem we can say, 0 k=1 µ(k)ν(x/k)x dx =

3
P∞ R ∞
k=1 0
µ(k)ν(x/k)x−s−1 dx).
P∞ s
For ℜ(s) > 1 we know that k=1 µ(k)/k = 1/ζ(s). Hence, for ℜ(s) > 1
1
1 − 21−s
 
1
Z
fµ xs−1 dx = (5)
0 x s
Now, since fµ (x) = 1 whenever x ∈ [1, 2) giving us for all ℜ(s) > 1
Z 1  
2 1
1 + fµ xs−1 dx = 0 (6)
0 x
P
Since, µ(k)ν(x/k) is always constant in any given [n, n + 1). So due to equation
(6) we have for all x ≥ 2 , fµ (x) = −1. (Properties of Dirichlet series).

Discussion: The derivation is not explained above. Notice, the integral in equa-
tion (6) above can be expanded as,
−1 − fµ (2) fµ (2) − fµ (3) fµ (3) − fµ (4)
+ + + ... = 0
2s 3s 4s
The uniqueness property implies, −1 − fµ (2) = fµ (2) − fµ (3) = ... = 0. Since
fn (x) is a step function, we get the result for x ≥ 2.

P∞
Theorem 3: If fn = O(nσ0 +ǫ ) where fn = | sup fn (x)|, then k=1 µ(k)/k s con-
verges for all ℜ(s) > σ0 .

Proof: If we set fn (x) = nk=1 µ(k)ν(x/k) we get for ℜ(s) > 0

P

Z 1 n
η(s) X µ(k) 1 − 21−s
 
2 1
1 + fn xs−1 dx = − (7)
0 x s k=1 k s s

Changing the integral we get

Z ∞ n
1 + fn (x) η(s) X µ(k) 1 − 21−s
dx = − (8)
2 xs+1 s k=1 k s s

Since, 1 + fn (x) = 0 for x ∈ [2, n), it follows from above,

  Z
n ∞ Z ∞
 η(s) X µ(k) 1 − 21−s  |1 + fn (x)| |1 + fn (x)| sup |1 + fn (x)|
− = dx = dx ≤

 s k s s xσ+1 xσ+1 σnσ

k=1
 2 n
(9)

4
If fn = O(nσ0 +ǫ ) then LHS converges to 0 for σ > σ0 as n → ∞, i.e,
 
 η(s) Xn 1−s 
µ(k) 1 − 2 
lim  − =0 (10)

n→∞  s
k=1
ks s 
which gives, for ℜ(s) > σ0
∞
η(s) X µ(k) 1 − 21−s
= (11)
s k=1 k s s
Now here η(s) 6= 0 for 1 > ℜ(s)  > σ0 . (This is because of, equation (9) with
 1−21−s 
η(s) = 0 gives LHS =  s . Ignoring the line ℜ(s) = 1, because it is known
that Möbius converges on that line.)

Since in equation (11), η(s) 6= 0, therefore ∞ s

P
k=1 µ(k)/k converges for ℜ(s) > σ0 .
This is because, we can see from the expression in equation (11), where the Möbius
sum exactly equals 1/ζ(s), which it wouldn’t if it did not converge for ℜ(s) > σ0 .

Theorem 4: ∞ s 1
P
k=1 µ(k)/k converges for ℜ(s) > 2
Proof:
For any σ > 0, σ 6= 1 the following is justified geometrically.
n Z n
X |µ(k)| dy n1−σ 1
≤ + 1 = − + 1 = O(n1−σ ) (12)
k σ
1 y σ 1 − σ 1 − σ
k=1

For σ = 1 we shall similarly have for any ǫ > 0, nk=1 |µ(k)|/k ≤ log(n)+1 = O(nǫ )
P

P∞
I. Assume 1 ≥ σ0 > 1/2 such that, k=1 µ(k)/k s converges only for ℜ(s) > σ0 .

II. Now consider for σ > 0,

Z ∞  Z ∞ ∞
fn (x)  |fn (x)| |fn (x)|
Z

 dx ≤ dx = dx

1 xσ+1  1 xσ+1 0 xσ+1
Z ∞ Pn
k=1 |µ(k)|ν(x/k)
≤ dx
0 xσ+1 (13)
n
η(σ) X |µ(k)|
=
σ kσ
k=1
1−σ
≤ cn

Assume σ0 > σ ′ > 1/2

1  ∞ fn (x) 
Z  Z ∞
1 |fn (x)| ′
′  dx ≤ σ′ dx ≤ cn1−σ −σ (14)
nσ
1 xσ+1 n 1 x σ+1

5
Now as n → ∞, equation (12) → 0 for all σ > 1 − σ ′ . If we consider
′
gn (x) = |fn (x)|/nσ

then the Mellin transform over gn (x) can be expressed as a Dirichlet series for
σ > 1 − σ ′ , because gn (x) is a step function (since fn (x) is a step function) and
the transform converges.
∞ ∞
gn (x) X an,k
Z
Dn = dx =
1 xσ+1 k=1
kσ

Since the Dirichlet series Dn → 0 for all σ > 1 − σ ′ as n → ∞, hence an,k → 0

as n → ∞ (by the uniqueness property of Dirichlet series, i.e., the series vanishes
identically). Giving us gn (1) → 0, gn (1) − gn (2) → 0, gn (2) − gn (3) → 0, and so
on... as n → ∞. Hence gn (x) → 0 as n → ∞, and therefore
′
fn = | sup fn (x)| = o(nσ )

P∞ s
III. Now considering theorem 3 and the result in II., we get k=1 µ(k)/k con-
verges for all ℜ(s) > σ ′ .

IV. But this contradicts our assumption that ∞ s

P
k=1 µ(k)/k converges
P∞ only fors
1 ′
ℜ(s) > σ0 > 2 , since σ0 > σ > 1/2. Therefore, we must have k=1 µ(k)/k
converges for all ℜ(s) > 12 , thereby validating the Riemann hypothesis.

References
[1] Peter Borwein, Stephen Choi, Brendan Rooney, Andrea Weirathmueller
”The Riemann Hypothesis: A Resource for the Afficionado and Virtuoso
Alike”.

[2] H.M. Edwards ”Riemann’s Zeta function”.

[3] E.C. Titchmarsh ”The theory of the Riemann Zeta function”.

[4] Julian Havil ”Gamma: Exploring Euler’s constant”.

[5] T.M. Apostol ”Introduction to Analytic Number Theory”

6
[6] A. Beurling, ”A closure problem related to the Riemann zeta function, Proc.
Natl. Acad.Sci. 41 (1955) 312314”

[7] Luis Báez Duarte, ”New versions of the Nyman-Beurling criterion for the
Riemann hypothesis” IJMMS 31:7 (2002) 387406 PII. S0161171202013248