Demonstration of Riemann Hypothesis

Diego Marin ∗

June 12, 2014

Abstract

We define an infinite summation which is proportional to the reverse of Riemann Zeta

function ζ(s). Then we demonstrate that such function can have singularities only for

Re s = 1/n with n ∈ N \ 0. Finally, using the functional equation, we reduce these

possibilities to the only Re s = 1/2.

∗dmarin.math@gmail.com

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Contents

1 Target 3

2 Strategy 4

3 Application of the Euler-MacLaurin formula 5

4 Calculating the limiting function 13

2

1 Target

Riemann hypothesis, proposed by Bernhard Riemann in the 1859, is a conjecture

that regards an apparently simple function of complex variable s. Such function,

called Riemann Zeta Function, is defined for Re s > 1 via the following summation

ζ(s) =+∞∑n=1

n−s

For every integer n, an unique decomposition as product of (powers of) prime

numbers exists. In this way

ζ(s) =+∞∏p=2

+∞∑j=0

p−sj =+∞∏p=2

1

1− p−s

where we have used the summation rule for the geometric series

+∞∑j=1

wj =1

1− w

for |w| < 1. Riemann Zeta function can’t have zeros in the convergence area,

because no term in the product can be equal to zero. Nevertheless an holomorphic

extension of ζ(s) can be defined over the entire complex plane C, with the exception

of s = 1. Such extension has infinite zeros corresponding to all negative even

integers; that is, ζ(s) = 0 when s is one of −2,−4,−6, . . .. These are the so-called

“trivial zeros”.

Again for the holomorphic extension, for s 6= 1 we can prove the functional

equation1

ζ(s) = 2sπs−1 sin(πs

2

)Γ(1− s)ζ(1− s)

1The first proof was given by Bernhard Riemann in the 10-page paper Ueber die Anzahl derPrimzahlen unter einer gegebenen Grsse (usual English translation: On the Number of PrimesLess Than a Given Magnitude) published in the November 1859 edition by Monatsberichte derKniglich Preuischen Akademie der Wissenschaften zu Berlin.

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Leaving out the zeros of sin(πs2

)which aren’t poles of Γ(1− s), i.e. for s = −2n,

n ∈ N, any other zero s0 must have a “mate” zero s′0 = 1− s0. Because there are

no zeros for Re s > 1, functional equation implies that there are no zeros also for

Re s < 0 (except for s = −2n). Other works have excluded also the presence of

zeros for Re s = 0 and Re s = 1. As consequence, all the non trivial zeros of ζ(s)

stay in the “critical strip” 0 < Re s < 1.

Riemann hypothesis conjectures that all the non-trivial zeros have real part

equal to 12. This is what we aim to demonstrate in the following.

2 Strategy

We start from the definition of ζ(s) as an infinite product of terms, one term for

every prime number p, running from 2 to +∞:

ζ(s) =+∞∏p=2

1

1− p−s(1)

The product converges for Re s > 1. From here we fix s = x+ iy with s ∈ C and

x, y ∈ R. So the convergence condition is x > 1. Out of convergence area, ζ(s) is

defined via holomorphic extension.

The derivative of a single term in the product (1) gives

d

ds

1

1− p−s= − 1

(1− p−s)2(log p p−s) = − 1

1− p−s· log p

ps − 1

Hence the derivative ζ ′(s) of ζ(s) results

ζ ′(s) =d

dsζ(s) = −

+∞∏p=2

1

1− p−s·

+∞∑p′=2

log p′

p′s − 1

and thenζ ′(s)

ζ(s)= −

+∞∑p=2

log p

ps − 1(2)

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For briefness we set C(s) = ζ′(s)ζ(s)

. The sum (2) converges for x > 1. Otherwise we

define C(s) as the holomorphic extension of (2), recognizing it by the label “H.e.”:

C(s) = −H.e.∞∑n=1

log pnpsn − 1

The uniqueness of the holomorphic extension ensures that C(s) = ζ′(s)ζ(s)

also for

x ≤ 1. At this point we rely upon the following evidence:

Any singularity of C(s) corresponds to a zero of ζ(s) and/or to a

singularity of ζ ′(s).

We can exclude that a zero of ζ ′(s) hides a zero of ζ(s): in fact, for any holomorphic

function f(s), if a point s0 exists which is a zero both for f(s) and f ′(s), we have

surely

f(s0)

f ′(s0)= lim

s→s0

∑∞k=1 fk(s− s0)k∑∞

k=1 kfk(s− s0)k−1= lim

s→s0

fk̂(s− s0)k̂

k̂fk̂(s− s0)k̂−1= lim

s→s0

s− s0

k̂= 0

where k̂ is the minor k for which a Taylor coefficient fk is 6= 0. Hence the zeros of

ζ(s) are singularities for C(s) also when they are zeros of both ζ(s) and ζ ′(s).

For this reason we’ll proceed with finding an ensemble of points which includes

all the singularities of C(s) and so, among them, all the zeros of ζ(s).

3 Application of the Euler-MacLaurin formula

We move from prime to integer numbers:

C(s) = −∞∑n=1

log pnpsn − 1

(3)

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where pn is the n-th prime number. For s 6= 0, no partial sum

CM1 (s) = −

M∑n=1

log pnpsn − 1

with M < +∞ has singularities. Hence, any function

CM(s) = −+∞∑n=M

log pnpsn − 1

has the same singularities of C(s) for s 6= 0. This permit us to work with CM(s)

in place of C(s), in such a way to exploit the freedom in choosing M . Consider

now the Euler-MacLaurin theorem at leading order:∣∣∣∣∣N∑l=M

f(l, s)−∫ N

M

f(w, s)dw

∣∣∣∣∣ ≤ FN(s) ∈ R+

FN(s) =

∫ N

M

dw

∣∣∣∣ ddwf(w, s)

∣∣∣∣for some f(w, a) analytic in w ∈ R and holomorphic in s ∈ C except at most for

isolated points, provided that in such points we have w /∈ N.

Consider now limN→∞ FN(s)

!= F (s). If F (s) exists (is finite) in a open set

x > A (with A some real number) except at most for isolated points, then the

Dominated Convergence Theorem ensures (for x > A) that

limN→∞

∣∣∣∣∣N∑l=M

f(l, s)−∫ N

M

f(w, s)dw

∣∣∣∣∣ ≤ F (s) (4)

Now it is possible that the sum and the integral in the left side converge only for

x > B with B > A. For A < x ≤ B we can separate the holomorphic extension of

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the summation (H.e.) from its divergent piece (D.p.):

∞∑l=M

f(l, s) = H.e.

∞∑l=M

f(l, s) +D.p.

∞∑l=M

f(l, s)

We can do the same for the integral:∫ ∞M

f(w, s)dw = H.e.

∫ ∞M

f(w, s)dw +D.p.

∫ ∞M

f(w, s)dw

Inserting them in (4) we obtain:∣∣∣∣∣H.e.∞∑l=M

f(l, s) +D.p.∞∑l=M

f(l, s)−H.e.∫ ∞M

f(w, s)dw −D.p.∫ ∞M

f(w, s)dw

∣∣∣∣∣ ≤ F (s)

Being F (s) finite (except for isolated points), it has to be true

D.p.∞∑l=M

f(l, s) = D.p.

∫ ∞M

f(w, s)dw

and so ∣∣∣∣∣H.e.∞∑l=M

f(l, s)−H.e.∫ ∞M

f(w, s)dw

∣∣∣∣∣ ≤ F (s)

for A < x ≤ B. Hence we can apply the Euler-Maclaurin formula not only

for a comparison between sum and integral, but also between their holomorphic

extensions, at least until F (s) is finite. Our case is∣∣∣∣∣∞∑

n=M

log pnpsn − 1

−∫ ∞M

dtlog p(t)

p(t)s − 1

∣∣∣∣∣ ≤ F (s)

Here p(t) is any analytic function which posses analytic inverse t(p) and satisfies

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p(n) = pn. Moreover

|∑−∫| ≤

∫ ∞M

dt

∣∣∣∣ ddt log p(t)

p(t)s − 1

∣∣∣∣≤

∫ ∞M

dt

∣∣∣∣dp(t)dt

[1

p(t)(p(t)s − 1)− sp(t)s−1 log p(t)

(p(t)s − 1)2

]∣∣∣∣ (5)

The right side converges for x > 0 with the exceptions of isolated points.

The function t(pn) returns the cardinality n of the prime number pn. This is

equivalent to return how many prime numbers exist that are less than or equal to

pn, i.e. t(p) = π(p), where π(p) is the prime-counting function. A holomorphic

definition was given by Riesel, Edwards and Derbyshire2:

π(p) = limT→+∞

1

2πi

∫ 2+iT

2−iT

ps

slog ζ(s)ds

π′(p) = limT→+∞

1

2πi

∫ 2+iT

2−iTps−1 log ζ(s)ds

For p→ +∞ we have

π(p) ∼ p

log p

π′(p) =dπ(p)

dp∼ 1

log p

(1− 1

log p

)∼ 1

log p

Before going any further, we prove that (the holomorphic extension of) the integral

in |∑−∫|,

H.e.

∫ +∞

pM

dpdt(p)

dp

log p

ps − 1= H.e.

∫ +∞

pM

dp π′(p)log p

ps − 1, (6)

2Riesel, H. “The Riemann Prime Number Formula” Prime Numbers and Computer Methodsfor Factorization, 2nd ed. Boston, MA: Birkhuser, pp. 50-52, 1994; Edwards, H. M. Riemann’sZeta Function. New York: Dover, 2001; Derbyshire, J. Prime Obsession: Bernhard Riemannand the Greatest Unsolved Problem in Mathematics. New York: Penguin, 2004.

8

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has no singularities. We introduce first a redefinition of the same integral which

permits to calculate it also for x ≤ 1. This is

1

2i sin(πs)

∫Cε

dp π′(p)log p (−p)−s

1− p−s, (7)

which is equivalent to (6) for x > 1, so giving the relative holomorphic extension.

The Cauchy principle ensures that integral maintains the same value for every

choice of ε (until the path doesn’t touch any pole of the integrated function); for

demonstration we find useful to work in the limit ε→ 0+.

The contribute at ∞ is null for x > 1. This is unique zone of interest to check

the equivalence between (6) and (7); so, at this aim, we can forget about it.

lim|p|→+∞

π′(p)log p (−p)−s

1− p−s= 0 for x > 1

Having said this, we proceed by calculating separately the contributes to the inte-

gral along Γ1, Γ3 and Γ2:∫Γ1

[ ] = limε→0+

∫ +∞

pM

dt π′(t+ iε)log(t+ iε) e−s log(t+iε)+iπs

1− e−s log(t+iε)

= eiπs∫ +∞

pM

dp π′(p)log p p−s

1− p−s

= eiπs∫ +∞

pM

dp π′(p)log p

ps − 1

∫Γ3

[ ] = limε→0+

∫ pM

+∞dt π′(t− iε) log(t− iε) e−s log(t−iε)−iπs

1− e−s log(t−iε)

= −e−iπs∫ +∞

pM

dp π′(p)log p p−s

1− p−s

= −e−iπs∫ +∞

pM

dp π′(p)log p

ps − 1

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For the integral in Γ2, we pose p = εeiθ + pM , dp = iεeiθdθ,

∫Γ2

[ ] = limε→0

∫ 3π4

− 3π4

dθπ′(εeiθ + pM) log(εeiθ + pM)(εeiθ + pM)−seiπsiεeiθ

1− (εeiθ + pM)−s

= limε→0

iε

∫ 3π4

− 3π4

dθπ′(pM) log(pM)(pM)−seiπseiθ

1− (pM)−s

= limε→0

iεπ′(pM) log(pM)eiπs

psM − 1

∫ 3π4

− 3π4

dθ eiθ

= limε→0

επ′(pM) log(pM)eiπs

psM − 1

[ei

3π4 − e−i

3π4

]= lim

ε→0iε

√2 π′(pM) log(pM)eiπs

psM − 1= 0

In the end:

1

2i sin(πs)

∫Cε

dp π′(p)log p (−p)−s

1− p−s=

1

2i sin(πs)

[∫Γ1

[ ] +

∫Γ2

[ ] +

∫Γ3

[ ]

]=

1

2i sin(πs)

(eiπs − e−iπs

)∫ +∞

pM

dp π′(p)log p

ps − 1

=1

2i sin(πs)2i sin(πs)

∫ +∞

pM

dp π′(p)log p

ps − 1

=

∫ +∞

pM

dp π′(p)log p

ps − 1

CVD. At this stage we can doubtless affirm that (7) is the holomorphic extension

of (6). Using the Cauchy’s integral theorem, the integral in (7) can be transformed

into a sum over the minimal circuitations around all the poles. These sit at p =

exp(

2πiks

), k ∈ Z. A single term is the following:

sin(πs)

∮ (k)

[ ] = limε→0+

iε

2i

∫ 2π

0

eiθdθ π′(e2πiks + εeiθ)

log(e2πiks + εeiθ)

(e2πiks + εeiθ)s − 1

eiπs =

= limε→0+

ε

2

∫ 2π

0

eiθdθ π′(e2πiks + εeiθ)

log(1 + εeiθ−2πiks ) + 2πk

s

e2πikss(1 + εeiθ−

2πiks )s − 1

eiπs =

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= limε→0+

ε

2

∫ 2π

0

eiθdθ π′(e2πiks + εeiθ)

log(1 + εeiθ−2πiks ) + 2πk

s

(1 + εeiθ−2πiks )s − 1

eiπs =

= limε→0+

ε

2

∫ 2π

0

eiθdθ π′(e2πiks + εeiθ)

εeiθ−2πiks + 2πk

s

sεeiθ−2πiks

eiπs =

= limε→0+

1

2

∫ 2π

0

eiθdθ π′(e2πiks )

2πks

seiθ−2πiks

eiπs =

=πk

s2π′(e

2πiks )e

2πiks

+iπs

∫ 2π

0

dθ =

=2π2k

s2π′(e

2πiks )e

2πiks

+iπs

By summing over all the poles:

sin(πs)H.e.

∫ +∞

pM

dp π′(p)log p

ps − 1= eiπs

+∞∑k=−∞

2π2k

s2π′(e

2πiks )e

2πkis

= eiπs+∞∑

k=−∞

2π2k

s2π′(e

2πiks )e

2πiks∗|s|2

=π

s2eiπs

∫ +∞

−∞log ζ(2 + iw)dw

+∞∑k=−∞

ke2πik(1+iw)s∗

|s|2 e2πiks∗|s|2

=π

s2eiπs

∫ +∞

−∞log ζ(2 + iw)dw

+∞∑k=1

k

[e

2πik(2+iw)s∗

|s|2 − e−2πik(2+iw)s∗

|s|2

]

Use now the summation rule

+∞∑k=1

keka =d

da

+∞∑k=1

eka =d

da

[1

1− ea− 1

]=

ea

(1− ea)2=

1

(e−a/2 − ea/2)2

The last term in the right is the correct value of summation only for Rea < 0.

Nevertheless, when we climb over the line Rea = 0, it gives the corresponding

holomorphic extension. Hence we can use the rule without care of convergence

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criterium:

=π

s2eiπs

∫ +∞

−∞log ζ(2 + iw)dw

1(eπi(2+iw)s∗|s|2 − e−

πi(2+iw)s∗|s|2

)2

For w → +∞ the integrand goes like

∼ π

s2eiπs log ζ(2 + iw) e

− 2πws∗|s|2

and so the integral converges at +∞ (remember that Re s∗ = x > 0). Similarly,

for w → −∞ the integrand goes like

∼ π

s2eiπs log ζ(2 + iw) e

2πws∗|s|2

and so the integral converges at −∞. Being know that ζ(2 + iw) has neither zeros

nor poles for w ∈ R, we can say that (the Holomorphic Extension of) the integral

in |∑−∫| has no singularities for s ∈ C \ R.

The inescapable conclusion is that all singularities of C(s) in the critical strip

emerge from the difference between it and the corresponding integral, i.e. they are

among the isolated points where F (s) = +∞.

4 Calculating the limiting function

Let recover the result (5):

|∑−∫| ≤

∫ +∞

M

dt

∣∣∣∣dp(t)dt

∣∣∣∣ ∣∣∣∣[ 1

p(t)(p(t)s − 1)− sp(t)s−1 log p(t)

(p(t)s − 1)2

]∣∣∣∣≤

∫ +∞

M

dtdp(t)

dtIt

∣∣∣[ ]∣∣∣− ∫ +∞

M

dtdp(t)

dt[1− It]

∣∣∣[ ]∣∣∣where It = 1 if dp(t)

dt≥ 0 and It = 0 otherwise. Use now dt dp(t)

dt= dp to achieve an

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advantageous change of variable:

≤∫ +∞

pM

dp It(p)

∣∣∣[ ]∣∣∣− ∫ +∞

pM

dp [1− It(p)]∣∣∣[ ]∣∣∣

≤∫ +∞

pM

dp It(p)

∣∣∣[ ]∣∣∣+

∫ +∞

pM

dp [1− It(p)]∣∣∣[ ]∣∣∣

≤∫ ∞pM

dp

∣∣∣∣[ 1

p(ps − 1)− sps−1 log p

(ps − 1)2

]∣∣∣∣≤

∫ ∞pM

dp

∣∣∣∣ps − 1− sps−1 log p

p(ps − 1)2

∣∣∣∣Now consider the following inequality:

|ps − 1|2 = (ps − 1)(ps∗ − 1) = ps+s

∗ − ps − ps∗ + 1

= p2x − 2pxcos(y log p) + 1

≥ p2x − 2px + 1 = (px − 1)2

For x > 0 we have also

|ps − 1− sps−1 log p|2 = (ps − 1− sps−1 log p)(ps∗ − 1− s∗ps∗−1 log p)

= p2x − 2pxcos(y log p) + 1−

−2xp2x−1 log p+ 2xpx−1 log p cos(y log p)−

−2ypx−1 log p sin(y log p) + |s|2p2x−2 log2 p

≤ p2x + 2px + 1 + 2xp2x−1 log p+ 2xpx−1 log p+

+2|y|px−1 log p+ (x2 + y2)p2x−2 log2 p

14

≤ p2x + 2px + 1 + 2xp2x−1 log p+ 2xpx−1 log p+

+2|y|px−1 log p+ (x2 + y2)p2x−2 log2 p+

+2x|y|p2x−2 log2 p+ 2|y|p2x−1 log p =

= (px + 1 + (x+ |y|)px−1 log p)2

Hence

|∑−∫| ≤

∫ ∞pM

dppx + 1 + (x+ |y|)px−1 log p

p(px − 1)2

≤ 2

x(1−px)− log(1−p−x)

x−

(x+|y|) log p((px−1)Φ(px, 1,− 1

x

)+x)

x2p(px − 1)

∣∣∣∣∣∞

pM

Φ(w, a, b) is the Lerch Transcendent3 which goes at +∞ in the first variable as

w−1; in our case as p−x. As consequence, the contribute at +∞ is null. Finally

|∑−∫| ≤ 2

x(pxM−1)+

log(1−p−xM )

x+

(x+|y|) log p((pxM−1)Φ(pxM , 1,− 1

x

)+x)

x2pM(pxM − 1)

The Lerch Transcendent Φ(w, a, b) has singularities (if a ∈ N\0) only for b ∈ −N\0.

In our case, a = 1 and so we have singularities for

x =1

nwith n ∈ N \ 0

This means that the zeros of ζ(s) in the strip 0 < x < 1 have to satisfy x = 1n

for

some n ∈ N \ 0. Moreover the functional equation

ζ(s) = 2sπs−1 sin(πs

2

)Γ(1− s)ζ(1− s)

reveals that if s0 is a non-trivial zero, then 1 − s0 is a zero too. Hence we search

3Φ(w, a, b) is usually defined as the holomorphic extension of Φ(w, a, b) = 1Γ(w)

∫ +∞0

ta−1e−bt

1−we−t dt

which works for {Re b > 0 ∧Rea > 0 ∧ |w| < 1} ∪ {Re b > 0 ∧Rea > 1 ∧ |w| = 1}.

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for two integers m,n such that

1

m= 1− 1

n,

but the unique solution of this equation is m = n = 2. Consequently, all the

non-trivial zeros of zeta must have real part equal to 1/2. CVD

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