Riemann zeta function has no zeros on ℜ⁡s=0,1

Riemann zeta function has no zeros on

ℜ

⁡

s

=

0

,

1

This article shows that the

Riemann zeta function

ζ

⁢

(

s

)

has no zeros along the lines

ℜ

⁡

s

=

0

or

ℜ

⁡

s

=

1

. That implies that all nontrivial zeros of

ζ

⁢

(

s

)

lie strictly within the

critical strip

0

<

ℜ

⁡

s

<

1

. As the article points out, this is known to be equivalent to one version of the

prime number theorem

.

It can in fact be shown that

ζ

⁢

(

s

)

≠

0

for any

s

=

σ

+

i

⁢

t

with

0

<

σ

<

1

if

σ

≥

1

-

c

log

⁡

(

|

t

|

+

1

)

for some constant

c

. By using the

functional equation

π

-

s

2

⁢

Γ

⁢

(

s

2

)

⁢

ζ

⁢

(

s

)

=

π

-

1

-

s

2

⁢

Γ

⁢

(

1

-

s

2

)

⁢

ζ

⁢

(

1

-

s

)

we have also that

ζ

⁢

(

σ

+

i

⁢

t

)

≠

0

if

σ

≤

c

log

⁡

(

|

t

|

+

1

)

Bounding the zeros of

ζ

⁢

(

s

)

away from

ℜ

⁡

s

=

0

,

1

leads to a version of the prime number theorem with more precise error terms.

Theorem 1

ζ

⁢

(

1

+

i

⁢

t

)

≠

0

for

t

∈

R

.

Proof.

Notice that for

θ

∈

ℂ

0

≤

2

⁢

(

1

+

cos

⁡

θ

)

2

=

2

⁢

cos

2

⁡

θ

+

4

⁢

cos

⁡

θ

+

2

=

3

+

4

⁢

cos

⁡

θ

+

cos

⁡

(

2

⁢

θ

)

(1)

If

σ

=

ℜ

⁡

s

>

1

, then

ζ

⁢

(

σ

+

i

⁢

t

)

=

∏

p

⁢

prime

(

1

-

p

-

σ

-

i

⁢

t

)

-

1

, so that

log

⁡

ζ

⁢

(

σ

+

i

⁢

t

)

=

-

∑

p

⁢

prime

log

⁡

(

1

-

p

-

σ

-

i

⁢

t

)

=

∑

p

⁢

prime

∑

m

=

1

∞

1

m

⁢

p

-

m

⁢

σ

-

i

⁢

m

⁢

t

and thus

log

⁡

|

ζ

⁢

(

σ

+

i

⁢

t

)

|

=

∑

p

⁢

prime

∑

m

=

1

∞

1

m

⁢

p

m

⁢

σ

⁢

cos

⁡

(

m

⁢

t

⁢

log

⁡

p

)

since the log of the

absolute value

is the

real part

of the log.

Using equation (

1

), we then have

3

⁢

log

⁡

ζ

⁢

(

σ

)

+

4

⁢

log

⁡

|

ζ

⁢

(

σ

+

i

⁢

t

)

|

+

log

⁡

|

ζ

⁢

(

σ

+

i

⁢

2

⁢

t

)

|

=

∑

p

⁢

prime

∑

m

=

1

∞

1

m

⁢

p

m

⁢

σ

⁢

(

3

+

4

⁢

cos

⁡

(

m

⁢

t

⁢

log

⁡

p

)

+

cos

⁡

(

2

⁢

m

⁢

t

⁢

log

⁡

p

)

)

≥

0

so that

ζ

⁢

(

σ

)

3

⁢

|

ζ

⁢

(

σ

+

i

⁢

t

)

|

4

⁢

|

ζ

⁢

(

σ

+

i

⁢

t

⋅

2

)

|

≥

1

⁢

for all

⁢

σ

>

1

,

t

∈

ℝ

(2)

But if

ζ

has a zero at

σ

+

i

⁢

t

0

, then

lim

σ

→

1

+

⁡

ζ

⁢

(

σ

)

3

⁢

|

ζ

⁢

(

σ

+

i

⁢

t

0

)

|

4

⁢

|

ζ

⁢

(

σ

+

i

⁢

2

⁢

t

)

|

=

0

since the first factor gives a pole (

http://planetmath.org/Pole

) of order 3 at

1

and the second factor gives a zero of order at least 4 at

1

+

i

⁢

t

0

. This contradicts equation (

2

).

Corollary 1

ζ

⁢

(

i

⁢

t

)

≠

0

for

t

∈

R

.

Proof.

Use the functional equation

π

-

s

2

⁢

Γ

⁢

(

s

2

)

⁢

ζ

⁢

(

s

)

=

π

-

1

-

s

2

⁢

Γ

⁢

(

1

-

s

2

)

⁢

ζ

⁢

(

1

-

s

)

and set

s

=

i

⁢

t

. The theorem implies that the RHS is nonzero, so the LHS is as well. Thus

ζ

⁢

(

s

)

≠

0

.

Title

Riemann zeta function has no zeros on

ℜ

⁡

s

=

0

,

1

Canonical name

RiemannZetaFunctionHasNoZerosOnReS01

Date of creation

2013-03-22 17:54:37

Last modified on

2013-03-22 17:54:37

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

5

Author

rm50 (10146)

Entry type

Theorem

Classification

msc 11M06

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