formulae for zeta in the critical strip

formulae for zeta in the critical strip

Let us use the traditional notation

s

=

σ

+

i

⁢

t

for the complex variable,
where

σ

and

t

are

real numbers

.

ζ

⁢

(

s

)

=

1

1

-

2

1

-

s

⁢

∑

n

=

1

∞

(

-

1

)

n

+

1

⁢

n

-

s

σ

>

0

(1)

ζ

⁢

(

s

)

=

1

s

-

1

+

1

-

s

⁢

∫

1

∞

x

-

[

x

]

x

s

+

1

⁢

𝑑

x

σ

>

0

(2)

ζ

⁢

(

s

)

=

1

s

-

1

+

1

2

-

s

⁢

∫

1

∞

(

(

x

)

)

x

s

+

1

⁢

𝑑

x

σ

>

-

1

(3)

where

[

x

]

denotes the largest

integer

≤

x

,
and

(

(

x

)

)

denotes

x

-

[

x

]

-

1

2

.

We will prove (

2

) and (

3

) with the help of this
useful lemma:

Lemma:

For integers

u

and

v

such that

0

<

u

<

v

:

∑

n

=

u

+

1

v

n

-

s

=

-

s

⁢

∫

u

v

x

-

[

x

]

x

s

+

1

⁢

𝑑

x

+

v

1

-

s

-

u

1

-

s

1

-

s

Proof:

If we can prove the special case

v

=

u

+

1

, namely

(

u

+

1

)

-

s

=

-

s

⁢

∫

u

u

+

1

x

-

[

x

]

x

s

+

1

⁢

𝑑

x

+

(

u

+

1

)

1

-

s

-

u

1

-

s

1

-

s

(4)

then the lemma will follow by

summing

a

finite sequence

of cases of
(

4

).
The integral in (

4

) is

∫

0

1

t

⁢

d

⁢

t

(

u

+

t

)

s

+

1

=

∫

0

1

(

u

+

t

)

-

s

⁢

𝑑

t

-

∫

0

1

u

⁢

(

u

+

t

)

-

s

-

1

⁢

𝑑

t

=

(

u

+

1

)

1

-

s

-

u

1

-

s

1

-

s

+

u

⁢

[

(

u

+

1

)

-

s

-

u

-

s

]

s

so the right side of (

4

) is

-

s

1

-

s

⁢

[

(

u

+

1

)

1

-

s

-

u

1

-

s

]

-

u

⁢

[

(

u

+

1

)

-

s

-

u

-

s

]

-

u

1

-

s

1

-

s

+

(

u

+

1

)

1

-

s

1

-

s

=

(

u

+

1

)

-

s

⁢

[

-

s

⁢

(

u

+

1

)

1

-

s

-

u

+

u

+

1

1

-

s

]

+

u

-

s

⁢

[

u

⁢

s

1

-

s

+

u

-

u

1

-

s

]

=

(

u

+

1

)

-

s

⋅

1

+

u

-

s

⋅

0

and the lemma is proved.

Now take

u

=

1

and let

v

→

∞

in the lemma, showing that
(

2

) holds for

σ

>

1

.
By the principle of analytic continuation, if
the integral in (

2

) is

analytic

for

σ

>

0

,
then (

2

) holds for

σ

>

0

.
But

x

-

[

x

]

is

bounded

, so the integral

converges
uniformly

on

σ

≥

ϵ

for any

ϵ

>

0

, and the claim
(

2

) follows.

We have

1

2

⁢

s

⁢

∫

1

∞

x

-

1

-

s

⁢

𝑑

x

=

1

2

Adding and subtracting this quantity from (

2

),
we get (

3

) for

σ

>

0

.
We need to show that

∫

1

∞

(

(

x

)

)

x

s

+

1

⁢

𝑑

x

is analytic on

σ

>

-

1

. Write

f

⁢

(

y

)

=

∫

1

y

(

(

x

)

)

⁢

𝑑

x

and integrate by parts:

∫

1

∞

(

(

x

)

)

x

s

+

1

⁢

𝑑

x

=

lim

x

→

∞

⁡

f

⁢

(

x

)

⁢

x

-

1

-

s

-

f

⁢

(

1

)

⁢

x

-

1

-

1

+

(

s

+

1

)

⁢

∫

1

∞

f

⁢

(

x

)

x

s

+

2

⁢

𝑑

x

The first two terms on the right are zero, and the integral

converges

for

σ

>

-

1

because

f

is bounded.

Remarks:

We will prove (

1

) in a later version of this entry.

Using

formula

(

3

), one can verify Riemann’s
functional equation in the strip

-

1

<

σ

<

2

.
By analytic continuation, it follows that the functional
equation holds everywhere.
One way to prove it in the strip is to decompose the
sawtooth

function

(

(

x

)

)

into a

Fourier series

, and
do a termwise integration.
But the proof gets rather technical, because that
series does not converge uniformly.

Title

formulae for zeta in the

critical strip

Canonical name

FormulaeForZetaInTheCriticalStrip

Date of creation

2013-03-22 13:28:14

Last modified on

2013-03-22 13:28:14

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

11

Author

mathcam (2727)

Entry type

Theorem

Classification

msc 11M99

Related topic

CriticalStrip

Related topic

ValueOfTheRiemannZetaFunctionAtS0

Related topic

AnalyticContinuationOfRiemannZeta

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