analytic continuation of Riemann zeta (using integral)

analytic continuation of Riemann zeta (using integral)

The

Riemann zeta function

can be analytically
continued to the whole

complex plane

minus the point 1
by means of an

integral

representation

. Remember that the zeta functon
is defined by the series

ζ

⁢

(

s

)

=

∑

n

=

1

∞

1

n

s

.

When

ℜ

⁡

s

>

1

, this series converges; furthermore, this convergence is
uniform on compact subsets of this half-plane, hence the series converges to
an

analytic function

on this half plane. However, the series diverges when
we have

ℜ

⁡

s

<

1

, so this series cannot be used to define the

zeta function

in the whole complex plane, which is why we must make an

analytic continuation

.

To make this

continuation

, we start by changing the variable in an integration:

n

s

⁢

∫

0

∞

e

-

n

⁢

x

⁢

x

s

-

1

⁢

𝑑

x

=

∫

0

∞

e

-

y

⁢

y

s

-

1

⁢

𝑑

y

=

Γ

⁢

(

s

)

This provides us with an integral representation of our summand.
Substituting this into the series, we find that

ζ

⁢

(

s

)

=

∑

n

=

1

∞

1

Γ

⁢

(

s

)

⁢

∫

0

∞

e

-

n

⁢

x

⁢

x

s

-

1

⁢

𝑑

x

=

1

Γ

⁢

(

s

)

⁢

∑

n

=

1

∞

∫

0

∞

e

-

n

⁢

x

⁢

x

s

-

1

⁢

𝑑

x

.

We note that

∑

n

=

1

∞

∫

0

∞

|

e

-

n

⁢

x

⁢

x

s

-

1

|

⁢

𝑑

x

=

∑

n

=

1

∞

∫

0

∞

e

-

n

⁢

x

⁢

x

|

s

-

1

|

⁢

𝑑

x

=

∑

n

=

1

∞

Γ

⁢

(

|

s

-

1

|

+

1

)

n

s

;

because the series converges, hence it is possible
to interchange integration and

summation

and
subsequently sum a geometric series.

ζ

⁢

(

s

)

=

1

Γ

⁢

(

s

)

⁢

∑

n

=

1

∞

∫

0

∞

e

-

n

⁢

x

⁢

x

s

-

1

⁢

𝑑

x

=

1

Γ

⁢

(

s

)

⁢

∫

0

∞

∑

n

=

1

∞

e

-

n

⁢

x

⁢

x

s

-

1

⁢

d

⁢

x

=

1

Γ

⁢

(

s

)

⁢

∫

0

∞

x

s

-

1

e

x

-

1

⁢

𝑑

x

As it stands, the integral representation we have is not of
much use for analytically continuing the zeta function
because the integral diverges when

ℜ

⁡

s

<

1

on account of
the fact that the integrand behaves like

x

-

s

when

x

is close to zero. However, it is possible to make use of the
theorem of Cauchy to move the path of integration away from
zero.

Given a

real number

r

>

0

, define the

contour

C

r

on the

Riemann surface

of

z

s

-

1

as follows:

C

r

passes from

+

∞

to

r

along a lift of the

real axis

, then continues
along the circle of radius

r

clockwise, and finally goes
from

r

to

+

∞

.

We now examine the integral over such a contour by breaking it
into three pieces.

∫

C

r

x

s

-

1

e

x

-

1

⁢

𝑑

x

=

∫

r

∞

x

s

-

1

e

x

-

1

⁢

𝑑

x

+

∫

|

x

|

=

r

x

s

-

1

e

x

-

1

⁢

𝑑

x

-

∫

r

∞

(

e

-

2

⁢

π

⁢

i

⁢

x

)

s

-

1

e

x

-

1

⁢

𝑑

x

.

We may estimate the third integral in

absolute value

like so:

|

∫

|

x

|

=

r

x

s

-

1

e

x

-

1

⁢

𝑑

x

|

≤

2

⁢

π

⁢

r

⁢

sup

|

x

|

=

r

⁡

|

x

s

-

1

e

x

-

1

|

The expression

x

/

(

e

x

-

1

)

represents an analytic function of

x

, and hence a bounded function of

x

in a neighborhood of

0

.
When

ℜ

⁡

s

>

1

, it happens that

lim

x

→

0

⁡

|

x

|

⁢

x

s

-

2

=

0

, so

lim

r

→

0

⁡

|

∫

|

x

|

=

r

x

s

-

1

e

x

-

1

⁢

𝑑

x

|

=

0

.

The third integral differs from the first integral by a phase, so
they may be combined by pulling out this common factor. When

ℜ

⁡

s

>

0

, we may take the limit as

r

approaches

0

after doing
so to obtain the following:

lim

r

→

0

⁡

∫

C

r

x

s

-

1

e

x

-

1

⁢

𝑑

x

=

(

1

+

e

2

⁢

π

⁢

i

⁢

(

1

-

s

)

)

⁢

∫

0

∞

x

s

-

1

e

x

-

1

⁢

𝑑

x

Since, aside from the

branch point

at

0

, the only singularities of our integrand occur at

multiples

of

2

⁢

π

⁢

i

, it follows from Cauchy’s theorem that

∫

C

a

x

s

-

1

e

x

-

1

⁢

𝑑

x

=

∫

C

b

x

s

-

1

e

x

-

1

⁢

𝑑

x

whenever

0

<

a

<

2

⁢

π

and

0

<

b

<

2

⁢

π

, which trivially implies that

lim

r

→

0

⁡

∫

C

r

x

s

-

1

e

x

-

1

⁢

𝑑

x

=

∫

C

r

x

s

-

1

e

x

-

1

⁢

𝑑

x

for any

r

between

0

and

2

⁢

π

. Therefore,

ζ

⁢

(

s

)

=

1

(

1

+

e

2

⁢

π

⁢

i

⁢

(

1

-

s

)

)

⁢

Γ

⁢

(

s

)

⁢

∫

C

π

x

s

-

1

e

x

-

1

⁢

𝑑

x

when

ℜ

⁡

z

>

1

. This integral converges for all complex

s

because
the

exponential

grows more rapidly than the power. Furthermore, this
integral defines an analytic function of

s

, so we have an analytic
continuation of the zeta function to the whole complex plane minus the point 1.

Title

analytic continuation of Riemann zeta (using integral)

Canonical name

AnalyticContinuationOfRiemannZetausingIntegral

Date of creation

2013-03-22 16:53:59

Last modified on

2013-03-22 16:53:59

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

21

Author

rspuzio (6075)

Entry type

Example

Classification

msc 30B40

Classification

msc 30A99

Related topic

EstimatingTheoremOfContourIntegral

Related topic

PeriodicityOfExponentialFunction

Related topic

AnalyticContinuationOfRiemannZeta

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