derivation of integral representations of Jacobi ϑ functions

derivation of integral representations of Jacobi

ϑ

functions

By rearranging the

Fourier series

of

cos

⁡

(

u

⁢

x

)

, one obtains the series

π

⁢

cos

⁡

(

u

⁢

x

)

2

⁢

u

⁢

sin

⁡

(

π

⁢

u

)

=

1

2

⁢

u

2

+

∑

n

=

1

∞

(

-

1

)

n

⁢

cos

⁡

(

n

⁢

x

)

u

2

-

n

2

This equation which is valid for all real values of

x

such that

-

π

≤

x

≤

π

and all non-integral

complex

values of

u

. By comparison with the

convergent series

∑

n

=

0

∞

1

/

n

2

, it follows that this series is absolutely convergent. Note that this series may be viewed as a Mittag-Leffler partial fraction expansion.

Let

y

be a positive real number. Multiply both by

2

⁢

u

⁢

e

-

y

⁢

u

2

and

integrate

.

∫

i

-

∞

i

+

∞

π

⁢

cos

⁡

(

u

⁢

x

)

⁢

e

-

y

⁢

u

2

sin

⁡

(

π

⁢

u

)

⁢

𝑑

v

=

2

⁢

∫

i

-

∞

i

+

∞

e

-

y

⁢

u

2

⁢

[

1

2

⁢

u

2

+

∑

n

=

0

∞

(

-

1

)

n

⁢

cos

⁡

(

n

⁢

x

)

u

2

-

n

2

]

⁢

u

⁢

𝑑

u

Because of the exponential, the integrand decays rapidly as

u

→

i

±

∞

provided that

ℜ

⁡

u

>

0

, and hence the integral converges absolutely. Make a change of variables

v

=

u

2

=

∫

P

e

-

y

⁢

v

⁢

[

1

2

⁢

v

+

∑

n

=

1

∞

(

-

1

)

n

⁢

cos

⁡

(

n

⁢

x

)

v

-

n

2

]

⁢

𝑑

v

The

contour

of integration

P

is a

parabola

in the complex

v

-plane,

symmetric about

the

real axis

with vertex at

v

=

-

1

, which encloses the real axis. Its equation is

ℜ

⁡

v

+

1

=

2

⁢

(

ℑ

⁡

v

)

2

Let

S

m

(

m

is an

integer

) be the

straight line

segment

joining the points

v

=

(

i

+

m

+

1

/

2

)

2

and

v

=

(

i

-

m

-

1

/

2

)

2

. Along this

line segment

, we may bound the integrand in

absolute value

as follows:

|

∑

n

=

1

∞

(

-

1

)

n

⁢

cos

⁡

(

n

⁢

x

)

v

-

n

2

|

≤

∑

n

=

1

∞

(

-

1

)

n

|

v

-

n

2

|

≤

∑

n

=

1

∞

(

-

1

)

n

|

v

m

-

n

2

|

where

v

m

=

m

2

+

m

-

3

/

4

is the point of

intersection

of

S

m

with the real axis. To proceed further, we break up the last

summation

into two parts.

Since the squares closest in absolute value to

v

m

are

m

2

and

(

m

+

1

)

2

=

m

2

+

2

⁢

m

+

1

, it follows that

|

v

m

-

n

2

|

≥

|

m

-

3

/

4

|

for all

m

,

n

. Hence, we have

∑

i

=

1

2

⁢

m

1

|

v

m

-

n

2

|

≤

2

⁢

m

m

-

3

/

4

≤

8

When

n

>

2

⁢

m

, we have

n

2

≥

(

2

⁢

m

+

1

)

2

=

4

⁢

m

2

+

4

⁢

m

+

1

>

4

⁢

m

2

+

4

⁢

m

-

3

=

4

⁢

v

m

. Hence,

|

n

2

-

v

m

|

>

3

⁢

n

2

/

4

and

∑

n

=

2

⁢

m

+

1

∞

1

|

v

m

-

n

2

|

<

4

3

⁢

∑

n

=

2

⁢

m

+

1

∞

1

n

2

<

4

3

⁢

∑

n

=

1

∞

1

n

2

=

2

⁢

π

9

Finally

1

/

(

2

⁢

v

m

)

<

1

/

2

since

v

m

>

1

when

m

≥

1

. Also,

|

e

-

y

⁢

v

|

=

e

-

y

⁢

ℜ

⁡

v

-

e

-

y

⁢

v

m

<

e

-

y

⁢

m

2

. From these observations, we conclude that

|

∫

S

m

e

-

y

⁢

v

⁢

[

1

2

⁢

v

+

∑

n

=

1

∞

(

-

1

)

n

⁢

cos

⁡

(

n

⁢

x

)

v

-

n

2

]

⁢

𝑑

v

|

<

e

-

y

⁢

m

2

⁢

(

1

+

8

+

2

⁢

π

9

)

⁢

∫

S

m

𝑑

v

=

(

4

⁢

m

+

2

)

⁢

(

9

+

2

⁢

π

9

)

⁢

e

-

y

⁢

m

2

Note that this quantity approaches 0 in the limit

m

→

∞

.

Let

P

m

be the arc of the parabola

P

bounded

by the

endpoints

of

S

m

. Together,

S

m

and

P

m

form a closed contour which encloses poles of the integrand. Hence, by the

residue theorem

, we have

∫

P

m

e

-

y

⁢

v

⁢

[

1

2

⁢

v

+

∑

n

=

1

∞

(

-

1

)

n

⁢

cos

⁡

(

n

⁢

x

)

v

-

n

2

]

⁢

𝑑

v

+

∫

S

m

e

-

y

⁢

v

⁢

[

1

2

⁢

v

+

∑

n

=

1

∞

(

-

1

)

n

⁢

cos

⁡

(

n

⁢

x

)

v

-

n

2

]

⁢

𝑑

v

=

2

⁢

π

⁢

i

⁢

∑

n

=

1

m

(

-

1

)

n

⁢

cos

⁡

(

n

⁢

x

)

⁢

e

-

n

2

⁢

y

Taking the limit

m

→

∞

we obtain

∫

P

e

-

y

⁢

v

⁢

[

1

2

⁢

v

+

∑

n

=

1

∞

(

-

1

)

n

⁢

cos

⁡

(

n

⁢

x

)

v

-

n

2

]

⁢

𝑑

v

=

2

⁢

π

⁢

i

⁢

(

1

2

+

∑

n

=

1

∞

(

-

1

)

n

⁢

cos

⁡

(

n

⁢

x

)

⁢

e

-

n

2

⁢

y

)

Going back to the beginning of the proof, where the integral on the left hand side was expressed as an integral with respect to

u

, we obtain

∫

i

-

∞

i

+

∞

π

⁢

cos

⁡

(

u

⁢

x

)

⁢

e

-

y

⁢

u

2

sin

⁡

(

π

⁢

u

)

⁢

𝑑

v

=

2

⁢

π

⁢

i

⁢

(

1

2

+

∑

n

=

1

∞

(

-

1

)

n

⁢

cos

⁡

(

n

⁢

x

)

⁢

e

-

n

2

⁢

y

)

Making a change of variables

x

=

2

⁢

z

,

y

=

-

i

⁢

π

⁢

τ

and tidying up some, we obtain

∫

i

-

∞

i

+

∞

cos

⁡

(

2

⁢

u

⁢

z

)

⁢

e

i

⁢

π

⁢

τ

⁢

u

2

sin

⁡

(

π

⁢

u

)

d

v

=

i

(

1

+

2

∑

n

=

1

∞

(

-

1

)

n

e

i

⁢

π

⁢

n

2

⁢

τ

cos

(

2

n

z

)

)

=

i

ϑ

4

(

z

|

τ

)

Because of the initial

assumption

about the Fourier series, we only know that this

formula

is valid when

τ

is

purely imaginary

with

strictly

positive

imaginary part

and

z

is real and

π

/

2

<

z

<

π

/

2

. However, we can use

analytic continuation

to extend the domain of its

validity

. On the one hand, the theta function on the right-hand side is

analytic

for all

z

and all

τ

such that

ℑ

⁡

τ

>

0

.

On the other hand, I claim that the integral on the left hand side is also an

analytic function

of

z

and

τ

whenever

ℑ

⁡

τ

>

0

. To validate this claim, we need to examine the behaviour of the integrand as

u

→

i

±

∞

. The contribution of the

denominator

is bounded;

|

1

sin

⁡

π

⁢

u

|

<

c

for some constant

c

whenever

ℑ

⁡

u

=

1

. The absolute value of the cosine in the numerator is easy to bound:

|

cos

⁡

(

2

⁢

u

⁢

z

)

|

≤

e

2

⁢

|

u

|

⁢

|

z

|

To bound the remaining term, let us examine the

argument

of the exponential carefully:

ℑ

⁡

(

τ

⁢

u

2

)

=

2

⁢

ℜ

⁡

τ

⁢

ℜ

⁡

u

+

ℑ

⁡

τ

⁢

(

ℜ

⁡

u

)

2

-

ℑ

⁡

τ

=

ℑ

⁡

τ

⁢

(

(

ℜ

⁡

u

+

ℜ

⁡

τ

ℑ

⁡

τ

)

2

-

1

-

(

ℜ

⁡

τ

ℑ

⁡

τ

)

2

)

Therefore, if

|

ℜ

⁡

u

|

>

1

+

3

⁢

|

ℜ

⁡

τ

|

/

(

ℑ

⁡

τ

)

, it will be the case that

ℑ

⁡

(

τ

⁢

u

2

)

≥

ℑ

⁡

τ

⁢

(

ℜ

⁡

u

)

2

/

9

, and so

|

e

i

⁢

π

⁢

τ

⁢

u

2

|

=

e

-

π

⁢

ℑ

⁡

(

τ

⁢

u

2

)

≤

e

-

π

⁢

ℑ

⁡

τ

⁢

(

ℜ

⁡

u

)

2

/

9

Taken together, the estimates of the last paragraph imply that

|

∫

i

+

R

i

+

∞

cos

⁡

(

2

⁢

u

⁢

z

)

⁢

e

i

⁢

π

⁢

τ

⁢

u

2

sin

⁡

(

π

⁢

u

)

|

<

c

⁢

∫

i

+

R

i

+

∞

e

2

⁢

|

u

|

⁢

|

z

|

-

π

⁢

ℑ

⁡

τ

⁢

(

ℜ

⁡

u

)

2

/

9

when

R

>

1

+

3

⁢

|

ℜ

⁡

τ

|

/

(

ℑ

⁡

τ

)

. If we impose the further conditions

R

>

180

⁢

|

z

|

π

⁢

ℑ

⁡

τ

R

2

>

180

⁢

|

z

|

π

⁢

ℑ

⁡

τ

,

it will be the case that

2

⁢

|

u

|

⁢

|

z

|

-

π

⁢

ℑ

⁡

τ

⁢

(

ℜ

⁡

u

)

2

/

9

<

2

⁢

ℜ

⁡

u

⁢

|

z

⁢

|

+

2

|

⁢

z

|

-

π

⁢

ℑ

⁡

τ

⁢

(

ℜ

⁡

u

)

2

/

9

<

(

2

⁢

ℜ

⁡

u

⁢

|

z

|

-

π

⁢

ℑ

⁡

τ

⁢

(

ℜ

⁡

u

)

2

/

180

)

+

(

2

⁢

|

z

|

-

π

⁢

ℑ

⁡

τ

⁢

(

ℜ

⁡

u

)

2

/

180

)

-

π

⁢

ℑ

⁡

τ

⁢

(

ℜ

⁡

u

)

2

/

10

<

-

π

⁢

ℑ

⁡

τ

⁢

(

ℜ

⁡

u

)

2

/

10

,

and hence

|

∫

i

+

R

i

+

∞

cos

⁡

(

2

⁢

u

⁢

z

)

⁢

e

i

⁢

π

⁢

τ

⁢

u

2

sin

⁡

(

π

⁢

u

)

⁢

𝑑

u

|

<

c

⁢

∫

i

+

R

i

+

∞

e

-

π

⁢

ℑ

⁡

τ

⁢

(

ℜ

⁡

u

)

2

/

10

⁢

𝑑

u

<

5

⁢

c

π

⁢

ℑ

⁡

τ

⁢

R

⁢

e

-

π

⁢

ℑ

⁡

τ

⁢

R

2

/

10

.

Likewise, under the same

restriction

on

R

,

|

∫

i

-

∞

i

-

R

cos

⁡

(

2

⁢

u

⁢

z

)

⁢

e

i

⁢

π

⁢

τ

⁢

u

2

sin

⁡

(

π

⁢

u

)

⁢

𝑑

u

|

<

c

⁢

∫

i

+

R

i

+

∞

e

-

π

⁢

ℑ

⁡

τ

⁢

(

ℜ

⁡

u

)

2

/

10

⁢

𝑑

u

<

5

⁢

c

π

⁢

ℑ

⁡

τ

⁢

R

⁢

e

-

π

⁢

ℑ

⁡

τ

⁢

R

2

/

10

.

Since the contour of integration is

compact

and the integrand is analytic in a

neighborhood

of the contour,

∫

i

-

R

i

+

R

cos

⁡

(

2

⁢

u

⁢

z

)

⁢

e

i

⁢

π

⁢

τ

⁢

u

2

sin

⁡

(

π

⁢

u

)

⁢

𝑑

u

will be an analytic function of

z

and

τ

. Suppose that

z

and

τ

are restricted to bounded regions of the

complex plane

and that, furthermore,

I

⁢

m

⁢

τ

is positive and bounded away from zero. Then the

inequalities

of the last paragraph imply that the integral converges uniformly as

R

→

∞

, and hence

∫

i

-

∞

i

+

∞

cos

⁡

(

2

⁢

u

⁢

z

)

⁢

e

i

⁢

π

⁢

τ

⁢

u

2

sin

⁡

(

π

⁢

u

)

⁢

𝑑

u

is an analytic function of

u

and

z

in the domain

ℑ

⁡

τ

>

0

.

Thus, by the fundamental theorem of analytic continuation, we may conclude that

∫

i

-

∞

i

+

∞

cos

⁡

(

2

⁢

u

⁢

z

)

⁢

e

i

⁢

π

⁢

τ

⁢

u

2

sin

⁡

(

π

⁢

u

)

d

v

=

i

(

1

+

2

∑

n

=

1

∞

(

-

1

)

n

e

i

⁢

π

⁢

n

2

⁢

τ

cos

(

2

n

z

)

)

=

i

ϑ

4

(

z

|

τ

)

throughout this domain.

Finally, integral representations of the remaining three theta functions may be easily obtained from this one by adding the appropriate half-quasiperiods to

z

.

Title

derivation of integral representations of Jacobi

ϑ

functions

Canonical

name

DerivationOfIntegralRepresentationsOfJacobivarthetaFunctions

Date of creation

2013-03-22 14:39:54

Last modified on

2013-03-22 14:39:54

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

26

Author

rspuzio (6075)

Entry type

Derivation

Classification

msc 33E05

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