On some mathematical equations concerning the functions zeta(s) and zeta(s,w) and some Ramanujan-type series for 1/pigreco . Mathematical connections with some equations concerning

8/8/2019 On some mathematical equations concerning the functions zeta(s) and zeta(s,w) and some Ramanujan-type series

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1

On some mathematical equations concerning the functions ( )s and ( )ws, and someRamanujan-type series for /1 . Mathematical connections with some equations concerningthe p-adic open string for the scalar tachyon field and the zeta strings.

Michele Nardelli 2,1

1 Dipartimento di Scienze della Terra

Universit degli Studi di Napoli Federico II, Largo S. Marcellino, 10

80138 Napoli, Italy

2 Dipartimento di Matematica ed Applicazioni R. Caccioppoli

Universit degli Studi di Napoli Federico II Polo delle Scienze e delle Tecnologie

Monte S. Angelo, Via Cintia (Fuorigrotta), 80126 Napoli, Italy

Abstract

In this paper, in the Section 1, we have described some equations concerning the functions ( )s

and ( )ws, . In this Section, we have described also some equations concerning a transformationformula involving the gamma and Riemann zeta functions of Ramanujan. Furthermore, we have

described also some mathematical connections with various theorems concerning the incomplete

elliptic integrals described in the Ramanujans lost notebook. In the Section 2, we have described

some Ramanujan-type series for /1 and some equations concerning the p-adic open string for thescalar tachyon field. In this Section, we have described also some possible and interesting

mathematical connections with some Ramanujans Theorems, contained in the first letter of

Ramanujan to G. H. Hardy. In the Section 3, we have described some equations concerning the zeta

strings and the zeta nonlocal scalar fields. In conclusion, in the Section 4, we have showed some

possible mathematical connections between the arguments above mentioned, the Palumbo-Nardelli

model and the Ramanujans modular equations that are related to the physical vibrations of the

bosonic strings and of the superstrings.

1. On some equations concerning some observations concerning the functions ( )s and( )ws, [1]

In the Mathematical Analysis there exist the proof of the following formula:

=)(s1

)]([)1(21

1

2

120

22

++

+

t

s

e

dttsArcTanSint

s (1.1)

that can be rewritten also as follows:


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2

( )( )( )

( ) ( )

++

+=

0 22/2

1

1

1

2

1

11

tansin2

sdt

et

tss

ts (1.1b)

where )(s represent the Riemann zeta function.

Now we analyze in greater detail the formula (1.1)

The function ( )s is represented from the series

( ) ...1

...3

1

2

11 +++++=

sssn

s , (1.2)

as the real part of the complex variable

is +

is greater than the unity. Under the same condition, we have still that

( )( )

=

0

1

1

1x

s

e

dxx

ss , (1.3)

what is easily proved by using the following equality

( )

=0

111 dxxe

sv

svx

s. (1.4)

It is from the expression (1.3) that Riemann reached by an ingenious application of the residue

calculus, extending ( )s across all the plan and discover such interesting properties of this function.

We set ( ) szzf = , where

( ) ( )

+=

tsttp

s

arctancos, 222 , ( ) ( )

+=

tsttq

s

arctansin, 222 . (1.5)

The following three conditions:

1 The function ( )zf is holomorphic for , for each t;

2 The condition ( ) 0lim 2 =+=

itfet

t

is verified uniformly for , however great ;

3 The function ( )zf is subject to the following condition: ( ) 0lim 2 =++

=itfe

t

;

are verified in the half-plane 0> ; assuming 1> , so that the series (1.2) converges, and by

1=m , we obtain, after the application of the following formula:

( ) ( ) ( )( )

+=

mm t

dt

e

tmqdfmfvf

0 2

1

,2

2

1

, (1.6)


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3

( ) ( ) ( )1

arctansin121

1

2

120

22

++

+=

t

s

e

dttst

ss

, (1.7)

that is the eq. (1.1), and, after the application of the following formula,

( ) ( ) ( )

=m

dttQdfvf

0

,2 , (1.8)

for2

1= ,

( ) ( )1

2arctansin4

12

1

220

22

1

+

+

=

t

ss

e

dttst

ss

. (1.9)

Another expression for ( )s is derived from the following formula:

( ) ( )+

=

i

im

dzzFzi

vf

2

sin2

1. (1.10)

By itz +==2

1,

2

1 , we find

( )( )[ ]

( )

+

+

=

0 2

2

1

2 2arctan1cos

4

1

1

4dt

ee

tst

ss

tt

s

. (1.11)

Starting from the relationship

( ) ...4

1

3

1

2

11

2

11

1++=

sssss

with the application of the following formula

( ) ( )( )

( ) +

=

=

n

m

i

i iziz

vdttQ

ee

dzzfvf

01 ,21 , (1.12)

with2

1,1 == m , we also find

( )( )

+

+

=

0

22

1

2arctancos

4

1

12

2dt

ee

tsts

tt

s

s

s

. (1.13)

It is easy to see that the integrals definite appearing in the above expressions are analytic functions,

holomorphic for any finite value of s . On the other hand, we deduce from the eq. (1.7), for 0=s ,

( )2

10 = ,


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4

then, subtracting the two members1

1

s, and making tend s to the unity, taking into account the

following equality:

dtet

tC

t

++=

0 22 1

1

12

2

1

, (1.14)

we obtain

( ) =

= 1

1lim

1 ss

s Cdt

et

tt

=+

+

0 22 1

1

12

2

1

, (1.15)

and finally, by differentiating with respect to s , putting 0=s and using the following equality

=

0 2

2log1

1arctan21

dt

et

t, (1.16)

we obtain

( )

=

=0 2

2log11

1arctan20'

dt

et

t. (1.17)

We note that there exist a mathematical connection between and2

15 = , i.e. the aurea

section, by the simple formula

2879,0arccos = , (1.18)

thence we have that

2879,0

1arccos = . (1.19)

We can rewrite the eq. (1.17) also as follows:

( )

=

=

0 2 2879,0

1arccos2log1

1

1arctan20'

dt

et

t. (1.19b)

Now we consider the function

( )( ) ( )

...2

1

1

11, +

++

++=

ssswww

ws , (1.20)

which reduces to ( )s for 1=w . We must replace our general formulas

( ) ( ) swzzf += , (1.21)

where


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5

( ) ( )[ ]

+++=

w

tstwtp

s

arctancos, 22

2, ( ) ( )[ ]

+++=

w

tstwtq

s

arctansin, 22

2. (1.22)

Assuming the real part of w positive, applying the following formula

( ) ( ) ( ) ( )

+=

mm t

dttmqe

dfmfvf0 2

,1

1221 , (1.23)

we obtain

( ) ( )

+++

=

0 2222

1

1

1arctansin2

21, dt

ew

tstw

w

s

wws

t

sss

, (1.24)

valid expression in the whole plan and shows that ( )ws, is a uniform function admitting forsingularity at finite distance, only the pole 1=s of residue 1. We conclude, on the other hand, for

0=s ,

( ) ww =21,0 , (1.25)

then, subtracting1

1

sand for s tending to the unity,

( )

= +++=

0 2221 1

12

2

1log

1

1,lim dt

etw

t

ww

sws

ts

, (1.26)

and finally, differentiating with respect to s and then for 0=s , we have that

( )

+

=

0 2

'

1

1arctan2log

2

1,0 dt

ew

twwww

ts . (1.27)

For the following equalities,

( ) ( )xJxxxx +

+= log

2

12loglog , (1.28)

( ) ( )xJ

x

xxDx '

2

1loglog += , (1.29)

these last two expressions on reduced respectively to( )( )ww

'and ( ) 2loglog w .

Here it is possible to obtain some mathematical connections with various theorems concerning the

incomplete elliptic integrals described in the Ramanujans lost notebook.

Let ( )qu denote the Rogers-Ramanujan continued fraction defined by

( )

...11

1

1

::

3

2

5/1

++

+

+

==

qq

q

qquu , 1


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6

and set 2quv = . Recall that ( )q is defined by

( ) ( )( ) ( )

( )

=

+

===0

2

222/13

;

;,:

n

nn

qq

qqqqqfq . (1.31)

Then

( )( )

( ) ( )( )

+

++=

2

232

5

5

251

251log5log

5

8

uv

uvvu

q

dq

q

q

. (1.32)

We note that ( ) 22

15236067977,1251

==+ , i.e. the aurea section multiplied by the

number 2, and that ( )

+==+ 2

2152360679777,3251 , i.e. the aurea ratio multiplied by

the number 2 and with the minus sign.

With ( )qf , ( )q , and ( )qu defined by

( ) ( ) ( ) ( ) ( ) ( )

=

====

n

iznnnzeqqqqqfqf 24/22/132 :;1,: , izeq 2= , 0Im >z , (1.33)

by (1.31) and (1.30), respectively, and with ( ) 2/15 += , we have that

( ) ( )( )( ) =

=

q

uddt

t

tftf

0

2/

cos 22/35

5224/3

2/51

sin51

125

( ) ( )( )( ) ( )( )

=

=

qfqfq qqq

dd3534/31 54/11/5tan2

0

/5tan2

0 22/122/35 sin51

15

sin51

1

. (1.34)

Let v be defined by the following expression

( )( ) ( )( ) ( )

3

53

15

::

==

qfqf

qfqfqqvv , (1.35)

and let ( ) 2/15 += . Then

( ) ( ) ( ) ( )( )

=

=

+

q

vv

vv ddttftftftf0

5/1tan2

1

111

5

1tan2

2

15531

2

21

sin25

91

1

5

1

( ) ( )( )( )( )

( )( )( )

( )( )

+

+

+

++

=

=2/

11

11

1

1tan2

53tan

11

1153tan

2251

5

3

31

1

5

511

sin16

151

1

4

1

sin81

11

1

9

1

vvvv

v

v

vv

vv dd . (1.36)

If v is defined by the following expression


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7

( )( ) ( )( ) ( )

4

72

14

::

==

qfqf

qfqfqqvv , (1.37)

and if 7

21613 +

=c , then

( ) ( ) ( ) ( )

+

=q c

v

vc

ddttftftftf0

cos

1

1cos

2

14721

1

sin232

132161

1

28

1

. (1.38)

Let

( ) ( )( ) ( )733

14323

qfqf

qfqfqv

= and

7

249 =c ,

Then

( ) ( ) ( ) ( )

++

=q c

x

xc

ddttftftftf0

cos

1221

221cos

2

4/714721

1

sin64

213321

12

. (1.39)

1.1On some equations concerning a transformation formula involving the gamma andRiemann zeta functions of Ramanujan.

In the Ramanujans lost notebook there is a claim that provides a beautiful series transformation

involving the logarithmic derivative of the gamma function and the Riemann zeta function. To state

Ramanujans claim, it will be convenient to use the familiar notation

( )( )( )

=

+

+=

=

0 1

11':

k kxkx

xx , (1.40)

where denotes Eulers constant. We also need to recall the following functions associated with

Riemanns zeta function ( )s . Let

( ) ( ) ( )sssss

+=

2

111: 2

1

.

Then Riemanns -function is defined by


8/32


9/32

9

log

0=

dxx

ee xx, 0, > . (1.48)

We now describe a proof of Theorem 1. Our first goal is to establish an integral representation for

the far left side of (1.42). Replacing z by n in (1.45) and summing on n ,


10/32

10

Letting log2

1=n and /2 tx = in (1.55), we deduce that

=

=+

+

0 0 /222

2

3 21

1

2

1

1

12

1

log2

1cos

4

1

2

11dttetedtt

tit

t tt

( )( ) ( ) ( )

+

+

=

0 2/222/2 21

1

111

/2dt

teteteetttt

. (1.56)

Hence, combining (1.54) and (1.56), in order to prove that the far left side of (1.42) equals the far

right side of (1.42), we see that it suffices to show that

( ) ( )

( )

=

+

=

+

0 0

2/

2

/

2/20

2

1

1

11

221

1du

u

e

ueu

dt

t

e

tet

u

u

t

t

, (1.57)

where we made the change of variable /2 tu = . In fact, more generally, we show that

( )( )

=

+

02

2log2

1

2

1

1

1adu

u

e

ueu

ua

u , (1.58)

so that if we set ( )2/1=a in (1.58), we deduce (1.57).Consider the integral, for 0>t ,

( ) ( ) ( )

++

=

+

+

=

0log

2

12log

2

1log

2

1log

22

11

1

1:,

a

tttttdu

u

ee

u

e

uetaF

tuuatu

u ,

(1.59)

where we applied (1.46) and (1.48). Upon the integration of (1.30), it is easily gleaned that, as

0t ,

( ) ttt loglog , (1.60)

where denotes Eulers constant. Using this in (1.59), we find, upon simplification, that, as 0t ,

( ) ( ) atttttaF log212log

21log, + . (1.61)

Hence,

( ) ( )ataFt

2log2

1,lim

0=

. (1.62)

Letting t approach 0 in (1.59), taking the limit under the integral sign on the right-hand side using

Lebesgues dominated convergence theorem, and employing (1.62), we immediately deduce (1.58).

As previously discussed, this is sufficient to prove the equality of the first and third expressions in

(1.42), namely,


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11

( )( )

= +

+

=

+

0 2

2

31 1

log2

1cos

4

1

2

11

2

2logdt

t

tit

tnn

. (1.63)

Lastly, using (1.63) with replaced by and employing the relation 1=

, we conclude that

( )( ) =

+

+

=

+

=

dtt

tit

tnn

2

2

031 1

log2

1cos

4

1

2

11

2

2log

( )=

+

+

=

dt

t

titt

2

2

03 1

/1log2

1cos

41

211

( )dt

t

tit

t2

2

03 1

log2

1cos

4

1

2

11

+

+

=

. (1.64)

Hence, the equality of the second and third expressions in (1.42) has been demonstrated, and so the

proof is complete.

2. On some Ramanujan-type series for /1 [2]

Ramanujans series representations for /1 depend upon Clausens product formulas forhypergeometric series and Ramanujans Eisenstein series

( )

= =

11241:

kk

k

q

kqqP , 1


12/32

12

{ } ( )

=

+

+=0

2/12

2515542

k

k

kAk

, (2.2)

( ) ( )

=

+=0 4

13201

8

kkk

kBk

. (2.3)

For 13=n ,if kA and kB , 0k , are defined by the formula (2.1b), then

( ) ( )kk

k

kBk

20 18

1232601

72

=

+=

. (2.4)

The last two identities was recorded by Ramanujan in the fundamental paper Modular equations

and approximations to (1914).

Proof of (2.2).

From Ramanujans second notebook (Notebooks 2 volumes 1957), we see that

( ) ( ) ( ) ( ) ( )( ) ( )( ){ }

=

=

++=

+1 1

55522

10

10

2

2

11324

1

130

11

k kk

k

k

k

qxqxqxqxqqq

kq

q

kq

( ) ( ) ( )( ) ( )( ){ }2/1

55 111 qxqxqxqx ++ . (2.5)

With the help of (2.1) we can rewrite (2.4) in the form

( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( )( ){ }++= 55522210 1132

15 qxqxqxqxqqqPqP

( ) ( ) ( )( ) ( )( ){ }2/1

55 111 qxqxqxqx ++ . (2.6)

Now we set 5/= eq and use, following Ramanujan, the following expressions

nn xx /11 = , nn znz =/1 , (2.7)

( ) ( )qxFqzz 212 ;1;2

1,

2

1:: =

== , (2.8)

to deduce that ( ) 55/1 1 xxqx == , ( ) 55 xqx = , and ( ) ( ) 5

525/2 55 zee == .

Thus, from (2.6), we find that

( ) ( ) ( ){ } ( )( ) =++= 5555255/252 1212

112355 xxxxzePeP

( ){ } ( )( )555525 1412

11435 xxxxz ++= . (2.9)

The eq. (2.9), putting ( )555 14 xxX = , can be rewritten also as follows


13/32

13

( ) ( ) ( ){ } ( )( ) =++= 5555255/252 1212

112355 xxxxzePeP

{ } ( )5525 12

135 XXz ++= . (2.9b)

But, the singular modulus 5x is given by

3

52

15

2

1

=x , (2.10)

so that

5495 =X and 255 =X . (2.11)

Thus, from (2.9), we find that

( ) ( ) ( ) 255/2522

151555 zePeP

+=

. (2.12)

Next, setting 5=n in the following equation

( ) ( )

nePenP

nn 6/22=+

, (2.13)

we find that

( ) ( )

565 5/252 =+ ePeP . (2.14)

Adding (2.12) and (2.14), we deduce that

( ) 25522

15

5

15

5

3zeP

++=

. (2.15)

Now, employing the following expression

=

=

=

0

23

2 ;1,1;2

1,

2

1,

2

1

k

k

kXAXFz ,2

10 x , (2.16)

in (2.15), we deduce the identity

( )

=

++=

0

5

52

2

15

5

15

5

3

k

k

kXAeP

. (2.17)



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14

( ) ( ) ( )

=

+=

0

2 1321k

k

nkn

nXAkxeP

, (2.18)

we find that

( ) ( ) ( ) ( ) ( )

=

=

+=+=

0 0

555

52 132521321k k

k

k

k

k XAkXAkxeP . (2.19)

Using (2.17) and (2.19), we arrive at (2.2).

Proof of (2.3).

Employing the following expression

( )

=

=

=

0

22

23

2 121

1;1,1;

4

3,

2

1,

4

1

21

1

k

k

k

kWB

xWF

xz ,

2221

2

10 4/1x , (2.20)

in (2.15), we find that

( )( )

( ) ( )

=

=

++=

++=

0 0

2

5

2

5

5

52 154

53

5

31

2

15

215

15

5

3

k k

k

k

kk

k

kWBWB

xeP

, (2.21)

where

2

1

1

2

5

5

5 =

=X

XW . (2.22)


( ) ( ) ( )

=

+++=

0

22

1

2/1131

k

k

nk

n

nnkn WBX

XXkeP

, (2.23)

we find that

( ) ( ) ( ) ( ) kkkk

k

n

k

k

n

nn WBkWBX

XXkeP

2

5

0

252 18

253

2

531

1

2/113

++=

+++=

=

. (2.24)

From (2.21) and (2.24), we readily arrive at (2.3). Thus, we complete the proof. The proof of (2.3b)

is similar.

2.1 On some equations concerning the p-adic open string for the scalar tachyon field. [3] [4]As a free action in p-adic field theory one can take the following functional

( ) =pQfDfdxfS (2.25)


15/32

15

where ( )xff = is a function RQf p : , dx is the Haar measure and D is the Vladimirov

operator or its generalizations. Boundary value problems for homogeneous solutions of nonlinear

equations of motion corresponding to the p-adic string,

pe = . (2.26)

Here is the dAlembert operator and the field and its argument are real-valued.

The dynamics of the open p-adic string for the scalar tachyon field is described by the non-linear

pseudodifferential equation

pp =21

, (2.27)

where 22211

...

=dxxt

, 0xt= , is the dAlembert operator and p is a prime number,

,...5,3,2=p . In what follows p is any positive integer. We consider only real solutions of

equations (2.27), since only real solutions have physical meaning. In the one-dimensional case (d =

1) we use the change

( ) ( )ptt ln2=

and write equation (2.27) in the following equivalent form:

pte =2

2

1

. (2.28)

Equation (2.28) is a non-linear integral equation of the following form:

( ) ( ) ( )

= tde pt

21 , Rt . (2.29)

Solutions of equation (2.29) are sought in the class of measurable functions ( )t such that

( ) ( ){ }21exp tCt for any 0> , Rt . (2.30)

The following boundary-value problems for the solutions of equation (2.29) have physical

meaning:

( ) 0lim =

tt

, ( ) 1lim =

tt

(2.31)

ifp is even, and

( ) 1lim =

tt

, ( ) 1lim =

tt

(2.32)

ifp is odd.

Assertion 1

If is a solution of equation (2.29) such that

( ) att

=lim ,


16/32

16

then 0=a or 1=a if p is even and 0=a or 1=a if p is odd, ( ) 0'lim = tp

t . If 0a ,

then ( ) 0'lim = tt .

We deduce from equation (2.29) the following chain of equalities:

( ) ( )[ ] ( ) ( ) ( )

=====

22 1lim

1limlimlim u

t

t

t

pp

t

p

teutdeatt

( )

== adueut u

t

2

lim1

, (2.34)

whence 0=a or 1=a ifp is even and 1,0 =a ifp is odd. Further, we have

( ) ( ) ( )( ) ( ) ( )

=== duueutdett u

t

t

t

p

t

22 1lim2

1lim2'lim

( )

===

= 00

22lim

2 22aduueaduueut uu

t

. (2.35)

If 0a , then ( ) 0'lim = tt , since

( ) ( ) ( ) ( ) ( ) 0'lim'lim'lim 11 ===

tpattpt

t

pp

t

p

t . (2.36)

We passed to the limit under the integral sign, using Lebesgues theorem and estimate (2.30). We

shall write ba if the integers a and b are both even or both odd, and ba if one of them iseven and the other is odd.

Hermite polynomials are defined to be the polynomials

( ) ( )22

1 xn

nxn

n edx

dexH

= , ,...,1,0=n (2.37)

whence ( ) 10 =xH , ( ) xxH 21 = , ( ) 242

2 = xxH , ( ) ,...1283

3 xxxH = . They form a complete

orthogonal system in the Hilbert space 12L , and

( ) ( )

== !21

22

1nxdxHH

n

nn . (2.38)

Any 12Lf can be expanded in Hermite polynomials:

( ) ( )( )

=

=0

1!2

,n

n

nn

n

xHHfxf in 12L , (2.39)

and the Parseval-Steklov equality holds:

( )

=

=0

2

1

2

1 !21,

nnn n

Hff . (2.40)


17/32

17

The expansion in powers ofx has the form

( ) =

=n

nmm

m

mnn xcnxH0

,! , ,...,1,0=n (2.41)

In particular, we have

!

2,

nc

n

nn = ,( )

!

10,2

nc

n

n

= ,

( )!

121,12

nc

n

n

=+ ,

( )( )!1

122,2

=

nc

n

n ,( )

( )!1314

3,12

=+

nc

n

n . (2.42)

The integral representation for the modified Hermite polynomials has the form:

( ) ( ) ( )

=

deHxVt

nn

22/22 , ,...1,0=n . (2.43)

Let 2/12Lf . It follows from (2.28) that1

2Lf ,

( )( )

( )

=

=

==0 0 !!2n n

nnn

nn

n

xVbxf

n

xHa in 12L . (2.44)

We denote by K the linear integral operator in equation (2.29):

( )( ) ( ) ( )

detK t21

. (2.45)

Lemma 1

The operator K assigns to every function ( )t f satisfying condition (2.30) an entire function

( )( )zKf with the estimate

( )( )

+

22 11

exp tyC

zKf

, iytz += . (2.46)

The proof follows immediately from (2.30):

( )( ) ( ) ( )

+==

deeC

deeCzKf ttyt 21222221

+

22 11

exp tyC

. (2.47)

Lemma 2

The operator K assigns to2Lf , 20


18/32

18

The proof follows from the Cauchy-Bunyakovskii inequality (applied to ( )( )zKf ) and the followingestimates:

( )( ) ( )

+

dzzefzKf

222/

2

12exp1 2

( ){ } ( ){ } =

+

+

2/1

222

2/1

22 2422exp1

242exp1

dttyfdtzf

( )

2/12

22

2

22exp

1

2exp

+=

dt

tyf . (2.49)

We can rewrite the above equation also as follows:

( )( ) ( ) =

+

dzzefzKf

222/

212exp

1 2

( )

2/12

22

2

22exp

1

2exp

+=

dt

tyf . (2.49b)

Lemma 3

The operator 22: LLK , 20


19/32

19

If 12Lf , then its image ( )( )tKf can be expanded in the Taylor series

( )( )

=

=0 !n

n

nn

tatKf , ( )

1, nn Hfa = , (2.52)

which converges uniformly on every compact set in R . If 2/12Lf , then

( )( )( )

=

=0 !2n

n

nn

n

tHbtKf in

1

2L , ( )1, nn HKfb = , (2.53)

and

( ) ( ) ,,,2/11 nn

VfHKf = ,...1,0=n . (2.54)

By lemma 3, the function ( )( )tKf is the trace of an entire function ( )( )zKf for 0=y . Hence, it canbe expanded in the Taylor series with the coefficients

( )( ) ( ) ( ) ( )( ) ( ) ( )

=

=

= === 00022

111

tn

tn

t

t

n

n

tn

n

dtHefdedt

dftKf

dt

d

( )( ) ( ) ( ) nnt

n

naHfdeHf ===

1,1

1 2

. (2.55)

Here we used equality (2.37). Further, if 2/12Lf , then (2.53) holds by (2.44), since1

2LKf by

Lemma 3. Equalities (2.54) can be proved as follows:

( ) ( )2/12/11

,,, nnn VfHKfHKf == . (2.56)

Here we used formula (2.43), which implies that nn HKV

= , where K is the operator adjoint to

K.

Let be a solution of equation (2.29) belonging to 12L , whence Kp

= . Putting ( )1

, nn Ha = ,

we deduce from (2.39) and (2.40) that

( )( )

=

=

0

!2n

n

nn

n

tHat in 12L ,

=

=

0

22

!2n

n

n

n

a . (2.57)

The function ( )tp is the trace of the entire function ( ) ( )( )zKzA = , for which (2.48) holds with1= :

( )2

1

zezA , Cz . (2.58)

By Lemma 4, it can be expanded in the Taylor series (2.52):

( )

=

=0 !n

n

n

p

n

tat . (2.59)

The integral equation (2.29) is equivalent to the following boundary-value problem for the heat

equation:


20/32

20

ttx uu4

1= , 10 x , Rt and, further, its

analytic continuation with respect to ( )tx, to the complex domain CT + , where +T is the righthalf-plane 0Re >=x .

Equation (2.29) takes the form

( ) ( ) ( )

=

det tq

212 , ,...2,1=q . (2.64)

If ( )t is a solution of equation (2.64), then ( )t and ( )0tt+ also are solutions of this equation

(for all 0t ).

Assertion 2

If ( )t is a solution of equation (2.64) such that (2.63) holds, then

( )( )

( )

+

12

121exp

124

112

2

2

xqx

qxxd

x

t

xq

q

q

q

(2.65)

for all ( )12/1 > qx .

We remember that if S is a measurable subset of Rn

with the Lebesgue measure, and and g are

measurable real- or complex-valued functions on S, then Hlder inequality is

( ) ( ) ( )( ) ( )( ) qS

qp

S

p

Sdxxgdxxfdxxgxf

/1/1

. (2.65b)

Denoting the left-hand side of inequality (2.65) by ( )txJ , , using the boundary conditions (2.61), theproperties of solutions of the heat equation and Holders inequality, we obtain the following chain

of relations for all ( )12/1 > qx :


21/32

21

( ) ( )( )

( )( )

( )

+

+=

dx

t

xd

x

t

xtxJ

q

1exp

1

1exp

1,

22

2

( )( )

( ) ( ) ( )( )

=

+

+

d

qxx

xqxt

qx

t

x 21

12exp

2exp

1

122

( )[ ] ( ) ( )

( )( )

q

q dqxx

xqxtJt

x

2

11

2

2

1

121

12exp

1

1

+

+=

, (2.66)

whence

( )

( )

( )( ) qqqxqx

qxx

xx

xJ

2

11

4

1

2

11

12

121

1

+

+

, (2.67)

which implies that (2.65) holds.Now we can rewrite the eq. (2.65) also as follows:

( )

( )

( )( )

+

+

qqq

xqx

qxx

xx

xJ

2

11

4

1

2

11

12

121

1

( )( )

( )

+

12

121exp

124

112

2

2

xqx

qxxd

x

t

xq

q

q

q

. (2.67b)

Corollary

For 1=x estimate (2.65) with ,...3,2=q takes the form

( ) ( ){ }

22

122exp

1 241

22

q

qdt q

q

. (2.68)

With regard the possible mathematical connections, we note that it is possible to obtain some

interesting and new relationships evidencing some Ramanujans Theorems. The first letter of

Ramanujan to G. H. Hardy, contain the bare statements of about 120 theorems, mostly formal

identities extracted from his note-books. We take, for the connections regarding this section, the

following identities:

( )

( ) ( )

+

+

++

+

=

++

++

+

++

0 2

2

2

2

12

1

2

11

2

1

2...

11

21

1

11

abba

abba

dx

a

x

b

x

a

x

b

x

, (2.69)

( )( )( ) ( )

+++++=

+++0106324222 ...12

1...111

1rrrr

dxxrxrx

, (2.70)


22/32

22

If 2 = , then

+=

+

0 2

4/1

0 2

4/1

141

141

22

dxe

xedx

e

xex

x

x

x

, (2.71)

5/242

2

15

2

55

...111

1

eee

+

+=

+++

, (2.72)

5/2

5

2/5

4/3

5452

2

15

12

1551

5

...111

1

eee

+

+

=+++

. (2.73)

We note that the eqs. (2.49b), (2.51b) and (2.68) can be related with the expression (2.69) and

(2.70). Indeed, we have:

( )( ) ( ) =

+

dzzefzKf

222/

212exp

1 2

( )

2/12

22

2

22exp

1

2

exp

+=

dt

tyf

( )

( ) ( )

+

+

++

+

=

++

++

+

++

0 2

2

2

2

12

1

2

11

2

1

2...

11

21

1

11

abba

abba

dx

a

x

b

x

a

x

b

x

, (2.74)

( )( ) ( ) =

+

dzzefzKf

222/

212exp

1 2

( )

2/12

22

222exp1

2exp

+=

dttyf

( )( )( ) ( )

+++++=

+++0106324222 ...12

1

...111

1

rrrrdx

xrxrx , (2.75)

( )( )

= dttKfeKf

t 22 2

( )

+

=

dedtef

t 22

22

42

21

( )

( ) ( )

+

+

++

+

=

+

+

++

+

++

0 2

2

2

2

12

1

2

11

2

1

2...

11

21

1

11

abba

abba

dx

a

x

b

x

a

x

b

x

, (2.76)


23/32

23

( )( )

= dttKfeKf

t 22 2

( )

+

=

dedtef

t 22

22

42

21

( )( )( ) ( )

+++++

=

+++0 106324222

...12

1

...111

1

rrrr

dx

xrxrx

, (2.77)

( ) ( ){ }

22

122exp

1 241

22

q

qdt q

q

( )

( ) ( )

+

+

++

+

=

++

++

+

++

0 2

2

2

2

12

1

2

11

2

1

2...

11

21

1

11

abba

abba

dx

a

x

b

x

a

x

b

x

, (2.78)

( ) ( ){ }

22

122exp

1 24 122

q

qdt q

q

( )( )( ) ( )

+++++=

+++0106324222 ...12

1

...111

1

rrrrdx

xrxrx . (2.79)

3. On some equations concerning the zeta strings and the zeta nonlocal scalar fields [5]

The exact tree-level Lagrangian for effective scalar field which describes open p-adic string

tachyon is

++

=

+

12

2

2 1

1

2

1

1

1 pp

pp

p

p

g

L , (3.1)

where p is any prime number,22

+= t is the D-dimensional dAlambertian and we adopt

metric with signature ( )++ ... . Now, we want to show a model which incorporates the p-adicstring Lagrangians in a restricted adelic way. Let us take the following Lagrangian


24/32

24

+

++=

==

1 1 1 1

1222 1

1

2

111

n n n n

n

nnnn

ngn

nCL

LL . (3.2)

Recall that the Riemann zeta function is defined as

( )

==1 1

11

n pss

pns , is += , 1> . (3.3)

Employing usual expansion for the logarithmic function and definition (3.3) we can rewrite (3.2) in

the form

( )

++

= 1ln

22

112

gL , (3.4)

where 1= 2220

2kkkr

, (3.5)

where ( ) ( ) ( )dxxek ikx

=~

is the Fourier transform of ( )x .

Dynamics of this field is encoded in the (pseudo)differential form of the Riemann zeta function.

When the dAlambertian is an argument of the Riemann zeta function we shall call such

string a zeta string. Consequently, the above is an open scalar zeta string. The equation of

motion for the zeta string is

( )( ) +> =

=

2

2

220 1

~

22

1

2 kkixk

Ddkk

ker

(3.6)

which has an evident solution 0= .

For the case of time dependent spatially homogeneous solutions, we have the following equation of

motion

( ) ( ) ( )

( )

( )tt

dkk

k

et ktikt

=

=

+>

1

~

22

1

2 00

2

0

2

2

0

0

. (3.7)

With regard the open and closed scalar zeta strings, the equations of motion are

( )( )

( )

=

=

n

n

nn

ixk

Ddkk

ke

1

2

12 ~

22

1

2

, (3.8)

( )( )

( )( )

( )

( )

+

+

+=

=

1

11

2

12

112

1~

42

1

4

2

n

n

nn

nixk

Dn

nndkk

ke

, (3.9)

and one can easily see trivial solution 0== .


25/32

25

The exact tree-level Lagrangian of effective scalar field , which describes open p-adic string

tachyon, is:

+

+

=+

122

2

1

1

2

1

1

2pm

p

D

p

p

p

p

p

p

g

mp

L , (3.10)

where p is any prime number,22

+= t is the D-dimensional dAlambertian and we adopt

metric with signature ( )++ ... , as above. Now, we want to introduce a model which incorporatesall the above string Lagrangians (3.10) with p replaced by Nn . Thence, we take the sum of all

Lagrangians nL in the form

+

=

++

=

++

==

1

12

1

2

2 1

1

2

1

1

2

n

nm

n n

D

nnnn

nn

n

n

g

mCCL n

L , (3.11)

whose explicit realization depends on particular choice of coefficients nC , masses nm and coupling

constants ng .

Now, we consider the following case

hnn

nC

+

=

2

1, (3.12)

where h is a real number. The corresponding Lagrangian reads

+

+= +

=

++

=

1

1

1

22 12

1 2

n

nh

n

hm

D

hn

nn

g

mL

(3.13)

and it depends on parameter h . According to the Euler product formula one can write

+

=

=p

hmn

hm

p

n2

2

21

2

1

1

. (3.14)

Recall that standard definition of the Riemann zeta function is

( ) +

=

==

1 1

11

n pss

pns , is += , 1> , (3.15)

which has analytic continuation to the entire complex s plane, excluding the point 1=s , where it

has a simple pole with residue 1. Employing definition (3.15) we can rewrite (3.13) in the form

++

+=

+

=

+

1

1

22 122

1

n

nhD

hn

nh

mg

mL

. (3.16)


26/32

26

Here

+ h

m22

acts as a pseudodifferential operator

( )

( )

( )dkkhm

kexh

m

ixk

D

~

22

1

22

2

2

+=

+

, (3.17)

where ( ) ( ) ( )dxxek ikx

=~

is the Fourier transform of ( )x .

We consider Lagrangian (3.16) with analytic continuations of the zeta function and the power series

+

+

1

1

nh

n

n , i.e.

++

+=

+

=

+

1

1

22 122

1

n

nhD

hn

nACh

mg

mL

, (3.18)

where AC denotes analytic continuation.Potential of the above zeta scalar field (3.18) is equal to hL at 0= , i.e.

( ) ( )

+=

+

=

+

1

12

2 12 n

nhD

hn

nACh

g

mV

, (3.19)

where 1h since ( ) =1 . The term with -function vanishes at ,...6,4,2 =h . The equationof motion in differential and integral form is

+

=

=

+

122 n

nhnAChm

, (3.20)

( )( )

+

=

=

+

12

2 ~

22

1

n

nh

R

ixk

DnACdkkh

m

ke

D

, (3.21)

respectively.

Now, we consider five values of h , which seem to be the most interesting, regarding the

Lagrangian (3.18): ,0=h ,1=h and 2=h . For 2=h , the corresponding equation of motion

now read:

( )

( )( )

( )

+=

=

DR

ixk

Ddkk

m

ke

m32

2

2

1

1~2

22

12

2

. (3.22)

This equation has two trivial solutions: ( ) 0=x and ( ) 1=x . Solution ( ) 1=x can be also

shown taking ( ) ( )( )Dkk 2~

= and ( ) 02 = in (3.22).For 1=h , the corresponding equation of motion is:

( )( )

( ) =

=

DR

ixk

Ddkk

m

ke

m22

2

21

~1

22

11

2

. (3.23)

where ( )12

11 = .

The equation of motion (3.23) has a constant trivial solution only for ( ) 0=x .For 0=h , the equation of motion is


27/32

27

( )( ) =

=

D

R

ixk

Ddkk

m

ke

m

1

~

22

1

2 2

2

2

. (3.24)

It has two solutions: 0= and 3= . The solution 3= follows from the Taylor expansion of theRiemann zeta function operator

( )( )( )

+=

122 2!

00

2 n

nn

mnm

, (3.25)

as well as from ( ) ( ) ( )kk D 32~

= .

For 1=h , the equation of motion is:

( ) ( ) ( ) =

+DR

ixk

D dkkm

ke

2

2

2

1ln2

1~122

1

, (3.26)

where ( ) =1 gives ( ) =1V .In conclusion, for 2=h , we have the following equation of motion:

( )( )

( )

=

+

DR

ixk

Ddw

w

wdkk

m

ke

0

2

2

2

2

1ln~2

22

1. (3.27)

Since holds equality

( ) ( )

===

1

0 1 2211ln

n ndw

w

w

one has trivial solution 1= in (3.27).

Now, we want to analyze the following case:2

2 1

n

nCn

= . In this case, from the Lagrangian (3.11),

we obtain:

+

+

=

121

22

1 2

222mmg

mL

D

. (3.28)

The corresponding potential is:

( )( )

2

124

731

=

g

mV

D

. (3.29)

We note that 7 and 31 are prime natural numbers, i.e. 16 n with n =1 and 5, with 1 and 5 that are

Fibonaccis numbers. Furthermore, the number 24 is related to the Ramanujan function that has 24

modes that correspond to the physical vibrations of a bosonic string. Thence, we obtain:


28/32

28

( )( )

2

124

731

=

g

mV

D ( )

++

+

42710

421110log

'

142

'

cosh

'cos

log42

'

'4

0

'

2

2

wtitwe

dxex

txw

anti

w

wt

wx

. (3.29b)

The equation of motion is:

( )[ ]( )2

2

221

11

21

2

+=

+

mm

. (3.30)

Its weak field approximation is:

022

12 22

=

+

mm

, (3.31)

which implies condition on the mass spectrum

22

12 2

2

2

2

=

+

m

M

m

M . (3.32)

From (3.32) it follows one solution for 02

>M at22

79.2 mM and many tachyon solutions when22 38mM < .

We note that the number 2.79 is connected with2

15 = and

2

15 += , i.e. the aurea section

and the aurea ratio. Indeed, we have that:

78,2772542,22

15

2

1

2

152

2

=

+

+.

Furthermore, we have also that:

( ) ( ) 79734,2179314566,0618033989,27/257/14 =+=+

4. Mathematical connections

We have the following new possible mathematical connections: between eqs. (1.7), (1.17) and

(1.19b) with eq. (3.6)

( ) ( ) ( )1

arctansin12

1

1

2

12

0

22

++

+=

t

s

e

dttst

s

s


29/32

29

( )

( ) +> =

=

2

2

220 1

~

22

1

2 kkixk

Ddkk

ker

, (4.1)

( )

=

=0

22log1

1

1arctan20'

dt

e

tt

( )

( ) +> =

=

2

2

220 1

~

22

1

2 kkixk

Ddkk

ker

, (4.2)

( )

=

=

0 2 2879,0

1arccos2log1

1

1arctan20'

dt

et

t

( )

( ) +> =

=

2

2

220 1

~

22

1

2 kkixk

Ddkk

ker

. (4.3)

We have further possible mathematical connections between eqs. (1.56) and (1.64) with eq. (3.6):

=

=

+

+

0 0 /222

2

3 21

1

2

1

1

12

1

log2

1cos

4

1

2

11dt

tetedt

t

tit

ttt

( )( ) ( ) ( )

+

+

=

0 2/222/2 21

1

111

/2dt

teteteetttt

( )

( ) +>

=

=

2

2

22

0 1

~

22

1

2kk

ixk

Ddkk

ker

, (4.4)

( )( ) =

+

+

=

+

=

dtt

tit

tnn

2

2

031 1

log2

1cos

4

1

2

11

2

2log

( )dt

t

tit

t2

2

03 1

log2

1cos

4

1

2

11

+

+

=

( )( )

+>

=

=

2

2

220 1

~

22

1

2 kk

ixk

D

dkkk

er

, (4.5)

With regard , we have these possible mathematical connections between eqs (4.4) and (4.5) with

some Ramanujans equations ((2.69) and (2.70)). Indeed:

=

=

+

+

0 0 /222

2

3 21

1

2

1

1

12

1

log2

1cos

4

1

2

11dt

tetedt

t

tit

ttt

( )( ) ( ) ( )

+

+

=

02/222/2

21

1

111

/2dt

teteteetttt


30/32

30

( )

( ) +> =

=

2

2

220 1

~

22

1

2 kkixk

Ddkk

ker

( )

( ) ( )

+

+

++

+

=

+

+

++

+

++

0 2

2

2

2

12

1

2

11

2

1

2...

11

21

1

11

abba

abba

dx

a

x

b

x

a

x

b

x

, (4.6)

=

=

+

+

0 0 /222

2

3 21

1

2

1

1

12

1

log2

1cos

4

1

2

11dt

tetedt

t

tit

ttt

( )( ) ( ) ( )

+

+

=

0 2/222/2 21

1

111

/2dt

teteteetttt

( )

( ) +> =

=

2

2

220 1

~22

12 kk

ixkD

dkkker

( )( )( ) ( )

+++++=

+++0106324222 ...12

1

...111

1

rrrrdx

xrxrx . (4.7)

We have also some possible mathematical connections with the eqs. (1.19b), (1.34), (1.36) and eq.

(3.6). Indeed:

( )

=

=

0 2 2879,0

1arccos2log1

1

1arctan20'

dt

et

t

( ) ( )

( )( ) ==

q

uddt

t

tftf

0

2/

cos 22/35

5224/3

2/51

sin51

125

( ) ( )( )( ) ( )( )

=

=

qfqfq qqq

dd3534/31 54/11/5tan2

0

/5tan2

0 22/122/35 sin51

15

sin51

1

( )

( ) +> =

=

2

2

220 1

~

22

1

2 kkixk

Ddkk

ker

, (4.8)

( )

== 0

2 2879,0

1arccos2log11

1arctan20' dtet t

( ) ( ) ( ) ( )( )

=

=

+

q

vv

vv ddttftftftf0

5/1tan2

1

111

5

1tan2

2

15531

2

21

sin25

91

1

5

1

( ) ( )( )( )( )

( )( )( )

( )( )

+

+

+

++

=

=2/

11

11

1

1tan2

53tan

11

1153tan

2251

5

3

31

1

5

511

sin16

151

1

4

1

sin81

11

1

9

1

vvvv

v

v

vv

vv dd

( )

( ) +>

=

=

2

2

22

0 1

~

22

1

2kk

ixk

Ddkk

ker

. (4.9)


31/32

31

In conclusion, we have also some possible mathematical connections between the eqs. (1.32) and

(1.39) with the following equation

++

+=

4

2710

4

21110log

142

8

3

, related to

and with the physical vibrations of the superstrings. Thence, we have:

( )( )

( ) ( )( )

+

++=

2

232

5

5

251

251log5log

5

8

uv

uvvu

q

dq

q

q

++

+=

4

2710

4

21110log

142

8

3

, (4.10)

( ) ( ) ( ) ( )

++

=q c

x

xc

ddttftftftf0

cos

1221

221cos

2

4/714721

1

sin64

213321

12

++

+=

4

2710

4

21110log

142

8

3

. (4.11)

Acknowledgments

I would like to thank Prof. Branko Dragovich of Institute of Physics of Belgrade (Serbia) for his

availability, useful discussion and friendship

References

[1] Bruce C. Berndt and Atul Dixit Transformation formula involving the gamma and

Riemann zeta functions in Ramanujans lost notebook H98230-07-01-0088

[2] Nayandeep Deka Baruah and Bruce C. Berndt Eisenstein series and Ramanujan-type

Series for1/ Math. Reviews Class.N: 33C05, 33E05, 11F11, 11R29

[3] B. Dragovich, A. Yu. Khrennikov, S. V. Kozyrev and I. V. volovich On p-adic


32/32

Mathematical Physics arXiv:0904.4205v1 Apr 2009

[4] V. S. Vladimirov On the equation of the p-adic open string for the scalar tachyon field

arXiv:math-ph/0507018v1 Jul 2005

[5] Branko Dragovich: Zeta Strings arXiv:hep-th/0703008v1 1 Mar 2007.

Documents

On some mathematical equations concerning the functions zeta(s) and zeta(s,w) and some Ramanujan-type series for 1/pigreco . Mathematical connections with some equations concerning