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 the p-adic open string for the scalar tachyon field and the zeta strings.

(1)

On some mathematical equations concerning the functions

ζ

( )

s and

ζ

( )

s,w and some Ramanujan-type series for 1/π . Mathematical connections with some equations concerning the p-adic open string for the scalar tachyon field and the zeta strings.

Michele Nardelli1,2

1Dipartimento di Scienze della Terra

Università degli Studi di Napoli Federico II, Largo S. Marcellino, 10 80138 Napoli, Italy

2Dipartimento di Matematica ed Applicazioni “R. Caccioppoli”

Università 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

ζ

( )

s and

ζ

( )

s,w . In this Section, we have described also some equations concerning a transformation formula 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 “Ramanujan’s lost notebook”. In the Section 2, we have described some Ramanujan-type series for 1/π and some equations concerning the p-adic open string for the scalar tachyon field. In this Section, we have described also some possible and interesting mathematical connections with some Ramanujan’s 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 Ramanujan’s 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

ζ

( )

s and

ζ

( )

s,w [1]

In the Mathematical Analysis there exist the proof of the following formula: ζ(s)= 1 )] ( [ ) 1 ( 2 1 1 2 1 2 0 2 2 − + + − +

_∫

∞ − _t s e dt t sArcTan Sin t s π (1.1) that can be rewritten also as follows:

(2)

( )

(

( )

)

( ) (

)

∫

∞ − + + ₋ − + = 0 2 /2 2 1 1 1 2 1 1 1 tan sin 2 s dt e t t s s t s _π

ζ

(1.1b)

where ζ(s) represent the Riemann zeta function. Now we analyze in greater detail the formula (1.1) The function

ζ

( )

s is represented from the series

( )

... 1 ... 3 1 2 1 1+ + + + + = _s _s _s n s ζ , (1.2)

as the real part of the complex variable

s≡ξ +iη

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

( ) ( )

∫

∞ − − Γ = 0 1 1 1 x s e dx x s s

ζ

, (1.3)

what is easily proved by using the following equality

( )

∫

∞ − − Γ = 0 1 1 1 dx x e s v s vx s . (1.4)

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

ζ

( )

s across all the plan and discover such interesting properties of this function. We set f

( )

z =z−s, where

( )

(

)

      + = − τ τ τ t t s t p s arctan cos , 2 2 2 _,

( )

(

)

      + − = − τ τ τ t t s t q s arctan sin , 2 2 2 _{. (1.5)}

The following three conditions:

1° The function f

( )

z is holomorphic for τ ≥α , for each t ; 2° The condition lim −2

(

+

)

=0

±∞

= e f it

t

t τ

π

is verified uniformly for α ≤τ ≤β, however great β; 3° The function f

( )

z is subject to the following condition: lim

∫

+∞ 2

(

+

)

=0

∞ − − ∞ = e f it t

τ

π τ ;

are verified in the half-plane τ >0; assuming

ξ

>1, so that the series (1.2) converges, and by 1

=

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

∑

( )

∫

( )

∫

( )

∞ _∞ _∞ − − + = m m t dt e t m q d f m f v f 0 2 ₁ , 2 2 1 π

τ

, (1.6)

(3)

( )

(

)

1 arctan sin 1 2 1 1 2 1 2 0 2 2 − + + − + =

_∫

∞ −s _t e dt t s t s s _π ζ , (1.7)

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

∑

( )

∫

( )

∫

( )

∞ _∞ _∞ − = m dt t Q d f v f α τ τ 2 0 α, , (1.8) for 2 1 = α ,

( )

(

)

1 2 arctan sin 4 1 2 1 2 2 0 2 2 1 +       ₊ − − =

_∫

∞ − − t s s e dt t s t s s _π

ζ

. (1.9)

Another expression for

ζ

( )

s is derived from the following formula:

∑

( )

∫

+∞

( )

∞ − ∞       − = i i m dz z F z i v f α α

π

2 sin 2 1 . (1.10) By = z= +it 2 1 , 2 1

α

, we find

( )

[

( )

]

(

)

∫

∞ ₋ − + −       + − = 0 2 2 1 2 cos 1arctan2 4 1 1 4 dt e e t s t s s t t s π π

π

ζ

. (1.11)

Starting from the relationship

( )

... 4 1 3 1 2 1 1 2 1 1 ₁ = − + − +      ₋ − s s s s

ζ

s

with the application of the following formula

∑

( ) ( )

∫

+∞

( )

∫

( )

∞ − ∞ − =− − − = − n m i i iz iz v dt t Q e e dz z f v f α α π π 2 0 1

α

, 1 , (1.12) with 2 1 , 1 = =

α

m , we also find

( )

∫

∞

(

₋

)

− − _ ₊     ₊ − = 0 2 2 1 2 arctan cos 4 1 1 2 2 dt e e t s t s _t _t s s s π π

ζ

. (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 s=0,

( )

2 1 0 =−

(4)

then, subtracting the two members 1 1 −

s , and making tend s to the unity, taking into account the following equality: dt e t t C

∫

∞ _t − + + = 0 2 2 ₁ 1 1 2 2 1 π , (1.14) we obtain _

( )

_= − − = ₁ 1 lim 1 s _s s ζ _t _e dt C t t − = + +

_∫

∞ 0 2 2 ₁ 1 1 2 2 1 π , (1.15)

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

∫

∞ = − − 0 2 ₁ log 2 1 arctan 2 1 _π dt π e t _t , (1.16) we obtain

( )

∫

∞ − =− − = 0 2 ₁ 1 log 2 1 arctan 2 0 ' π ζ _π dt e t _t . (1.17)

We note that there exist a mathematical connection between

π

and

2 1 5− =

φ , i.e. the aurea section, by the simple formula

arccosφ =0,2879π , (1.18) thence we have that

2879 , 0 1 arccos ⋅ =

φ

π

. (1.19)

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

( )

∫

∞       ⋅ − = − − = 0 2 ₀_,₂₈₇₉ 1 arccos 2 log 1 1 1 arctan 2 0 '

φ

ζ

π dt e t _t . (1.19b)

Now we consider the function

( )

(

) (

2

)

... 1 1 1 1 , + + + + + = s s s w w w w s

ζ

, (1.20)

which reduces to

ζ

( )

s for w=1. We must replace our general formulas f

( ) (

z = z+w

)

−s, (1.21)

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( ) (

[

)

]

      + + + = − w t s t w t p s

τ

, 2 2 2cos arctan _,

( ) (

[

)

]

      + + + − = − w t s t w t q s

τ

, 2 2 2sin arctan _{. (1.22)} Assuming the real part of w positive, applying the following formula

∑

( )

∫

( )

∫

( )

∞ _∞ _∞ − − + = m m t q m t dt e d f m f v f 0 2 ₁ , 1 2 2 1 π τ τ , (1.23) we obtain

( )

∫

∞

(

)

− − − −       + + + − = 0 2 2 2 2 1 1 1 arctan sin 2 2 1 , dt e w t s t w w s w w s _t s s s π

ζ

, (1.24)

valid expression in the whole plan and shows that

ζ

( )

s,w is a uniform function admitting for singularity at finite distance, only the pole s=1 of residue 1. We conclude, on the other hand, for

0 = s ,

( )

w = −w 2 1 , 0 ζ , (1.25) then, subtracting 1 1 −

s and for s tending to the unity,

( )

∫

∞ = =− + + + −    − − 0 2 2 2 1 ₁ 1 2 2 1 log 1 1 , lim dt e t w t w w s w s _t s ζ π , (1.26)

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

( )

∫

∞ − + −       ₋ = 0 2 ' 1 1 arctan 2 log 2 1 , 0 dt e w t w w w w _t s π ζ . (1.27)

For the following equalities,

( )

x x  x−x+J

( )

x      − + = Γ log 2 1 2 log log π , (1.28)

( )

J

( )

x x x x Dx ' 2 1 log logΓ = − + , (1.29)

these last two expressions on reduced respectively to

( )

w w Γ Γ

− ' and logΓ

( )

w −log 2π . Here it is possible to obtain some mathematical connections with various theorems concerning the incomplete elliptic integrals described in the “Ramanujan’s lost notebook”.

Let u

( )

q denote the Rogers-Ramanujan continued fraction defined by

( )

... 1 1 1 1 : : 3 2 5 / 1 + + + + = = q q q q q u u , q <1, (1.30)

(6)

and set v=u

( )

q2 . Recall that

ψ

( )

q is defined by

( )

∑

( )

(

( )

)

∞ = ∞ ∞ + ₌ = = 0 2 2 2 2 / 1 3 ; ; , : n n n q q q q q q q f q

ψ

. (1.31) Then

∫

( )

(

)

_       + − − + + = 2 3 ₂2 5 5 2 5 1 2 5 1 log 5 log 5 8 uv uv v u q dq q q

ψ

. (1.32) We note that

(

)

2 2 1 5 236067977 , 1 2 5

1+ − = = − × , i.e. the aurea section multiplied by the

number 2, and that

(

)

_

      × + − = − = + − 2 2 1 5 2360679777 , 3 2 5

1 , i.e. the aurea ratio multiplied by

the number 2 and with the minus sign. With f

( )

−q ,

ψ

( )

q , and u

( )

q defined by

( )

(

)

∑

( )

∞ −∞ = − ∞ − ₌ ₌ − = − − = − n iz n n n z e q q q q q f q f : , 2 1 3 1/2 ; : 2π /24η , q=e2πiz, Imz>0, (1.33) by (1.31) and (1.30), respectively, and with ε =

(

5+1

)

/2, we have that

( )

(

)

∫

− = − = − − − − q u d dt t t f t f 0 2 / cos 5 3/2 2 5 2 2 4 / 3 2 / 5 1 sin 5 1 1 2 5 π ε

_ε

_ϕ

ϕ

( )

(

)

( )

(

)

∫

− − − ₋ ₋

∫

− ₋ − = − = qf q f q q q q d d 3 5 3 4 / 3 1₅ _/ 1 1/4 5 tan 2 0 / 5 tan 2 0 1/2 2 2 2 / 3 5 sin 5 1 1 5 sin 5 1 1 ψ ψ ϕ ϕ ε ϕ ϕ ε . (1.34)

Let v be defined by the following expression

( )

( ) ( )

( )

3 5 3 15 : : _      − − − − = = q f q f q f q f q q v v , (1.35)

and let ε =

(

5+1

)

/2. Then

( )

( ) ( ) ( )

( ) = − = − − − −

∫

−−         − + − − q v v v v d dt t f t f t f t f 0 5 / 1 tan 2 1 11 1 5 1 tan 2 2 15 5 3 1 2 2 1 sin 25 9 1 1 5 1

_ϕ

ϕ

( )

(

₍ ₎

₍

)(

₎

)

( ) ( )

(

)

(

)(

)

∫

        + − − + + − −         + + − − − − − − − − − − − − = − = /2 1 1 1 1 1 1 tan 2 5 3 tan 1 1 1 1 5 3 tan 2 2 5 1 5 3 3 1 1 5 5 1 1 sin 16 15 1 1 4 1 sin 81 1 1 1 9 1 π ε ε ε ε ε ε ε ε ε ε

ϕ

v v v v v v v v v v d d . (1.36)

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( )

( ) ( )

( )

4 7 2 14 : : _      − − − − = = q f q f q f q f q q v v , (1.37) and if 7 2 16 13+ = c , then

∫

( )

( ) ( ) ( )

∫

− − _      − + − − = − − − − q c v v c d dt t f t f t f t f 0 cos 1 1 cos 2 14 7 2 1 1 sin 2 32 13 2 16 1 1 2 8 1

_ϕ

ϕ

. (1.38) Let

( ) ( )

( )

3

( )

7 3 14 3 2 3 q f q f q f q f q v − − − − = and 7 2 4 9− = c , Then

∫

( )

( ) ( ) ( )

∫

− −         + + − − − − = − − − − q c x x c d dt t f t f t f t f 0 cos 1 2 2 1 2 2 1 cos 2 4 / 7 14 7 2 1 1 sin 64 2 13 32 1 1 2

ϕ

. (1.39)

1.1On some equations concerning a transformation formula involving the gamma and

Riemann zeta functions of Ramanujan.

In the Ramanujan’s 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 Ramanujan’s claim, it will be convenient to use the familiar notation

( )

∑

∞ =       + − + − − = Γ Γ = 0 1 1 1 ' : k k x k x x x γ ψ , (1.40)

where γ denotes Euler’s constant. We also need to recall the following functions associated with Riemann’s zeta function

ζ

( )

s . Let

ξ

( ) ( )

s s

π

s s

ζ

( )

s      ₊ Γ − = − 2 1 1 1 : 2 1 .

(8)

      ₊ =       Ξ t it 2 1 2 1 : 2 1 _ξ . Theorem 1 Define

( )

x x x x log 2 1 :=ψ + − φ . (1.41)

If

α

and β are positive numbers such that αβ =1, then

(

)

( )

=       + − =       + −

_∑

∞ = ∞ =1 2 1 2 log 2 2 log n n n n

φ

β

πβ

γ

β

α

φ

α

πα

γ

α

∫

∞ +            − + Γ       Ξ − = 0 2 2 3 ₁ log 2 1 cos 4 1 2 1 1 dt t t it t

α

π

, (1.42)

where γ denotes Euler’s constant and Ξ

( )

x denotes Riemann’s Ξ-function.

Although Ramanujan does not provide a proof of (1.42), he does indicate that (1.42) “can be deduced from”

(

x

)

x

) (

nx

)

dx

(

1 n

)

logn

)

2 1 2 cos log 1 0 + − = + −

∫

∞ψ π ψ . (1.43)

We have that, for t≠0,

∑

∞ =       + − − = + 1 2 2 2 2 1 1 1 1 2 1 4 1 n t t e t n t π . (1.44)

We find that, for Rez>0,

( )

∫

∞

(

)(

)

− + − = 0 2 2 2 ₁ 2 _t e z t tdt z _π φ . (1.45)

We require Binet’s integral for logΓ

( )

z , i.e., for Rez>0, thence

( )

∫

∞ −       − + − + + −       − = Γ 0 ₁ 1 1 2 1 2 log 2 1 log 2 1 log dt t e e t z z z z zt t

π

. (1.46) We find that

∫

∞ ₋  − =      ₋ − 0 1 1 1 _γ dx e x e x x , (1.47)

(9)

µ

ν

ν µ log 0 = −

∫

∞ − − dx x e e x x , µ,ν >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

α

in (1.45) and summing on n, 1≤n<∞, we find, by absolute convergence, that

( )

(

)(

)

(

)

( )

∑

∞

∑∫

∫

∑

= ∞ = ∞ ∞ ∞ = + − − = − + − = 1 1 0 0 1 2 2 2 2 2 2 2 2 / 1 1 2 1 2 n n n t t n t e tdt e n t tdt n

α

φ

π π . (1.49)

Invoking (1.44) in (1.49), we see that

∑

( )

∫

(

)

∞ = ∞       ₋ ₊ − − − = 1 0 / 2 2 2 1 2 1 1 1 1 2 n t t dt t e e n π α απ α φ _π _π _α . (1.50)

Next, setting x=2πt in (1.47), we readily find that

∫

∞ −       − − = 0 2 2 1 2 dt t e e t t π π

π

γ

. (1.51) By Frullani’s integral (1.48),

∫

∞

(

)

− − =       = − 0 2 / 2 log / 1 2 log

πα

α

π

π α dt t e e t t . (1.52)

Combining (1.51) and (1.52), we arrive at

(

)

dt t e e t t

∫

∞ −       − − = − 0 / 2 1 2 2 log α π

π

πα

γ

. (1.53)

Hence, from (1.40) and (1.43), we deduce that

(

)

_{( )}

(

)

∫

∑

∞ ∞ − ∞ =       + − − − −       − − =       + − 0 0 2 2 / / 2 1 2 1 2 1 1 1 1 2 1 2 2 1 2 2 log dt t e e dt t e e n _t _t t t n

π

α

π

α

φ

α

πα

γ

α

_π α _π _π _α

(

)

(

)(

)

∫

∞ −        − − − − − = 0 / 2 / 2 2 2 1 1 2 1 t dt e e e e t t t t t

α

π

α

α π α π π . (1.54)

Now, for n real, we have that

∫

∞

∫

∞ = +      − + Γ       Ξ = +             Ξ      − − Γ      − + Γ 0 2 2 0 2 2 1 cos 4 1 2 1 1 cos 2 1 4 1 4 1 dt t nt it t dt t nt t it it dx xe e xe exen n xe n n

∫

∞ − _     − −       − − = − 0 3 1 1 1 1 1 1

π

. (1.55)

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Letting logα 2 1 =

n and x=2

π

t/

α

in (1.55), we deduce that

∫

∞

∫

∞  =      ₋ −       ₋ − − = +            − + Γ       Ξ − 0 2 0 2 2 / 2 3 ₁ ₂ 1 2 1 1 1 2 1 log 2 1 cos 4 1 2 1 1 dt t e t e dt t t it t _t _t

π

α

π

α

π

α

π

π π α

(

)(

) (

)

(

)

∫

∞        − − + − + − − − = 0 2 / 2 2 2 / ₁ ₂ 2 1 1 1 1 / 2 dt t e t e t e e t t t t

π

α

π

α π π π α π . (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 / 2 ₂ 0 1 1 1 1 2 2 1 1 du u e u e u dt t e t e t u u t t π α α π

π

_α

α

, (1.57)

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

∫

∞

(

)

( )

− − =       + − − 0 2 ₂log 2 1 2 1 1 1 a du u e u e u ua u

π

, (1.58)

so that if we set a=1/

( )

2

π

in (1.58), we deduce (1.57). Consider the integral, for t>0,

( )

∫

∞ − − −

( )

 + −

( )

+      − − Γ =       ₋ +       + − − = ₀ log 2 1 2 log 2 1 log 2 1 log 2 2 1 1 1 1 : , a t t t t t du u e e u e u e t a F tu ua tu u

π

, (1.59)

where we applied (1.46) and (1.48). Upon the integration of (1.30), it is easily gleaned that, as 0

→ t ,

logΓ

( )

t ≈−logt−

γ

t, (1.60)

where γ denotes Euler’s constant. Using this in (1.59), we find, upon simplification, that, as t →0,

F

( )

a t t t t t

( )

loga 2 1 2 log 2 1 log , ≈−γ − + − π − . (1.61) Hence, F

( )

a t

( )

a t ₂log 2π 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 Lebesgue’s 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,

(11)

(

)

∑

( )

∫

∞ ∞ = +            − + Γ       Ξ − =       + − 0 2 2 3 1 1 log 2 1 cos 4 1 2 1 1 2 2 log dt t t it t n n

α

π

α

φ

α

πα

γ

α

. (1.63)

Lastly, using (1.63) with

α

replaced by β and employing the relation αβ =1, we conclude that

( )

= +            − + Γ       Ξ − =       + −

∫

∑

∞ ∞ = dt t t it t n n 2 2 0 3 1 1 log 2 1 cos 4 1 2 1 1 2 2 log β π β φ β πβ γ β

( )

= +            − + Γ       Ξ − =

_∫

∞ dt t t it t ₂ 2 0 3 ₁ / 1 log 2 1 cos 4 1 2 1 1 α π

( )

dt t t it t ₂ 2 0 3 ₁ log 2 1 cos 4 1 2 1 1 +            − + Γ       Ξ − =

_∫

∞ α π . (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/π [2]

Ramanujan’s series representations for 1/π depend upon Clausen’s product formulas for hypergeometric series and Ramanujan’s Eisenstein series

( )

∑

∞ = − − = 11 24 1 : k k k q kq q P , q <1. (2.1)

More precisely, but briefly, by combining two different relations between P

( )

q and

( )

n

q

P , for certain positive integers n, along with a Clausen formula, we can obtain series representations for

π

/ 1 . Theorem 1

For n=5, if A and k B , k k≥0, are defined by the following formula

₃ 3 ! 2 1 : k A k k       = , ₃ ! 4 3 2 1 4 1 : k B k k h k                   = , ₃ ! 6 5 2 1 6 1 : k C k k k k                   = , (2.1b) then

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∑

{

} (

)

∞ = + − − + = 0 2 / 1 2 2 5 1 5 5 4 2 k k k A k π , (2.2)

∑

( ) (

)

∞ = + − = 0 4 1 3 20 1 8 k k k k B k π . (2.3)

For n=13, if A and k B , k k≥0, are defined by the formula (2.1b), then

( ) (

)

_k _k k k B k ₂ 0 18 1 23 260 1 72

∑

∞ = + − = π . (2.4)

The last two identities was recorded by Ramanujan in the fundamental paper “Modular equations and approximations to π(1914)”.

Proof of (2.2).

From Ramanujan’s second notebook (Notebooks – 2 volumes – 1957), we see that

∑

( )

{

( )

(

( )

)

(

( )

)

}

∞ = ∞ = × − − + + = − − − + 1 1 5 5 5 2 2 10 10 2 2 1 1 3 2 4 1 1 30 1 1 k k k k k k q x q x q x q x q q q kq q kq

_φ

{

( )

(

( )

)

(

( )

)

}

2 / 1 5 5 1 1 1+ x q x q + −x q −xq × . (2.5)

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

( ) ( )

10 − 2 = 2

( )

2

( )

5

{

3+

( )

5 +

(

1−

( )

)

(

1−

( )

5

)

}

× 2 1 5Pq P q φ qφ q x q xq x q xq

{

( )

(

( )

)

(

( )

)

}

2 / 1 5 5 1 1 1+ x q xq + −x q −xq × . (2.6)

Now we set q=e−π/ 5 and use, following Ramanujan, the following expressions 1−x_n =x₁_/_n, z₁_/_n = nz_n, (2.7) z z

( )

q ₂F₁ ;1;x 2

( )

q 2 1 , 2 1 : : =φ      = = , (2.8)

to deduce that x

( )

q = x₁_/₅ =1−x₅, x

( )

q5 =x₅, and

φ

2

( )

e−π/ 5 = 5

φ

2

( )

e−π 5 = 5z₅. Thus, from (2.6), we find that

( ) (

−2 5 − −2 / 5

)

= ₅2

{

+ ₅

(

− ₅

)

}

(

1+2 ₅

(

1− ₅

)

= 2 1 1 2 3 5 5Pe π Pe π z x x x x ₅2

{

₅

(

₅

)

}

(

1 4 ₅

(

1 ₅

)

2 1 1 4 3 5z + x −x + x −x = . (2.9)

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( ) (

−2 5 − −2 / 5

)

= ₅2

{

+ ₅

(

− ₅

)

}

(

1+2 ₅

(

1− ₅

)

= 2 1 1 2 3 5 5Pe π Pe π z x x x x

{

5

} (

5

)

2 5 1 2 1 3 5z + X + X = . (2.9b)

But, the singular modulus x is given by 5

3 5 2 1 5 2 1         ₋ − = x , (2.10) so that X5 =9−4 5 and X5 = 5−2. (2.11)

Thus, from (2.9), we find that

( ) (

)

(

)

52 5 / 2 5 2 2 1 5 1 5 5 5Pe Pe _z       ₋ + = − − − π π . (2.12)

Next, setting n=5 in the following equation

( ) (

)

π π π n e P e nP −2 n + −2 / n = 6 , (2.13) we find that

( ) (

)

π π π 6 5 5Pe−2 5 +Pe−2 / 5 = . (2.14)

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

( )

2 5 ₅2 2 1 5 5 1 5 5 3 z e P _       ₋ ⋅ + + = −

π

π . (2.15)

Now, employing the following expression

∑

∞ = =       = 0 2 3 2 ; 1 , 1 ; 2 1 , 2 1 , 2 1 k k kX A X F z , 2 1 0≤x≤ , (2.16)

in (2.15), we deduce the identity

( )

∑

∞ = −         ₋ ⋅ + + = 0 5 5 2 2 1 5 5 1 5 5 3 k k kX A e P

π

π . (2.17)

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( )

(

) (

∑

)

∞ = − ₌ ₋ ₊ 0 2 1 3 2 1 k k n k n n X A k x e P π , (2.18) we find that

( )

(

) (

∑

)

(

)

∑

(

)

∞ = ∞ = − ₌ ₋ ₊ ₌ ₋ ₊ 0 0 5 5 5 5 2 1 3 2 5 2 1 3 2 1 k k k k k kX k A X A k x e P π . (2.19)

Using (2.17) and (2.19), we arrive at (2.2). Proof of (2.3).

Employing the following expression

∑

( )

∞ = − − =       − − = 0 2 2 2 3 2 1 2 1 1 ; 1 , 1 ; 4 3 , 2 1 , 4 1 2 1 1 k k k k W B x W F x z ,       ₋ ₋ ≤ ≤ 1 2 2 2 2 1 0 x 1/4 , (2.20) in (2.15), we find that

( )

₍

₎

∑

∞

( )

∑

( )

= ∞ = − ₋ ₌ ₊ + ₋         ₋ ⋅ − + + = 0 0 2 5 2 5 5 5 2 1 5 4 5 3 5 3 1 2 1 5 2 1 5 1 5 5 3 k k k k k k k k W B W B x e P

π

π , (2.21) where 2 1 1 2 5 5 5 = ₋ = X X W . (2.22)

Next, setting n=5in the following equation

( )

∑

( ) (

)

∞ = − − + + + − = 0 2 2 1 2 / 1 1 3 1 k k n k n n n k n W B X X X k e P π , (2.23) we find that

( )

(

)

( )

k _k k k k n k k n n n _B _W _k _B_W X X X k e P ₅2 0 2 5 2 1 8 2 5 3 2 5 3 1 1 2 / 1 1 3 ₋         ₊ + = − − + + + =

_∑

∞ = − π . (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

( )

=

∫

p

Q fDfdx

f

(15)

where f = f

( )

x is a function f :Q_p →R, 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,

e Φ=Φp. (2.26)

Here is the d’Alembert operator and the field Φ 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 p p2 Φ=Φ 1 , (2.27) where 2 2 2 1 1 −...−∂ − ∂ − ∂

= t x x_d , t= x0, is the d’Alembert 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

ϕ

( )

t =Φ

(

t 2lnp

)

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

e ∂t

ϕ

=

ϕ

p

2

2 1

. (2.28)

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

∫

∞ ( )

( )

∞ − − − _d ₌ _t e t

ϕ

τ

ϕ

p

π

τ 2 1 , t∈R. (2.29)

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

ϕ

( )

t such that

ϕ

( )

t ≤Cexp

{

(

1−

ε

)

t2

}

for any ε >0, t∈R. (2.30)

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

( )

=0 −∞ → t t ϕ , limt→∞ϕ

( )

t =1 (2.31) if p is even, and lim

( )

=−1 −∞ → t t ϕ , limt→∞ϕ

( )

t =1 (2.32) if p is odd. Assertion 1

If ϕ is a solution of equation (2.29) such that

( )

t a

t→∞ϕ =

(16)

then a=0 or a=1 if p is even and a=0 or a=±1 if p is odd, lim_t_→_∞

( )

ϕp '

( )

t =0. If a≠0, then lim_t_→_∞

ϕ

'

( )

t =0.

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

( )

[

( )

]

∫

∞

( )

∫

(

)

∞ − ∞ ∞ − − ∞ → − − ∞ → ∞ → ∞ → = = = = − = 2 2 1 lim 1 lim lim lim u t t t p p t p t

ϕ

t

ϕ

t a

π

ϕ

τ

e d

τ

π

ϕ

t u e τ

∫

∞

(

)

∞ − − ∞ → − = = t u e u du a t 2 lim 1

_ϕ

π

, (2.34)

whence a=0 or a=1 if p is even and a=0,±1 if p is odd. Further, we have

( )

∫

( )(

)

( )

∫

∞

(

)

∞ − − ∞ ∞ − →∞ − − ∞ → ∞ → t =− t− e d =− t−uue du= u t t t p t 2 2 1 lim 2 1 lim 2 ' lim

ϕ

π

τ

ϕ

π

ϕ

τ

∫

(

)

∫

∞ ∞ − − − ∞ ∞ − →∞ − =− =− ⋅ = − = 2 lim t u ue u2du 2 a ue u2du 2 a 0 0 t

ϕ

π

. (2.35)

If a≠0, then limt_→_∞

ϕ

'

( )

t =0, since

lim

( )

'

( )

= lim 1

( ) ( )

' = 1lim '

( )

=0

∞ → − − ∞ → ∞ → t p t t pa t t p p t p t

ϕ

. (2.36)

We passed to the limit under the integral sign, using Lebesgue’s theorem and estimate (2.30). We shall write a≡b if the integers a and b are both even or both odd, and a≠b if one of them is even and the other is odd.

Hermite polynomials are defined to be the polynomials

( ) ( )

1 2 _n x2 n x n n e dx d e x H = − − , n=0,1,..., (2.37) whence H₀

( )

x =1, H₁

( )

x =2x,

( )

4 2 2 2 x = x −

H , H₃

( )

x =8x3−12x,.... They form a complete orthogonal system in the Hilbert space L , and 12

∫

∞

( ) ( )

∞ − = = 1 2 ! 2 2 1 H x d x n Hn n

µ

n . (2.38)

Any f ∈L1₂ can be expanded in Hermite polynomials:

( )

∑

(

) ( )

∞ = = 0 1 ! 2 , n n n n n x H H f x f in 1 2 L , (2.39)

and the Parseval-Steklov equality holds:

∑

(

)

∞ = 2 2 1 ,H f f . (2.40)

(17)

The expansion in powers of x has the form

( )

∑

≡ = = n n m m m m n n x n c x H 0 , ! , n=0,1,..., (2.41) In particular, we have ! 2 , n c n n n = ,

( )

! 1 0 , 2 n c n n − = ,

( )

! 1 2 1 , 1 2 n c n n − = + ,

(

( )

)

! 1 1 2 2 , 2 − − − = n c n n ,

( )

(

1

)

! 3 1 4 3 , 1 2 − − − = + n c n n . (2.42)

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

( )

∫

∞

( )

( ) ∞ − − − =

τ

π

H e τ d x Vn n t 2 2 / 2 2 , n=0,1,.... (2.43)

Let f ∈L1₂/2. It follows from (2.28) that f ∈L1₂,

∑

( )

∑

( )

∞ = ∞ = = = 0 2 ! 0 ! n n n n n n n n x V b x f n x H a in 1 2 L . (2.44)

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

( )( )

∫

∞ ( )

( )

∞ − − − ≡ →

ϕ

τ

π

ϕ

τ d e t K 1 t 2 . (2.45) Lemma 1

The operator K assigns to every function f

( )

t satisfying condition (2.30) an entire function

( )( )

Kf z with the estimate

( )( )

            − + ≤ 2 2 1 1 exp y t C z Kf

ε

, z=t+iy. (2.46)

The proof follows immediately from (2.30):

( )( )

_∫

∞ ( ) ( )

_∫

∞ − ∞ ∞ − + − − − − − ₌ ₌ ≤

τ

π

τ

π

τ ετ τ τ ε d e e C d e e C z Kf 1 1 2 t 2 y2 t2 2 2t             − + 2 2 1 1 exp y t C

ε

. (2.47) Lemma 2

The operator K assigns to f ∈Lα2, 0<α <2, an entire function

( )( )

Kf z with the estimate

( )( )

(

)

      − + − ≤ −1/4 2 2 2 exp 2 y t f z Kf α α α α , z=t+iy. (2.48)

(18)

The proof follows from the Cauchy-Bunyakovskii inequality (applied to

( )( )

Kf z ) and the following estimates:

( )( )

( )

≤             ₋ − + − ≤

_∫

∞ ∞ − −

_τ

α

_τ

τ

π

ατ d z z e f z Kf /2 2 2 2 1 2 exp 1 2

(

)

{

}

{

(

)

}

_ =      − − + − ≤       − − + ≤

_∫

∞ ∞ − ∞ ∞ − 2 / 1 2 2 2 2 / 1 2 2 2 4 2 2 exp 1 2 4 2 exp 1

_τ

_α

_τ

π

τ

α

τ

π

α α z t d f y t t d f

(

)

2 / 1 2 2 2 2 2 2 exp 1 2 exp                       − − −       − + =

_∫

∞ ∞ −

α

τ

α

τ

π

α

α d t t y f . (2.49)

We can rewrite the above equation also as follows:

( )( )

( )

=             − − + − ≤

_∫

∞ ∞ − −

_τ

α

_τ

τ

π

f e ατ z z d z Kf /2 2 2 2 1 2 exp 1 2

(

)

2 / 1 2 2 2 2 2 2 exp 1 2 exp                       − − −       − + =

_∫

∞ ∞ −

α

τ

α

τ

π

α

α d t t y f . (2.49b) Lemma 3 The operator K:Lα₂ →Lβ₂, 0<α <2, α α β − > 2 2 , is bounded, and _β

α

_α

β

α

f Kf 4 / 1 2 2 2 2 −       − − ≤ , f ∈Lα₂. (2.50)

We prove the lemma by writing the following chain of equalities and inequalities for f ∈Lα₂:

∫

∞

( )( )

∫

( )

∫

( )

( ) ( ) ∞ − ∞ ∞ − ∞ ∞ − + − − − + − − ₌ _≤ = e Kf t dt e f e d dt Kf t t t 2 2 2 / 1 2 / 2 2 2 2 2 1 2 2

τ

π

β

π

β

β β α τ α τ τ β

∫

( )

∫

∞

( )

∫

( ) ∞ − ∞ ∞ − + − − − ∞ ∞ − + − ₌ ≤ e t f

τ

e d

τ

e t d

τ

dt

π

β

π

β ατ 2 ατ 4τ 2 2 2 2 2 1

∫

∞

∫

( )

(

)

∞ − ∞ ∞ − − − −       − − + − − − = = 2 2 1/2 2 4 2 2 2 2 1 2 2 α τ α α β α

_α

β

αβ

α

β

τ

α

β

π

f e dt e d f t . (2.51)

This equation can be rewritten also as follows:

∫

∞

( )( )

∞ − − = e Kf t dt Kf 2_β βt2 2

π

β

_∫

∞

_∫

( ) ∞ − ∞ ∞ − − −       − − + − =

τ

α

β

π

α ατ β α e dt e d f t2 ₂ 2 2 4 2 2 1 . (2.51b) Lemma 4

(19)

If 1 2

L

f ∈ , then its image

( )( )

Kf t can be expanded in the Taylor series

( )( )

∑

∞ = = 0 ! n n n n t a t Kf , a_n =

(

f,H_n

)

₁, (2.52) which converges uniformly on every compact set in R . If 1/2

2 L f ∈ , then

( )( )

∑

( )

∞ = = 0 2 ! n n n n n t H b t Kf in 1 2 L , b_n =

(

Kf,H_n

)

₁, (2.53) and

(

Kf,H_n

) (

₁= f,V_n

)

₁_/₂, n=0,1,.... (2.54)

By lemma 3, the function

( )( )

Kf t is the trace of an entire function

( )( )

Kf z for y=0. Hence, it can be expanded in the Taylor series with the coefficients

( )( )

∫

∞

( )

∫

( )( )

( )

(

)

∞ − ∞ ∞ − = − − = − − =0 = 0 = − − 0= 2 2 1 1 1 t n t n t t n n t n n d t H e f d e dt d f t Kf dt d

_τ

π

τ

π

τ τ

( )( ) ( )

(

n

)

n t n n a H f d e H f − − = = =

_∫

∞ ∞ − − 1 , 1 1 2

τ

π

. (2.55)

Here we used equality (2.37). Further, if f ∈L12/2, then (2.53) holds by (2.44), since

1 2

L Kf ∈ by Lemma 3. Equalities (2.54) can be proved as follows:

(

Kf,H_n

)

₁ =

(

f,K∗H_n

)

₁_/₂ =

(

f,V_n

)

₁_/₂. (2.56)

Here we used formula (2.43), which implies that V_n =K∗H_n, where K is the operator adjoint to ∗ K .

Let ϕ be a solution of equation (2.29) belonging to 1 2

L , whence ϕp =Kϕ

. Putting a_n =

(

ϕ,H_n

)

₁, we deduce from (2.39) and (2.40) that

( )

∑

( )

∞ = = 0 2 ! n n n n n t H a t ϕ in 1 2 L ,

∑

∞ = = 0 2 2 ! 2 n n n n a

_ϕ

. (2.57)

The function ϕp

( )

t is the trace of the entire function A

( ) ( )( )

z = K

ϕ

z , for which (2.48) holds with 1 = α :

( )

2 1 z e z A ≤

ϕ

, z∈C. (2.58) By Lemma 4, it can be expanded in the Taylor series (2.52):

( )

∑

∞ = = 0 ! n n n p n t a t

ϕ

. (2.59)

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

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u_x u_tt 4 1

= , 0<x≤1, t∈R, (2.60) u

( ) ( )

0,t =

ϕ

t , u

( )

1,t =

ϕ

p

( )

t , t∈R. (2.61)

Let us note that if there is an interpolating function, it can be represented by Poisson’s formula for equation (2.60):

( )

∫

∞

( )

(

)

∞ − _     ₋ − =

ϕ

τ

π

x d t x t x u 2 exp 1 , , 0<x≤1. (2.62) If ϕ such that

ϕ

( )

t ≤Cexp t

{ }

ε

2 for any ε >0, t∈R, (2.63)

then formula (2.62) gives its analytic continuation to the domain x>1, t∈R and, further, its analytic continuation with respect to

( )

x,t to the complex domain T+×C, where T+is the right half-plane Reζ =x>0.

Equation (2.29) takes the form

( )

∫

∞

( )

( ) ∞ − − − = ϕτ τ π ϕ τ d e t t q 1 2 2 , q=1,2,.... (2.64)

If

ϕ

( )

t is a solution of equation (2.64), then

ϕ

( )

−t and ϕ

(

t+t₀

)

also are solutions of this equation (for all t ). ₀

Assertion 2

If

ϕ

( )

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

∫

∞

( )

(

)

(

)

∞ − − − − − − − + ≤       ₋ − 1 2 1 2 1 exp 1 2 4 1 1 2 2 2 x qx q x x d x t x q q q q τ τ τ ϕ π (2.65) for all x>1/

(

2q−1

)

.

We remember that if S is a measurable subset of Rn with the Lebesgue measure, and ƒ and g are measurable real- or complex-valued functions on S, then Hölder inequality is

( ) ( )

(

( )

)

(

( )

)

q S q p S p S f x g x dx f x dx g x dx / 1 / 1

∫

≤ . (2.65b)

Denoting the left-hand side of inequality (2.65) by J

( )

x,t , using the boundary conditions (2.61), the properties of solutions of the heat equation and Holder’s inequality, we obtain the following chain of relations for all x>1/

(

2q−1

)

:

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( )

(

)

(

)

( )

(

)

∫

₋∞_∞

∫

₋∞_∞ ≤       + − − + =       ₋ − ≡

ϕ

τ

π

τ

ϕ

π

x d t x d x t x t x J q 1 exp 1 1 exp 1 , 2 2 2

(

)

( )

(

)

(

) (

)

 =     + − − − −       ₋ − + ≤

_∫

∞ ∞ −

τ

ϕ

π

x qx d x qx t qx t x 1 2 1 2 exp 2 exp 1 1 2 2

(

)

[ ]

(

) (

)(

)

q q _d q x x x qx t J t x 2 1 1 2 2 1 1 2 1 1 2 exp 1 1 − ∞ ∞ −               − + − − − − + =

π

_∫

τ

π

, (2.66) whence

( )

(

)

(

)(

)

q q q x qx q x x x x x J 2 1 1 4 1 2 1 1 1 2 1 2 1 1 − −         − − − + + ≤

π

, (2.67)

which implies that (2.65) holds.

Now we can rewrite the eq. (2.65) also as follows:

( )

(

)

(

)(

)

 ⇒       − − − + + ≤ − − _q q q x qx q x x x x x J 2 1 1 4 1 2 1 1 1 2 1 2 1 1

π

∫

∞

( )

(

)

(

)

∞ − − − − − − − + ≤       ₋ − ⇒ 1 2 1 2 1 exp 1 2 4 1 1 2 2 2 x qx q x x d x t x q q q q

τ

ϕ

π

. (2.67b) Corollary

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

∫

∞

( )

{

(

)

}

∞ − − − − − ≤ − − 2 2 1 2 2 exp 1 4 2 1 2 2 q q d t q q

τ

ϕ

π

. (2.68)

With regard the possible mathematical connections, we note that it is possible to obtain some interesting and new relationships evidencing some Ramanujan’s 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:

(

)

( )

(

)

∫

∞ + − Γ       + Γ Γ       ₋ ₊ Γ + Γ       ₊ Γ =       + +       + + ⋅       +       + + 0 2 2 2 2 1 2 1 2 1 1 2 1 2 ... 1 1 2 1 1 1 1 a b b a a b b a dx a x b x a x b x

π

, (2.69)

∫

∞

(

)(

)

(

)

+ + + + + = + + + 0 2 2 2 4 2 ₂₁ 3 6 10 _... 1 ... 1 1 1 1 r r r r dx x r x r x

π

, (2.70)

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If

αβ

=

π

2 , then _       − + =         − +

_∫

∞ − −

_∫

∞ − − 0 2 4 / 1 0 2 4 / 1 1 4 1 1 4 1 2 2 dx e xe dx e xe x x x x π β π α

β

α

, (2.71) 2 /5 4 2 2 1 5 2 5 5 ... 1 1 1 1 π π π e e e         ₊ −         ₊ = + + + − − , (2.72) 2 / 5 5 2 / 5 4 / 3 5 4 5 2 2 1 5 1 2 1 5 5 1 5 ... 1 1 1 1 π π π e e e                   + −         −         ₋ + = + + + − − . (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:

( )( )

( )

=             ₋ − + − ≤

_∫

∞ ∞ − −

_τ

α

_τ

τ

π

ατ d z z e f z Kf /2 2 2 2 1 2 exp 1 2

(

)

2 / 1 2 2 2 2 2 2 exp 1 2 exp                       − − −       − + =

_∫

∞ ∞ −

α

τ

α

τ

π

α

α d t t y f

(

)

( )

(

)

∫

π

, (2.74)

( )( )

( )

=             − − + − ≤

_∫

∞ ∞ − −

_τ

α

_τ

τ

π

ατ d z z e f z Kf /2 2 2 2 1 2 exp 1 2

(

)

2 / 1 2 2 2 2 2 2 exp 1 2 exp                       − − −       − + =

_∫

∞ ∞ −

α

τ

α

τ

π

α

α d t t y f

(

)(

)

(

)

∫

∞ = ₊ ₊ ₊ ₊ ₊ + + + 0 2 2 2 4 2 ₂₁ 3 6 10 _... 1 ... 1 1 1 1 r r r r dx x r x r x

π

, (2.75)

∫

∞

( )( )

∞ − − = e Kf t dt Kf 2_β βt2 2

π

β

∫

( ) ∞ ∞ − ∞ ∞ − − −       − − + − =

τ

α

β

π

α ατ β α e dt e d f t2 ₂ 2 2 4 2 2 1

(

)

( )

(

)

∫

∞ + − Γ     + Γ Γ       ₋ ₊ Γ + Γ       ₊ Γ =     +       + + ⋅     +       + + 0 2 2 2 2 1 1 2 1 1 2 1 2 ... 1 2 1 1 1 1 a b b a a b b a dx x b x x b x

π

, (2.76)

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∫

∞

( )( )

∞ − − = e Kf t dt Kf 2_β βt2 2

π

β

_∫

∞

_∫

( ) ∞ − ∞ ∞ − − −       − − + − =

τ

α

β

π

α ατ β α e dt e d f t2 ₂ 2 2 4 2 2 1

(

)(

)

(

)

∫

∞ = ₊ ₊ ₊ ₊ ₊ + + + 0 2 2 2 4 2 ₂₁ 3 6 10 _... 1 ... 1 1 1 1 r r r r dx x r x r x

π

, (2.77)

∫

∞

( )

{

(

)

}

∞ − − − − − ≤ − − 2 2 1 2 2 exp 1 4 2 1 2 2 q q d t q q

τ

ϕ

π

(

)

( )

(

)

∫

π

, (2.78)

∫

∞

( )

{

(

)

}

∞ − − − − − ≤ − − 2 2 1 2 2 exp 1 4 2 1 2 2 q q d t q q

τ

ϕ

π

(

)(

)

(

)

∫

∞ = ₊ ₊ ₊ ₊ ₊ + + + 0 2 2 2 4 2 ₂₁ 3 6 10 _... 1 ... 1 1 1 1 r r r r dx x r x r x

π

. (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 ϕ which describes open p-adic string tachyon is       + + − − = −2 +1 2 2 1 1 2 1 1 1 p p p p p p g

ϕ

L _{, (3.1)}

where p is any prime number, =−∂2_t +∇2 is the D-dimensional d’Alambertian and we adopt metric with signature

(

−+...+

)

. Now, we want to show a model which incorporates the p-adic string Lagrangians in a restricted adelic way. Let us take the following Lagrangian