# Some New Inequalities for the Generalized ðœ– – Gamma, Beta and Zeta Functions

DOI : 10.17577/IJERTV1IS9511

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#### Some New Inequalities for the Generalized – Gamma, Beta and Zeta Functions

R. Suryanarayana And Ch.Gopala Rao

Dept. of Mathematics, GMR Institute of Technology,Rajam-532127, Srikakulam, A.P, India

Abstract

In this paper, we establish some properties and inequalities for the – generalized functions which are – Gamma function, Beta function and Zeta function and has

, = , Re(x )> 0,Re (y) > 0 (1.5)

+

And -Zeta function as

=0 +

, = 1 , , > 0, s > 1 (1.6) The function , satisfies the equality

given some identities which they satisfy. This inequality

leads to new inequalities involving the Beta, Gamma and Zeta functions and a large family of functions. The gamma

, = 1 1 1

0

which follows

1 1

(1.7)

and Beta functions belong to the category of the special

, = 1 , . (1.8)

transcendental functions and we will see that some mathematical constants are occurring in its study.

We mention that lim , (, ) and – Zeta

function is a generalization of Hurwitz Zeta function

, = 1 which is a generalization of the

Keywords: – generalized Gamma function , Beta function and – Zeta function.

1. Introduction

The generalized – Gamma function as

=0 +

Reimann Zeta function

=1

= 1 . The motivation to study properties of

generalized Gamma and – Beta functions is the fact

=lim

! 1

,

,k>0,x

(1.1)

that ,

appears in the combinatorics of creation and

where ,

is the – Pochammer symbol and is given by

annihilation operators[3]. Recently M. Mansour [4]

,

(1.2)

=x(x+ )(x+2 )…(x+(n-1) ),x , ,n +

determined the generalized Gamma function by a combination of some functional equations.

In this paper , we use the definitions of the above

It is obvious that 1, where is known as Gamma function. Also for Re (x) > 0, it holds

0

= 1 (1.3)

And it follows that

1

generalized functions to prove a formula for 2 which is a generalization of the Legendre duplication formula for and to prove inequalities for the function , , for x,y, > 0 and x + y and the

product 1 , for 0 < x, < 1.we also give

= . (1.4)

In this paper [1],[2],[3] introduced the – Beta function

, as

monotonicity properties for = , where

, = and , for s 2.

We mention that using (1.4) the following inequalities hold:

= 1 ,>0,a (1.9)

= 1( 1)!, >0,n (1.10) = 1, >0, (1.11)

2 1 2 !

holds. It is known [1] that () is completely monotonic for x > 0, so from (2.4) it follows the desired result.

Remark 2.1. (i) From (2.3) it follows that is

logarithmic convex on (0, ) which is proved in [2], (ii) Theorem 2.1 is a generalization of the known [1] result

( 2 + 1 2)= 2 2 ! , >0,n (1.12)

also , using (1.5) and (1.8) the following equalities hold:

+

+ , = , , , +

that the function () is completely monotonic.

Result 2.1. For x > 0 the function , = log

satisfies the differential equation

=

+

, x,y, > 0. (1.13)

22

, + 2

2 , +

3 , =

, = 1 and , = 1, x ,y, > 0. (1.14)

(2.5)

, = 1 , , a,b, > 0 (1.15)

Proof: From (2.1) taking the first and second derivative of

, with respect to , we obtain

, = 1 1 !2 , > 0, n (1.16)

1 1

21 !

, =

2 + 2 / (2.6)

2 , = 2 3 + 1 + / +

2. The Function

3

3

2

2

(/) (2.7)

Theorem2.1: let x, > 0 and be the logarithmic derivative of . Then the function () is completely monotonic.

Proof: From (1.4) , we get

3

From (2.3), (2.6) and (2.7), we get (2.5)

Theorem2.2: The function satisfies the equality

2 1

2 = 2 + /2 (2.8)

Log = 1 + log / or by setting

, = Log , we obtain

For x with Re(x) > 0.

Proof: From (1.7) it follows that

, = 1 + log / (2.1)

1 1 1

1

, = 0

1

We get , , = 1 + / (2.2)

1+

2 1 2

1

Or by setting t =

2 , , =

22 1

0 1

We remind that / = . from (2.2)

taking the derivative with respect to x, we have

or by setting 2 = ,

1 1 11

1

we obtain , =

22 1

0 2

1

=

2 = 1 (/) (2.3)

3 , = 1 (/)

2

1

22 1

, =

, Or

1

22 1 2

, = 1

2

By induction, we obtain +1 , = 1 (/)

22 1

(2.9)

+

2

Or if we call = , , then the equation

from (1.9) for a=1/2 , we get = , since 1 =

= 1 (/) (2.4)

2 2

, from (2.9) and (1.5) , we get the equality (2.8).

Remark 2.3. Theorem 2.2 is a generalization of the legendre duplication formula of (x).

Repeating the same , we get (3.4) ,

since s (s+1) .(s+n-1) = ,1.

In[2]it was proved that

3. The Function ,

Theorem3.1 (i) Let x, > 0 and s > 1. Then the positive

2 , = 1

=0 + 2

From (1.6) for s + 2 and (3.7), we get

(3.7)

function x, s decreases with respect to x and also

2

, = , 2 (3.8)

decreases with respect to . (ii) let x > 0 and s > 1. Then the positive function x, s decreases with respect to s for x > 1 and > 0, 0 and increases with respect to s for > 0, 0 < < 1/ and 0 < x < 1 – .

Proof: From (1.6) we obtain

Differentiating (3.7) with respect to x and using (3.1) for s = 2, we get

3 , = 1 2 2 , 3

and 4 , = 1 23! , 4

By induction, we obtain (3.5). The equation (3.6) follows

, =

1 , , > 0, s >1

from the definition(1.6), since

=0 + +1

Or

, = 1

+ 1

= 1 + + , .

, = – , + 1 (3.1)

+ +

=0

, =

= -s

, > 0,

4. Inequalities For , ,

=0 + +1

=1 + +1

s >1 (3.2)

Then (3.1) and (3.2) , prove the theorem 3.1(i)also the

Theorem 4.1: le x,y, > 0 and x + y . Then the function B x, y satisfies the inequalitie

definition(1.6) gives

+

2

2

+

< B x, y < 122

(4.1)

, =

ln (+ )

(3.3)

+

+

=0 +

Lemma 4.1: The function B(x,y) satisfies the

If x >1 then x > 1- , for , > 0 thus ln(x + ) > 0 so

from (3.3) it follows that the function , decreases

inequalities

22( + ) < B(x, y) < 122( + ) , x,y > 0, x+y 1 (4.2)

with s > 1 and if 0 < < 1/ and 0 < x < 1- then ln(x

+ 1

+1

+ ) < 0 from (3.3) it follows that the function ,

increases with s > 1.

Result 3.1: Let x > 0, > 0 and s > 1. Then the function

, satisfies the identities:

Proof: The function B(x,y) is defined [1] by the integral B(x,y) = 1 1 1 1

0

Which can be written as

, = 1

, + (3.4)

B(x,y)= 1/2 1 1 1 + 1

1 1 1

,1

0 1/2

, = 1 , , 2 (3.5)

(4.3)

1 !

And + , = , 1/ (3.6)

If 0 < t < Â½ then t < 1- t , so that the following inequalities

1 1

Proof: From (3.1) we obtain

2 , = , + 1 = 1 2 + 1 , + 2

hold 2 +2 < 2 1 1 1 <

0

0 0

1/2 1 +2 (4.4)

1/2

and if Â½ < t < 1 then 1 t < t, so that the following inequalities hold

(4.8)

Proof: By setting y = + 1 x instead of y in (4.1) we

1

1/2

1 +2 < 1

1

1 1 1 <

+2

obtain

211/ < B x, + 1 < 1 211/ (4.9)

1/2

(4.5)

Using (1.5) the inequalities (4.9) become

From (4.3), using the inequalities (4.4) and (4.5) and evaluating the integrals on the left and right side of the above inequalities , we obtain the inequalities (4.2) .

Proof of theorem 4.1:By setting x / and y/ , instead of x

211/ < +1 < 1 211/ (4.10)

+1

From (1.4) we obtain

+ 1 = (1 ) 1

11

and + 1 = 1 = 1/ .

and y respectively in (4.2) and taking in account the relation (1.8) we get the inequalities (4.1).

Corollary 4.1: Let x,y, > 0. Then the function B x, y

satisfies the inequalities:

From (4.10) using the above equalities we obtain the inequalities (4.8).

References

+

21

+

< B x, y < 121

(4.6)

1. M.Abramowitz, I.A. Stegun (Eds) Hand book of

1 +

Or 2

1 +

< B x, y < 12

(4.7)

mathematical functions with formulae and mathematical tables, Applied Math. Series, Vol.55,4th edition with

Proof: The above inequalities follow from (4.1) by setting x + (or y + ) instead of x (or y) and taking in account relations (1.13).

Corollary 4.2: Let 0 < x < 1 and 0 < < 1. Then the

following inequalities for the product 1

corrections. Nat. Ber. Of Standards, Washington(1965).

2. R.Diaz and E.Pariguan , On hypergeometric functions and k- Pochhammer symbol, arXiv:math. CA/0405596v2(2005).

3. R.Diaz and E.Pariguan, Quantum symmetric functions

holds

2 11/ 1/ 2

< 1 <

1

11/ 1

1/ 2 1

,arXiv : math. QA/0312494(2003)

4. M. Mansour, Determining the k-generalized Gamma

function by functional equations , Int. J. Contemp. Math. Sciences, 4, (2009), 1037-1042.

1

1

5. M. Abramowitz and I. tegun, Handbook of Mathematical Functions, Dover, New York, (1964) .