 Open Access
 Total Downloads : 354
 Authors : R. Suryanarayana, Ch. Gopala Rao
 Paper ID : IJERTV1IS9511
 Volume & Issue : Volume 01, Issue 09 (November 2012)
 Published (First Online): 29112012
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
Some New Inequalities for the Generalized – Gamma, Beta and Zeta Functions
R. Suryanarayana And Ch.Gopala Rao
Dept. of Mathematics, GMR Institute of Technology,Rajam532127, 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.

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+(n1) ),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 + / +

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+n1) = ,1.
In[2]it was proved that

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)

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

R.Diaz and E.Pariguan, Quantum symmetric functions
holds
2 11/ 1/ 2
< 1 <
1
11/ 1
1/ 2 1
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M. Mansour, Determining the kgeneralized Gamma
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1
1

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