Scholarly article on topic 'A Double Inequality for Gamma Function'

A Double Inequality for Gamma Function Academic research paper on "Mathematics"

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Academic research paper on topic "A Double Inequality for Gamma Function"

Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2009, Article ID 503782,7 pages doi:10.1155/2009/503782

Research Article

A Double Inequality for Gamma Function

Xiaoming Zhang1 and Yuming Chu2

1 Haining Radio and TV University, Haining 314400, Zhejiang, China

2 Department of Mathematics, Huzhou Teachers College, Huzhou 313000, Zhejiang, China

Correspondence should be addressed to Yuming Chu, chuyuming2005@yahoo.com.cn Received 12 June 2009; Revised 21 August 2009; Accepted 30 August 2009 Recommended by Ramm Mohapatra

Using the Alzer integral inequality and the elementary properties of the gamma function, a double inequality for gamma function is established, which is an improvement of Merkle's inequality.

Copyright © 2009 X. Zhang and Y. Chu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

1. Introduction

For real and positive values of x, the Euler gamma function r and its logarithmic derivative f, the so-called psi function, are defined by

respectively. For extensions of these functions to complex variables and for basic properties, see [1].

Recently, the gamma function has been the subject of intensive research, many remarkable inequalities for r can be found in literature [2-21]. In particular, the ratio (r(s)/r(r))(s > r > 0) have attracted the attention of many mathematicians and physicists. Gautschi [22] first proved that

i s r(n +1) .

n < r(„ + c) < exp [(1 - s)v(n +1)]

for 0 < s < 1 and n = 1,2,3.

Journal of Inequalities and Applications A strengthened upper bound was given by Erber [23]:

r(n + 1) < 4(n + s)(n + 1)

r(n + s) 4n + (s + 1)2 In [24], KeCkiC and VasiC established the following double inequality for b > a > 0:

1 b r(b) bb-(1/2) b

-e < -b < b „,^ea-b.

aa-1 r(a) aa-(1/2)

In [25], Kershaw obtained

exp [(1 - s)f(^x + s1/2)j

r(x +1) < F(XXTs) <exp

(1 - s)y(x + 2 (s + 1)

x + - s 2

r(x + 1) < —-- <

r(x + s)

x -1+(s+4)

for x > 0 and 0 < s < 1.

The generalized logarithmic mean Lp(a, b) of order p of two positive numbers a and b with a = b is defined by

L„(a, b) =

bp+i - ap+i

(p + 1)(b - a) b - a log b - log a'

1/ (b-a)

, p= - 1, p / 0, p = -1,

p = 0.

It is well known that Lp(a, b) is strictly increasing with respect to p for fixed a and b.

If we denote A(a,b) = L1 (a,b) = (a + b)/2,1(a,b) = Lc(a, b) = (1/e)(bb/aa)1/(b-a} ,L(a,b) = L-1(a,b) = (b - a)/(log b - log a), and G(a,b) = L-2(a,b) = Vab the arithmetic mean, identric mean, logarithmic mean, and geometric mean of a and b with a =/ b, respectively, then

min{a,b} < G(a,b) < L(a,b) < I(a,b) < A(a,b) < max{a,b}. (1.7)

In 1996, Merkle [26] established

A(f (a),f(b)) < logmb - 0»r(a) (A(a, b))

for a,b > 0 with a = b.

It is the aim of this paper to present the new upper and lower bounds of inequality (1.8) in terms of I and L.

2. Lemmas

In order to establish our main result we need several lemmas, which we present in this section.

Lemma 2.1 (see [27, page 2670]). If x > 0, then

f(x) > x + 2X2 ■ (11)

Lemma 2.2 (see [28]). Let f e C[a,b] be a strictly increasing function. If 1/f 1 is strictly convex (or concave, resp.), then

b - a J a

[ f (t)dt > (or <,resp)f (L(a,b))■ (2.2)

Here, f 1 is the inverse of f. Lemma 2.3. If x > 0, then

0 < 2f'(x)+ xy"(x) <x■ (2.3)

Proof. It is well known that logr(x) = -jx + ^O=i [x/k - log(1 + (x/k))] - logx, where j = 0577 215 ■■■ is the Euler constant. Then, we have

V'(x) = X --3 (2.4)

k=0 (k + x)

From (2.4) and (2.5), we get

2w'(x) + xw"(x) = Y,-5- > 0,

^ W ^ X k=1 (k + x)3

y 2k 2f '(x) + xf"(x) = y

k=1 (k + x)3

< k=1 (k - 1 + x)(k + x)(k + 1 + x)

(k - 1 + x)(k + x) (k + x)(k + 1 + x) 1

k=1 (k - 1 + x)(k + x)

= y /J___L_

k=1\k - 1 + x k + x

Lemma 2.4. Suppose that b > a > 0 and f : [a, b] ^ R is a twice differentiable function. If f'(x) > 0 and 2f '(x) + xf"(x) > (or <, resp.) 0 for x e [a, b], then there exists the inverse function f-1 of f and 1/f-1 is strictly convex (or concave, resp.).

Proof. The existence of f-1 can be derived from f '(x) > 0 directly. Next, let y = f (x), then simple computation yields

f '(x)( f -1(y))' = 1, f "(x)[(f-1 (y))']2 + f '(x)(f -1(y))" = 0,

2[(f-1(y))']2 (f-1(y))''

J-1^)/ (/-1(y))3 (f-1(y)Y

From (2.7) and x = f ~l(y), we get

"_ 2/' (x) + xf "(x)

f-1( y)) x3J <(x))

Therefore, the strict convexity (or concavity, resp.) of 1/f 1 follows from (2.8) and the assumed condition 2f'(x) + xf"(x) > (or <, resp.) 0. □

3. Main Result

Theorem 3.1. For all a,b> 0 with a = b, one has

f^ < ^^ <fLiaMU .o8 ™ <3.1)

Proof. Without loss of generality, we assume that b > a > 0. From <2.4) and Lemma 2.3, together with Lemma 2.4, we clearly see that f is strictly increasing and 1/f-1 is strictly convex on [a,b]. Then, Lemma 2.2 leads to

(f <t)dt > f<L<a, b)). <3.2)

Therefore, the left-side inequality in <3.1) follows from <3.2).

Next, for x e [a,b], let g<x) = f<x) - logx. Then, Lemmas 2.1 and 2.3 lead to

g'<x) = f'<x) - x > 2x2 > 0, <3.3)

2g'<x) + xg"<x) = 2f'<x) + xf"<x) - x< 0. <3.4)

From <3.3) and <3.4), together with Lemma 2.4, we clearly see that g<x) is strictly increasing and 1/g-1 is strictly concave on [a,b]. Then, Lemma 2.2 implies

[ (f<t) - log t)dt < f<L<a, b)) - logL<a,b). <3.5)

Therefore, the right-side inequality in <3.1) follows from <3.5).

To compare the bounds in Theorem 3.1 with that in <1.8), we have the following two remarks. □

Remark 3.2. The lower bound in Theorem 3.1 is greater than that in <1.8), that is, f<L<a, b)) > A<f<a),f<b)) for a,b > 0 with a / b. In fact, for any b > a > 0 and x e [a,b], Lemmas 2.1 and 2.3 lead to

f'<x) + xf"<x) < -< 0. <3.6)

From <3.6) and [29], we know that f<x) is a strictly geometric-arithmetic concave function on [a,b], hence, we get

f <G<a,b)) > A(f<a),f<b)). <3.7)

Since f is strictly increasing and G<a, b) < L<a, b), so we have

f <L<a,b)) > f <G<a,b)).

Inequalities (3.7) and (3.8) show that f(L(a, b)) > A(f(a), f(b)) for a,b> 0 with a f b.

Remark 3.3. The upper bound in Theorem 3.1 is less than that in (1.8), that is, f (L(a,b)) + log I (a,b) - log L(a,b) < f (A(a,b)). In fact, for any b > a > 0 and x e [a,b], (3.3) and L(a,b) < I(a,b) imply

f(L(a,b)) - log L(a,b) <f (I(a,b)) - log I (a,b). (3.9)

On the other hand, the monotonicity of f and I (a, b) < A(a, b) leads to

f (I(a, b)) <f (A(a,b)). (3.10)

From (3.9) and (3.10), we get

f (L(a,b))+ log I (a, b) - log L(a,b) <f (A(a,b)). (3.11)

Acknowledgments

The authors wish to thank the anonymous referee for the very careful reading of the manuscript and fruitful comments and suggestions. This research is partly supported by N S Foundation of China under Grants 60850005 and 10771195, and N S Foundation of Zhejiang Province under Grants D7080080 and Y607128.

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