A New Proof of an Inequality for the Logarithm of the Gamma Function and Its Sharpness

Mansour Mahmoud

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A New Proof of an Inequality for the Logarithm of the Gamma Function and Its Sharpness

Mansour Mahmoud1,2,

1Department of Mathematics, Faculty of Science, King Abdulaziz University, P. O. Box 80203, Jeddah 21589, Saudi Arabia
2Department of Mathematics, Faculty of Science, Mansoura University, ansoura 35516, Egypt

Abstract

In the paper, the author shows that the partial sums are alternatively larger and smaller than the generalized Euler’s harmonic numbers with sharp bounds, where γ is the Euler's constant, are the Bernoulli numbers and ψ is the digamma function.

Cite this article:

  • Mahmoud, Mansour. "A New Proof of an Inequality for the Logarithm of the Gamma Function and Its Sharpness." Turkish Journal of Analysis and Number Theory 2.4 (2014): 147-151.
  • Mahmoud, M. (2014). A New Proof of an Inequality for the Logarithm of the Gamma Function and Its Sharpness. Turkish Journal of Analysis and Number Theory, 2(4), 147-151.
  • Mahmoud, Mansour. "A New Proof of an Inequality for the Logarithm of the Gamma Function and Its Sharpness." Turkish Journal of Analysis and Number Theory 2, no. 4 (2014): 147-151.

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1. Introduction

Euler, in his "Institutiones calculi differentialis" [10], introduced the concept of inexplicable functions. These functions were appeared originally as functions in the positive integers of one symbol or more. He presented the following examples of the inexplicable functions:

where there is no necessity for x to be an integer. The first function generalizes the factorial function and the second one generalizes the harmonic numbers

(1)

which are partial sums of the harmonic series. The function can be defined by the definite integral

(2)

We will call the function , Euler's generalized Harmonic numbers. The recurrence relation of is given by

(3)

and its reflection relation is

The multiplication formula is given by

The function is related to the Euler's constant γ by the relation

and its relation with the digamma function (the logarithmic derivative of the gamma function) is

(4)

There are many other generalizations of the harmonic numbers all of them depend only on the positive integers see [3,7,8,11,12,13,17,30,31]. Also, there are many estimations of the harmonic numbers see [4,5,6,9,18,22,25,26,27,28,32,33].

A function f is said to be completely monotonic on an interval J if

(5)

If the inequality (5) is strict and all then f is said to be strictly completely monotonic on the interval J.

In [1], Alzer proved that the functions

and

For are completely monotonic on (0,∞). This means that the functions and are also are completely monotonic on (0,∞) see [15]. These complete monotonicity have been repeated in [16] and [23]. These results can also be found in the survey [29]. From the complete monotonicity of and , we get the following double inequality

(6)

In this paper, we will present a new proof of the double inequality (6), using Artin's technique [2] and we will use a method due to Mortici [24] to prove that the bounds in (6) are the best possible. Also, we will provide new proof of the complete monotonicity of the two Functions and .

2. Main Results

Theorem 1.

For x>0,

(7)

with sharp bounds.

Proof

Consider the function

(8)

Then using the recurrence relation (3), we get

If we define the following function

then

We can express the function by the integral representation

and hence

By considering the known discontinuous function

(9)

we get

The oscillating of the function and provided us by the existence of the integration

(10)

In [2], Artin introduced the functions

(11)
(12)

Where is the Fourier series [20, 21] of the function . He showed that the series is absolutely and uniformly convergent for all v with n=2,3,… and the series is uniformly convergent in every closed interval has no integer for n=1. Also,

(13)

for all v when n=2,3,… and for nonintegral v when n=1.

Now by repeated the integrations by parts of (10), we obtain

(14)

But

(15)

If n is even, the has the same sign as for all v and thus the integral (15) has this sign. Also,

(16)

Then the signs of alternate between minus and plus. Then the function lies between any two successive partial sums

(17)

In other words, for every n, there exists a number satisfies

(18)

and hence

(19)

Now, we will prove the sharpness of the bounds in (19). By the definition of the asymptotic expansion [14], the expansion of a function of the form

satisfies for every fixed k, that

Then

(20)

If we have other constants have the property that for all

etc. These inequalities give us that

(21)

the relations (20) and (21), gives us that

(22)

Then the choice of the constants in the inequality (7) is the best one. To complete our results, we need to prove that the constant 1/2 in the the function can not be improved by any method whatsoever. Consider the function

then

Now, let

then

The The function will be increasing if

and the function is increasing function with . So, the best choice of A is 1/2. Also, the function is increasing with limit tends to zero as x→∞, then

Hence

As n→∞, we get

with sharp bound. Now, consider the function

then

Let

then

The function will be increasing if

and the function is increasing function with . So, the best choice of B is 1/12. Also, the function is increasing with limit tends to zero as , then

Hence

As , we get

Hence

with sharp bounds.

As a special case we get the following result [19]

Corollary 2.1.

For any natural number ,

(23)

with sharp bounds.

Now, we will present a new proof for the complete monotonicity of the functions and .

Lemma 2.2

For m=0,1,2,…, the functions

and

are completely monotonic on (0,∞).

Proof

Using the relations (15), (17) and (16) at n=4m+2, we get

and hence

(24)

But for m=0,1,2,3,…, the has the same sign as for all v and thus the integral (24) has this sign. Using the relation (11), we obtain

and hence

But

then

Similarly, we can prove the complete monotonicity for by replacing n by 4m+4.

Remark 1.

Series (17) is divergent so we can not take the limit as n tends to ∞ but the relation (18) provided us by approximations of the function or any finite order. For example, if n=5 we obtain the following approximation:

Remark 2.

The formula (10) implies the the function is completely monotonic on (0,∞), that is

Acknowledgement

The author is very grateful to Prof. Feng Qi at Tianjin Polytechnic University in China for valuable comments and useful suggestions, which greatly helped improve the presentation and the quality of the paper.

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