COMPLETE MONOTONICITY OF A DIFFERENCE BETWEEN THE EXPONENTIAL AND TRIGAMMA FUNCTIONS
COMPLETE MONOTONICITY OF A DIFFERENCE BETWEEN THE EXPONENTIAL AND TRIGAMMA FUNCTIONS
The Pure and Applied Mathematics. 2014. May, 21(2): 141-145
• Received : April 06, 2014
• Accepted : May 12, 2014
• Published : May 31, 2014 PDF e-PUB PubReader PPT Export by style
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FENG, QI
COLLEGE OF MATHEMATICS, INNER MONGOLIA UNIVERSITY FOR NATIONALITIES, TONGLIAO CITY, INNER MONGOLIA AUTONOMOUS REGION, 028043, CHINAEmail address:qifeng618@gmail.com,qifeng618@hotmail.com,qifeng618@qq.comURL: http://qifeng618.wordpress.com
XIAO-JANG, ZHANG
THE 59TH MIDDLE SCHOOL, JIANXI DISTRICT, LUOYANG CITY, HENAN PROVINCE, 471000, CHINAEmail address:xiao.jing.zhang@qq.com

Abstract
In the paper, by directly verifying an inequality which gives a lower bound for the first order modified Bessel function of the first kind, the authors supply a new proof for the complete monotonicity of a difference between the exponential function e 1/t and the trigamma function 𝜓′( t ) on (0, ∞).
Keywords
1. INTRODUCTION
In [3, Lemma 2] , the inequality PPT Slide
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on (0, ∞) was discovered and employed, where 𝜓( t ) denotes the digamma function PPT Slide
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and Γ is the classical Euler gamma function which may be defined for ℜ( z ) > 0 by PPT Slide
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The functions 𝜓′( z ) and 𝜓′′( z ) are respectively called the trigamma function and the tetragamma function. As a whole, the derivatives 𝜓 (k) ( z ) for k ∈ {0} ∪ PPT Slide
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are called polygamma functions.
An infinitely differentiable function f defined on an interval I is said to be a completely monotonic function on I if it satisfies PPT Slide
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for all k ∈ {0} ∪ PPT Slide
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on I . Some properties of the completely monotonic functions can be found in, for example, [2 , 8] .
In [5, Theorem 3.1] and [6, Theorem 1.1] , the following theorem was proved by three methods totally.
Theorem 1.1. The function PPT Slide
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is completely monotonic on (0, ∞) and PPT Slide
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The second main result of the paper  is [6, Theorem 1.2] which has been referenced in [4, Section 1.2] and [5, Lemma 2.1] as follows.
Theorem 1.2. For k ∈ {0} ∪ PPT Slide
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and z ≠ 0, let PPT Slide
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For ℜ( z ) > 0, the function Hk ( z ) has the integral representations PPT Slide
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and PPT Slide
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where the hypergeometric series PPT Slide
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for bi ∉ {0, −1, −2, ... }, the shifted factorial ( a ) 0 = 1 and PPT Slide
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for n > 0 and any real or complex number a, and the modified Bessel function of the first kind PPT Slide
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for ν PPT Slide
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and z PPT Slide
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.
When k = 0, the integral representations (1.6) and (1.7) may be written as PPT Slide
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and PPT Slide
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for ℜ( z ) > 0. Hence, by the well known formula PPT Slide
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for ℜ( z ) > 0 and n PPT Slide
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, see [1, p. 260, 6.4.1] , the function h ( t ) defined by (1.3) has the following integral representation PPT Slide
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Proposition 1.3 (Hausdorff-Bernstein-Widder Theorem [8, p. 161, Theorem 12b] ).
A necessary and sufficient condition that f ( x ) should be completely monotonic for 0 < x < ∞ is that PPT Slide
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where α ( t ) is non-decreasing and the integral converges for 0 < x < ∞.
Combining the complete monotonicity in Theorem 1.1 and the integral representation (1.14) with the necessary and sufficient condition in Proposition 1.3, it was revealed in  that PPT Slide
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Replacing PPT Slide
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by t in (1.16) yields [6, Theorem 1.3] below.
Theorem 1.4. For t > 0, we have PPT Slide
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We note that the complete monotonicity in Theorem 1.1 is the basis of the inequality (1.17) and some results in the subsequent papers [4 , 5] .
The aim of this paper is, with the help of the integral representation (1.14) but without using Proposition 1.3, to supply a new proof of Theorems 1.1 and 1.4 in a converse direction with that in [4 , 5 , 6] . In other words, Theorem 1.4 will be firstly and straightforwardly proved, and then Theorem 1.1 will be done.
2. A NEW PROOF OF THEOREMS 1.1 AND 1.4
By the definition of the modified Bessel function Iν ( z ) in (1.10), it is easy to see that PPT Slide
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Hence, in order to prove (1.16), it suffices to show PPT Slide
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which is equivalent to PPT Slide
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Consequently, the proof of the inequality (1.16), that is, Theorem 1.4, is thus complete.
Substituting the inequality (1.16) into the integral representation (1.14) leads to h ( t ) > 0 and for k PPT Slide
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on (0, ∞). As a result, the function h ( t ) is completely monotonic on (0, ∞).
The limit (1.4) follows immediately from taking t → ∞ on both sides of the integral representation (1.14). Theorem 1.1 is thus proved.
Remark 2.1. The inequality (2.1) is equivalent to
An immediate differentiation yields
• Q′ (u) =eu(u2− 4u+ 6) − 2 (u+ 3),
• Q′′ (u) =eu(u2− 2u+ 2) − 2,
• Q′′′ (u) =u2eu.
Since Q ′′′ ( u ) and Q ′′ (0) = 0, it follows that Q ′′ ( u ) > 0 on (0, ∞). Owing to Q ′(0) = 0 and Q ′′ ( u ) > 0, it is derived that Q ′ ( u ) > 0. Finally, since Q (0) = 0, the function Q ( u ) is positive on (0, ∞). This gives an alternative proof of the inequality (2.1).
Remark 2.2. This is a slightly modified version of the preprint  .
References