| ||Computer Modeling in Engineering & Sciences || |
Complete Monotonicity of Functions Related to Trigamma and Tetragamma Functions
1Faculty of Science, Mathematics Department, Mansoura University, Mansoura, 35516, Egypt
2Faculty of Science, Mathematics Department, Jeddah University, Jeddah, 21589, Saudi Arabia
3Faculty of Science, Mathematics Department, King Abdulaziz University, Jeddah, 21589, Saudi Arabia
*Corresponding Author: Mansour Mahmoud. Email: firstname.lastname@example.org
Received: 11 April 2021; Accepted: 20 October 2021
Abstract: In this paper, we study the completely monotonic property of two functions involving the function and deduce the double inequality which improve some recent results, where is the logarithmic derivative of the Gamma function. Also, we deduce the completely monotonic degree of a function involving .
Keywords: Trigamma function; tetragamma function; completely monotonic function; completely monotonic degree; inequality
The Euler’s gamma function is defined  by the improper integral
and the psi or digamma function is defined by the logarithmic derivative of gamma function, that is
The two derivatives and are respectively called the trigamma and tetragamma functions.
Kirchhoff was the first to apply the polygamma functions in the field of physics  and in Feynman calculations arised several series containing polygamma functions . Recently, Wilkins et al.  use the digamma function, as well as new variants of the digamma function, as a new family of basis functions in mesh-free numerical methods for solving partial differential equations. The polygamma functions and their inequalities have many interesting applications in physics, statistics and applied mathematics by giving estimates and approximations to the values of some special functions .
A function f(x) is said to be completely monotonic  (Chapter XIII) on an interval J if the derivatives f(m)(x) exist on J for all and
According to the Bernstein–Widder theorem  the function f(x) is completely monotonic on if and only if there is a function satisfying that
where the integral converges for and is non-decreasing and bounded, that is, f(x) is completely monotonic on if and only if it is a Laplace transform of a non-decreasing and bounded measure .
In 2004, Alzer  presented the inequality
In 2013, Guo et al.  proved the complete monotonicity on of the function
In , Zhao et al. proved the complete monotonicity on of the functions
They also presented the double inequality
The lower bound of inequality (2) and the bound of inequality (1) are not included each other.
In 2015, Qi  proved the complete monotonicity on of the function
If and only if and so its negative if and only if . Hence, he deduced the two-sided inequality
If and only if and and posed a two-sided inequality of the function on x > 0. For and , inequality (3) recovers the lower bound and refines the upper bound of inequality (2).
The survey  presented several results about the function , its divided difference forms, its variants and its q-analogous including inequalities, positivity, generalizations, (logarithmically) complete monotonicity and applications. Also, the authors pose several open problems. For more results related to this function, we refer to [13–25], and the literature listed therein.
Let the function f (x) be completely monotonic for and denote . If the function is completely monotonic on x > 0 if and only if , then the number is called the completely monotonic degree [16,26] of f (x) with respect to x > 0 and is denoted by . This concept can help to measure completely monotonic functions more precisely.
Qi  proved that
and Guo et al.  proved that
where C(x) is a polynomial of positive coefficients of degree twenty one. For more results related to properties of completely monotonic degree, we refer to [12,16,28–32] and the references therein.
Our first aim will be to establish the double inequality
which improves the upper bound of inequality (3) for x > 0 and the lower bound of inequality (3) for . Also, we proved the complete monotonicity of the following two functions on :
The second aim of this paper is to compute the completely monotonic degree of a function involving .
For proving our main results, we need the following lemmas.
Lemma 2.1 The function
is completely monotonic on and
Proof. Using the integral representations formulas 
Using the inequalities 3n > n2 for , we get
Also, by using the inequalities for and for , we have
and hence an > 0 for . Furthermore, Mathematica software computation shows that is a polynomial in v with all positive coefficients. Then an > 0 for and hence for v > 0 and consequently the function L(x) is completely monotonic in . Now the function L(x) is decreasing, and using the asymptotic expansion 
where the mth Bernoulli number Bm is defined by 
We obtain and hence L(x) > 0 for x > 0.
Lemma 2.2 The function
is completely monotonic on and
Proof. Using the formulas (6) and (7), we obtain
By using that for we deduce that
For , and hence
Hence bn < 0 for . Using Mathematica, we can see that all coefficients of the polynomial are positive. Then bn < 0 for , hence u(v) > 0 for v > 0, and consequently the function U(x) is completely monotonic in . Now the function U(x) is decreasing, and using the asymptotic expansion (9), we obtain and hence U(x) > 0 for x > 0.
3 Main Results
Now we begin to prove our main results.
Theorem 3.1 The functions
are completely monotonic on and
Proof. Using recursion formula 
The two functions and L(x) are completely monotonic on and the product of two completely monotonic functions is also completely monotonic, then the difference fl(x) − fl(x + 1) is completely monotonic on , and hence the function fl(x) is also completely monotonic on , see . Now using the formula (9) and its derivative formula
We get and hence fl(x) > 0 for x > 0, that is
Now, using the recursion formula (16), we get
The two functions and U(x) are completely monotonic on , and hence the functions fu(x) − fu(x + 1), and fu(x) are completely monotonic on . Now using the formulas (9) and (17), we have , and then fu(x) > 0 for x > 0, that is
The proof of Theorem 1 is complete.
Remark 1. The upper bound of inequality (15) is better than the upper bound of inequality (3) for x > 0. Also, the lower bound of inequality (15) is better than the lower bound of inequality (3) for .
Theorem 3.2. The completely monotonic degree of L(x) on is 1.
Proof. Using the integral representation (8), we get
Now using the inequalities
we obtain that kn > 0 for . Furthermore, Mathematica computation shows that is a polynomial in v with all positive coefficients. Then kn > 0 for , and hence m(v) > 0 for v > 0 and consequently the function x L(x) is completely monotonic in . Hence we have
If we suppose that is completely monotonic on , then the function is decreasing, that is
Using the relation 
and hence we get
Combining (18) and (19) completes the proof.
The main conclusions of this paper are stated in Theorems 3.1 and 3.2. Concretely speaking, the authors proved the completely monotonic property of two functions involving the sum of the Trigamma square and Tetragamma functions, derived a new double inequality for this sum and deduced the completely monotonic degree of a function involving the Trigamma function.
Funding Statement: The authors received no specific funding for this study.
Conflicts of Interest: The authors declare that they have no conflicts of interest to report regarding the present study.
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