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 Computer Modeling in Engineering & Sciences

DOI: 10.32604/cmes.2022.016927

ARTICLE

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: mansour@mans.edu.eg
Received: 11 April 2021; Accepted: 20 October 2021

Abstract: In this paper, we study the completely monotonic property of two functions involving the function Δ(x)=[ψ(x)]2+ψ(x) and deduce the double inequality x2+3x+33x4(2x+1)2<Δ(x)<625x2+2275x+50433x4(50x+41)2,x>0 which improve some recent results, where ψ(x) is the logarithmic derivative of the Gamma function. Also, we deduce the completely monotonic degree of a function involving ψ(x).

Keywords: Trigamma function; tetragamma function; completely monotonic function; completely monotonic degree; inequality

1  Introduction

The Euler’s gamma function is defined [1] by the improper integral

Γ(x)=0evvx1dv,x>0

and the psi or digamma function is defined by the logarithmic derivative of gamma function, that is

ψ(x)=Γ(x)Γ(x).

The two derivatives ψ(x) and ψ(x) are respectively called the trigamma and tetragamma functions.

Kirchhoff was the first to apply the polygamma functions ψ(n)(x) in the field of physics [2] and in Feynman calculations arised several series containing polygamma functions [3]. Recently, Wilkins et al. [4] 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 [5].

A function f(x) is said to be completely monotonic [6] (Chapter XIII) on an interval J if the derivatives f(m)(x) exist on J for all m0 and

(1)mf(m)(x)0,xJ,m0.

According to the Bernstein–Widder theorem [7] the function f(x) is completely monotonic on x0 if and only if there is a function μ(v) satisfying that

f(x)=0exvdμ(v),

where the integral converges for x0 and μ(v) is non-decreasing and bounded, that is, f(x) is completely monotonic on x0 if and only if it is a Laplace transform of a non-decreasing and bounded measure μ(v).

In 2004, Alzer [8] presented the inequality

Δ(x)>α(x)900x4(x+1)10,x>0 (1)

where

Δ(x)=[ψ(x)]2+ψ(x)

and

α(x)=450+3600x+13290x2+29700x3+44101x4+45050x5+31865x6+15370x7+4840x8+900x9+75x10.

In 2013, Guo et al. [9] proved the complete monotonicity on (0,) of the function

K(x)=Δ(x)α(x)900x4(x+1)10.

In [10], Zhao et al. proved the complete monotonicity on (0,) of the functions

A(x)=x2+1212x4(x+1)2+Δ(x)

and

B(x)=x+1212x4(x+1)Δ(x).

They also presented the double inequality

x2+1212x4(x+1)2<Δ(x)<x+1212x4(x+1),x>0. (2)

The lower bound of inequality (2) and the bound of inequality (1) are not included each other.

In 2015, Qi [11] proved the complete monotonicity on (0,) of the function

Δ(x)x2+βx+1212x4(x+1)2

If and only if β0 and so its negative if and only if β4. Hence, he deduced the two-sided inequality

x2+δx+1212x4(x+1)2<Δ(x)<x2+λx+1212x4(x+1)2,x>0 (3)

If and only if δ0 and λ4 and posed a two-sided inequality of the function Δ(x) on x > 0. For δ=0 and λ=4, inequality (3) recovers the lower bound and refines the upper bound of inequality (2).

The survey [12] presented several results about the function Δ(x), 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 [1325], and the literature listed therein.

Let the function f (x) be completely monotonic for x(0,) and denote limxf(x)=f(). If the function xβ[f(x)f()] is completely monotonic on x > 0 if and only if β[0,α], then the number αR is called the completely monotonic degree [16,26] of f (x) with respect to x > 0 and is denoted by degcmx[f(x)]=α. This concept can help to measure completely monotonic functions more precisely.

Qi [27] proved that

degcmx[Δ(x)]=4

and Guo et al. [26] proved that

degcmx[Δ(x)C(x)1800x2(x+1)10(x+2)10]=1,

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 [12,16,2832] and the references therein.

Our first aim will be to establish the double inequality

x2+3x+33x4(2x+1)2<Δ(x)<625x2+2275x+50433x4(50x+41)2,x>0

which improves the upper bound of inequality (3) for x > 0 and the lower bound of inequality (3) for x>132(9+849)1.1918. Also, we proved the complete monotonicity of the following two functions on (0,):

Δ(x)x2+3x+33x4(2x+1)2

and

625x2+2275x+50433x4(50x+41)2Δ(x).

The second aim of this paper is to compute the completely monotonic degree of a function involving ψ(x).

2  Lemmas

For proving our main results, we need the following lemmas.

Lemma 2.1 The function

L(x)=ψ(x)12x21xx22[x2+3x+33x4(2x+1)2(x+1)2+3(x+1)+33(x+1)4(2x+3)2] (4)

is completely monotonic on (0,) and

ψ(x)>k(x)6x2(x+1)4(2x+1)2(2x+3)2,x>0, (5)

where

k(x)=96x9+816x8+3040x7+6536x6+8968x5+8157x4+4962x3+1977x2+477x+54.

Proof. Using the integral representations formulas [33]

1xs=1Γ(s)0vs1exvdv,s,x>0 (6)

and

(1)mψ(r)(x)=0vmexvev1dv,mN,x>0, (7)

we have

L(x)=0e3v/224(ev1)(v)exvdv,x>0, (8)

where

(x)=2ev/2(v315v2+92v246)+2e3v/2(v315v2+104v252)63v+12e5v/2e2v(7v+36)+ev(70v+528)492=n=5anvn2nn!

with

an=1627(3n27)n389(17×3n189)n2+(35×2n+17×22n1+4136×3n3520)n+12(11×2n+214×3n+13×4n+5n+41).

Using the inequalities 3n > n2 for n1, we get

an[1]=1627(3n27)n389(17×3n189)n2=n2[3n(16n271369)16n+168]>n2[n2(16n271369)16n]>8n327(2n251n54)>0,n27.

Also, by using the inequalities 12(54)n>7n2+708 for n19 and 2nn for n1, we have

an[2]=(35×2n+17×22n1+4136×2n3520)n+12(11×2n+214×4n+13×4n+5n+41)>[12×5n22n(7n2+708)]+[2n(587n+528)520n]>0,n19

and hence an > 0 for n27. Furthermore, Mathematica software computation shows that n=526anvn2nn! is a polynomial in v with all positive coefficients. Then an > 0 for n5 and hence (v)>0 for v > 0 and consequently the function L(x) is completely monotonic in (0,). Now the function L(x) is decreasing, and using the asymptotic expansion [33]

ψ(x)1x+12x2+m=2Bmxm+1,x, (9)

where the mth Bernoulli number Bm is defined by [34]

m=0Bmm!tm=tet1,|t|<2π,

We obtain limxL(x)=0 and hence L(x) > 0 for x > 0.

Lemma 2.2 The function

U(x)=x22[625x2+2275x+50433x4(50x+41)2625(x+1)2+2275(x+1)+50433(x+1)4(50x+91)2]+1x+12x2ψ(x) (10)

is completely monotonic on (0,) and

ψ(x)<x22[625x2+2275x+50433x4(50x+41)2625(x+1)2+2275(x+1)+50433(x+1)4(50x+91)2]+1x+12x2,x>0. (11)

Proof. Using the formulas (6) and (7), we obtain

U(x)=0e91v/5067818264(ev1)u(v)exvdv,x>0 (12)

where

u(v)=e91v/50(5651522v367613182v2+354726096v491362936)+e41v/50(5651522v3+67613182v2286907832v+491773100)1681e2v(14391v+40100)410164e141v/501183(100737v+415700)+2ev(71681571v+279590600)=n=5bn2n125nvn753571n!

with

bn=(54017153140041×2n+225n709777372049200×41n+327338438907600×91n18229840278741×100n)n+10000(19789528031×41n3500263931×91n)n26028568(1397953×2n+225n+1+674081×4n25n+1122943275×41n+122840734×91n+102541×141n)20500000(753571×41n68921×91n)n3.

Now

bn[1]=(18229840278741×100n+54017153140041×2n+225n)n=52n2nn(18229840278741×2n+216068612560164)<0,n5.

By using that (9141)n>19789528031n70977737204.923500263931n32733843890.76 for n10 we deduce that

bn[2]=(709777372049200×41n+327338438907600×91n)n+10000(19789528031×41n3500263931×91n)n2<0,n10.

For n24, n3<(32)n and hence

bn[3]=6028568(1397953×2n+225n+1+674081×4n25n+1122943275×41n+122840734×91n+102541×141n)20500000(753571×41n68921×91n)n3<23n[18839275 41n(4917731 2n102500 3n)507967893251 8n25n+1+753571×2n+1(1397953×50n+161420367×91n)+1681×3n(105062500×91n45967831×94n)]<0forn26.

Hence bn < 0 for n26. Using Mathematica, we can see that all coefficients of the polynomial n=525(bn2n125nvn753571n!) are positive. Then bn < 0 for n5, hence u(v) > 0 for v > 0, and consequently the function U(x) is completely monotonic in (0,). Now the function U(x) is decreasing, and using the asymptotic expansion (9), we obtain limxU(x)=0 and hence U(x) > 0 for x > 0.

3  Main Results

Now we begin to prove our main results.

Theorem 3.1 The functions

fl(x)=Δ(x)x2+3x+33x4(2x+1)2 (13)

and

fu(x)=625x2+2275x+50433x4(50x+41)2Δ(x) (14)

are completely monotonic on (0,) and

x2+3x+33x4(2x+1)2<Δ(x)<625x2+2275x+50433x4(50x+41)2,x>0. (15)

Proof. Using recursion formula [33]

ψ(x+1)=ψ(x)+1x,x>0 (16)

we get

fl(x)fl(x+1)=ψ(x)ψ(x+1)+(x+1)2+3(x+1)+33(x+1)4(2(x+1)+1)2+{ψ(x)+ψ(x+1)}{ψ(x)ψ(x+1)}x2+3x+33x4(2x+1)2=2x2L(x).

The two functions 2x2 and L(x) are completely monotonic on (0;) and the product of two completely monotonic functions is also completely monotonic, then the difference fl(x) − fl(x + 1) is completely monotonic on (0;), and hence the function fl(x) is also completely monotonic on (0;), see [11]. Now using the formula (9) and its derivative formula

ψ(x)1x21x3m=2(m+1)Bmxm+2,x (17)

We get limxfl(x)=0 and hence fl(x) > 0 for x > 0, that is

Δ(x)>x2+3x+33x4(2x+1)2,x>0.

Now, using the recursion formula (16), we get

fu(x)fu(x+1)=ψ(x+1)ψ(x)+625x2+2275x+50433x4(50x+41)2+{ψ(x+1)+ψ(x)}{ψ(x+1)ψ(x)}625(x+1)2+2275(x+1)+50433(x+1)4(50(x+1)+41)2=2x2U(x).

The two functions 2x2 and U(x) are completely monotonic on (0;), and hence the functions fu(x) − fu(x + 1), and fu(x) are completely monotonic on (0;). Now using the formulas (9) and (17), we have limxfu(x)=0, and then fu(x) > 0 for x > 0, that is

Δ(x)<625x2+2275x+50433x4(50x+41)2,x>0.

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 x>132(9+849)1.1918.

Theorem 3.2. The completely monotonic degree of L(x) on (0,) is 1.

Proof. Using the integral representation (8), we get

L(x)=1x0e3v/248(ev1)2m(v)exvdv,x>0,

where

m(v)=8e3v/2(v318v2+122v344)4ev/2(v318v2+122v338)+ev(385v+2722)4e5v/2(v318v2+134v350)+e3v(7v+22)27(7v+50)e2v(203v+1394)=n=5kn2n1vn3375n!

with

kn=22n(82944n2+9450000)[(54)n68.5125n+940.958.2944n2+945]+6n(1728n3+1528416n)[157.5n+148517.28n3+15284.16n(56)n]+2n(2598750n+18373500)+103(16n3480n2+4856n18576)×[3n216n32592n2+8964n912616n3480n2+4856n18576].

Now using the inequalities

68.5125n+940.958.2944n2+945<(54)nn1,157.5n+148517.28n3+15284.16n>(56)nn27,216n32592n2+8964n912616n3480n2+4856n18576<3nn15

we obtain that kn > 0 for n27. Furthermore, Mathematica computation shows that n=526kn2n1vn3375n! is a polynomial in v with all positive coefficients. Then kn > 0 for n5, and hence m(v) > 0 for v > 0 and consequently the function x L(x) is completely monotonic in (0,). Hence we have

degcmx[L(x)]1. (18)

If we suppose that xμL(x) is completely monotonic on (0,), then the function xμL(x) is decreasing, that is

μxL(x)L(x)=x[ψ(x)ϕ(x)]ψ(x)ϕ(x),

where

ϕ(x)=k(x)6x2(x+1)4(2x+1)2(2x+3)2

with

k(x)=96x9+816x8+3040x7+6536x6+8968x5+8157x4+4962x3+1977x2+477x+54.

Using the relation [35]

ψ(r)(x)=(1)r+1r!i=01(i+x)r+1,x>0;r=1,2,3,...

We get

limx0[xψ(x)1x]=0andlimx0[x2ψ(x)+2x]=0.

Also,

limx0[xϕ(x)1x]=12andlimx0[x2ϕ(x)+2x]=12.

Then

μ[x2ψ(x)+2x][x2ϕ(x)+2x][xψ(x)1x][xϕ(x)1x]1asx0

and hence we get

degcmx[L(x)]1. (19)

Combining (18) and (19) completes the proof.

4  Conclusions

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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