View of On Inequalities for the Ratio of v-Gamma and v-Polygamma Functions

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https://doi.org/10.34198/ejms.13123.121131

On Inequalities for the Ratio of v-Gamma and v-Polygamma Functions

˙Inci Ege

Department of Mathematics, Aydın Adnan Menderes University, Aydın, Turkey e-mail: [email protected]

Abstract

In this paper, the author presents some double inequalities involving a ratio of v-Gamma and v-polygamma functions. The approach makes use of the log-convexity property ofv-Gamma function and the monotonicity property of v-polygamma function. Some of the results also give generalizations and extensions of some previous results.

1 Introduction and Preliminaries

Logarithmically convex (log-convex) functions are of interest in many areas of mathematics. They have been found to play an important role and have many applications, especially in the theory of special functions, [3, 6, 9, 15].

LetC ∈Rbe a convex set. The functionf :C →[0,∞) is said to be convex on C if it satisfies the inequality

f(cr+ (1−c)t)≤c f(r) + (1−c)f(t), r, t∈C, 0≤c≤1.

Also, a function f :C →[0,∞) is said to be logarithmically convex (log-convex) if logf is convex or equivalently it satisfy the inequality

f(cr+ (1−c)t)≤(f(r))^c(f(t))^1−c, r, t∈C, 0≤c≤1.

Note that, a log-convex function is convex, and a family of log-convex functions is closed under both addition and multiplication.

Received: April 27, 2023; Revised & Accepted: May 16, 2023; Published: May 20, 2023 2020 Mathematics Subject Classification: 33B15, 26D07, 26A48.

Keywords and phrases: Gamma function, v-Gamma function, v-polygamma function,

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Perhaps, the most known and used of the special functions is Euler’s Gamma function. It is defined by

Γ(t) = Z ∞

0

x^t−1e^−xdx, t >0.

The Psi (or digamma) function is defined as the logarithmic derivative of the Gamma function, and the polygamma function is defined as the rth order derivative of the digamma function.

The subject of present new inequalities including the Gamma function has attracted the attention of many mathematicians. For example, Shabani, in [13]

proved the following inequalities:

Γ(a+b)^c

Γ(d+e)^f ≤ Γ(a+bt)^c

Γ(d+et)^f ≤ Γ(a)^c

Γ(d)^f, t∈[0,1] (1) wherea, b, c, d, eand f are real numbers such that a+bt >0,d+et >0,a+bt≤ d+et,ef ≥bc >0 andψ(a+bt)>0 or ψ(d+et)>0.

Also, Vinh and Ngoc, in [14] proved the following inequalities:

n

Y

i=1

Γ(1 +α_i) Γ(β+

n

X

i=1

α_i)

≤

n

Y

i=1

Γ(1 +α_it) Γ(β+

n

X

i=1

α_it)

≤ 1

Γ(β), t∈[0,1].

where β≥1,α_i >0,n∈N.

Some new extensions of the Gamma function and including inequalities of them have been given by many researchers, [1, 5, 7, 8, 11, 12, 13]. Recently, a new one-parameter deformation of the classical Gamma function is introduced as a v-analogue (v-deformation or v-generalization) of the Gamma function, [4]. It is defined as

Γ_v(t) = Z ∞

0

x v

_v^t−1

e^−xdx, t, v >0.

Note that when v= 1, we have Γ_v(t) = Γ(t). The logarithmic derivative of Γ_v is calledv-digamma orv-psi function and denoted byψv and therth order derivative

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of ψ_v is calledv-polygamma function. The series representations are given in [4]

as

ψ_v(t) =−lnv+γ

v −1

t +

∞

X

n=1

1

nv − 1

t+nv

, (2)

and

ψ_v^(r)(t) = (−1)^r+1r! +

∞

X

n=0

1

(t+nv)^r+1. (3)

The main aim of the present study is to give some new generalized inequalities including the functions Γvandψv^(r)by using similar methods, used in [10, 13]. This method is based on some monotonicity properties of certain functions associated with v-Gamma andv-polyamma functions.

2 Main Results

Theorem 1. Let 0 < ai+

m

X

j=1

bjt ≤ di +

l

X

k=1

ekt,

l

X

k=1

ek

! fi ≥





m

X

j=1

bj



ci,

ai, ci, di, fi>0, and





m

X

j=1

bj



ci>0 for i= 1,2, . . . n. If

ψ_v



a_i+

m

X

j=1

b_jt



>0 or ψ_v d_i+

l

X

k=1

e_kt

!

>0 (4)

then the function

F(t) =

n

Y

i=1

Γv



ai+

m

X

j=1

bjt





ci

Γv di+

l

X

k=1

e_kt

!^fi

is decreasing on [0,∞) and the following inequalities bounding a ratio of the v-Gamma function hold forv >0:

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n

Y

i=1

Γ_v



a_i+

m

X

j=1

b_j





ci

Γ_v d_i+

l

X

k=1

e_k

!fi ≤F(t)≤

n

Y

i=1

Γ_v(a_i)^cⁱ

Γ_v(d_i)^fⁱ, t∈[0,1], (5)

and

F(t)≤

n

Y

i=1

Γ_v



a_i+

m

X

j=1

b_j





ci

Γv di+

l

X

k=1

e_k

!^fi, t∈(1,∞). (6)

Proof. LetG(t) = lnF(t). Then

G(t) =

n

X

i=1

ciln Γv



ai+

m

X

j=1

bjt



−

n

X

i=1

filn Γv di+

l

X

k=1

ekt

! ,

and G⁰(t) =





m

X

j=1

bj





n

X

i=1



ciψv



ai+

m

X

j=1

bjt







−

l

X

k=1

ek

! _n X

i=1

"

fiψv di+

l

X

k=1

ekt

!#

=

n

X

i=1









m

X

j=1

bj



ciψv



ai+

m

X

j=1

bjt



−

l

X

k=1

ek

!

fiψv di+

l

X

k=1

ekt

!

.

Let ψ_v



a_i+

m

X

j=1

b_jt



 >0 for i = 1,2, . . . n. Since from the equation (2) we have that ψv is an increasing function on [0,∞) we get

ψv



ai+

m

X

j=1

bjt



≤ψv di+

l

X

k=1

ekt

! ,

(5)

so we have ψ_v d_i+

l

X

k=1

e_kt

!

>0. Then





m

X

j=1

b_j



c_iψ_v



a_i+

m

X

j=1

b_jt



 ≤





m

X

j=1

b_j



c_iψ_v d_i+

l

X

k=1

e_kt

!

≤

l

X

k=1

e_k

!

f_iψ_v d_i+

l

X

k=1

e_kt

! .

This proves the inequality





m

X

j=1

bj



ciψv



ai+

m

X

j=1

bjt



−

l

X

k=1

e_k

!

fiψv di+

l

X

k=1

e_kt

!

≤0 (7)

fori= 1,2, . . . n. Now, letψv di+

l

X

k=1

ekt

!

>0. Then ifψv



ai+

m

X

j=1

bjt



>0, we have the inequality (7) with the above discussion.

Ifψ_v



a_i+

m

X

j=1

b_jt



≤0 with ψ_v d_i+

l

X

k=1

e_kt

!

>0, we can write

l

X

k=1

e_k

!

fiψv di+

l

X

k=1

e_kt

!

≥





m

X

j=1

bj



ciψv di+

l

X

k=1

e_kt

!

≥





m

X

j=1

bj



ciψv



ai+

m

X

j=1

bjt



,

and the inequality (7) again follows for i= 1,2, . . . n. Then we haveG⁰(t)≤0. It implies thatGand soF is decreasing on [0,∞). Then for everyt∈[0,1], we have

F(1)≤F(t)≤F(0), and for t∈(1,∞) we have

F(t)≤F(1), and the inequalities (5) and (6) are valid.

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Theorem 2. Let F be a function given in the Theorem 1, where 0< ai+

m

X

j=1

bjt≤di+

l

X

k=1

ekt,





m

X

j=1

bj



ci≥

l

X

k=1

ek

!

fi, ai, ci, di, fi >0, and

l

X

k=1

e_kfi >0fori= 1,2, . . . n. Ifψv



ai+

m

X

j=1

bjt



<0orψv di+

l

X

k=1

e_kt

!

<0, then the function F is decresing for t ∈ [0,∞) and the inequalities (5) and (6) hold.

Proof. It can be proved by using the similar way in the Theorem 1.

Now, we give the following remarks by using the Theorems 1 and 2.

Remark 3. If in the Theorem 1 we take n = m = l = v = 1, we obtain the inequality (1).

Remark 4. If we take F : [0,∞) → R as any differentiable log-convex function instead of Γv satisfying the conditions in the Theorems 1 and 2, the results of the Theorems still hold, since the logarithmic convexity ofF on[0,∞)implies that its logarithmic derivative is an increasing function on [0,∞). For example, one can take the log-convex function Γor any log-convex generalizations such as Γ_k, Γ_q,k, Γ_p,q,k functions instead of Γ_v, [1, 2, 5].

Now, we give some inequalities including thev-polygamms functions.

Theorem 5. Let 0< a+bt≤c+dt, a, c, α, β >0,αb >0 and

H(t) = h

ψ_v^(r)(a+bt)iα

h

ψv^(r)(c+dt) iβ.

If αb≤βd and r= 2k+ 1, k∈N∪ {0}, then H is decreasing function on [0,∞) and the following inequalities hold for t∈[0,1]:

h

ψv^(r)(a+b) iα

h

ψ_v^(r)(c+d)iβ ≤H(t)≤ h

ψv^(r)(a) iα

h

ψ_v^(r)(c)iβ, (8)

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and if αb≥βd and r= 2k, k∈N∪ {0}, then H is increasing function on [0,∞) and the inequalities (8) are reversed.

Proof. LetI(t) = lnH(t). Then I⁰(t) = αbψ^(r+1)v (a+bt)

ψv^(r)(a+bt)

−βdψv^(r+1)(c+dt) ψv^(r)(c+dt)

= αb ψ^(r+1)v (a+bt)ψv^(r)(c+dt)−βd ψv^(r+1)(c+dt)ψv^(r)(a+bt) ψ^(r)_v (a+bt)ψ_v^(r)(c+dt) . Now, by using equation (3) observe that we haveψv^(r)>0 forr= 2k+1,k∈N∪{0}

and ψ_v^(r) < 0 for r = 2k, k ∈ N∪ {0}. Firstly, let αb ≤ βd and r = 2k+ 1, k ∈ N∪ {0}. Then ψ^(r)v (a+bt) > 0, ψv^(r)(c+dt) > 0, ψv^(r+1)(a+bt) < 0 and ψv^(r+1)(c+dt) < 0, ψv^(r) is decreasing and ψv^(r+1) is increasing. Then we have ψv^(r)(a+bt)≥ψv^(r)(c+dt) and ψ^(r+1)v (a+bt)≤ψ^(r+1)v (c+dt). Then we have

ψ^(r)_v (c+dt)ψ_v^(r+1)(a+bt) ≤ ψ_v^(r)(c+dt)ψ_v^(r+1)(c+dt)

≤ ψ_v^(r)(a+bt)ψ_v^(r+1)(c+dt).

Now, using the condition αb≤βdwe get

αb ψ_v^(r)(c+dt)ψ_v^(r+1)(a+bt) ≤ αb ψ^(r)_v (c+dt)ψ_v^(r+1)(c+dt)

≤ αb ψ^(r)_v (a+bt)ψ_v^(r+1)(c+dt)

≤ βd ψ_v^(r)(a+bt)ψ_v^(r+1)(c+dt), that is

αb ψ^(r)_v (c+dt)ψ^(r+1)_v (a+bt)−βd ψ^(r)_v (a+bt)ψ_v^(r+1)(c+dt)≤0,

so we have I⁰(t) ≤ 0. That implies that I is decreasing on [0,∞). Hence H is decreasing on [0,∞). Then fort∈[0,1] we have

H(1)≤H(t)≤H(0), and the inequalities (8) follow.

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Let αb ≥ βd and r = 2k, k ∈ N∪ {0}. Then we have ψ^(r)v (a+bt) < 0, ψv^(r)(c+dt)<0,ψv^(r+1)(a+bt)>0 and ψv^(r+1)(c+dt)>0,ψ^(r)v is increasing and ψv^(r+1) is decreasing. Then we have

ψ^(r+1)_v (c+dt)ψ_v^(r)(a+bt) ≤ ψ_v^(r+1)(c+dt)ψ^(r)_v (c+dt)

≤ ψ_v^(r+1)(a+bt)ψ^(r)_v (c+dt), and using the condition αb≥βdwe get

αb ψ_v^(r+1)(a+bt)ψ^(r)_v (c+dt) ≥ αb ψ^(r+1)_v (c+dt)ψ_v^(r)(c+dt)

≥ αb ψ^(r+1)_v (c+dt)ψ_v^(r)(a+bt)

≥ βd ψ_v^(r+1)(c+dt)ψ_v^(r)(a+bt),

so we have I⁰(t) ≥ 0, means that I is increasing for t ∈ [0,∞). Hence H is increasing on [0,∞). Then the reverse of the inequality (8) is valid.

Theorem 6. Let0< ai+

m

X

j=1

bjt≤ci+

l

X

k=1

dkt, ai, ci, αi, βi >0fori= 1,2, . . . n, r ∈N∪ {0} and

J(t) =

n

Y

i=1



ψv^(r)



a_i+

m

X

j=1

b_jt









αi

"

ψv^(r) c_i+

l

X

k=1

d_kt

!#^βi .

If





m

X

j=1

bj



αi <0 and

l

X

k=1

d_k

!

βi >0 for i= 1,2, . . . n. ThenJ is increasing on [0,∞) and the following inequalities hold on [0,1]:

n

Y

i=1

h

ψv^(r)(a_i)iαi

h

ψ^(r)v (ci)

iβi ≤J(t)≤

n

Y

i=1



ψ^(r)_v



a_i+

m

X

j=1

b_j









αi

"

ψ_v^(r) c_i+

l

X

k=1

d_k

!#βi , (9)

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and if





m

X

j=1

b_j



α_i > 0 and

l

X

k=1

d_k

!

β_i < 0 for i = 1,2, . . . n. Then J is increasing function and the inequalities (9) are reversed.

Proof. LetK(t) = lnJ(t). Then

K⁰(t) =

n

X

i=1











m

X

j=1

b_j



α_i ψ^(r+1)v



a_i+

m

X

j=1

b_jt





ψv^(r)



a_i+

m

X

j=1

b_jt





−

t

X

k=1

d_k

! β_i

ψv^(r+1) ci+

l

X

k=1

dkt

!

ψv^(r) ci+

l

X

k=1

dkt

!





 .

Now, observe that

ψ_v^(r+1)



a_i+

m

X

j=1

b_jt





ψv^(r)



ai+

m

X

j=1

bjt





< 0 and

ψv^(r+1) c_i+

l

X

k=1

d_kt

!

ψ^(r)v ci+

l

X

k=1

d_kt

! < 0.

For the case





m

X

j=1

b_j



α_i < 0 with

l

X

k=1

d_k

!

β_i > 0 for i = 1,2, . . . n, we get K⁰(t)>0. Then we haveJ is increasing on [0,∞) and the inequalities (9) follow on [0,1]. Now, for the second case





m

X

j=1

bj



αi > 0 with

l

X

k=1

dk

!

βi < 0 for i= 1,2, . . . n, we getJ is decreasing and then the inequalities (9) are reversed on [0,1].

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