Logarithmically Complete Monotonicity of Certain Ratios Involving the k -Gamma Function
aa r X i v : . [ m a t h . G M ] F e b Logarithmically Complete Monotonicity of Certain RatiosInvolving the k -Gamma Function Kwara Nantomah and Li Yin Department of Mathematics, University for Development Studies, Navrongo Campus,P. O. Box 24, Navrongo, UE/R, Ghana (E-mail: [email protected]) Department of Mathematics, Binzhou University, Binzhou City, Shandong Province,P.O.Box 256603 China (E-mail: yinli [email protected])
Submitted to RGN Publications on: 17-October-2018, Accepted on: 26-November-2018
Abstract
In this paper, we prove logarithmically complete monotonicity properties of certain ratiosof the k -gamma function. As a consequence, we deduce some inequalities involving the k -gamma and k -trigamma functions. k -gamma function, k -polygamma function, logarithmicallycompletely monotonic function, inequalityArticle type: (Research Article) The k -gamma function (also known as the k -analogue or k -extension of the classical Gammafunction) was defined by D´ıaz and Pariguan [5] for k > x ∈ C \ k Z asΓ k ( x ) = lim n →∞ n ! k n ( nk ) xk − ( x ) n,k = Z ∞ t x − e − tkk dt, (1)where ( x ) n,k = x ( x + k )( x + 2 k ) . . . ( x + ( n − k ) is the Pochhammer k -symbol. It satisfiesthe basic properties Γ k ( x + k ) = x Γ k ( x ) , Γ k ( k ) = 1 , Γ k ( nk ) = k n − ( n − , n ∈ N . k -analogue of the Gauss multiplication formula is given as [15]Γ k ( mx ) = m mxk − k m − (2 π ) − m m − Y s =0 Γ k (cid:18) x + skm (cid:19) , m ≥ , (2)and by letting x = km , one obtains the identity m − Y s =1 Γ k (cid:18) skm (cid:19) = k − m (2 π ) m − √ m , m ≥ . (3)The logarithmic derivative of the k -gamma function, which is called the k -digamma function,is defined as (see [6], [7], [12], [13] ) ψ k ( x ) = ddx ln Γ k ( x ) = ln k − γk − x + ∞ X n =1 xnk ( nk + x )= ln k − γk + ∞ X n =0 (cid:18) nk + k − nk + x (cid:19) , (4)and satifies the properties ψ k ( x + k ) = 1 x + ψ k ( x ) ,ψ k ( k ) = ln k − γk , where γ is the Euler-Mascheroni’s constant. The k -polygamma function of order r is defined as[7] ψ ( r ) k ( x ) = d r dx r ψ k ( x ) = ( − r +1 ∞ X n =0 r !( nk + x ) r +1 , r ∈ N . (5)Also, it is well known in the literature that the integral r ! x r +1 = Z ∞ t r e − xt dt, (6)holds for x > r ∈ N = N ∪ { } . See for instance [1, p. 255].We recall that a function h is said to be completely monotonic on an interval I , if h hasderivatives of all order and satisfies ( − r h ( r ) ( z ) ≥ , (7)for all z ∈ I and r ∈ N . If inequality (7) is strict, then f is said to be strictly completelymonotonic on I . In particular, each completely monotonic function is positive, decreasing andconvex. 2 positive function h is said to be logarithmically completely monotonic on an interval I , if h satisfies [14] ( − r [ln h ( z )] ( r ) ≥ , (8)for all z ∈ I and r ∈ N . If inequality (8) is strict, then h is said to be strictly logarithmicallycompletely monotonic on I . It has been established that, if a function is logarithmicallycompletely monotonic, then it is completely monotonic [14]. However, the converse is notnecessarily true.Completely monotonic functions are frequently encounted in various aspects of mathematics.They are particularly useful in the theory of inequalities, in probability theory and in potentialtheory. There exists an extensive literature on this subject. See for instance [2], [4], [10], [11],[17], [16] and the related references therein.In [9] it was shown that the functions F ( x ) = Γ(2 x ) x Γ ( x ) and G ( x ) = Γ(2 x )Γ ( x ) , are strictly logarithmically convex and strictly logarithmically concave respectively on (0 , ∞ ).In [3], the author established a more deeper results by proving that F ( x ) and 1 /G ( x ) are strictlylogarithmically completely monotonic on (0 , ∞ ). Then in [8], the authors generalized the resultsof [3] by proving the following results.Let F and G be defined for an integer m ≥ x ∈ (0 , ∞ ) as F ( x ) = Γ( mx ) x m − Γ m ( x ) and G ( x ) = Γ( mx )Γ m ( x ) . Then F ( x ) and 1 /G ( x ) are strictly logarithmically completely monotonic on (0 , ∞ ).In this paper the main objective is to extend the results of [8] to the k -gamma function. Webegin with the following auxiliary results. Lemma 2.1.
The k -polygamma function satisfies the identity ψ ( r ) k ( mx ) = 1 m r +1 m − X s =0 ψ ( r ) k (cid:18) x + skm (cid:19) , m ≥ , (9) where r ∈ N . This may be called the multiplication formula for the k -polygamma function.Proof. This follows easily from (2). 3 emma 2.2.
The k -digamma and k -polygamma functions satisfy the following integral representations. ψ k ( x ) = ln k − γk + Z ∞ e − kt − e − xt − e − kt dt (10)= ln k − γk + Z t k − − t x − − t k dt, (11) ψ ( r ) k ( x ) = ( − r +1 Z ∞ t r e − xt − e − kt dt (12)= − Z (ln t ) r t x − − t k dt. (13) Proof.
By using (4) in conjunction with (6), we obtain ψ k ( x ) = ln k − γk + ∞ X n =0 Z ∞ (cid:16) e − kt − e − xt (cid:17) e − nkt dt = ln k − γk + Z ∞ (cid:16) e − kt − e − xt (cid:17) ∞ X n =0 e − nkt dt = ln k − γk + Z ∞ e − kt − e − xt − e − kt dt, which proves (10), and by change of variables, we obtain (11). Representations (12) and (13)respectively follow directly from (10) and (11). Lemma 2.3.
For t > and n ∈ N , we have n X s =1 e − stn +1 − ne − t > . (14) Proof.
Note that e − ut > e − t for all 0 < u < t >
0. Then we have n X s =1 e − stn +1 = e − n +1 t + e − n +1 t + e − n +1 t + · · · + e − nn +1 t > e − t + e − t + e − t + · · · + e − t = ne − t , which completes the proof. We are now in position to prove the main results of this paper.4 heorem 3.1.
Let F and G be defined for an integer m ≥ , k > and x ∈ (0 , ∞ ) as F ( x ) = Γ k ( mx ) x m − Γ mk ( x ) and G ( x ) = Γ k ( mx )Γ mk ( x ) . Then(a) F ( x ) is strictly logarithmically completely monotonic on (0 , ∞ ) ,(b) /G ( x ) is strictly logarithmically completely monotonic on (0 , ∞ ) .Proof. By repeated differentiations and applying (9), we obtain(ln F ( x )) ( r ) = m r ψ ( r − k ( mx ) − mψ ( r − k ( x ) + ( − r ( m − r − x r = m − X s =0 ψ ( r − k (cid:18) x + skm (cid:19) − mψ ( r − k ( x ) + ( − r ( m − r − x r , for r ∈ N . Then by applying (6) and (12), we obtain( − r (ln F ( x )) ( r ) = Z ∞ " m − X s =0 e − skm t − m + ( m − − e − kt ) t r − e − xt − e − kt dt = Z ∞ " m − X s =1 e − skm t − ( m − e − kt t r − e − xt − e − kt dt> , which follows from Lemma 2.3. This completes the proof of (a). Similarly, we obtain (cid:18) ln 1 G ( x ) (cid:19) ( r ) = mψ ( r − k ( x ) − m r ψ ( r − k ( mx )= mψ ( r − k ( x ) − m − X s =0 ψ ( r − k (cid:18) x + skm (cid:19) , for r ∈ N . This implies that( − r (cid:18) ln 1 G ( x ) (cid:19) ( r ) = Z ∞ " m − m − X s =0 e − skm t t r − e − xt − e − kt dt = Z ∞ " m − X s =1 − m − X s =0 e − skm t t r − e − xt − e − kt dt = Z ∞ m − X s =1 (cid:16) − e − skm t (cid:17) t r − e − xt − e − kt dt> , which completes the proof of (b). 5 orollary 3.2. Let m ≥ be an integer and k > . Then the inequality k m − ( m − < Γ k ( mx )Γ mk ( x ) < x m − ( m − , (15) holds if x ∈ ( k, ∞ ) and reverses if x ∈ (0 , k ) .Proof. The conclusions of Theorem 3.1 imply that F ( x ) is strictly decreasing while G ( x ) isstrictly increasing. Then for x ∈ ( k, ∞ ), we have F ( x ) < F ( k ) which gives the right hand sideof (15). Also for x ∈ ( k, ∞ ) we have G ( x ) > G ( k ) yielding the left hand side of (15). The prooffor the case where x ∈ (0 , k ) follows the same procedure and so we omit the details. Corollary 3.3.
Let m ≥ be an integer and k > . Then the inequality Γ k ( mx )Γ mk ( x ) < x m − m , (16) holds for x ∈ (0 , ∞ ) .Proof. Since F ( x ) is strictly decreasing on (0 , ∞ ), it follows easily that F ( x ) < F (0) = lim x → F ( x ) = 1 m , which gives (16). Corollary 3.4.
Let m ≥ be an integer and k > . Then the inequality m m − X s =0 ψ ′ k (cid:18) x + skm (cid:19) < ψ ′ k ( x ) < m m − X s =0 ψ ′ k (cid:18) x + skm (cid:19) + m − mx , (17) holds for x ∈ (0 , ∞ ) .Proof. Infering from Theorem 3.1, it follows that F ( x ) is strictly logarithmically convex while G ( x ) is strictly logarithmically concave. In this way,(ln F ( x )) ′′ = m − X s =0 ψ ′ k (cid:18) x + skm (cid:19) − mψ ′ k ( x ) + m − x > , which yields the right hand side of (17). Also,(ln G ( x )) ′′ = m − X s =0 ψ ′ k (cid:18) x + skm (cid:19) − mψ ′ k ( x ) < , which gives the left hand side of (17). This completes the proof.6 Conclusion
We have established logarithmically complete monotonicity properties of certain ratios of the k -gamma function. As a consequence, we derived some inequalities involving the k -gamma andthe k -trigamma functions. The established results could trigger a new research direction in thetheory of inequalities and special functions. Acknowledgement
The authors are grateful to the anonymous referees for thorough reading of the manuscript.
Competing Interests
The authors declare that they have no competing interests.
Authors’ Contributions
All the authors contributed significantly in writing this article. The authors read and approvedthe final manuscript.
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