arXiv:1511.07156v2 [math.CA] 17 Dec 2015 ucin Γ( functions tions, ψ 1 Γ (1) S (2010): MSC xesv plctosi ahmtclsciences. mathematical in applications extensive 3 lim (3) 2 Γ (2) and The 4 Γ (4) and ( k keywords: ti elkonta h lsia ue am ucinmay function gamma Euler classical the that well-known is It The OAIHIAL OPEEYMNTNI FUNCTIONS MONOTONTIC COMPLETELY LOGARITHMICALLY EAE THE RELATED ) ( q x − for ) q q gmafunction, -gamma ie eut eeaie hoesdet le n egand Berg and Alzer to due theorems generalized results given nqaiisfrthe for inequalities oooi ucin.Orfntoshv ncmo htthey that common in have functions the Our of functions. monotonic Abstract. am ucinΓ function gamma − nlgeo sdfie [] p 9-9]for 493-496] pp. [[5], defined is Γ of analogue k x ∈ opeeymntncfntos oaihial complet Logarithmically functions, monotonic Completely ) q ψ , − N am and gamma r aldteplgmafntos ti omnkoldet knowledge common is It functions. polygamma the called are ( nti ae epeetsvrlnwcasso logarithmica of classes new several present we paper this In x and ) q ( q q x − − q ( = ) ψ q ( q am n the and gamma AM N THE AND GAMMA x − q ( ( k z − q (1 = ) iam ucin Inequalities function, digamma −→ ) a h olwn ai properties: basic following the has ) iam ucin.A napiain fti eut,some results, this of applications an As functions. digamma ( q IHAPPLICATIONS WITH x 1 for ) − − Γ( q 1) Γ − ( q z 1 x HLDMEHREZ KHALED 1. ( q k − = ) = ) z ) x ∈ lim = ) 1 Introduction q − q − N x x q Z ( iam ucin r salse.Sm fthe of Some established. are functions digamma x ( j Y q 0 2 x r udmna n motn n aemuch have and important and fundamental are ∞ − =0 −→ ∞ − 1) 1 1)( t 2 1 1 1 j Y x x ∞ + =0 − − − − Γ 2) 1 q e q 1 1 q Γ − j ( j − − + +1 z 1 q t q Γ( = ) dt, x ( − q q z , − − IAM FUNCTIONS DIGAMMA ) > x 0 . ( ( j j < q < + +1) x z by 0 .P hnadF Qi. F. and Chen C.-P. ) ) > q , , r endi terms in defined are 1 edfie by defined be , 1 . l completely lly l oooi func- monotonic ely a special hat 2 KHALED MEHREZ
The q−digamma function ψq, the q−analogue of the psi or digamma function ψ is defined by ′ Γq(x) ψq(x)= Γq(z) ∞ qk+x = − ln(1 − q)+ln q 1 − qk+x k=0 (5) X∞ qkx = − ln(1 − q)+ln q 1 − qk k X=1 ∞ e−xt = − ln(1 − q) − dγ (t), 1 − e−t q Z0 for 0 (6) γq(t)= δ(t + k ln q), 0 (n) (−1)n ln f(x) (x) ≥ 0, for all x ∈ I and n ∈ N. A positive function f is said to be logarithmically convex on an interval I, or simply log-convex, if its natural logarithm ln f is convex, that is, for all x,y ∈ I and λ ∈ [0, 1] we have − f (λx + (1 − λ)y) ≤ [f(x)]λ [f(y)]1 λ . LOGARITHMICALLY COMPLETELY MONOTONTIC FUNCTIONS 3 We note that every logarithmically completely monotonic function is log-convex. From the definition of ψq(x), direct differentiation, and the induction we get n kn+1qkx (8) (−1)nψ(n+1)(x) = (ln q)n+2 > 0, q 1 − qk k X=1 ′ which implies that the function ψq is strictly completely monotonic function on (0, ∞), for q ∈ (0, 1). The relation (4) and the definition of the q−digamma function (5) give 2x − 3 (9) ψq(x)= ψ 1 (x)+ ln q, q 2 ′ for q > 1. Thus, implies that the function ψq(x) is strictly completely monotonic in (0, ∞) for q > 1, and consequently the function ψq(x) is strictly completely monotonic on (0, ∞). It is the aim of this paper to provide several new classes of logarithmically completely monotonic functions. The functions we study have in common that they are defined in terms of q−gamma and q−digamma functions. In the next section we collect some lemmas. Our monotonicity theorems are stated and proved in sections 3. 2. Useful lemmas We begin this section with the following useful lemmas which are needed to completes the proof of the main theorems. The following monotonicity theorem is proved in [2]. Lemma 1. Let n be a natural number and c be a real number. The function xc|ψn(x)| is decreasing on (0, ∞) if and only if c ≤ n. Lemma 2. [14] Let f, g : [a, b] −→ R two continuous functions which are differentiable on ′ f ′ (a, b). Further, let g 6= 0 on (a, b). If g′ is increasing (or decreasing) on (a, b), then the f(x)−f(a) f(x)−f(b) functions g(x)−g(a) and g(x)−g(b) are also increasing (or decreasing) on (a, b). The next lemma is given in [1, 16]. Lemma 3. The function ψq, q> 0 has a uniquely determined positive zero, which we denoted by x0 = x0(q) ∈ (1, 2). A proof for the following lemma can be found in [8]. ′ 1 Lemma 4. Let f be a positive function. If f is completely monotonic function, then f is logarithmically completely monotonic function. The following inequality for completely monotonic functions is due to Kimberling [12]. Lemma 5. If a function f, defined on (0, ∞), is continuous and completely monotonic and maps (0, ∞) into (0, 1), then ln f is super-additive, that is for all x,y > 0 we have ln f(x)+ln f(y) ≤ ln f(x + y) or f(x)f(y) ≤ f(x + y). 4 KHALED MEHREZ 3. The main results (Γ(x))2 In 1997, Merkle [15] proved that the function Γ(2x) is log-convex on (0, ∞). Recently, Alzer and Berg [4], presented a substantial generalization. They established that the function (Γ(ax))α (10) , 0 < b < a, (or 0 < a < b) (Γ(bx))β is completely monotonic on (0, ∞), if and only if, α ≤ 0, (or α ≥ 0) and αa = βb. The main objective of the next Theorem extend and generalize this results. An application of Γ2(x) their result leads to sharp upper and lower bounds for q in terms of the ψ -function. Γq(2x) q α (Γq (ax)) Theorem 1. Let 0 α (Γq(ax)) Proof. Let q ∈ (0, 1). Assume that the function β is logarithmically completely (Γq(bx)) monotonic on (0, ∞). By definition, this gives us that for all real x> 0, ′ α (Γq(ax)) (11) ln β = αaψq(ax) − βbψq(bx) ≤ 0. " (Γq(bx)) # Now we prove α ≥ 0. By contradiction, suppose that α< 0 such that αa = βb. Since the function ψq(x) is strictly increasing on (0, ∞) we have (12) αa(ψq(ax) − ψq(bx)) > 0, for x> 0 and 0 < a < b. On the other hand, by inequality (11) we obtain that (13) αa(ψq(ax) − ψq(bx)) ≤ 0. Inequalities (12) and (13) implies that the constant α ≥ 0. On the other hand, using the fact lim ψq(x)= − ln(1 − q), x−→+∞ for q ∈ (0, 1), and equality (11) we obtain (αa − βb)(− ln(1 − q)) ≤ 0. and consequently (αa − βb) ≤ 0. On the other hand, using the definition we get α (2) (Γq(ax)) 2 (1) 2 (1) ln β = αa ψq (ax) − βb ψq (bx) ≥ 0. " (Γq(bx)) # If x −→ 0+, we obtain (14) αa2 − βb2 ≥ 0, 1 1 since 0 < b < a we have αa ≥ βb, this inequality and inequality (14) leads to αa = βb. LOGARITHMICALLY COMPLETELY MONOTONTIC FUNCTIONS 5 α (Γq(ax)) Now we show that the function β is logarithmically completely monotonic on (Γq(bx)) (0, ∞) for α, β ≥ 0, 0 < a < b such that αa = βb. Let n = 1, since the function ψq(x) is strictly increasing on (0, ∞) we have ′ α (Γq(ax)) (15) (−1) ln β = αa (ψq(bx) − ψq(ax)) ≥ 0. (Γq(bx)) ! For n ≥ 1, we get α (n+1) n+1 (Γq(ax)) n n (n) n (n) (16) (−1) ln β = (−1) αa b ψq (bx) − a ψq (ax) . (Γq(bx)) ! ′ N Since the function ψq(x) is strictly completely monotonic on (0, ∞), we have for n ∈ and q > 0 we have (n) n+1 (n) (17) |ψq (x)| = (−1) ψq (x). Thus α (n+1) n+1 n (Γq(ax)) n n n (n) n (n) (−1) x ln β = (−1) αax b ψq (bx) − a ψq (ax) (Γq(bx)) ! n n n (n) n n (n) = (−1) αa (xb) (−1) |ψq (bx)|− (ax) (−1) |ψq (ax)| n (n) n (n) = αa (ax) |ψq (ax)|− (bx) |ψq (bx)| , ! and the last expression is nonnegative by Lemma 1. Hence, for q ∈ (0, 1) and n ∈ N α (n) n (Γq(ax)) (−1) ln β ≥ 0. (Γq(bx)) ! The proof of Theorem 1 is complete. α (Γq(ax)) Corollary 1. Let q > 1 and 0 < a < b. If, α ≥ 0 and αa = βb., Then the function β (Γq(bx)) is logarithmically completely monotonic on (0, ∞). Proof. Follows immediately by Theorem 1 and equality (9). Corollary 2. Let q > 0 and 0 < a < b, the following inequalities β α (Γq(bx1)) (Γq(ax)) (18) exp [αa(x − x1) (ψq(ax1) − ψq(bx1))] ≤ α β ≤ 1, (Γq(ax1)) (Γq(bx)) holds for all α, β ≥ 0 such that αa = βb. and x>x1 > 0 In particular, the following inequalities holds true for every integer n ≥ 1 : ln q Γ2(n) (19) exp 2q(n − 1) ≤ q ≤ 1. 1 − q Γ (2n) q 6 KHALED MEHREZ Proof. Let q > 0 and 0 < a < b. We suppose that α, β ≥ 0 such that αa = βb and define the function hα,β(q; x) by β α (Γq(bx1)) (Γq(ax)) hα,β(q; x)= α β (Γq(ax1)) (Γq(bx)) where 0 < x1 < x, and Hα,β,q(x) = ln hα,β(q; x). Since the function hα,β(q; x) is logarith- mically completely monoyonic on (0, ∞) for α ≥ 0 and αa = βb, we conclude that the ′ (hα,β(q;x)) logarithmic derivative is increasing on (0, ∞). By Lemma 2 we deduce that the hα,β(q;x) H q x function α,β ( ; ) is increasing for all 0 < x < x. By l’Hospital’s rule and (11) it is easy x−x1 1 to deduce that H (q; x) α,β αa ψ ax ψ bx , −→lim = ( q( 1) − q( 1)) x x1 x − x1 from which follows the right side inequality of (18). As hα,β(q; x) is logarithmically completely monotonic on (0, ∞), we deduce that hα,β(q; x) is decreasing on (0, ∞). The following inequality hold true for all 0 hα,β(q; x) ≤ hα,β(q; x1) = 1, we conclude the left side inequality of (18). Taking α = b = 2, β = a = 1 and x1 = 1 in (18) and using the recurrence formula of ψq [[9], p. 1245, Theorem 4.4] n−1 x − − d q (20) ψ(n 1)(x + 1) − ψ(n 1)(x)= − ln q q q dxn−1 1 − qx we obtain the inequalities (19). The main purpose of the next Theorem is to present monotonicity properties of the function 2 1 2 2x 1 Γ 2 x + β(1 − q )q g q x q 2 ψ x , (21) β( ; )= exp 2x + q(2 ) 1+ q Γ 2 (x + 1) 2(1 − q ) " q # where q ∈ (0, 1) and x> 0. It is worth mentioning that Ai-Jun Li and Chao-Ping Chen [13] considered the function 2 1 Γ x + 1 β (22) g (x)= 2 exp + ψ(2x) β 2 Γ(x + 1) 2x " # which is a special case of the function gβ(q; x) on letting q −→ 1 and proved that gβ(x) is 13 logarithmically completely monotonic on (0, ∞) if β ≥ 12 . The objective of this Theorem is to generalize this result. Theorem 2. Let q ∈ (0, 1). The function gβ(q; x) defined by (21) is logarithmically com- −13 ln q pletely monotonic on (0, ∞) if β ≥ 6(1−q2) . Proof. It is clear that 1 β(1 − q2)q2x ln g (q; x)=2lnΓ 2 (x + ) − 2lnΓ 2 (x +1)+ ψq(2x)+ − ln(q + 1). β q 2 q 2(1 − q2x) LOGARITHMICALLY COMPLETELY MONOTONTIC FUNCTIONS 7 Using the q−analogue of Legendre’s duplication formula (??) we get 1 1 1 1 β(1 − q2)q2x ln g (q; x)=2lnΓ 2 (x + ) − 2lnΓ 2 (x +1)+ ψ 2 (x)+ ψ 2 (x + )+ . β q 2 q 2 q 2 q 2 2(1 − q2x) In view of (20) and (5) we obtain that n (n) n (n−1) 1 (n−1) 1 (n) 1 (n) 1 (−1) (ln gβ(q; x)) = (−1) 2ψq2 (x + ) − 2ψq2 (x +1)+ ψq2 (x)+ ψq2 (x + ) " 2 2 2 2 2 β(1 − q ) (n) (n) − ψq2 (x + 1) − ψq2 (x) 4 ln q # ∞ 1 e−xt = Φ (t)dγ 2 (t) 2 et − 1 β,q q Z0 where 2 2 −β(1 − q ) t t β(1 − q ) Φ (t)= − 1 te + (4 − t)e 2 + t − 4 β,q 2 ln q 2 ln q ∞ ∞ tn+1 −β(1 − q2) 1 tn β(1 − q2) = − 1 − + 4 + t n! 2 ln q 2n 2nn! 2 ln q n=0 n=1 X∞ X∞ tn+1 −β(1 − q2) 1 tn+1 = − 1 − + 4 − 2t n! 2 ln q 2n 2n+1(n + 1)! n=1 n=0 X∞ X tn+1 −β(1 − q2) 1 1 = − 1 − + n! 2 ln q 2n (n + 1)2n−1 n=1 X 1 1 1 −13 ln q Since the max 2n + (n+1)2n−1 = 12 , we conclude that β ≥ 6(1−q2) . The proof is com- ! pleted. Theorem 3. Let q > 0, the function 1 is Logarithmically completely monotonic on ψq (x) (x0, ∞). ′ Proof. Since the function ψq is completely monotonic function on (0, ∞), and the function ψq is increasing on (0, ∞) and a uniquely determined zero on (0, ∞) by Lemma 3. We conclude that the function ψ (x) > 0 for all x>x , and consequently the function 1 q 0 ψq(x) is Logarithmically completely monotonic on (x0, ∞), by Lemma 4. Corollary 3. Let q > 0 and a> 1. The following inequality 1 1− 1 x 1 (23) [ψ (x)] a [ψ (y)] a ≤ ψ + 1 − y q q q a a holds for all x>x0 and y>x0. In particular, the following inequality holds a a−1 [ψq(u + 1)] (24) [ψq(2)] ≤ ψq (a(x − 1) + 2) 8 KHALED MEHREZ −2 for all a> 1 and u> a + 1. Proof. Let q ∈ (0, 1),a> 1,x>x0 and y>x0. By theorem 3 we obtain that the function 1 is log-convex on (x , ∞). Thus, ψq(x) 0 1 1 x y [ψ (x)] p [ψ (y)] q ≤ ψ + , q q q p q 1 1 a where p > 1, q> 1, p + q = 1. If p = a and q = a−1 . Then we get the inequality (23). Now, let y = 2 and x = a(u − 1) + 2 we obtain the inequality (24). Remark 1. Replacing u by n ∈ N and a by 2 in inequality (24) and using the identity ln q ψ (n +1) = γ − ln qH q 1 − q q n,q 1−q where γq = ln q ψq(1) is the q−analogue of the Euler-Mascheroni constant [17] and Hn,q is the q-analogue of Harmonic number is defined by [18] as n qk Hn,q = , n ∈ N, 1 − qk Xk=1 we obtain ln q 2 (25) ψ2(2)ψ (2n) ≤ γ − ln qH , n ∈ N. q q 1 − q q n,q Theorem 4. Let q > 0 The function Γq(x) is logarithmically completely monotonic on (0,x0). So, the following inequality (26) Γq(x + 1)Γq(y + 1) ≤ Γq(x + y + 2) holds for all x,y ∈ (0, 1). Proof. Proving by induction that n (n) (−1) (ln Γq(x)) ≥ 0, for all n ∈ N. For n = 1, we get (1) (−1)(ln Γq(x)) = ψq(x) ≥ 0, for all x ∈ (0,x0). Suppose that k (k) (−1) (ln Γq(x)) ≥ 0, for all 1 ≤ k ≤ n and x ∈ (0,x0). ′ Since the function ψq is completely monotonic on (0, ∞), for q > 0 we get n+1 (n+1) n (n+1) (−1) (ln Γq(x)) = (−1) ψq (x) ≥ 0. We note that every logarithmically completely monotonic function is completely mono- tonic [7], and the function Γq(x) ≤ 1 if and only if 1 ≤ x ≤ 2. The above results imply that the function Γq(x) is completely monotonic on (0,x0) and maps (1, 2) into (0, 1), applying Lemma 5, we conclude the asserted inequality (26) . LOGARITHMICALLY COMPLETELY MONOTONTIC FUNCTIONS 9 References [1] H. Alzer, A. Z. Grinsphan, Inequalities for gamma and q−gamma functions, J. of App. Theory, 144 (2007) 6783. [2] H. Alzer, Mean-value inequalities for the polygamma functions, Aequat. Math. 61 (2001), 151161. [3] H. Alzer, A.Z. Grinshpan, Inequalities for the gamma and q-gamma functions, Journal of Approxima- tion Theory 144 (2007) 6783. [4] H. Alzer and C. Berg, Some classes of completely monotonic functions II, Ramanujan J (2006) 11: 225248. 445460. [5] G. E. Andrews, R. A. Askey, and R. Roy, Special Functions, Cambridge University Press, Cambridge, 1999 [6] R. Askey, The q−Gamma and q−Beta functions, Applicable Analysis: An International Journal, 2013. [7] C. Berg, Integral representation of some functions related to the gamma function, Mediterranean Journal of Mathematics 1(4) (2004) 433439. [8] J. El Kamel, K. Mehrez, A Function Class of strictly positive definite and logarithmically completely monotonic functions related to the modified Bessel functions. arXiv:1205.1112v1. [9] B.-N. Guo, F. Qi, Properties and applications of a functions involving exponential functions, Commun. Pure Appl. Anal. 8 (2009). [10] M. E. H. Ismail and M. E. Muldoon, Inequalities and monotonicity properties for gamma and q-gamma functions, available online at http://arxiv.org/abs/1301.1749. [11] C. Krattenthaler, H.M. Srivastava, Summations for basic hypergeometric series involving a q-analogue of the digamma function, Computers and Mathematics with Applications 32 (2) (1996) 7391. [12] C.H. Kimberling, A probabilistic interpretation of complete monotonicity, Aequationes Math. 10 (1974) 152164. [13] A.-J. Li, C.-P. Chen, Some completely monotonic functions involving the Gamma and Polygamma functions, J. Korean Math. Soc. 45 (2008), No. 1, pp. 273287. [14] K. Mehrez, Redheffer type inequalities for modified Bessel functions, Arab. J. Math. Sci, 2015; avaible online at: http://dx.doi.org/10.1016/j.ajmsc.2015.03.001 [15] M. Merkle, On log-convexity of a ratio of gamma functions, Univ. Beograd. Publ. Elektrotehn. Fak. Ser. Mat. 8 (1997), 114119. [16] A. Salem, Ceratin class of approximation for the q−digamma function,Rocky Mountain Journal of Mathematics, ( To appear). [17] A. Salem, Some Properties and Expansions Associated with q-Digamma Func- tion, Quaestiones math- ematicae, (To appear) [18] C. Wei, Q. Gu, q-Generalizations of a family of harmonic number identities, Advances in Applied Mathematics, Vol.45 (2010) 24-27. Khaled Mehrez. Departement´ de Mathematiques´ ISSAT Kasserine, Universite´ de Kairouan, Tunisia. E-mail address: [email protected]