Some completely monotonic functions involving the q -gamma function
aa r X i v : . [ m a t h . C A ] N ov SOME COMPLETELY MONOTONIC FUNCTIONS INVOLVING THE q -GAMMA FUNCTION PENG GAO
Abstract.
We present some completely monotonic functions involving the q -gamma function thatare inspired by their analogues involving the gamma function. Introduction
The q -gamma function is defined for a complex number z and q = 1 byΓ q ( z ) = ( ( q ; q ) ∞ ( q z ; q ) ∞ (1 − q ) − z , < q < ( q − ; q − ) ∞ ( q − z ; q − ) ∞ ( q − − z q z ( z − , q > . (1.1)where the product ( a ; q ) ∞ is defined by( a ; q ) ∞ = ∞ Y n =0 (1 − aq n ) . In what follows we restrict our attention to positive real numbers x . We note here [17] the limitof Γ q ( x ) as q → − gives back the well-known Euler’s gamma function:lim q → − Γ q ( x ) = Γ( x ) = Z ∞ t x e − t dtt . It’s then easy to see using (1.1) that lim q → Γ q ( x ) = Γ( x ). For historical remarks on gamma and q -gamma functions, we refer the reader to [17], [2] and [3].There exists an extensive and rich literature on inequalities for the gamma and q -gamma functionsof positive real numbers. For the recent developments in this area, we refer the reader to the articles[14], [2]-[4], [20] and the references therein. Many of these inequalities follow from the monotonicityproperties of functions which are closely related to Γ (resp. Γ q ) and its logarithmic derivative ψ (resp. ψ q ) as ψ ′ and ψ ′ q are completely monotonic functions on (0 , + ∞ ) (see [15], [4]). Here werecall that a function f ( x ) is said to be completely monotonic on ( a, b ) if it has derivatives of allorders and ( − k f ( k ) ( x ) ≥ , x ∈ ( a, b ) , k ≥
0. We further note that Lemma 2.1 of [7] asserts that f ( x ) = e − h ( x ) is completely monotonic on an interval if h ′ is. Following [13], we call such functions f ( x ) logarithmically completely monotonic.We note here that lim q → ψ q ( x ) = ψ ( x ) (see [18]), hence in what follows we also write Γ ( x ) forΓ( x ) and ψ ( x ) for ψ q ( x ). Thus we may also regard the gamma function as a q -gamma functionwith q = 1 and in this manner, many completely monotonic functions involving Γ q ( x ) and ψ q ( x )are inspired by their analogues involving Γ( x ) and ψ ( x ). It is our goal in this paper to presentsome completely monotonic functions involving Γ q , ψ q that are motivated by this point of view. Inthe remaining part of this introduction, we briefly mention the motivations for our results in thepaper. Mathematics Subject Classification.
Primary 33D05.
Key words and phrases.
Completely monotonic function, q -gamma function. In [16], Kershaw proved the q = 1 case of the following result for 0 < s < , x > e (1 − s ) ψ q ( x + s / ) < Γ q ( x + 1)Γ q ( x + s ) < e (1 − s ) ψ q ( x +( s +1) / . (1.2)A result of Ismail and Muldoon [14] establishes the second inequality in (1.2) for 0 < q <
1. In [7],Bustoz and Ismail showed that when q = 1, the function (0 < s < x Γ q ( x + s )Γ q ( x + 1) e (1 − s ) ψ q ( x +( s +1) / is completely monotonic on (0 , + ∞ ). In [12], it is shown that the result of Bustoz and Ismail alsoholds for any q > a ( q, s ) and b ( q, s ) such that thefollowing inequalities hold for all x > , < q = 1 , < s < e (1 − s ) ψ q ( x + a ( q,s )) < Γ q ( x + 1)Γ q ( x + s ) < e (1 − s ) ψ q ( x + b ( q,s )) . (1.3)We shall determine the best possible values of a ( q, s ) and b ( q, s ) in Section 3. Another resultgiven in Section 3 is motivated by the following result of Alzer and Batir [5], who showed that thefunction ( x > , c ≥ G c ( x ) = ln Γ( x ) − x ln x + x −
12 ln(2 π ) + 12 ψ ( x + c )is completely monotonic if and only if c ≥ / − G c ( x ) is completely monotonic if and only if c = 0. We shall present a q -analogue in Section 3 for G ′ c ( x ).Muldoon [19] studied the monotonicity property of the function h α ( x ) = x α Γ( x )( e/x ) x . He showed that h α ( x ) is logarithmically completely monotonic on (0 , + ∞ ) for α ≤ /
2. We pointout here that as was shown in [6, Theorem 3.3], 1 / h α ( x ). In [12, Proposition 4.1], it is shown that if one defines for α ≥ f α ( x ) = − ln Γ( x ) + ( x −
12 ) ln x − x + 112 ψ ′ ( x + α ) , then f ′ α ( x ) is completely monotonic on (0 , + ∞ ) if α ≥ / − f ′ α ( x ) is completely monotonicon (0 , + ∞ ) if α = 0. As was pointed out in [12], this implies a result of Alzer [1, Theorem 1]. InSection 3, we shall establish a q -analogue of the above result.It’s shown in the proof of Theorem 2.2 in [8] that for x > < q < ψ ′ q ( x + 1) < ln(1 /q ) q x − q x . The q = 1 analogue of inequality (1.5) is ψ ′ ( x +1) ≤ /x , which reminds us the following asymptoticexpansion [4, (1.5)] for the derivatives of ψ ( x ):(1.6) ( − n +1 ψ ( n ) ( x ) = ( n − x n + n !2 x n +1 + O (cid:18) x n +2 (cid:19) , n ≥ , x → + ∞ . We note that Lemma 2.2 of [11] asserts that for fixed n ≥ , a ≥
0, the function f a,n ( x ) = x n ( − n +1 ψ ( n ) ( x + a ) is increasing on [0 , + ∞ ) if and only if a ≥ /
2. It follows from this and(1.6) that we have ψ ′ ( x + 1 / ≤ /x and this suggests that inequality (1.5) would still hold if onereplaces ψ ′ q ( x + 1) with ψ ′ q ( x + 1 / OME COMPLETELY MONOTONIC FUNCTIONS INVOLVING THE q -GAMMA FUNCTION 3 Lemmas
The following lemma gathers a few results on Γ q and ψ q . Equality (2.1) below is given in [3,(2.7)] and the rest can be easily derived from (1.1) and (2.1). Lemma 2.1.
For < q < , x > , ψ q ( x ) = − ln(1 − q ) + ln q ∞ X n =1 q nx − q n , (2.1) ln Γ q ( x + 1) = ln Γ q ( x ) + ln 1 − q x − q , (2.2) ψ q ( x + 1) = ψ q ( x ) − (ln q ) q x − q x , (2.3) ψ ′ q ( x + 1) = ψ ′ q ( x ) − (ln q ) q x (1 − q x ) . (2.4)Our next lemma is a result in [21]: Lemma 2.2.
For positive numbers x = y and real number r , we define E ( r, x, y ) = (cid:18) r · x r − y r ln x − ln y (cid:19) /r , r = 0; E (0 , x, y ) = √ xy. Then the function r E ( r, x, y ) is strictly increasing on R . Lemma 2.3.
Let < q < , then for any integer n ≥ , ln q − q n + 1 n − ln q − n (ln q ) q n/ − q n ) < , (2.5) ln q − q n + 1 n − ln q − n (ln q ) − q n ) > . (2.6) Proof.
On setting ln q n = x , it is easy to see that inequality (2.5) follows from f ( x ) < x < f ( x ) = 6 x (1 + e x ) + 12(1 − e x ) − x e x/ . As f ′′ ( x ) = 6 xe x/ ( e x/ − − x/ − x /
24) and it is easy to see that there is a unique solution x ∈ ( −∞ ,
0) of the equation e x/ − − x/ − x /
24 = 0, it follows that f ′′ ( x ) > x < x and f ′′ ( x ) < x < x <
0. One then deduces easily via the expression of f ′ ( x ) and the observation f ′ (0) = 0 that f ′ ( x ) > x <
0. It follows from this and f (0) = 0 that f ( x ) < x < g ( x ) > x <
0, where g ( x ) = 6 x (1 + e x ) + 12(1 − e x ) − x . As g ′′ ( x ) = 6 x ( e x − > x < g ′ (0) = 0, we see that g ′ ( x ) < x < g (0) = 0 that g ( x ) > x < (cid:3) Main Results
We first determine the best possible value for a ( q, s ) in (1.3). For this, for any q > , t > s > I ψ q ( s, t ) as the integral ψ q mean of s and t : I ψ q ( s, t ) = ψ − q (cid:18) t − s Z ts ψ q ( u ) du (cid:19) . (3.1)Then we have the following result: PENG GAO
Theorem 3.1.
For every q > , x > , t > s > , we have ψ q (cid:0) x + I ψ q ( s, t ) (cid:1) < t − s Z ts ψ q ( x + u ) du, where the constant I ψ q ( s, t ) is best possible.Proof. We note that the case q = 1 of the assertion of the theorem is already established in [9,Thereom 4]. The general case can be established similarly, on noting that the function x I ψ q ( x + s, x + t ) − x is increasing by Theorem 4 of [10], in view that ψ ′ q is completely monotonic on (0 , + ∞ ). Onconsidering the case x → + , we see immediately that the constant I ψ q ( s, t ) is best possible andthis completes the proof. (cid:3) On setting t = 1 in Theorem 3.1, we readily deduce the following result concerning the bestpossible value a ( q, s ) in (1.3): Corollary 3.1.
Let q > and < s < . The first inequality of (1.3) holds for all x > with thebest possible value a ( q, s ) = I ψ q ( s, , where I ψ q is defined as in (3.1) . Now to determine the best possible value for b ( q, s ) in (1.3), we note that it is easy to see onconsidering the case x → + ∞ that the best possible value for b ( q, s ) is (1 + s ) / q >
1. When0 < q <
1, we have the following result:
Theorem 3.2.
Let < q < and < s < . Let b ( q, s ) = ln q s − q ( s −
1) ln q ln q . For x > , let f q,s,c ( x ) = ln Γ q ( x + 1) − ln Γ q ( x + s ) − (1 − s ) ψ q ( x + c ) , where c > . Then − f q,s,c ( x ) is completely monotonic on (0 , + ∞ ) if and only if c ≥ b ( q, s ) .Proof. We have, using (2.1), that f ′ q,s,b ( q,s ) ( x ) = ψ q ( x + 1) − ψ q ( x + s ) − (1 − s ) ψ ′ q ( x + b ( q, s ))= ln q ∞ X n =1 q nx − q n (cid:16) q n − q ns − (1 − s )(ln q n ) q nb ( q,s ) (cid:17) . We want to show q n − q ns − (1 − s )(ln q n ) q nb ( q,s ) ≤
0, which is equivalent to E s − ( n ( s − , q, ≥ q b ( q,s ) − , where E is defined as in Lemma 2.2. It also follows from Lemma 2.2 that E s − ( n ( s − , q, ≥ E s − ( s − , q,
1) = q b ( q,s ) − . We then deduce that f ′ q,s,c ( x ) is completely monotonicon (0 , + ∞ ) when c ≥ b ( q, s ). This together with the observation that lim x → + ∞ f q,s,c ( x ) = 0 impliesthe “if” part of the assertion of the theorem.To show the “only if” part of the assertion of the theorem, we use (2.2) and (2.3) to deduce that f q,s,c ( x + 1) − f q,s,c ( x ) = ln 1 − q x +1 − q x + s + (1 − s ) ln q q x + c − q x + c . If we set z = q x and consider the Taylor expansion of the above expression at z = 0, then the firstorder term is: ( q s − q + (1 − s ) (ln q ) q c ) z. Note that the expression in the parenthesis above is < c < b ( q, s ) as it is 0 when c = b ( q, s ).This implies that f q,s,c ( x + 1) < f q,s,c ( x ) when x is large enough and this shows that − f q,s,c ( x ) OME COMPLETELY MONOTONIC FUNCTIONS INVOLVING THE q -GAMMA FUNCTION 5 can’t be completely monotonic on (0 , + ∞ ) when c < b ( q, s ) and this completes the proof of the“only if” part of the assertion of the theorem. (cid:3) Theorem 3.2 now allows us to determine the best possible value of b ( q, s ) in (1.3) when 0 < q < Corollary 3.2.
Let < q < and < s < . The inequality Γ q ( x + 1)Γ q ( x + s ) < e (1 − s ) ψ q ( x + b ( q,s )) (3.2) holds for all x > with the best possible value b ( q, s ) given as in the statement of Theorem 3.2.Proof. Using the same notions in the proof of Theorem 3.2, we see from the proof of Theorem 3.2that f ′ q,s,b ( q,s ) ( x ) > x >
0, which implies the strict inequality in (3.2). To show b ( q, s ) is bestpossible, we note that in the proof of Theorem 3.2, we’ve shown that f q,s,c ( x + 1) − f q,s,c ( x ) < x large enough if c < b ( q, s ). It follows that f q,s,c ( x + k ) − f q,s,c ( x ) < k when x is large enough and c < b ( q, s ). On letting k → + ∞ , we see immediately that thisimplies that − f q,s,c ( x ) <
0, so that inequality (3.2) fails to hold with b ( q, s ) being replaced by any c < b ( q, s ) and this completes the proof. (cid:3) We note here that Corollary 3.2 refines a result of Ismail and Muldoon in [14], mentioned inthe introduction of this paper, where b ( q, s ) is replaced by (1 + s ) / b ( q, s ) ≤ (1 + s ) /
2, as it follows from E ( s − , q, ≤ E (0 , q, q → − , b ( q, s ) → (1 + s ) / q = 1.Our next result is a q -analogue of the result of Alzer and and Batir [5] mentioned in Section 1. Theorem 3.3.
Let < q < be fixed. Let c ≥ . Let a q = ( q − − ln q ) / (ln q ) . The function g q,c ( x ) = ψ q ( x ) − ln 1 − q x − q + a q ψ ′ q ( x + c ) is completely monotonic on (0 , + ∞ ) if and only if c = 0 .Proof. We have, using (2.1), that g q,c ( x ) = ln q ∞ X n =1 q nx − q n (cid:18) − q n n ln q + a q (ln q n ) q nc (cid:19) . On setting t = − ln q n , we have t ≥ − ln q and the expression in the parenthesis above when c = 0can be rewritten as 1 − − e − t t − a q t = 1 t ( − t + e − t − a q t ) := h q ( t ) /t. It suffices to show h q ( t ) ≤ t ≥ − ln q . For this, note that h q ( − ln q ) = 0 and that h ′ q ( t ) = 1 − a q t − e − t , h ′′ q ( t ) = − a q + e − t . We have (ln q ) h ′′ q ( − ln q )2 = q (ln q ) q + 1 − q, and the right-hand side expression above is easily seen to be < < q <
1. As h (3) q ( t ) < t ≥ − ln q , we conclude that h ′′ q ( t ) < t ≥ − ln q . It’s also easy to see that h ′ q ( − ln q ) < h ′ q ( t ) < t ≥ − ln q and this implies h q ( t ) ≤ t ≥ − ln q , whichcompletes the proof of the “if” part of the assertion of the theorem. PENG GAO
For the “only if” part of the assertion of the theorem, note that we have by (2.3) and (2.4), g q,c ( x + 1) − g q,c ( x ) = − (ln q ) q x − q x − ln 1 − q x +1 − q x − a q (ln q ) q x + c (1 − q x + c ) . If we set z = q x and consider the Taylor expansion of the above expression at z = 0, then the firstorder term is:( − ln q + q − − a q (ln q ) q c ) z = ( h q ( − ln q ) + a q (ln q ) − a q (ln q ) q c ) t > , if c >
0. This implies that g q,c ( x + 1) > g q,c ( x ) when x is large enough and this shows that g q,c ( x )can’t be completely monotonic on (0 , + ∞ ) when c > (cid:3) Similar to Theorem 3.3, one can prove the following result, whose proof we leave to the reader.
Theorem 3.4.
Let < q < be fixed. Let c ≥ . The function x ψ q ( x ) − ln 1 − q x − q + 12 ψ ′ q ( x + c ) is completely monotonic on (0 , + ∞ ) if c = 0 and its negative is completely monotonic on (0 , + ∞ ) if c ≥ / . Related to the function given in (1.4), we have the following q -analogue: Theorem 3.5.
Let < q < be fixed, the functions − ψ q ( x ) + ln (cid:16) − q x − q (cid:17) + (ln q ) q x − q x ) + 112 ψ ′′ q ( x + 1 / , (3.3) ψ q ( x ) − ln (cid:16) − q x − q (cid:17) − (ln q ) q x − q x ) − ψ ′′ q ( x )(3.4) are completely monotonic on (0 , + ∞ ) .Proof. The function given in (3.3) being completely monotonic on (0 , + ∞ ) follows from (2.5) and(2.1). As by (2.1), we have ψ q ( x ) − ln (cid:16) − q x − q (cid:17) − (ln q ) q x − q x ) − ψ ′′ q ( x + 1 /
2) = ∞ X n =1 (cid:16) ln q − q n + 1 n − ln q − n (ln q ) q n/ − q n ) (cid:17) q nx . Similarly, the function given in (3.4) being completely monotonic on (0 , + ∞ ) follows from (2.6)and (2.1). (cid:3) Our next result is motivated by (1.5) and (1.6):
Theorem 3.6.
Let < q < be fixed, the functions ψ ′ q ( x ) − (ln q ) q x (1 − q )(1 − q x ) − (ln q ) q x (1 + q )(1 − q x ) , (3.5) − ψ ′ q ( x + 1 /
2) + (ln q ) q x +1 / (1 − q )(1 − q x )(3.6) are completely monotonic on (0 , + ∞ ) .Proof. To show the function given in (3.5) is completely monotonic on (0 , + ∞ ), we note that q x (1 − q x ) = ∞ X n =1 nq nx . OME COMPLETELY MONOTONIC FUNCTIONS INVOLVING THE q -GAMMA FUNCTION 7 Using this and (2.1), we can recast (3.5) as ψ ′ q ( x ) − (ln q ) q x (1 − q )(1 − q x ) − (ln q ) q x (1 + q )(1 − q x ) = (ln q ) (cid:16) ∞ X n =1 nq nx − q n − − q ∞ X n =1 q nx − q x q ∞ X n =1 nq nx (cid:17) = (ln q ) ∞ X n =2 (cid:16) n − q n − − q − n −
11 + q (cid:17) q nx = (ln q ) ∞ X n =2 (cid:16) u n ( q )(1 − q n )(1 − q )(1 + q ) (cid:17) q nx , where u n ( q ) = n (1 − q ) − (1 + q )(1 − q n ) − ( n − − q )(1 − q n ) = n (1 − q )( q + q n ) − q (1 − q n ) . It suffices to show that u n ( q ) ≥ n ≥
2, 0 < q <
1, or equivalently, n (1 + q n − ) ≥ − q n − q = 2 n − X i =0 q i . It is easy to see that the function q P n − i =0 q i − nq n − is an increasing function of 0 < q ≤ q = 1, we see that it implies u n ( q ) ≥ n ≥ < q < , + ∞ ), we use (2.1) to get ψ ′ q ( x + 1 / − (ln q ) q x +1 / (1 − q )(1 − q x ) = (ln q ) ∞ X n =1 (cid:16) nq n/ − q n − q / − q (cid:17) q nx . It suffices to show that nq n/ − / ≤ (1 − q n ) / (1 − q ) = P n − i =0 q i for 0 < q <
1. This follows by notingthat 2 P n − i =0 q i = P n − i =0 ( q i + q n − i − ) and that q i + q n − i − ≥ q n/ − / by the arithmetic-geometricinequality and this completes the proof. (cid:3) Corollary 3.3.
Let < q < be fixed, then for x > , we have ψ ′ q ( x + 1 / ≤ (ln q ) q x +1 / (1 − q )(1 − q x ) . The above inequality follows readily from Theorem 3.6 on considering the value of the functiongiven in (3.6) as x → + ∞ . As it’s easy to see that − (ln q ) q / < − q when 0 < q <
1, the aboveinequality gives a refinement of inequality (1.5).
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