- Email: [email protected]
Monotonicity properties of the gamma function
Applied Mathematics Letters 20 (2007) 778–781 www.elsevier.com/locate/aml
Monotonicity properties of the gamma function Horst Alzer a,∗ , Necdet Batir b a Morsbacher Str. 10, 51545 Waldbr¨ol, Germany b Department of Mathematics, Faculty of Arts and Sciences, Yuzuncu Yil University, 65080, Van, Turkey
Received 18 May 2006; received in revised form 16 July 2006; accepted 1 August 2006
Abstract Let G c (x) = log 0(x) − x log x + x −
1 1 log(2π) + ψ(x + c) 2 2
(x > 0; c ≥ 0).
We prove that G a is completely monotonic on (0, ∞) if and only if a ≥ 1/3. Also, −G b is completely monotonic on (0, ∞) if and only if b = 0. An application of this result reveals that the best possible nonnegative constants α, β in √ √ 1 1 2π x x exp −x − ψ(x + α) < 0(x) < 2π x x exp −x − ψ(x + β) (x > 0) 2 2 are given by α = 1/3 and β = 0. c 2006 Elsevier Ltd. All rights reserved.
Keywords: Gamma function; Psi function; Complete monotonicity; Inequalities
1. Introduction A function f : (0, ∞) → R is said to be completely monotonic if f has derivatives of all orders and (−1)n f (n) (x) ≥ 0
for x > 0 and n = 0, 1, 2, . . . .
(1.1)
If f is nonconstant and completely monotonic, then the inequality in (1.1) is strict; see [1]. Completely monotonic functions have remarkable applications in various fields, like, for instance, probability theory and numerical analysis. The most important facts on these functions are collected in [2, Chapter IV]. We also refer the reader to [3] and the list of references given therein. It is the aim of this note to present monotonicity properties of the function G c (x) = log 0(x) − x log x + x −
1 1 log(2π ) + ψ(x + c) 2 2
(x > 0; c ≥ 0).
∗ Corresponding author.
E-mail addresses: [email protected] (H. Alzer), necdet [email protected] (N. Batir). c 2006 Elsevier Ltd. All rights reserved. 0893-9659/$ - see front matter doi:10.1016/j.aml.2006.08.026
(1.2)
H. Alzer, N. Batir / Applied Mathematics Letters 20 (2007) 778–781
779
Here, 0 denotes Euler’s gamma function, Z ∞ e−t t x−1 dt (x > 0), 0(x) = 0
and ψ is the logarithmic derivative of 0, known as the psi or digamma function, Z ∞ −t e − e−xt 0 0 (x) = −γ + dt (γ = Euler’s constant). ψ(x) = 0(x) 1 − e−t 0 In the next section we present all nonnegative parameters a, b such that G a and −G b are completely monotonic. An application of our result leads to sharp upper and lower bounds for 0(x) in terms of the ψ-function. More precisely, we determine the smallest number α and the largest number β such that the double inequality √ √ 1 1 x x 2π x exp −x − ψ(x + α) < 0(x) < 2π x exp −x − ψ(x + β) (1.3) 2 2 is valid for all x > 0. 2. Main result The following monotonicity theorem holds. Theorem. Let a, b ≥ 0. The function G a , as defined in (1.2), is completely monotonic if and only if a ≥ 1/3. Also, −G b is completely monotonic if and only if b = 0. Proof. Let x > 0 and c ≥ 0. Differentiation gives 1 1 1 G 0c (x) = ψ(x) − log x + ψ 0 (x + c) and G 00c (x) = ψ 0 (x) − + ψ 00 (x + c). 2 x 2 Applying the integral formulas Z ∞ Z ∞ 1 tn e−xt dt, ψ (n) (x) = (−1)n+1 e−xt = dt (x > 0; n ∈ N) x 1 − e−t 0 0 (see [4, p. 260]), we obtain the representation Z ∞ ∆c (t) 00 e−xt G c (x) = dt, 1 − e−t 0
(2.1)
where t 2 −ct e . 2 Moreover, from the asymptotic formulas 1 1 1 log 0(x) ∼ x − log x − x + log(2π ) + + ···, 2 2 12x 1 n! 1 (n) n−1 (n − 1)! − + ψ(x) ∼ log x − + · · · , ψ (x) ∼ (−1) + · · · 2x xn 12x 2 2x n+1 ∆c (t) = −1 + t + e−t −
(x → ∞; n ∈ N)
(see [4, pp. 257,259,260]), we get the limit relations lim G c (x) = lim G 0c (x) = 0.
x→∞
(2.2)
x→∞
Let a ≥ 1/3. A short calculation yields for t > 0: ∆a (t) ≥ −1 + t + e−t −
∞ X t 2 −t/3 δk e = e−t 2 k! k=4
k t 3
780
H. Alzer, N. Batir / Applied Mathematics Letters 20 (2007) 778–781
with δk = (k − 1)(3k − 9k2k−3 ). Since δk > 0 for k ≥ 4, it follows that ∆a is positive on (0, ∞). Applying (2.1) gives that G a00 is completely monotonic. Also, from (2.2) we obtain G a0 (x) < 0
and
G a (x) > 0,
which reveals that G a is completely monotonic. Conversely, if G a is completely monotonic, then we get for x > 0: 0 < x G a (x) =
1 x + x H (x) − [log x − ψ(x + a)], 12 2
(2.3)
where 1 1 1 log x + x − log(2π ) − . H (x) = log 0(x) − x − 2 2 12x
Since lim x H (x) = 0
x→∞
and
we conclude from (2.3) that 1 1 1 0≤ + a− 12 2 2
lim x[log x − ψ(x)] =
x→∞
or a ≥
1 , 2
1 . 3
We have for t > 0: ∆0 (t) = −
∞ (k − 1)(k − 2) k e−t X t < 0, 2 k=3 k!
so that (2.1) yields that −G 000 is completely monotonic. Using (2.2) we obtain −G 00 (x) < 0
and
− G 0 (x) > 0.
Thus, −G 0 is completely monotonic. We assume that −G b (with b > 0) is completely monotonic. Then G b is negative on (0, ∞). But, this contradicts lim G b (x) = ∞.
x→0
The proof of the theorem is complete.
We are now in a position to provide the sharp constants α, β in (1.3). Corollary. Let α, β ≥ 0 be real numbers. For all x > 0 we have √ √ 1 1 x x 2π x exp −x − ψ(x + α) < 0(x) < 2π x exp −x − ψ(x + β) 2 2 with the best possible constants α = 1/3 and β = 0. Proof. From the theorem we obtain for x > 0: G 0 (x) < 0 < G 1/3 (x), which is equivalent to (2.4) with α = 1/3 and β = 0. If the left-hand side of (2.4) is valid, then we have 0 < x G α (x)
(x > 0).
(2.4)
781
H. Alzer, N. Batir / Applied Mathematics Letters 20 (2007) 778–781
As shown in the proof of the theorem this leads to α ≥ 1/3. And, if we assume that there exists a positive number β such that the right-hand side of (2.4) holds for all x > 0, then we obtain √ 1 lim 0(x) ≤ 2π exp − ψ(β) . x→0 2 A contradiction! Hence, the best possible constants in (2.4) are given by α = 1/3 and β = 0.
Remarks. (1) Upper and lower bounds for the ratio 0(x + 1)/ 0(x + s) in terms of the psi function or its derivative can be found in [5–8]. (2) A detailed bibliography on inequalities for the gamma and related functions is given in the survey paper [9]. Acknowledgements We thank the referees for helpful comments. References J. Dubourdieu, Sur un th´eor`eme de M.S. Bernstein relatif a´ la transformation de Laplace–Stieltjes, Compos. Math. 7 (1939) 96–111. D.V. Widder, The Laplace Transform, Princeton Univ. Press, Princeton, 1941. H. Alzer, C. Berg, Some classes of completely monotonic functions, Ann. Acad. Sci. Fenn. 27 (2002) 445–460. M. Abramowitz, I.A. Stegun (Eds.), Handbook of Mathematical Functions with Formulas and Mathematical Tables, Dover, New York, 1965. H. Alzer, Some gamma function inequalities, Math. Comp. 60 (1993) 337–346. J. Bustoz, M.E.H. Ismail, On gamma function inequalities, Math. Comp. 47 (1986) 659–667. D. Kershaw, Some extensions of W. Gautschi’s inequalities for the gamma function, Math. Comp. 41 (1983) 607–611. M. Merkle, Convexity, Schur-convexity and bounds for the gamma function involving the digamma function, Rocky Mountain J. Math. 28 (1998) 1053–1066. [9] W. Gautschi, The incomplete gamma function since Tricomi, in: Tricomi’s Ideas and Contemporary Applied Mathematics, Atti Convegni Lincei, vol. 147, Accad. Naz. Lincei, Rome, 1998, pp. 207–237. [1] [2] [3] [4] [5] [6] [7] [8]
Monotonicity properties of the gamma function Horst Alzer a,∗ , Necdet Batir b a Morsbacher Str. 10, 51545 Waldbr¨ol, Germany b Department of Mathematics, Faculty of Arts and Sciences, Yuzuncu Yil University, 65080, Van, Turkey
Received 18 May 2006; received in revised form 16 July 2006; accepted 1 August 2006
Abstract Let G c (x) = log 0(x) − x log x + x −
1 1 log(2π) + ψ(x + c) 2 2
(x > 0; c ≥ 0).
We prove that G a is completely monotonic on (0, ∞) if and only if a ≥ 1/3. Also, −G b is completely monotonic on (0, ∞) if and only if b = 0. An application of this result reveals that the best possible nonnegative constants α, β in √ √ 1 1 2π x x exp −x − ψ(x + α) < 0(x) < 2π x x exp −x − ψ(x + β) (x > 0) 2 2 are given by α = 1/3 and β = 0. c 2006 Elsevier Ltd. All rights reserved.
Keywords: Gamma function; Psi function; Complete monotonicity; Inequalities
1. Introduction A function f : (0, ∞) → R is said to be completely monotonic if f has derivatives of all orders and (−1)n f (n) (x) ≥ 0
for x > 0 and n = 0, 1, 2, . . . .
(1.1)
If f is nonconstant and completely monotonic, then the inequality in (1.1) is strict; see [1]. Completely monotonic functions have remarkable applications in various fields, like, for instance, probability theory and numerical analysis. The most important facts on these functions are collected in [2, Chapter IV]. We also refer the reader to [3] and the list of references given therein. It is the aim of this note to present monotonicity properties of the function G c (x) = log 0(x) − x log x + x −
1 1 log(2π ) + ψ(x + c) 2 2
(x > 0; c ≥ 0).
∗ Corresponding author.
E-mail addresses: [email protected] (H. Alzer), necdet [email protected] (N. Batir). c 2006 Elsevier Ltd. All rights reserved. 0893-9659/$ - see front matter doi:10.1016/j.aml.2006.08.026
(1.2)
H. Alzer, N. Batir / Applied Mathematics Letters 20 (2007) 778–781
779
Here, 0 denotes Euler’s gamma function, Z ∞ e−t t x−1 dt (x > 0), 0(x) = 0
and ψ is the logarithmic derivative of 0, known as the psi or digamma function, Z ∞ −t e − e−xt 0 0 (x) = −γ + dt (γ = Euler’s constant). ψ(x) = 0(x) 1 − e−t 0 In the next section we present all nonnegative parameters a, b such that G a and −G b are completely monotonic. An application of our result leads to sharp upper and lower bounds for 0(x) in terms of the ψ-function. More precisely, we determine the smallest number α and the largest number β such that the double inequality √ √ 1 1 x x 2π x exp −x − ψ(x + α) < 0(x) < 2π x exp −x − ψ(x + β) (1.3) 2 2 is valid for all x > 0. 2. Main result The following monotonicity theorem holds. Theorem. Let a, b ≥ 0. The function G a , as defined in (1.2), is completely monotonic if and only if a ≥ 1/3. Also, −G b is completely monotonic if and only if b = 0. Proof. Let x > 0 and c ≥ 0. Differentiation gives 1 1 1 G 0c (x) = ψ(x) − log x + ψ 0 (x + c) and G 00c (x) = ψ 0 (x) − + ψ 00 (x + c). 2 x 2 Applying the integral formulas Z ∞ Z ∞ 1 tn e−xt dt, ψ (n) (x) = (−1)n+1 e−xt = dt (x > 0; n ∈ N) x 1 − e−t 0 0 (see [4, p. 260]), we obtain the representation Z ∞ ∆c (t) 00 e−xt G c (x) = dt, 1 − e−t 0
(2.1)
where t 2 −ct e . 2 Moreover, from the asymptotic formulas 1 1 1 log 0(x) ∼ x − log x − x + log(2π ) + + ···, 2 2 12x 1 n! 1 (n) n−1 (n − 1)! − + ψ(x) ∼ log x − + · · · , ψ (x) ∼ (−1) + · · · 2x xn 12x 2 2x n+1 ∆c (t) = −1 + t + e−t −
(x → ∞; n ∈ N)
(see [4, pp. 257,259,260]), we get the limit relations lim G c (x) = lim G 0c (x) = 0.
x→∞
(2.2)
x→∞
Let a ≥ 1/3. A short calculation yields for t > 0: ∆a (t) ≥ −1 + t + e−t −
∞ X t 2 −t/3 δk e = e−t 2 k! k=4
k t 3
780
H. Alzer, N. Batir / Applied Mathematics Letters 20 (2007) 778–781
with δk = (k − 1)(3k − 9k2k−3 ). Since δk > 0 for k ≥ 4, it follows that ∆a is positive on (0, ∞). Applying (2.1) gives that G a00 is completely monotonic. Also, from (2.2) we obtain G a0 (x) < 0
and
G a (x) > 0,
which reveals that G a is completely monotonic. Conversely, if G a is completely monotonic, then we get for x > 0: 0 < x G a (x) =
1 x + x H (x) − [log x − ψ(x + a)], 12 2
(2.3)
where 1 1 1 log x + x − log(2π ) − . H (x) = log 0(x) − x − 2 2 12x
Since lim x H (x) = 0
x→∞
and
we conclude from (2.3) that 1 1 1 0≤ + a− 12 2 2
lim x[log x − ψ(x)] =
x→∞
or a ≥
1 , 2
1 . 3
We have for t > 0: ∆0 (t) = −
∞ (k − 1)(k − 2) k e−t X t < 0, 2 k=3 k!
so that (2.1) yields that −G 000 is completely monotonic. Using (2.2) we obtain −G 00 (x) < 0
and
− G 0 (x) > 0.
Thus, −G 0 is completely monotonic. We assume that −G b (with b > 0) is completely monotonic. Then G b is negative on (0, ∞). But, this contradicts lim G b (x) = ∞.
x→0
The proof of the theorem is complete.
We are now in a position to provide the sharp constants α, β in (1.3). Corollary. Let α, β ≥ 0 be real numbers. For all x > 0 we have √ √ 1 1 x x 2π x exp −x − ψ(x + α) < 0(x) < 2π x exp −x − ψ(x + β) 2 2 with the best possible constants α = 1/3 and β = 0. Proof. From the theorem we obtain for x > 0: G 0 (x) < 0 < G 1/3 (x), which is equivalent to (2.4) with α = 1/3 and β = 0. If the left-hand side of (2.4) is valid, then we have 0 < x G α (x)
(x > 0).
(2.4)
781
H. Alzer, N. Batir / Applied Mathematics Letters 20 (2007) 778–781
As shown in the proof of the theorem this leads to α ≥ 1/3. And, if we assume that there exists a positive number β such that the right-hand side of (2.4) holds for all x > 0, then we obtain √ 1 lim 0(x) ≤ 2π exp − ψ(β) . x→0 2 A contradiction! Hence, the best possible constants in (2.4) are given by α = 1/3 and β = 0.
Remarks. (1) Upper and lower bounds for the ratio 0(x + 1)/ 0(x + s) in terms of the psi function or its derivative can be found in [5–8]. (2) A detailed bibliography on inequalities for the gamma and related functions is given in the survey paper [9]. Acknowledgements We thank the referees for helpful comments. References J. Dubourdieu, Sur un th´eor`eme de M.S. Bernstein relatif a´ la transformation de Laplace–Stieltjes, Compos. Math. 7 (1939) 96–111. D.V. Widder, The Laplace Transform, Princeton Univ. Press, Princeton, 1941. H. Alzer, C. Berg, Some classes of completely monotonic functions, Ann. Acad. Sci. Fenn. 27 (2002) 445–460. M. Abramowitz, I.A. Stegun (Eds.), Handbook of Mathematical Functions with Formulas and Mathematical Tables, Dover, New York, 1965. H. Alzer, Some gamma function inequalities, Math. Comp. 60 (1993) 337–346. J. Bustoz, M.E.H. Ismail, On gamma function inequalities, Math. Comp. 47 (1986) 659–667. D. Kershaw, Some extensions of W. Gautschi’s inequalities for the gamma function, Math. Comp. 41 (1983) 607–611. M. Merkle, Convexity, Schur-convexity and bounds for the gamma function involving the digamma function, Rocky Mountain J. Math. 28 (1998) 1053–1066. [9] W. Gautschi, The incomplete gamma function since Tricomi, in: Tricomi’s Ideas and Contemporary Applied Mathematics, Atti Convegni Lincei, vol. 147, Accad. Naz. Lincei, Rome, 1998, pp. 207–237. [1] [2] [3] [4] [5] [6] [7] [8]