SOME GAMMA FUNCTION INEQUALITIES NECDET BATIR Abstract. In this paper we derive various new inequalities for the gamma function. Our results improve, refine and extend some previously known in- equalities. 1. INTRODUCTION The gamma Γ and psi ψ (or digamma )functions are defined as Z ∞ Γ0(x) Γ(x) = ux−1e−udu , ψ(x) = 0 Γ(x) for positive real numbers x, respectively. For basic properties of these functions refer to [7, Chapter I]. We call the derivatives ψ0, ψ00, ψ000, .... as polygamma functions. The gamma function play a key role in the proof of Stirling’s formula n! √ lim √ = 2π, n→∞ nne−n n see [10, Theorem 20], and has many applications in many fields of science such as special functions, number theory, statistics and physics. In particular, in the recent past many authors have studied these functions and they obtained many interesting inequalities for them, see([2], [3], [4], [5], [6], [8], [9], [12], [14], [15], [16], [17]). In this paper we aim at presenting several new inequalities for the gamma function. In what follows c = 1.462632144968362 ... denotes the only positive zero of ψ-function(see[1, p.259,(6.3.19)]. In order to prove our main results we need to present some lemmas. Lemma 1.1. We have Γ(x) √ lim = 2π. x→∞ exp ψ(x)eψ(x) − eψ(x) Proof. Using Stirling’s formula we get Γ(x) √ xxe−x lim = 2π lim √ . x→∞ exp(ψ(x)eψ(x) − eψ(x)) x→∞ x exp(ψ(x)eψ(x) − eψ(x)) In [9] the author of this paper proved that lim x − eψ(x) = 1/2. x→∞ 2000 Mathematics Subject Classification. Primary: 33B15; Secondary: 26D07. Key words and phrases. gamma function, digamma function, Gautschi inequality and inequalities. 1 2 NECDET BATIR Thus we have Γ(x) lim x→∞ exp(ψ(x)eψ(x) − eψ(x)) r2π 1 = exp lim x log x − log x − ψ(x)eψ(x) . e x→∞ 2 In [5] it was proved that lim [x(log x − ψ(x))] = 1/2. x→∞ Using this limit we obtain that Γ(x) lim x→∞ exp(ψ(x)eψ(x) − eψ(x)) √ 1 = 2π exp lim ψ(x)(x − eψ(x)) − log x . x→∞ 2 In [5, Eq.(2.2)], it is given 1 1 log x − < ψ(x) < log x − , (x > 0). x 2x Using these bounds, we obtain 1 1 (x log x − 1)(1 − e−1/(2x))− log x < ψ(x)(x − eψ(x)) − log x 2 2 1 1 < (x log x − )(1 − e−1/x) − log x. 2 2 It is not difficult to show that the limits of both of the bounds go to 0 when x tends to ∞, from which we get the desired limit. Lemma 1.2. For x > 0 we have Z ∞ x x2 x √ log Γ(t)dt = − + log 2π − (1 − x) log Γ(x) − log G(x), 0 2 2 2 where G is Barnes G-function defined by the infinite product 1 G(z + 1) = (2π)z/2 exp − [(1 + γ)z2 + z] 2 ∞ Y z k z2 × 1 + exp −z + k 2k k=1 and γ is Euler-Mascheroni constant given by SOME GAMMA FUNCTION INEQUALITIES 3 Xn 1 γ = lim − log n = 0.577215664.... n→∞ k=1 k Barnes G-function satisfies the functional equation G(1) = 1 and G(x + 1) = Γ(x)G(x), see [11]. Let I be an interval, s, t ∈ I and f : I → R continuous and strictly monotone. Then by mean value theorem for integration there exists a unique ξ ∈ [s, t] for which 1 Z t f(u)du = f(ξ). t − s s ξ is called integral f-mean of s and t, and denoted by Z t −1 1 If = If (s, t) = f f(u)du . t − s s If has the following property: Lemma 1.3. 1. Let f ∈ C(2)(I), f is increasing, f 0 is decreasing and f 00/f 0 is increasing. Then we have that x → If (x + s, x + t) − x is increasing. This is a known result and proved by Elezovic and Pecaric[13, Theorem 4]. 2. MAIN RESULTS In this section we collect our main results. Our first result gives new bounds in terms of psi function for the gamma function. Theorem 2.1. The following inequalities for the gamma function hold: For x ≥ c α exp ψ(x)eψ(x) − eψ(x) < Γ(x) < β exp ψ(x)eψ(x) − eψ(x) , √ where α = eΓ(c) = 2.4073190705. and β = 2π = 2.5066282746. are best possi- ble constants and for 0 < a ≤ x ≤ c eΓ(c) exp[P (x)] ≤ Γ(x) < Γ(a) exp[P (x) − P (a)], where P (t) = eψ(t)(ψ(t) − 1). Proof. Set Γ(x) u(x) = . exp(ψ(x)eψ(x) − eψ(x)) Differentiation gives Γ(x)ψ(x)(1 − ψ0(x)eψ(x)) u0(x) = . exp ψ(x)eψ(x) − eψ(x) Since ψ0(x)eψ(x) < 1 by [9, Lemma 1.2] for x > 0, ψ(x) ≥ 0 for x ≥ c and ψ(x) ≤ 0 4 NECDET BATIR for x ≤ c, u √is strictly increasing on [c, ∞) and strictly decreasing on (0, c] . Since lim u(x) = 2π by Lemma 1.1, we have for x ≥ c x→∞ √ eΓ(c) = u(c) ≤ u(x) < lim u(x) = 2π, x ≥ c. x→∞ This proves the first part of Theorem 2.1. Similarly if 0 < a ≤ x ≤ c we have u(c) < u(x) ≤ u(a).Using the definition of u, we prove the second part of Theorem 2.1. The author of present paper[9] proved the inequalities for the gamma function: Γ(x + 1/2) > Γ(c)xxe−x+1, x ≥ 1/2 and Γ(x + c − 1) ≤ Γ(c)xkxe−kx, x ≥ c − 1 where k = 6eγ /π2 = 1.08... and γ is Euler constant. The following theorem extend, refine and improve these inequalities. Theorem 2.2. For x > 1 the following inequalities hold ϕ(1) Γ(x + c − 1) < Γ(c) xxe1−x (2.1) and ϕ(x) Γ(x + 1/2) > Γ(c) xxe1−x , (2.2) and for 0 < a ≤ x ≤ 1 ϕ(x) Γ(x + c − 1) < Γ ψ−1(log x) a−axxea−x (2.3) and ϕ(a) Γ(x + a + ψ−1(log a) − 1) ≥ Γ ψ−1(log a) a−axxea−x , (2.4) where 1 ϕ(x) = . (2.5) xψ0 (ψ−1(log x)) Proof. Applying mean value theorem to log (Γ(x)) on the interval (u, u + 1), there exists a θ0 such that 0 < θ0 = θ0(u) < 1 and ψ(u + θ0(u)) = log u. (2.6) It is easy to deduce from this identity that ψ(u) ψ(u) θ0 e = u − e . (2.7) SOME GAMMA FUNCTION INEQUALITIES 5 Integrating both sides of (2.6) on the interval 1 ≤ u ≤ x we get Z x ψ(u + θ0(u))du = x log x − x + 1. (2.8) 1 Inducing the change of variable u = eψ(t) here and using (2.6), we obtain Z x+θ0(x) ψ(t)ψ0(t)eψ(t)dt = x log x − x + 1, x > 1. (2.9) c In [9, Lemma 1.1], it was proved that (ψ0(x))2 + ψ00(x) > 0 for x > 0, so that mapping t → ψ0(t)eψ(t) is strictly increasing on (0, ∞). Since ψ(t) ≥ 0 for t ≥ c we conclude from (2.9) that Z x+θ0(x) 0 xψ (x + θ0(x)) ψ(t)dt > x log x − x + 1, x > 1 c or 0 −1 x log x − x + 1 ≤ xψ ψ (log x) [log (Γ(x + θ0(x)) − log (Γ(c))] , x > 1 Simplifying this we get for x > 1 x 1−xϕ(x) Γ(x + θ0(x)) > Γ(c) x e . (2.10) In [9] it is proved that x → θ0(x) is strictly increasing on (0, ∞) and lim θ0(x) = x→∞ 1/2. Hence we obtain for x > 1 ϕ(x) Γ(c) xxe1−x ≤ Γ(x + 1/2), proving (2.2). From (2.9) we write 0 x log x − x + 1 ≥ ψ (c) [log (Γ(x + θ0(x)) − log (Γ(c))] , x > 1 or x 1−xϕ(1) Γ(c) x e ≥ Γ(x + θ0(x)), x > 1. (2.11) Since x + θ0(x) > x + θ0(1) = x + c − 1 for x > 1, we conclude for x > 1 that ϕ(1) Γ(c) xxe1−x ≥ Γ(x + c − 1). This proves (2.1). Now we let 0 < a ≤ x ≤ 1. Integrating both sides of (2.6) on the interval a ≤ u ≤ x (x ≤ 1), we find that Z x ψ(u + θ0(u))du = x log x − x − a log a + a. a 6 NECDET BATIR Making the change of variable u = eψ(t) here, we get Z x+θ0(x) (−ψ(t))ψ0(t)eψ(t)dt = −x log x + x + a log a − a. (2.12) a+θ0(a) Since −ψ(t) ≥ 0 for t≤ c, we get from (2.12) that Z x+θ0(x) −x log x + a log a − a ≤ xψ0 ψ−1(log x) (−ψ(t))dt. a+θ0(a) Simplifying this identity we get for 0 < a ≤ x ≤ 1 −1 x a−x −aϕ(x) Γ ψ (log a) x e a ≥ Γ(x + θ0(x)). (2.13) Since x + θ0(x) ≥ x + θ0(1) = x + c − 1 this proves (2.3). Similarly, from (2.12) we get Z x+θ0(x) −x log x + a log a − a ≥ aψ0 ψ−1(log a) (−ψ(t))dt a+θ0(a) or −1 x a−x −aϕ(a) Γ(x + θ0(x) ≥ Γ ψ (log a) x e a .