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A REFINEMENT OF A DOUBLE INEQUALITY FOR THE GAMMA FUNCTION 1 1 FENGQIANDBAI-NIGUO 0 2 Abstract. Inthepaper,wepresentamonotonicityresultofafunctioninvolv- n ingthegammafunctionandthelogarithmicfunction,refineadoubleinequal- a ityforthegammafunction,andimprovesomeknownresultsforboundingthe J gammafunction. 5 2 ] A 1. Introduction C In [10], the following double inequality was complicatedly procured: For x h. (0,1), ∈ t x2+1 x2+2 a <Γ(x+1)< , (1.1) m x+1 x+2 where Γ(x) stands for the classical Euler’s gamma function which may be defined [ for x>0 by 2 ∞ v Γ(x)= tx−1e−tdt. (1.2) Z 5 0 9 The aim of this paper is to simply and concisely generalize, refine and sharpen 4 the double inequality (1.1). 1 Our main results may be stated as the following theorem. . 1 Theorem 1. The function 0 0 lnΓ(x+1) 1 (1.3) ln(x2+1) ln(x+1) : − v is strictly increasing on (0,1), with the limits i X lnΓ(x+1) r lim =γ (1.4) a x→0+ ln(x2+1) ln(x+1) − and lnΓ(x+1) lim =2(1 γ). (1.5) x→1− ln(x2+1) ln(x+1) − − As a result, the double inequality x2+1 α x2+1 β <Γ(x+1)< (1.6) (cid:18) x+1 (cid:19) (cid:18) x+1 (cid:19) 2010 Mathematics Subject Classification. Primary26A48,33B15;Secondary 26D15. Key words and phrases. monotonicity, generalization, refinement, sharpening, inequality, gammafunction,Descartes’SignRule,openproblem,conjecture. The first author was supported in part by the Science Foundation of Tianjin Polytechnic University. ThispaperwastypesetusingAMS-LATEX. 1 2 F.QIANDB.-N.GUO holds on (0,1) if and only if α 2(1 γ) and β γ, where γ =0.57 stands for ≥ − ≤ ··· Euler-Mascheroni’s constant. Consequently, the double inequality (x x )2+1 α⌊x⌋−1 −⌊ ⌋ (x i)<Γ(x+1) (cid:20) x x +1 (cid:21) − −⌊ ⌋ iY=0 (x x )2+1 β⌊x⌋−1 < −⌊ ⌋ (x i) (1.7) (cid:20) x x +1 (cid:21) − −⌊ ⌋ iY=0 holds for x (0, ) N if and only if α 2(1 γ)and β γ, where x represents ∈ ∞ \ ≥ − ≤ ⌊ ⌋ the largest integer less than or equal to x. In Section 2, we cite three lemmas for proving in Section 3 Theorem 1. In Section 4, we compare Theorem 1 with several known results and pose some open problems and conjectures. 2. Lemmas In order to prove Theorem 1, we need the following lemma which can be found in [3], [18, pp. 9–10,Lemma 2.9], [19, p. 71, Lemma 1] or closely-relatedreferences therein. Lemma 1. Let f and g be continuouson [a,b] anddifferentiable on (a,b)suchthat g′(x) = 0 on (a,b). If f′(x) is increasing (or decreasing) on (a,b), then so are the 6 g′(x) functions f(x)−f(b) and f(x)−f(a) on (a,b). g(x)−g(b) g(x)−g(a) We also need the following elementary conclusions. Lemma 2. For x (0,1), we have ∈ x4+4x3 2x2 4x 3<0, − − − x2+1 (x 1) x2+2x 1 (x+1) x2+1 ln >0, − − − x+1 (cid:0) (cid:1) (cid:0) (cid:1) x6+6x5 3x4 16x3 21x2 6x 1<0, − − − − − x5+5x4 2x3 8x2 7x 1<0, − − − − 5x7+34x6+27x5 62x4 205x3 198x2 83x 6<0. − − − − − Proof. For our own convenience,denote the functions above by h (x) for 1 i 5 i ≤ ≤ on [0,1] in order. By Descartes’ Sign Rule, the function h (x) has just one possible positive root. 1 Since h (1)= 4 and h (2)=29, the function h (x) is negative on [0,1]. 1 1 1 − A straightforwardcalculation gives d h (x) (x 1)h (x) 2 1 = − , dx(cid:20)(x+1)(x2+1)(cid:21) (x+1)2(x2+1)2 so the function h2(x) is strictly increasing on [0,1]. Due to h (0) = 1, it is (x+1)(x2+1) 2 derived that h (x)>0 on (0,1). 2 Since h (1)= 40, h (3)=1304, h (1)= 12, 3 3 4 − − h (2)=49, h (1)= 488, h (2)=84, 4 5 5 − A REFINEMENT OF AN INEQUALITY FOR THE GAMMA FUNCTION 3 using Descartes’ Sign Rule again yields the negativity of the functions h (x) for i 3 i 5 on (0,1). The proof of Lemma 2 is complete. (cid:3) ≤ ≤ For our own convenience, we also recite the following double inequality for polygamma functions ψ(k)(x) on (0, ). ∞ Lemma 3. The double inequality (k 1)! k! (k 1)! k! − + <( 1)k+1ψ(k)(x)< − + (2.1) xk 2xk+1 − xk xk+1 holds for x>0 and k N. ∈ For the proof of the inequality (2.1), please refer to [5, p. 131], [6, p. 223, Lemma 2.3], [7, p. 107, Lemma 3], [8, p. 853], [12, p. 55, Theorem 5.11], [13, p. 1625], [16, p. 79], [17, p. 2155, Lemma 3] and closely-related references therein. 3. Proof of Theorem 1 Now we are in a position to prove our main results in Theorem 1. It is easy to see that lnΓ(x+1) 1 lnΓ(x+1) ln x−1 Γ(x+1) = x−1 = ln(x2+1)−ln(x+1) x−11lnxx2++11 ln px−1 xx2++11 q (3.1) ln x−1 Γ(x+1) ln 0−1 Γ(0+1) f(x) f(0) = − = − , lnpx−q1 xx2++11 −ln 0−qp1 002++11 g(x)−g(0) where x2+1 f(x)=ln x−1 Γ(x+1) and g(x)=ln x−1 r x+1 p on [0,1]. Easy computation and simplification yield f′(x) (x+1) x2+1 [(x 1)ψ(x+1) lnΓ(x+1)] = − − g′(x) (x 1)(cid:0)(x2+2x(cid:1) 1) (x+1)(x2+1)lnx2+1 − − − x+1 and d f′(x) (1 x) x4+4x3 2x2 4x 3 q(x) = − − − − , dx(cid:20)g′(x)(cid:21) (x 1) x2+(cid:0)2x 1 (x+1) x2+1(cid:1) lnx2+1 2 − − − x+1 where (cid:2) (cid:0) (cid:1) (cid:0) (cid:1) (cid:3) (x+1) x2+1 q(x)=lnΓ(x+1) (x 1)ψ(x+1) − − − x4+4x3 (cid:0)2x2 4(cid:1)x 3 − − − x2+1 (x 1) x2+2x 1 (x+1) x2+1 ln ψ′(x+1). ×(cid:20) − − − x+1 (cid:21) (cid:0) (cid:1) (cid:0) (cid:1) Further computation and simplification give (1 x) x2+2x 1 +(x+1) x2+1 lnx2+1 q′(x)= − − x+1 q (x), (cid:0) (x4+4x3(cid:1) 2x2 4x(cid:0) 3)2 (cid:1) 1 − − − where q (x)=2 x6+6x5 3x4 16x3 21x2 6x 1 ψ′(x+1) 1 − − − − − (cid:0) +(x+1) x2+1 x4+4x(cid:1)3 2x2 4x 3 ψ′′(x+1) − − − (cid:0) (cid:1)(cid:0) (cid:1) 4 F.QIANDB.-N.GUO and satisfies q′(x)=12 x5+5x4 2x3 8x2 7x 1 ψ′(x+1) 1 − − − − (cid:0)+ x4+4x3 2x2 4x 3 3(cid:1)3x2+2x+1 ψ′′(x+1) − − − (cid:0) (cid:1)(cid:2) (cid:0) +(x+1(cid:1)) x2+1 ψ′′′(x+1) . (cid:0) (cid:1) (cid:3) By virtue of Lemmas 2 and 3, we obtain 2 3 q′(x)< x4+4x3 2x2 4x 3 (x+1) x2+1 + 1 − − − (cid:26) (cid:20)(x+1)3 (x+1)4(cid:21) (cid:0) (cid:1) (cid:0) (cid:1) 1 2 3 3x2+2x+1 + − (cid:20)(x+1)2 (x+1)3(cid:21)(cid:27) (cid:0) (cid:1) 1 1 +12 x5+5x4 2x3 8x2 7x 1 + − − − − (cid:20)x+1 2(x+1)2(cid:21) (cid:0) (cid:1) 5x7+34x6+27x5 62x4 205x3 198x2 83x 6 = − − − − − (x+1)3 <0 on [0,1]. So the function q (x) is strictly decreasing on [0,1]. Since 1 q (0)= 2ψ′(1) 3ψ′′(1)=3.922 1 − − ··· and π2 q (1)=80 1 16ψ′′(2)= 45.128 , 1 (cid:18) − 6 (cid:19)− − ··· the function q (x) has a unique zero on (0,1), and so is the function q′(x). As a 1 result, the function q(x) has a unique minimum on (0,1). Because of 1 π2 q(0)= 3γ = 0.028 3(cid:18) 6 − (cid:19) − ··· and q(1) = 0, we obtain that q(x) < 0 on (0,1). Combining this with Lemma 2 leads to d f′(x) >0 dx(cid:20)g′(x)(cid:21) on (0,1), which means that the function f′(x) is strictly increasing on (0,1). Fur- g′(x) thermore, from Lemma 1 and the equation (3.1), it follows that the function (1.3) is strictly increasing on (0,1). By L’Hospital’s rule, we have lnΓ(x+1) (x+1) x2+1 ψ(x+1) lim = lim x→0+ ln(x2+1) ln(x+1) x→0+ x(cid:0)2+2x(cid:1) 1 − − = ψ(1) − =γ and lnΓ(x+1) (x+1) x2+1 ψ(x+1) lim = lim x→1− ln(x2+1) ln(x+1) x→1− x(cid:0)2+2x(cid:1) 1 − − =2ψ(2) =2(1 γ). − A REFINEMENT OF AN INEQUALITY FOR THE GAMMA FUNCTION 5 Hence, the double inequality (1.6) and its sharpness follow. Thedoubleinequality(1.7)maybededucedfrom(1.6)andtherecurrentformula Γ(x+1)=xΓ(x) for x>0. The proof of Theorem 1 is complete. 4. Remarks Inthissection,wecompareTheorem1withsomeknownresultsandposeseveral open problems and conjectures. Remark 1. It is clear that the double inequality (1.6) refines the double inequality (1.1). Moreover,the inequality (1.6) may be rearrangedas 1 x2+1 2(1−γ) 1 x2+1 γ <Γ(x)< , x (0,1). (4.1) x(cid:18) x+1 (cid:19) x(cid:18) x+1 (cid:19) ∈ Remark 2. In [1, p. 145, Theorem 2], it was obtained that if x (0,1), then ∈ xα(x−1)−γ <Γ(x)<xβ(x−1)−γ (4.2) with the best possible constants 1 π2 α=1 γ =0.42278 and β = γ =0.53385 , (4.3) − ··· 2(cid:18) 6 − (cid:19) ··· and if x (1, ), then (4.2) holds with the best possible constants ∈ ∞ 1 π2 α= γ and β =1. (4.4) 2(cid:18) 6 − (cid:19) In [2, p. 780, Corollary], the following conclusion was established: Let α and β be nonnegative real numbers. For x>0, we have 1 1 √2πxxexp x ψ(x+α) <Γ(x)<√2πxxexp x ψ(x+β) (4.5) (cid:20)− − 2 (cid:21) (cid:20)− − 2 (cid:21) with the best possible constants α= 1 and β =0. 3 In [9, p. 3, Theorem 5], among other things, it was demonstrated that for x ∈ (0,1] we have xx[1−lnx+ψ(x)] xx[1−lnx+ψ(x)] <Γ(x) . (4.6) ex ≤ ex−1 By the well-known software Mathematica Version 7.0.0, we can show that (1) thedoubleinequalities(4.1)and(4.2)arenotincludedeachotheron(0,1), (2) when x>0 is smaller, the double inequalities (4.1) is better than (4.2), (3) the double inequality (4.1) improves (4.5) on (0,1), (4) the left-hand side inequality in (4.1) refines the correspondingone in (4.6), (5) the right-hand side inequalities in (4.1) and (4.6) are not contained each other, (6) whenx>0is smaller,the right-handside inequality in(4.1)is better than the corresponding one in (4.6). Remark 3. In[4,Corollary1.2,Theorem1.4andTheorem1.5],thefollowingsharp inequalities for bounding the gamma function were obtained: For x>0, we have x+1/2 x+1/eγ 1 1 √2 x+ e−x Γ(x+1) eγ/eγ x+ e−x, (4.7) (cid:18) 2(cid:19) ≤ ≤ (cid:18) eγ(cid:19) 6 F.QIANDB.-N.GUO x+1/2 x+1/2 x+1/2 x+1/2 √2e Γ(x+1)<√2π (4.8) (cid:18) e (cid:19) ≤ (cid:18) e (cid:19) and 1 4 √2x+1xxexp x+ <Γ(x+1) (cid:26)−(cid:20) 6(x+3/8) − 9(cid:21)(cid:27) 1 < π(2x+1)xxexp x+ . (4.9) (cid:26)−(cid:20) 6(x+3/8)(cid:21)(cid:27) p By the software Mathematica Version 7.0.0, we can reveal that (1) the double inequalities (1.6) and (4.7) do not include each other on (0,1), (2) the right-hand side inequality in (1.6) is better than the one in (4.8) on (0,1), (3) theleft-handsideinequalitiesin(1.6)and(4.8)arenotincludedeachother on (0,1), (4) the lower bound in (1.6) improves the corresponding one in (4.9), but the right-hand side inequalities in (1.6) and (1.6) do not contain each other on (0,1). Remark 4. It is clear that when x N the inequality (1.7) becomes equality. This ∈ shows us that for x > 1 the double inequality (1.7) is better than those double inequalities listed in the above Remarks 2 and 3. Remark 5. In[15,Theorem1],amongotherthings,itwasprovedthatthefunction lnΓ(x+1) F(x)= (4.10) xln(2x) is both strictly increasing and strictly concave on 1, . By L’Hospital Rule and 2 ∞ the double inequality (2.1) for k =1, we obtain (cid:0) (cid:1) ψ(x+1) lim F(x)= lim = lim xψ′(x+1) =1, x→∞ x→∞1+ln(2x) x→∞ (cid:2) (cid:3) soitfollowsthatΓ(x+1)<(2x)xon 1, ,whichisnotbetterthantheright-hand 2 ∞ side inequality in (1.6) on 1,1 . (cid:0) (cid:1) 2 (cid:0) (cid:1) Remark 6. By similar argumentto the proofof Theorem1, we may provethat the function lnΓ(x+1) (4.11) ln(x2+6) ln(x+6) − is strictly decreasing on (0,1). Consequently, x2+6 6γ x2+6 7(1−γ) <Γ(x+1)< , x (0,1). (4.12) (cid:18) x+6 (cid:19) (cid:18) x+6 (cid:19) ∈ Motivating by monotonic properties of the functions (1.3) and (4.11), we pose the following open problem: What is the largest number λ > 1 (or the smallest number λ<6 respectively) for the function lnΓ(x+1) (4.13) ln(x2+λ) ln(x+λ) − to be strictly increasing (or decreasing respectively) on (0,1)? Remark 7. Finally, we pose the following conjectures. A REFINEMENT OF AN INEQUALITY FOR THE GAMMA FUNCTION 7 (1) Thefunction(1.3)isstrictlyincreasingnotonlyon(0,1)butalsoon(0, ). ∞ (2) For τ >0, the function lnΓ(x) , x=1 ln(x2+τ) ln(x+τ) 6 (4.14)  (1+τ)γ,− x=1 − is strictly increasing with respect to x (0, ). ∈ ∞ (3) Recall from [11, Chapter XIII], [20, Chapter 1] or [21, Chapter IV] that a function f is completely monotonic on an interval I if f has derivatives of all orders on I and 0 ( 1)nf(n)(x)< (4.15) ≤ − ∞ for x I and n 0. 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Vondraˇcek, Bernstein Functions, de Gruyter Studies in Mathematics37,DeGruyter,Berlin,Germany,2010.7 [21] D.V.Widder,The Laplace Transform,PrincetonUniversityPress,PrincetonNJ,1946.7 (F.Qi)DepartmentofMathematics,CollegeofScience,TianjinPolytechnicUniver- sity,Tianjin City, 300160,China E-mail address: [email protected], [email protected], [email protected] URL:http://qifeng618.wordpress.com (B.-N.Guo) School of Mathematicsand Informatics,Henan Polytechnic University, JiaozuoCity, HenanProvince,454010,China E-mail address: [email protected], [email protected]