A refinement of a double inequality for the gamma function

0.12 MB·English

by  Feng Qi

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 inequal-
ity (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. We conjecture that the function
∈ ≥
lnx
, x=1
h(x)=ln(1+x2) ln(1+x) 6 (4.16)
−
2, x=1
is completely monotonic on (0, ).
∞
Remark 8. For the history, backgrounds, origins, developments of bounding the
gamma function, please refer to the expository and survey article [12] and plenty
of references therein.
Remark 9. This paper is a revised version of the preprint [14].
Acknowledgements. Theauthorsappreciateanonymousrefereesfortheirhelpful
and valuable comments on this paper.
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(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]