Journal of Inequalities and Applications

J Inequal Appl. 2017; 2017(1): 210.
Published online 2017 September 8.
PMCID: PMC5591381

# On rational bounds for the gamma function

## Abstract

In the article, we prove that the double inequality

$x2+p0x+p0<Γ(x+1)

holds for all x ∈ (0, 1), we present the best possible constants λ and μ such that

$λ(x2+9/5)x+9/5≤Γ(x+1)≤μ(x2+p0)x+p0$

for all x ∈ (0, 1), and we find the value of x in the interval (0, 1) such that Γ(x + 1) > (x2 + 1/γ)/(x + 1/γ) for x ∈ (0, x) and Γ(x + 1) < (x2 + 1/γ)/(x + 1/γ) for x ∈ (x, 1), where Γ(x) is the classical gamma function, $γ=limn→∞(∑k=1n1/k−logn)=0.577…$ is Euler-Mascheroni constant and p0γ/(1 − γ) = 1.365….

Keywords: gamma function, psi function, rational bound, completely monotonic function

## Introduction

For x > 0, the classical Euler gamma function Γ(x) and its logarithmic derivative, the so-called psi function ψ(x) [1] are defined by

$Γ(x)=∫0∞tx−1e−tdt,ψ(x)=Γ′(x)Γ(x),$

respectively.

A real-valued function f is said to be completely monotonic [2] on an interval I if f has derivatives of all orders on I and (−1)nf(n)(x) ≥ 0 for all n ≥ 0 and x ∈ I. The well-known Bernstein theorem [3] states that a function f on [0, ∞) is completely monotonic if and only if there exists a bounded and non-decreasing function ω(t) such that $f(x)=∫0∞e−xtdω(t)$ converges for all x ∈ [0, ∞).

Recently, the gamma function have attracted the attention of many researchers. In particular, many remarkable inequalities and properties for Γ(x) can be found in the literature [414].

Due to Γ(x + 1) = xΓ(x) and Γ(n + 1) = n, we will only need to focus our attention on Γ(x + 1) with x ∈ (0, 1). Gautschi [15] proved that the double inequality

$n1−s<Γ(n+1)Γ(n+s)
1.1

holds for all s ∈ (0, 1) and n ∈ ℕ.

Inequality (1.1) was generalized and improved by Kershaw [16] as follows:

$(x+s2)1−s<Γ(x+1)Γ(x+s)

for all x > 0 and s ∈ (0, 1).

Elezović, Giordano and Pečarić [17] established the double inequality

$(12+14+x)1−xxx<Γ(x+1)<21−xxx$
1.2

for the gamma function being valid for all x ∈ (0, 1), and asked for ‘other bounds for the gamma function in terms of elementary functions’.

Ivády [18] provided the bounds for gamma function in terms of very simple rational functions as follows:

$x2+1x+1<Γ(x+1)
1.3

for all x ∈ (0, 1). Inequality (1.3) can be regarded as a simple estimation of the value of the gamma function.

In [19], Zhao, Guo and Qi proved that the function

$x→Q(x)=logΓ(x+1)log(x2+1)−log(x+1)$

is strictly increasing on (0, 1). The monotonicity of Q(x) on the interval (0, 1) and the facts that Q(0+) = γ and Q(1) = 2(1 − γ) lead to the conclusion that

$(x2+1x+1)2(1−γ)<Γ(x+1)<(x2+1x+1)γ$
1.4

for all x ∈ (0, 1), where $γ=limn→∞(∑k=1n1/k−logn)=0.577…$ is the Euler-Mascheroni constant.

Let

$L1(x)=x2+1x+1,L2(x)=(x2+1x+1)2(1−γ),$
1.5

$U1(x)=x2+2x+2,U2(x)=(x2+1x+1)γ.$
1.6

Then we clearly see that

L1(x) < L2(x)
1.7

for all x ∈ (0, 1), and numerical computations show that

$U1(1/4)=0.916…>U2(1/4)=0.910…,U1(1/8)=0.948…>U2(1/8)=0.942….$
1.8

Motivated by (1.3)-(1.8), it is natural to ask what the better parameters p and q on the interval (1, 2) are such that the double inequality

$x2+px+p<Γ(x+1)

holds for all x ∈ (0, 1). The main purpose of the article is to deal with this questions. Some complicated computations are carried out using the Mathematica computer algebra system.

## Lemmas

In order to establish our main results we need several lemmas, which we present in this section.

### Lemma 2.1

See [20, Theorem 1.25]

Let −∞ < a < b < ∞, fg:[ab] → ℝ be continuous on [ab] and differentiable on (ab), and g(x) ≠ 0 on (ab). If f(x)/g(x) is increasing (decreasing) on (ab), then so are the functions

$f(x)−f(a)g(x)−g(a),f(x)−f(b)g(x)−g(b).$

If f(x)/g(x) is strictly monotone, then the monotonicity in the conclusion is also strict.

### Lemma 2.2

See [21, Lemma 7]

Let n ∈ ℕ and m ∈ ℕ ∪ {0} with n > m, ai ≥ 0 for all 0 ≤ i ≤ n, anam > 0 and

$Pn(t)=−∑i=0maiti+∑i=m+1naiti.$

Then there exists t0 ∈ (0, ∞) such that Pn(t0) = 0, Pn(t) < 0 for t ∈ (0, t0) and Pn(t) > 0 for t ∈ (t0, ∞).

### Lemma 2.3

See [22, Corollary 3.1]

The inequality

$ψ(x+1)<192log(x2+x+43)+4592log(x2+x+1445)$

holds for all x > 0.

### Lemma 2.4

See [23, Corollary 3.3(ii)]

The double inequality

$(x+12)(x2+x+2321)x4+2x3+177x2+107x+1235<ψ′(x+1)<(x+12)(x2+x+π215(π2−9))x4+2x3+7π2−605(π2−9)x2+2π2−155(π2−9)x+15(π2−9)$

holds for all x > 0.

### Lemma 2.5

The inequalities

$1x+12−112(x+12)3≤ψ′(x+1)≤1x+12,$
2.1

$−1(x+12)2≤ψ″(x+1)≤−1(x+12)2+14(x+12)4,$
2.2

$2(x+12)3−1(x+12)5≤ψ‴(x+1)≤2(x+12)3$
2.3

hold for all x > −1/2.

### Proof

Let x > −1/2, and R1(x) and R2(x) be defined by

R1(x) = ψ(x + 1)−log(x + ½),
2.4

$R2(x)=−ψ(x+1)+log(x+12)+124(x+12)2,$
2.5

respectively. Then making use of the well-known formulas

$ψ(x)=∫0∞(e−tt−e−xt1−e−t)dt,logx=∫0∞e−t−e−xttdt$

we get

$R1(x)=∫0∞(e−tt−e−(x+1)t1−e−t)dt−∫0∞e−t−e−(x+1/2)ttdt=∫0∞(1t−e−t/21−e−t)e−(x+1/2)tdt=∫0∞(1t−12sinh(t/2))e−(x+1/2)tdt=∫0∞sinh(t/2)−t/2tsinh(t/2)e−(x+1/2)tdt,$
2.6

$R2(x)=−∫0∞(1t−12sinh(t/2))e−(x+1/2)tdt+124∫0∞te−(x+1/2)tdt=∫0∞((t2−24)sinh(t/2)+12t24tsinh(t/2))e−(x+1/2)tdt,$
2.7

where sinh(t) = (et − et)/2 is the hyperbolic sine function.

Note that

$6tsinh(t/2)(t2−24)sinh(t/2)+12t24tsinh(t/2)=∑n=3∞2(n−2)(2n+1)(2n−3)!(t2)2n−1>0,$
2.8

$sinh(t/2)−t2>0$
2.9

for t > 0.

It follows from (2.6)-(2.9) and the Bernstein theorem for complete monotonicity property that the two functions R1(x) and R2(x) are completely monotonic on the interval (−1/2, ∞).

Therefore, Lemma 2.5 follows easily from (2.4), (2.5) and the complete monotonicity of R1(x) and R2(x) on the interval (−1/2, ∞) together with the facts that

$[log(x+12)](n)=(−1)n−1(n−1)!(x+12)n,[1(x+12)2](n)=(−1)n(n+1)!(x+12)n+2.$

### Lemma 2.6

The double inequality

$(x2+1x+1)2/(q+1)
2.10

holds for all x ∈ (0, 1) and q > 1.

### Proof

Let x ∈ (0, 1), q > 1, and H1(x) and H2(x) be defined by

H1(x) = log(x2q) − log(xq),  H2(x) = log(x2 + 1) − log(x + 1),
2.11

respectively. Then simple computations lead to

$limx→0+H1(x)H2(x)=1q,limx→1−H1(x)H2(x)=2q+1,$
2.12

H1(0+) = H2(0+) = 0,
2.13

H1(1) = H2(1) = 0,
2.14

$H1′(x)H2′(x)=(x+1)(x2+1)(x2+2qx−q)(x+q)(x2+q)(x2+2x−1),[H1′(x)H2′(x)]′=(q−1)△(x,q)(x+q)2(x2+q)2(x2+2x−1)2,$
2.15

where

$△(x,q)=[8x5+4x4+(1−x)3(4x2+3x+1)]q2+4x2[2x4+4x3+(1−x)2]q−x4(x4−4x3−6x2−4x+1)>△(x,1)=(1−x)x7+3x7+10x6+(2x3−1)2+24x5+8x4+2x2(2x−1)2>0.$
2.16

From (2.15) and (2.16) we clearly see that $H1′(x)/H2′(x)$ is strictly increasing on $(0,2−1)∪(2−1,1)$. We assert that the function H1(x)/H2(x) is strictly increasing on (0, 1). Indeed, if $x∈(0,2−1)$, then $H2′(x)≠0$, and Lemma 2.1 and (2.13) together with the monotonicity of $H1′(x)/H2′(x)$ on $(0,2−1)$ lead to the conclusion that H1(x)/H2(x) is strictly increasing on $(0,2−1)$; if $x∈(2−1,1)$, then $H2′(x)≠0$, and Lemma 2.1 and (2.14) together with the monotonicity of $H1′(x)/H2′(x)$ on $(2−1,1)$ lead to the conclusion that H1(x)/H2(x) is strictly increasing on $(2−1,1)$.

Therefore, Lemma 2.6 follows easily from (2.11) and (2.12) together with the monotonicity of the function H1(x)/H2(x) on (0, 1).

Let p > 0, x ∈ (0, 1), and f(px), f1(px), f2(px) and f3(px) be defined by

$f(p,x)=logΓ(x+1)−log(x2+px+p),$
2.17

$f1(p,x)=∂f(p,x)∂x=ψ(x+1)−2xx2+p+1x+p,$
2.18

$f2(p,x)=∂2f(p,x)∂x2=ψ′(x+1)+4x2(x2+p)2−2x2+p−1(x+p)2,$
2.19

$f3(p,x)=∂3f(p,x)∂x3=ψ″(x+1)−16x3(x2+p)3+12x(x2+p)2+2(x+p)3.$
2.20

### Lemma 2.7

Let f2(px) be defined by (2.19). Then

f2(p, 1/3) < 0
2.21

for p ∈ [8/5, 9/5].

### Proof

From (2.19) and the second inequality in Lemma 2.4 we have

$f2(p,1/3)=[ψ′(x+1)+4x2(x2+p)2−2x2+p−1(x+p)2]x=1/3<[(x+12)(x2+x+π215(π2−9))x4+2x3+7π2−605(π2−9)x2+2π2−155(π2−9)x+15(π2−9)]x=1/3+[4x2(x2+p)2−2x2+p−1(x+p)2]x=1/3=15(23π2−180)2(152π2−1,179)−9(162p3+171p2+24p−1)(3p+1)2(9p+1)2.$
2.22

$(15(23π2−180)2(152π2−1,179)−9(162p3+171p2+24p−1)(3p+1)2(9p+1)2)′=54(729p4+1,215p3+243p2−27p−8)(3p+1)3(9p+1)3>0$
2.23

for p ∈ [8/5, 9/5].

From (2.22) and (2.23) we get

$f2(p,1/3)<[15(23π2−180)2(152π2−1,179)−9(162p3+171p2+24p−1)(3p+1)2(9p+1)2]p=9/5=15(23π2−180)2(152π2−1,179)−2,167,0651,893,376=−0.047…<0.$

### Lemma 2.8

Let f2(px) be defined by (2.19). Then

f2(9/5, 7/50) > 0.
2.24

### Proof

From (2.19) and the first inequality in Lemma 2.4 we have

$f2(9/5,7/50)=[ψ′(x+1)+4x2(x2+p)2−2x2+p−1(x+p)2]p=9/5,x=7/50>[(x+12)(x2+x+2321)x4+2x3+177x2+107x+1235+4x2(x2+p)2−2x2+p−1(x+p)2]p=9/5,x=7/50=84,826,873,256,410,10015,239,152,138,614,823,989>0.$

### Lemma 2.9

Let f1(px) be defined by (2.18). Then

f1(9/5, x) < 0

for x ∈ (7/50, 1/3).

### Proof

It follows from Lemma 2.3 and (2.18) that

$f1(9/5,x)<192log(x2+x+43)+4592log(x2+x+1445)−2xx2+9/5+1x+9/5.$
2.25

$[192log(x2+x+43)+4592log(x2+x+1445)−2xx2+9/5+1x+9/5]′=h(x)2(5x+9)2(5x2+9)2(3x2+3x+4)(45x2+45x+14),$
2.26

where

$h(x)=168,750x9+1,029,375x8+3,923,625x7+7,884,000x6+9,344,775x5+5,316,100x4+203,355x3−2,426,940x2−544,401x+118,017,h(7/50)=−406,357,216,255,013156,250,000,000<0,$
2.27

$h′(x)=1,518,750x8+8,235,000x7+27,465,375x6+47,304,000x5+46,723,875x4+21,264,400x3+610,065x2−4,853,880x−544,401,$
2.28

$h′(1/3)=−120,000,368243<0.$
2.29

From Lemma 2.2, (2.28) and (2.29) we know that h(x) is strictly decreasing on (7/50, 1/3), then (2.27) leads to the conclusion that h(x) < 0 for x ∈ (7/50, 1/3).

Therefore,

$f1(9/5,x)<[192log(x2+x+43)+4592log(x2+x+1445)−2xx2+9/5+1x+9/5]x=7/50=192log11,1977,500+4592log10,59122,500+159,550441,253=−0.0026…<0$

for x ∈ (7/50, 1/3) follows from (2.25) and (2.26) together with h(x) < 0 for x ∈ (7/50, 1/3).

### Lemma 2.10

Let p ∈ [3/2, 2] and f3(px) be defined by (2.20). Then there exists η(p) ∈ (0, 1) such that f3(px) < 0 for x ∈ (0, η(p)) and f3(px) > 0 for x ∈ (η(p), 1).

### Proof

Let

$g(p,x)=(x2+p)3f3(p,x)=(x2+p)3ψ″(x+1)+2(x2+p)3(x+p)3−4x3+12px.$
2.30

g(p, 0+) = p3ψ(1) + 2,  g(p, 1) = (p+1)3ψ(2) + 12p − 2,
2.31

$∂g(p,x)∂x=(x2+p)3ψ‴(x+1)+6x(x2+p)2ψ″(x+1)−6(x2+p)3(x+p)4+12x(x2+p)2(x+p)3−12x2+12p.$
2.32

It follows from the first inequalities in (2.2) and (2.3) together with the identity ψ(n)(x + 1) = ψ(n)(x) + (−1)nn! /xn+1 that

$ψ″(x+1)=ψ″(x+2)−2(x+1)3≥−1(x+3/2)2−2(x+1)3,$
2.33

$ψ‴(x+1)=ψ‴(x+2)+6(x+1)4≥2(x+3/2)3−1(x+3/2)5+6(x+1)4.$
2.34

From (2.32)-(2.34) we have

$∂g(p,x)∂x≥(x2+p)3[2(x+3/2)3−1(x+3/2)5+6(x+1)4]+6x(x2+p)2[−1(x+3/2)2−2(x+1)3]−6(x2+p)3(x+p)4+12x(x2+p)2(x+p)3−12x2+12p=2g1(p,x)(2x+3)5(x+1)4(x+p)4,$
2.35

where

$g1(p,x)=∑k=06bkxk−∑k=716bkxk$
2.36

with

$b0=(785p4+1,458p2−729)p3,b1=2(1,375p4+679p3+5,346p2+2,916p−1,944)p3,b2=(3,992p5+5,093p4+32,250p3+41,310p2+2,106p−729)p2,b3=2(1,544p5+3,955p4+27,270p3+60,214p2+30,186p+1,701)p2,b4=(1,352p6+6,520p5+56,049p4+187,562p3+184,880p2+38,070p+2,187)p,b5=2(160p6+1,510p5+16,743p4+78,961p3+128,815p2+54,729p+5,832)p,b6=32p7+752p6+8,088p5+44,529p4+135,746p3+123,972p2+19,926p−729,b7=−80p6+3,000p5+47,018p4+106,872p3+27,156p2+9,720p+5,346,b8=2,904p5+60,104p4+245,744p3+259,818p2+99,246p+17,334,b9=864p5+31,572p4+204,464p3+360,816p2+195,790p+34,398,b10=96p5+8,896p4+96,576p3+272,304p2+216,076p+46,726,b11=1,264p4+26,944p3+126,120p2+151,960p+44,884,b12=64p4+4,096p3+35,712p2+69,752p+30,400,b13=256p3+5,664p2+20,320p+14,180,b14=384p2+3,424p+4,336,b15=256p+784,b16=16,g1(p,1)=−199,181−732,767p−813,801p2−48,835p3+408,665p4+189,699p5+24,733p6+12,319p7,$
2.37

$g1(3/2,1)=90,546,875128>0.$
2.38

From Lemma 2.2, (2.37) and (2.38) we clearly see that

g1(p, 1) > 0
2.39

for p ∈ [3/2, 1].

Making use of Lemma 2.2 again, and (2.36) and (2.39) together with the facts that bk > 0 for p ∈ [3/2, 1] and k = 0, 1, 2, …, 16 we know that g1(px) > 0 for p ∈ [3/2, 1] and x ∈ (0, 1). Then inequality (2.35) leads to the conclusion that the function x → g(px) is strictly increasing on (0, 1) for p ∈ [3/2, 2].

From (2.2) and the identity ψ(n)(x) = ψ(n)(x + 1) + (−1)n+1n! /xn+1 we get

$−1(x+1/2)2≤ψ″(x+1)=ψ″(x+2)−2(x+1)3≤−1(x+3/2)2+14(x+3/2)4−2(x+1)3.$
2.40

Taking x = 1 in the first inequality of (2.40) and x = 0 in the second inequality of (2.40), one has

$ψ″(2)≥−49,ψ″(1)≤−19481.$
2.41

It follows from (2.31) and (2.41) that

$g(p,0+)≤−19481×(32)3+2=−7312<0,$
2.42

$g(p,1−)≥−49(p+1)3+12p−2$
2.43

for p ∈ [3/2, 2].

Note that

$[−49(p+1)3+12p−2]′=43(p+4)(2−p).$
2.44

Inequality (2.43) and equation (2.44) imply that

$g(p,1−)≥−49(32+1)3+12×32−2=16318>0$
2.45

for p ∈ [3/2, 2].

Therefore, Lemma 2.10 follows easily from (2.30), (2.42), (2.45) and the monotonicity of the function x → g(px) on the interval (0, 1).

### Lemma 2.11

Let p ∈ [8/5, 9/5] and f2(px) be defined by (2.19). Then there exist η1(p), η2(p) ∈ (0, 1) with η1(p) < η2(p) such that f2(px) > 0 for x ∈ (0, η1(p)) ∪ (η2(p), 1) and f2(px) < 0 for x ∈ (η1(p), η2(p)).

### Proof

It follows from (2.19) that

$f2(p,0+)=π26p2(p−6(π2+6)+6π2)(p+6(π2+6)−6π2)>0,$
2.46

$f2(p,1−)=(π2−6)p2+2(π2−12)p+π26(p+1)2>0$
2.47

for p ∈ [8/5, 9/5].

From Lemma 2.10 and [8/5, 9/5] ⊂ [3/2, 2] we know that there exists η(p) ∈ (0, 1) such that the function x → f2(px) is strictly decreasing on (0, η(p)) and strictly increasing on (η(p), 1). Then Lemma 2.7 leads to the conclusion that

f2(pη(p)) ≤ f2(p, 1/3) < 0.
2.48

Therefore, there exist η1(p) ∈ (0, η(p)) and η2(p) ∈ (η(p), 1) such that f2(px) > 0 for x ∈ (0, η1(p)) ∪ (η2(p), 1) and f2(px) < 0 for x ∈ (η1(p), η2(p)) follow from (2.46)-(2.48) and the piecewise monotonicity of the function x → f2(px) on the interval (0, 1).

## Main results

### Theorem 3.1

Let p > 0 and p0γ/(1 − γ) = 1.365…. Then the inequality

$Γ(x+1)>x2+px+p$
3.1

holds for all x ∈ (0, 1) if and only if p ≤ p0, and the inequality

$Γ(x+1)≤μ(x2+p0)x+p0$
3.2

holds for all x ∈ (0, 1) if and only if μ ≥ μ0, where

$μ0=(x0+p0)Γ(x0+1)(x02+p0)=1.027…$
3.3

and x0 = 0.346… is the unique solution of the equation

$ψ(x+1)−2xx2+p0+1x+p0=0$
3.4

on the interval (0, 1).

### Proof

If inequality (3.1) holds for all x ∈ (0, 1), then p ≤ p0 follows easily from

$limx→1−logΓ(x+1)−log(x2+px+p)1−x=−ψ(2)+11+p=γ−p1+p≥0.$

Next, we prove that inequality (3.1) holds for all x ∈ (0, 1) and pp0 and (3.2) holds for all x ∈ (0, 1) if and only if μ ≥ μ0.

Let f(px), f1(px), f2(px) be defined by (2.17)-(2.19) and

$g(x)=(x2+p0)2x2f2(p0,x)=(x2+p0)2x2ψ′(x+1)−2p0x2−(x2+p0)2x2(x+p0)2+2.$
3.5

f(p0, 0+) = f(p0, 1) = 0,
3.6

$f1(p0,0+)=1−γ−γ2γ>0,f1(p0,1−)=0,$
3.7

$g(0+)=−∞,g(1−)=(π2−6)p02+2(π2−12)p0+π26>0,$
3.8

$g′(x)=(x2+p0)2ψ″(x+1)x2−2(p02−x4)ψ′(x+1)x3−2p0[x4−4x3−6p0x2−2p0(3p0+1)x−(2p0+1)p02](x+p0)3x3.$
3.9

It follows from the second inequality in (2.1) and the first inequality in (2.2) together with (3.9) that

$g′(x)≥−(x2+p0)2x2(x+1/2)2−2(p02−x4)(x+1/2)x3−2p0[x4−4x3−6p0x2−2p0(3p0+1)x−(2p0+1)p02](x+p0)3x3=2x3(2x+1)2(x+p0)3g1(x),$

where

$g1(x)=2x8+2(3p0+1)x7+2p0(3p0−1)x6+2p0(p02−3p0+6)x5−p0(10p02−18p0−15)x4−2p0(p0+2)(2p02−7p0−1)x3−2p02(5p02−11p0−7)x2−2p02(p0+1)(3p02−4p0−1)x−p03(2p02−2p0−1).$

It is easy to verify that all the coefficients of the polynomial g1(x) are positive, which implies that g(x) is strictly increasing on (0, 1), then from (3.5) and (3.8) we know that there exists η ∈ (0, 1) such that the function f1(p0x) is strictly decreasing on (0, η) and strictly increasing on (η, 1).

It follows from (2.18) and (3.7) together with the piecewise monotonicity of the function f1(p0x) on the interval (0, 1) that there exists x0 ∈ (0, 1) such that f(p0x) is strictly increasing on (0, x0) and strictly decreasing on (x0, 1) and x0 is the unique solution of equation (3.4) on the interval (0, 1).

Therefore, the desired results follow easily from (2.17), (3.3), (3.6) and the piecewise monotonicity of the function f(p0x) on the interval (0, 1) together with the fact that the function p → (x2p)/(xp) is strictly increasing.

Numerical computations show that x0 = 0.346… and $μ0=(x0+p0)Γ(x0+1)/(x02+p0)=1.027…$.

### Theorem 3.2

The inequality

$Γ(x+1)>x2+1γx+1γ$

holds for all x ∈ (0, x), and its reverse inequality

$Γ(x+1)

holds for all x ∈ (x, 1), where x = 0.385… is the unique solution of the equation

$Γ(x+1)−x2+1γx+1γ=0$

on the interval (0, 1).

### Proof

Let f(px), f1(px) and f2(px) be, respectively, defined by (2.17), (2.18) and (2.19). Then simple computations lead to

$f(1γ,0+)=f(1γ,1−)=0,$
3.10

$f1(1γ,0+)=0,f1(1γ,1−)=1−γ−γ21+γ>0.$
3.11

From Lemma 2.11 and 1/γ = 1.732… ∈ [8/5, 9/5] we know that there exist η1(1/γ), η2(1/γ) ∈ (0, 1) with η1(1/γ) < η2(1/γ) such that f1(1/γx) is strictly increasing on (0, η1(1/γ)) ∪ (η2(1/γ), 1) and strictly decreasing on (η1(1/γ), η2(1/γ)). We claim that

f1(1/γη2(1/γ)) < 0.
3.12

Indeed, if f1(1/γη2(1/γ)) ≥ 0, then the piecewise monotonicity of the function f1(1/γx) on the interval (0, 1) and (3.11) lead to the conclusion that f(1/γx) is strictly increasing on (0, 1), which contradicts (3.10).

It follows from (3.11) and (3.12) together with the piecewise monotonicity of the function f1(1/γx) on the interval (0, 1) that there exist $η1∗(1/γ)∈(η1(1/γ),η2(1/γ))$ and $η2∗(1/γ)∈(η2(1/γ),1)$ such that f(1/γx) is strictly increasing on $(0,η1∗(1/γ))∪(η2∗(1/γ),1)$ and strictly decreasing on $(η1∗(1/γ),η2∗(1/γ))$.

Therefore, Theorem 3.2 follows easily from (2.17) and (3.10) together with the piecewise monotonicity of f(1/γx) on (0, 1). Numerical computations show that x = 0.385….

### Theorem 3.3

The double inequality

$λ(x2+95)x+95≤Γ(x+1)
3.13

holds for all x ∈ (0, 1) with the best possible constant

$λ=(5τ0+9)Γ(τ0+1)5τ02+9=0.991…,$
3.14

where τ0 = 0.719… is the unique solution of the equation

$ψ(x+1)−2xx2+95+1x+95=0$
3.15

on the interval (0, 1).

### Proof

Let f(px), f1(px) and f2(px) be, respectively, defined by (2.17), (2.18) and (2.19). Then simple computations lead to

$f(95,0+)=f(95,1−)=0,$
3.16

$f1(95,0+)=59−γ<0,f1(95,1−)=914−γ>0.$
3.17

It follows from Lemma 2.11 that there exist η1(9/5), η2(9/5) ∈ (0, 1) with η1(9/5) < η2(9/5) such that f2(9/5, x) > 0 for x ∈ (0, η1(9/5)) ∪ (η2(9/5), 1) and f2(9/5, x) < 0 for x ∈ (η1(9/5), η2(9/5)), and f1(9/5, x) is strictly increasing on (0, η1(9/5)) ∪ (η2(9/5), 1) and strictly decreasing on (η1(9/5), η2(9/5)). Then Lemmas 2.7-2.9 lead to the conclusion that η1(9/5) ∈ (7/50, 1/3) and

f1(9/5, η1(9/5)) < 0.
3.18

From (2.18), (3.17), (3.18) and the piecewise monotonicity of f1(9/5, x) on (0, 1) we clearly see that there exists τ0 such that τ0 is the unique solution of equation (3.15) on the interval (0, 1), and f(9/5, x) is strictly decreasing on (0, τ0) and strictly increasing on (τ0, 1).

Equation (3.16) and the piecewise monotonicity of the function f(9/5, x) on the interval (0, 1) lead to the conclusion that

f(9/5, τ0) ≤ f(9/5, x) < 0
3.19

for all x ∈ (0, 1).

Therefore, inequality (3.13) holds for all x ∈ (0, 1) follows from (2.17) and (3.19). We clearly see that the parameter λ given by (3.14) is the best possible constant such that the first inequality in (3.13) holds for all x ∈ (0, 1). Numerical computations show that τ0 = 0.719… and λ = 0.991….

### Remark 3.4

From Theorems 3.1 and 3.3 we clearly see that the double inequality

$x2+p0x+p0<Γ(x+1)
3.20

holds for all x ∈ (0, 1) with p0γ/(1 − γ) = 1.365… and p1 = 9/5, the constant p0 appears to be the best possible, but this is not true for p1, and a slightly smaller value for p1 is possible. Unfortunately, we cannot find the best possible constant p1 in the article; we leave this as an open problem for the reader.

### Remark 3.5

From the monotonicity of the function p ↦ (x2p)/(xp) we clearly see that both the upper and lower bounds for Γ(x + 1) given in (3.20) are better than that given in (1.3), and the first (second) inequality in Theorem 3.2 is the improvement of the first (second) inequality in (1.3) for x ∈ (0, x) (x ∈ (x, 1)), where x = 0.385… is given by Theorem 3.2.

### Remark 3.6

From Lemma 2.6, γγ2 < 1, 1/γ > 1, γ/(1 − γ) > 1 and (x2 + 1)/(x + 1) < 1 for x ∈ (0, 1) one has

$x2+γ1−γx+γ1−γ>(x2+1x+1)2(1−γ),(x2+1x+1)γ>x2+1γx+1γ>(x2+1x+1)2γ/(1+γ)>(x2+1x+1)2(1−γ).$

Therefore, the lower bound for Γ(x + 1) given in (3.20) is better than that given in (1.4), the first inequality in Theorem 3.2 is an improvement of the first inequality in (1.4) for x ∈ (0, x) and the second inequality in Theorem 3.2 is an improvement of the second inequality in (1.4) for x ∈ (x, 1), where x = 0.385… is given by Theorem 3.2.

### Remark 3.7

It is not difficult to verify that

$minx∈(0,1)(x2+1x+1)γ=[2(2−1)]γ=0.897…,minx∈(0,1)(21−xxx)=2e−2/e=0.958…,x2+95x+95<0.89$

for x ∈ (0.44, 0.45) and

$x2+95x+95<0.95$

for x ∈ (θ0θ1), where $θ0=(0.95−0.5425)/2=0.106…$ and $θ1=(0.95+0.5425)/2=0.843…$. Therefore, the upper bound (x2 + 9/5)/(x + 9/5) for Γ(x + 1) given in (3.20) is better than that given in (1.4) for x ∈ (0.44, 0.45), and it is also better than that given in (1.2) for x ∈ (θ0θ1).

### Remark 3.8

Let

$L3(x)=(12+14+x)1−xxx,L4(x)=x2+γ1−γx+γ1−γ.$

Then numerical computations show that

$L3(1/8)=0.846…

Therefore, there exists δ ∈ (0, 1/8) such that the lower bound for Γ(x + 1) given in (3.20) is better than that given in (1.2) for x ∈ (δ, 1/8 + δ) ∪ (1/4 − δ, 1/4 + δ) ∪ (3/8 − δ, 3/8 + δ) ∪ (1/2 − δ, 1/2 + δ) ∪ (5/8 − δ, 5/8 + δ) ∪ (3/4 − δ, 3/4 + δ) ∪ (7/8 − δ, 7/8 + δ).

## Results and discussion

In this paper, we provide the accurate bounds for the classical gamma function in terms of very simple rational functions, which can be used to estimate the value of the gamma function in the area of engineering and technology.

## Conclusion

In the article, we present several very simple and practical rational bounds for the gamma function, which can be regarded as a simple estimation of the value of the gamma function. The given results are improvements of some well-known results.

## Acknowledgements

The research was supported by the Natural Science Foundation of China (Grants Nos. 61673169, 61374086, 11371125, 11401191) and the Tianyuan Special Funds of the National Natural Science Foundation of China (Grant No. 11626101).

## Footnotes

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Publisher’s Note

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## Contributor Information

Zhen-Hang Yang, moc.361@mkhzy.

Wei-Mao Qian, moc.621@779166mwq.

Yu-Ming Chu, moc.621@5002gnimuyuhc.

Wen Zhang, moc.liamg@18new.gnahz.

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