Inequalities, asymptotic expansions and completely monotonic functions related to the gamma function

Applied Mathematics and Computation 250 (2015) 514–529

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Inequalities, asymptotic expansions and completely monotonic functions related to the gamma function Chao-Ping Chen a,⇑, Richard B. Paris b a b

School of Mathematics and Informatics, Henan Polytechnic University, Jiaozuo City 454000, Henan Province, China School of Engineering, Computing and Applied Mathematics, University of Abertay, Dundee DD1 1HG, UK

a r t i c l e

i n f o

a b s t r a c t In this paper, we present some completely monotonic functions and asymptotic expansions related to the gamma function. Based on the obtained expansions, we provide new bounds     for Cðx þ 1Þ=C x þ 12 and C x þ 12 . Ó 2014 Published by Elsevier Inc.

Key words: Gamma function Psi function Polygamma functions Asymptotic expansion Completely monotonic functions Inequality

1. Introduction A function f is said to be completely monotonic on an interval I if it has derivatives of all orders on I and satisfies the following inequality:

ð1Þn f

ðnÞ

ðxÞ P 0 ðx 2 I; n 2 N0 :¼ N [ f0g;

N :¼ f1; 2; 3; . . .gÞ:

ð1:1Þ

Dubourdieu [13, p. 98] pointed out that, if a non-constant function f is completely monotonic on I ¼ ða; 1Þ, then strict inequality holds true in (1.1). See also [16] for a simpler proof of this result. It is known (Bernstein’s Theorem) that f is completely monotonic on ð0; 1Þ if and only if

f ðxÞ ¼

Z

1

ext dlðtÞ;

0

where l is a nonnegative measure on ½0; 1Þ such that the integral converges for all x > 0 (see [48, p. 161]). The main properties of completely monotonic functions are given in [[48], Chapter IV]. We also refer to [4], where an extensive list of references on completely monotonic functions can be found. Euler’s gamma function:

CðxÞ ¼

Z

1

t x1 et dt;

x > 0;

0

is one of the most important functions in mathematical analysis and has applications in many diverse areas. The logarithmic derivative of the gamma function:

wðxÞ ¼

C0 ðxÞ ; CðxÞ

⇑ Corresponding author. E-mail addresses: [email protected] (C.-P. Chen), [email protected] (R.B. Paris). http://dx.doi.org/10.1016/j.amc.2014.11.010 0096-3003/Ó 2014 Published by Elsevier Inc.

C.-P. Chen, R.B. Paris / Applied Mathematics and Computation 250 (2015) 514–529

515

is known as the psi (or digamma) function. The derivatives of the psi function wðxÞ: n

wðnÞ ðxÞ :¼

d fwðxÞg; dxn

n 2 N;

are called the polygamma functions. In this paper, we present some completely monotonic functions and asymptotic expansions related to the gamma func  tion. Based on the obtained expansions, we provide new bounds for Cðx þ 1Þ=Cðx þ 12Þ and C x þ 12 . The numerical values given in this paper have been calculated via the computer program MAPLE 13. 2. Lemmas The Bernoulli polynomials Bn ðxÞ and Euler polynomials En ðxÞ are defined by the generating functions 1 X text tn ¼ Bn ðxÞ et  1 n¼0 n!

and

1 X 2ext tn ¼ En ðxÞ : et þ 1 n¼0 n!

The rational numbers Bn ¼ Bn ð0Þ and integers En ¼ 2n En ð1=2Þ are called Bernoulli and Euler numbers, respectively. It follows from Problem 154 in Part I, Chapter 4, of [39] that 2m X X B2j B2j 2j t t 2mþ1 t < t 1þ < t 2j ; e 1 2 ð2jÞ! ð2jÞ! j¼1 j¼1

ð2:1Þ

for t > 0 and m 2 N0 . The inequality (2.1) can be also found in [17,40]. Lemma 1 presents an analogous result to (2.1). Lemma 1. For x > 0 and m 2 N, 2mþ1 X j¼2

2m ð1  22j ÞB2j x2j1 2 x X ð1  22j ÞB2j x2j1 < x 1þ < ; 2 j¼2 j ð2j  1Þ! e þ 1 j ð2j  1Þ!

ð2:2Þ

where Bn ðn 2 N0 Þ are the Bernoulli numbers. Proof. The noted Boole’s summation formula [45, p. 17]) states for k 2 N that

f ð1Þ ¼

k1  1X Ej ð1Þ  ðjÞ 1 ðjÞ f ð1Þ þ f ð0Þ þ 2 j¼0 j! 2ðk  1Þ!

Z

1

f

ðkÞ

ðtÞEk1 ðtÞdt;

0

which can be written for m 2 N as

f ð1Þ  f ð0Þ ¼

Z 1 m  X E2j1 ð1Þ  ð2j1Þ 1 ð2j1Þ ð2mÞ f ð1Þ þ f ð0Þ þ f ðtÞE2m1 ðtÞdt: ð2j  1Þ! ð2m  1Þ! 0 j¼1

ð2:3Þ

Applying formula (2.3) to f ðtÞ ¼ ext , we obtain



ex

m 2 x X E2j1 ð1Þ 2j1 x x2m1 þ1 ¼ x þ x 2 j¼2 ð2j  1Þ! e þ 1 ð2m  1Þ! þ1

Z

1

ext E2m1 ðtÞdt:

0

It is well known (see [1, p. 804]) that

E2mþ1 ð1  tÞ ¼ E2mþ1 ðtÞ and E2mþ1

  1 ¼ 0: 2

Noting that

E4m1 ðtÞ > 0;

E4mþ1 ðtÞ < 0 for 0 < t < 1=2;

m ¼ 1; 2; . . . ;

we deduce for x > 0 that

Z

1

ext E4m1 ðtÞdt ¼

0

Z

1=2



 ext  exð1tÞ E4m1 ðtÞdt < 0;

1=2



 ext  exð1tÞ E4mþ1 ðtÞdt > 0:

0

and

Z 0

1

ext E4mþ1 ðtÞdt ¼

Z 0

ð2:4Þ

516

C.-P. Chen, R.B. Paris / Applied Mathematics and Computation 250 (2015) 514–529

Combining these with (2.4), we immediately obtain that for x > 0 and m 2 N, 2mþ1 X



j¼2

2m X E2j1 ð1Þ 2j1 2 x E2j1 ð1Þ 2j1 x < x 1þ < x : ð2j  1Þ! e þ1 2 ð2j  1Þ! j¼2

ð2:5Þ

Noting that

En ð1Þ ¼

2ð2nþ1  1Þ Bnþ1 ; nþ1

n 2 N;

the inequality (2.5) can be written as (2.2). The proof of Theorem 1 is complete. h The inequality (2.2) can also be written for x > 0 and m 2 N0 as 2mþ1 X j¼1

2m X ð1  22j ÞB2j x2j1 2 ð1  22j ÞB2j x2j1 < x 1< ; j ð2j  1Þ! e þ 1 j ð2j  1Þ! j¼1

ð2:6Þ

i.e.,

ð1Þ

mþ1

m X 2 ð1  22j ÞB2j x2j1 1 x e þ1 j ð2j  1Þ! j¼1

! > 0:

ð2:7Þ

Lemma 2 (10). Let r – 0 be a given real number and ‘ P 0 be a given integer. The gamma function has the following asymptotic expansion:

Cðx þ 1Þ 

1 X pffiffiffiffiffiffiffiffiffixx bj 1þ 2px e xj j¼1

!x‘ =r x ! 1;

;

ð2:8Þ

where the coefficients bj  bj ð‘; rÞ ðj 2 NÞ are given by

bj  bj ð‘; rÞ ¼

 kj X rk1 þk2 þþkj  B2 k1  B3 k2 Bjþ1  ; k1 !k2 !    kj ! 1  2 23 jðj þ 1Þ

ð2:9Þ

summed over all nonnegative integers kj satisfying the equation

ð1 þ ‘Þk1 þ ð2 þ ‘Þk2 þ    þ ðj þ ‘Þkj ¼ j: Lemma 3. Let r – 0 be a given real number and ‘ P 0 be a given integer. The gamma function has the following asymptotic expansion:

!x‘ =r  pffiffiffiffiffiffiffi x 1 X cj 1 x  2p C xþ 1þ ; 2 e xj j¼1 

x ! 1;

ð2:10Þ

where the coefficients cj  cj ð‘; rÞ ðj 2 NÞ are given by

cj ¼

X ðrÞk1 þk2 þþkj k1 !k2 !    kj !

ð1  21 ÞB2 12

!k1

ð1  23 ÞB4 34

!k 2 

ð1  212j ÞB2j ð2j  1Þð2jÞ

!k j ;

ð2:11Þ

summed over all nonnegative integers kj satisfying the equation

ð1 þ ‘Þk1 þ ð3 þ ‘Þk2 þ    þ ð2j þ ‘  1Þkj ¼ j: Proof. The following asymptotic expansion can be found [27, p. 32]

  1 pffiffiffiffiffiffiffi X B2j ð12Þ 1  x ln x  x þ ln 2p þ ln C x þ x12j ; 2jð2j  1Þ 2 j¼1 It is well known (see [1, p. 805]) that

Bn

  1 ¼ ð1  21n ÞBn ; 2

n 2 N0 ;

x ! 1:

ð2:12Þ

C.-P. Chen, R.B. Paris / Applied Mathematics and Computation 250 (2015) 514–529

517

and then the expansion (2.12) can be rewritten as

! m X Cðx þ 1Þ ð1  212k ÞB2k pffiffiffiffiffiffiffi 2 x ¼ exp ; þ R ðxÞ m 2kð2k  1Þx2k1 2pðx=eÞ k¼1

x ! 1;

ð2:13Þ

where Rm ðxÞ ¼ Oð1=x2mþ1 Þ. Further, we have

Cðx þ 12Þ

!r=x‘

! m X rð1  212k ÞB2k ¼e exp 2kð2k  1Þx2kþ‘1 k¼1 2 3 ! !2 m Y rð1  212k ÞB2k 1 rð1  212k ÞB2k rRm ðxÞ=x‘ 4 þ ¼e 1þ þ   5 2! 2kð2k  1Þx2kþ‘1 2kð2k  1Þx2kþ‘1 k¼1 rRm ðxÞ=x‘

pffiffiffiffiffiffiffi 2pðx=eÞx

¼ erRm ðxÞ=x

‘

1 X 1 X

1 X



k1 ¼0 k2 ¼0 1

rð1  2 ÞB2 12

 

km

1 k !k !    km ! 1 2 ¼0

!k 1

rð1  23 ÞB4 34

!k 2 

rð1  212m ÞB2m ð2m  1Þð2mÞ

!k m

1 : xð1þ‘Þk1 þð3þ‘Þk2 þþð2mþ‘1Þkm

ð2:14Þ

On the other hand, from (2.13) it follows that for any positive integer m,

Cðx þ 12Þ

pffiffiffiffiffiffiffi 2pðx=eÞx

!r=x‘ ¼1þ

m X cj j¼1

xj

þ Oð1=xmþ1 Þ;

ð2:15Þ

for some real numbers c1 ; . . . ; cm . Equating coefficients of equal powers of x in (2.14) and (2.15), we see that

cj ¼

X ðrÞk1 þk2 þþkj k1 !k2 !    kj !

ð1  21 ÞB2 12

!k 1

ð1  23 ÞB4 34

!k 2

ð1  212j ÞB2j  ð2j  1Þð2jÞ

!k j ;

summed over all nonnegative integers kj satisfying the equation

ð1 þ ‘Þk1 þ ð3 þ ‘Þk2 þ    þ ð2j þ ‘  1Þkj ¼ j: This completes the proof of Lemma 3. h Lemma 4. Let r – 0 be a given real number and ‘ P 0 be a given integer. The following asymptotic expansion holds: 1 X pj Cðx þ 1Þ pffiffiffi    x 1 þ xj C x þ 12 j¼1

!x‘ =r ;

x ! 1;

ð2:16Þ

where the coefficients pj  pj ð‘; rÞ ðj 2 NÞ are given by

pj ¼

X r k1 þk2 þþkj ð22  1ÞB2 k1 !k2 !    kj ! 1  1  22

!k1

ð24  1ÞB4

!k 2 

2  3  24

ð22j  1ÞB2j jð2j  1Þ22j

!kj ;

ð2:17Þ

summed over all nonnegative integers kj satisfying the equation

ð1 þ ‘Þk1 þ ð3 þ ‘Þk2 þ    þ ð2j þ ‘  1Þkj ¼ j: Proof. From (3.7) we obtain the following asymptotic expansion:

!

 1  X Cðx þ 1Þ pffiffiffi 1 B2j ; x exp 1   Cðx þ 12Þ 22j jð2j  1Þx2j1 j¼1

x ! 1:

ð2:18Þ

Write (2.18) as

!

m X Cðx þ 1Þ ð22k  1ÞB2k pffiffiffi  exp þ Rm ðxÞ ; 2k 2k1 xCðx þ 12Þ k¼1 kð2k  1Þ2 x

x ! 1;

ð2:19Þ

518

C.-P. Chen, R.B. Paris / Applied Mathematics and Computation 250 (2015) 514–529

where Rm ðxÞ ¼ Oð1=x2mþ1 Þ. Further, we have

Cðx þ 1Þ pffiffiffi xCðx þ 12Þ

!r=x‘ ¼e

¼e

rRm ðxÞ=x‘

rRm ðxÞ=x‘

exp m Y

2

rð22k  1ÞB2k

k¼1

kð2k  1Þ22k x2kþ‘1

kð2k  1Þ22k x2kþ‘1



1 X 1 X



1 rð22k  1ÞB2k þ 2! kð2k  1Þ22k x2kþ‘1

!2

3 þ   5

1 X

1 k !k !    km ! 1 2 k1 ¼0k2 ¼0 km ¼0 !k1 !k2 rð22  1ÞB2 rð24  1ÞB4 ‘

!

rð22k  1ÞB2k

41 þ

k¼1

¼ erRm ðxÞ=x

!

m X

1  1  22

2  3  24



rð22m  1ÞB2m mð2m  1Þ22m

!km 

1 : xð1þ‘Þk1 þð3þ‘Þk2 þþð2mþ‘1Þkm

ð2:20Þ

On the other hand, from (2.19) it follows that for any positive integer m,

Cðx þ 1Þ pffiffiffi xCðx þ 12Þ

!r=x‘ ¼1þ

m X pj j¼1

xj

þ Oð1=xmþ1 Þ;

ð2:21Þ

for some real numbers p1 ; . . . ; pm . Equating coefficients of equal powers of x in (2.20) and (2.21), we see that

X r k1 þk2 þþkj ð22  1ÞB2 pj ¼ k1 !k2 !    kj ! 1  1  22

!k 1

ð24  1ÞB4

!k2 

2  3  24

ð22j  1ÞB2j jð2j  1Þ22j

!kj ;

summed over all nonnegative integers kj satisfying the equation

ð1 þ ‘Þk1 þ ð3 þ ‘Þk2 þ    þ ð2j þ ‘  1Þkj ¼ j: This completes the proof of Lemma 4. h Lemma 5. For t – 0,

1 1 7 2 31 4 127 6 cosh t 1 1 7 2 31 4 þ  t þ t  t < < þ  t þ t : 6048 172800 coshð2tÞ  1 2t2 12 240 6048 2t 2 12 240

ð2:22Þ

Proof. We only prove the second inequality in (2.22). The proof of the first inequality in (2.22) is analogous. By using the power series expansion of cosh t, we find that



 1 X 1 1 7 2 31 4 an coshð2tÞ  1 þ  t þ t ð Þ  cosh t ¼ t 2n 2 12 240 6048 ð2n þ 2Þ! 2t n¼4

with

  104 10513 11117 2 1577 3 4303 4 31 5 62 6 þ nþ n  n  n  n þ n  2ðn þ 1Þð2n þ 1Þ 3 1260 1890 1260 1890 63 189   1005649 1355243 382391 125897 155 62 ¼ 22n4 339 þ ðn  4Þ þ ðn  4Þ2 þ ðn  4Þ3 þ ðn  4Þ4 þ ðn  4Þ5 þ ðn  4Þ6 1260 1890 1260 1890 21 189

an ¼ 22n4

 2ðn þ 1Þð2n þ 1Þ > 22n4  339  2ðn þ 1Þð2n þ 1Þ > 0 for n P 4: Hence, the second inequality in (2.22) holds. This completes the proof of Lemma 5. h

3. Completely monotonic functions It is known that [45, p. 64] that n t t X B2j 2j  1 þ ¼ t þ ð1Þn t 2nþ2 mn ðtÞ; et  1 2 j¼1 ð2jÞ!

n P 0;

ð3:1Þ

C.-P. Chen, R.B. Paris / Applied Mathematics and Computation 250 (2015) 514–529

519

where

mn ðtÞ ¼

1 X

2 2

ðt 2 þ 4p2 k Þð2pkÞ

k¼1

2n

> 0:

It is easy to see that (3.1) implies (2.1). Binet’s first formula [44, p. 16] states that

ln CðxÞ ¼

   pffiffiffiffiffiffiffi Z 1  t 1 t ext ln x  x þ ln 2p þ x  1 þ dt; 2 et  1 2 t2 0

x > 0:

ð3:2Þ

Combining (3.1) with (3.2), Xu and Han [46] deduced in 2009 that for every m 2 N0 , the function

"

#   m pffiffiffiffiffiffiffi X 1 B2j ln x þ x  ln 2p  ln CðxÞ  x  2 2jð2j  1Þx2j1 j¼1

m

Rm ðxÞ ¼ ð1Þ

ð3:3Þ

is completely monotonic on ð0; 1Þ. For m ¼ 0, the complete monotonicity property of Rm ðxÞ was proved by Muldoon [38]. Alzer [2] first proved in 1997 that Rm ðxÞ is completely monotonic on ð0; 1Þ. In 2006, Koumandos [17] proved the double inequality (2.1), and then used (2.1) and (3.2) to give the proof of the complete monotonicity property of Rm ðxÞ. In 2009, Koumandos and Pedersen [18, Theorem 2.1] strengthened this result. Based on the inequality (2.2), in this section we prove that for every m 2 N0 , the function

"

!





#

m X Cðx þ 1Þ 1 1 B2j  ln x  ; 1  2j 1 2 jð2j  1Þx2j1 Cðx þ 2Þ 2 j¼1

F m ðxÞ ¼ ð1Þm ln

ð3:4Þ

is completely monotonic on ð0; 1Þ. This result is similar to the complete monotonicity property of Rm ðxÞ in (3.3). Theorem 1. For every m 2 N0 , the function F m ðxÞ, defined by (3.4), is completely monotonic on ð0; 1Þ. Proof. The logarithm of the gamma function has the following integral representation (see [1, p. 258]):

ln CðxÞ ¼

Z

1

0



et  ext dt ðx  1Þet  ; t 1  et

x > 0:

ð3:5Þ

By using (3.5) and the following representations:

ln x ¼

Z

1

et  ext dt; t

0

x > 0;

in [1, p. 230, 5.1.32] and

1 1 ¼ xr CðrÞ

Z

1

tr1 ext dt;

x > 0 and r > 0;

0

in [1, p. 255, 6.1.1], we find that

F m ðxÞ ¼ ð1Þm

"Z

1

 

0

¼

1 2

Z

1

#   Z 1 m  X B2j 1 1 ext 1 2j2 xt þ dt  1  t e dt et=2 þ 1 2 t 22j jð2j  1Þ! 0 j¼1

ð1Þmþ1 km ðtÞ

0

ext dt; t

where

km ðtÞ ¼

 2j1 m X 2 ð1  22j ÞB2j t  1  : et=2 þ 1 jð2j  1Þ! 2 j¼1

By (2.7), we have ð1Þmþ1 km ðtÞ > 0 for t > 0 and m 2 N0 . From (3.6) we obtain that for every m 2 N0 ,

ð1Þn F ðnÞ m ðxÞ ¼

1 2

Z

1

ð1Þmþ1 km ðtÞt n1 ext dt > 0;

0

for x > 0 and n 2 N0 . The proof of Theorem 1 is complete. h From the inequality ð1Þn F ðnÞ m ðxÞ > 0 for x > 0 and m; n 2 N0 , we obtain the following.

ð3:6Þ

520

C.-P. Chen, R.B. Paris / Applied Mathematics and Computation 250 (2015) 514–529

Corollary 1 (i) Let m 2 N0 . Then for x > 0,

!  2m  X pffiffiffi 1 B2j Cðx þ 1Þ pffiffiffi < x exp 1  2j < x exp 2j1 jð2j  1Þx Cðx þ 12Þ 2 j¼1

2mþ1 X

1

j¼1

(ii) Let m; n 2 N. Then for x > 0,

1



22j

! B2j : jð2j  1Þx2j1

ð3:7Þ

     2m  X X 1 2B2j ð2j þ n  2Þ! 1 ðn  1Þ! 2m1 1 2B2j ð2j þ n  2Þ! n ðn1Þ ðn1Þ þ 1  2j < ð1Þ w ðx þ 1Þ  w x þ < 1  : x2jþn1 2 2xn x2jþn1 ð2jÞ! 2 22j ð2jÞ! j¼1 j¼1 ð3:8Þ By using the above results, we present below inequalities and integral representations for the constant

p.

The problem of finding new and sharp inequalities for the gamma function C and, in particular, for the Wallis ratio

Cðn þ 1Þ 1 ð2nÞ!!   ¼ pffiffiffiffi  ; C n þ 12 p ð2n  1Þ!!

n 2 N;

ð3:9Þ

has attracted the attention of many researchers (see [11,19,20,33] and references therein). Some inequalities for found in, for example, [15,21,34,35]. Here, we employ the special double factorial notation as follows:

p can be

ð2nÞ!! ¼ 2  4  6    ð2nÞ ¼ 2n n!;

  1 ; ð2n  1Þ!! ¼ 1  3  5    ð2n  1Þ ¼ p1=2 2n C n þ 2 0!! ¼ 1;

ð1Þ!! ¼ 1

(see [1, p. 258]). Very recently, Lin [21, Theorem 2.4] proved that for all n 2 N,



2   ð2nÞ!! 1 1 1 1 17 31

Setting x ¼ n in (3.7), we obtain the estimate for the constant



ð2nÞ!! ð2n  1Þ!!

2

1 exp  n

2mþ1 X

1

j¼1

1



22j

2B2j jð2j  1Þn2j1

ð3:10Þ

p:

!


ð2nÞ!! ð2n  1Þ!!

2

!  2m  X 1 1 2B2j : exp  1  2j n jð2j  1Þn2j1 2 j¼1 ð3:11Þ

Obviously, (3.11) is a generalization of (3.10). Formula (3.6) gives the following integral representation:

!





m X Cðx þ 1Þ 1 1 B2j 1  ln x  1  2j ¼ 1 2j1 2 2 jð2j  1Þx Cðx þ 2Þ 2 j¼1

ln

Z

1 0

 2j1 ! xt m X 2 ð1  22j ÞB2j t e  1  dt; et=2 þ 1 2 jð2j  1Þ! t j¼1 ð3:12Þ

which implies

    2j1 ! Z m  m X 1 1 X 1 B2j 1 1 2 ð1  22j ÞB2j t ext dt  wðx þ 1Þ  w x þ þ 1  2j ¼  1  2j 2 2x j¼1 2 0 et=2 þ 1 jð2j  1Þ! 2 2 jx j¼1

ð3:13Þ

for x > 0 and m 2 N0 . Formulas (3.12) and (3.13) can provide integral representations for the constant p. For example, setting ðx; mÞ ¼ ð1=2; 0Þ in (3.12) yields

Z

1

 1

0

eu

 u p 2 e du ¼ ln : þ1 u 2

ð3:14Þ

Setting ðx; mÞ ¼ ð1=4; 1Þ in (3.13) yields

Z 0

1



 2 u u=2 e  1 þ du ¼ 4  p: eu þ 1 2

ð3:15Þ

Many formulas exist for the representation of p, and a collection of these formulas is listed in [42,43]. For more history of

p see [6,7,14].

C.-P. Chen, R.B. Paris / Applied Mathematics and Computation 250 (2015) 514–529

521

Very recently, Mortici et al. [36] proved some completely monotonic functions and inequalities associated with the ratio of gamma functions. 4. Asymptotic expansions Stirling’s formula

n! 

pffiffiffiffiffiffiffiffiffinn 2pn ; e

n 2 N :¼ f1; 2; . . .g;

ð4:1Þ

has many applications in statistical physics, probability theory and number theory. Actually, it was first discovered in 1733 by the French mathematician Abraham de Moivre (1667–1754) in the form

n!  constant 

pffiffiffi nðn=eÞn ;

when he was studying the Gaussian distribution and the central limit theorem. Afterwards, the Scottish mathematician pffiffiffiffiffiffi ffi James Stirling (1692–1770) found the missing constant 2p when he was trying to give the normal approximation of the binomial distribution. Stirling’s series for the gamma function is given (see [1, p. 257, Eq. (6.1.40)]) by 1 X pffiffiffiffiffiffiffiffiffixx B2m Cðx þ 1Þ  2px exp e 2mð2m  1Þx2m1 m¼1

! ¼

  pffiffiffiffiffiffiffiffiffixx 1 1 1 1 ; exp  þ  þ    2p x e 12x 360x3 1260x5 1680x7

ð4:2Þ

as x ! 1, where Bn ðn 2 N0 Þ are the Bernoulli numbers. The following asymptotic formula is due to Laplace:

Cðx þ 1Þ 

 pffiffiffiffiffiffiffiffiffixx  1 1 139 571 1þ þ   þ    2p x e 12x 288x2 51840x3 2488320x4

ð4:3Þ

as x ! 1 (see [1, p. 257, Eq. (6.1.37)]). The expression (4.3) is sometimes incorrectly called Stirling’s series (see [12, pp. 2-3]). Stirling’s formula is in fact the first approximation to the asymptotic formula (4.3). Stirling’s formula has attracted much interest of from many mathematicians and has motivated a large number of research papers concerning various generalizations and improvements; see, for example, [8,9,22–26,29–32,37] and the references cited therein). See also an overview in [28]. Windschitl (see [5, p. 128] and [49]) noted that for x > 8, the approximation

Cðx þ 1Þ 

x=2 pffiffiffiffiffiffiffiffiffixx  1 1 x sinh þ ; 2p x e x 810x6

ð4:4Þ

gives at least eight decimal places of the gamma function. The formula (4.4) yields

Cðx þ 1Þ ¼

x=2    pffiffiffiffiffiffiffiffiffixx  1 1 x sinh 1þO 5 ; 2p x e x x

x ! 1:

ð4:5Þ

Inspired by (4.4) and (4.5), Alzer [3] proved in 2009 that for all x > 0,

x=2  x=2   pffiffiffiffiffiffiffiffiffixx  pffiffiffiffiffiffiffiffiffixx  1 a 1 b x sinh 1 þ 5 < Cðx þ 1Þ < 2px x sinh 1þ 5 ; 2px e x e x x x

ð4:6Þ

with the best possible constants a ¼ 0 and b ¼ 1=1620. In 2014, Lu et al. [24] extended Windschitl’s formula as follows:

Cðn þ 1Þ 

 n=2 pffiffiffiffiffiffiffiffiffinn  1 a7 a9 a11 2pn n sinh þ 7 þ 9 þ 11 þ    ; e n n n n

ð4:7Þ

where

a7 ¼

1 ; 810

a9 ¼ 

67 ; 42; 525

a11 ¼

19 ;...: 8505

ð4:8Þ

However, the authors did not give the general formula for the coefficients aj (j P 7) in (4.7). Subsequently, Chen [9] gave a recurrence relation formula for determining the coefficients of nj (j 2 N) in (4.7). Also in [9], Chen developed Windschitl’s approximation formula to produce the new asymptotic expansion:

 pffiffiffiffiffiffiffiffiffixx  1 Cðx þ 1Þ  2px x sinh e x

1 X

x=2þ

j¼0

rj xj

;

x ! 1;

and provided a recurrence relation for determining the coefficients rj in (4.9).

ð4:9Þ

522

C.-P. Chen, R.B. Paris / Applied Mathematics and Computation 250 (2015) 514–529

Smith [41, Eq. (43)] presented the following analogous result to (4.5):



C xþ

 pffiffiffiffiffiffiffi x  x=2    1 x 1 1 ¼ 2p ; 2x tanh 1þO 5 2 e 2x x

x ! 1:

ð4:10Þ

Let r – 0 be a given real number and ‘ P 0 be a given integer. We here determine the coefficients aj  aj ð‘; rÞ and dj  dj ð‘; rÞ (for j 2 N) such that

Cðx þ 1Þ 

1 pffiffiffiffiffiffiffiffiffixx 1 X aj x sinh þ 2px e x j¼1 xj

!x‘ =r ;

and

!x‘ =r  pffiffiffiffiffiffiffi x 1 1 x 1 X dj  2p C xþ 2x tanh þ ; 2 e 2x j¼1 xj 

as x ! 1. Theorem 2. Let r – 0 be a given real number and ‘ P 0 be a given integer. The gamma function has the following asymptotic expansion: 1 pffiffiffiffiffiffiffiffiffixx 1 X aj Cðx þ 1Þ  2px x sinh þ e x j¼1 xj

!x‘ =r

;

x ! 1;

ð4:11Þ

with the coefficients aj  aj ð‘; rÞ ðj 2 NÞ given by

a2j1 ¼ b2j1 ð‘; rÞ;

a2j ¼ b2j ð‘; rÞ 

1 ; ð2j þ 1Þ!

ð4:12Þ

where bj ð‘; rÞ ðj 2 NÞ can be calculated using (2.9). Proof. The Maclaurin expansion of sinh t with t ¼ 1=x gives

x sinh

1 X 1 1 ¼1þ ; x ð2j þ 1Þ!x2j j¼1

x – 0:

ð4:13Þ

In view of (2.8) and (4.13), we can let

Cðx þ 1Þ pffiffiffiffiffiffiffiffiffi 2pxðx=eÞx

!r=x‘  x sinh

1 1 X aj  ; x xj j¼1

x ! 1;

ð4:14Þ

where aj ðj 2 NÞ are real numbers to be determined. It follows that 1 X bj j¼1

xj



1 X j¼1

1 X 1 aj  ; 2j ð2j þ 1Þ!x xj j¼1

x ! 1;

ð4:15Þ

where bj  bj ð‘; rÞ ðj 2 NÞ are given in (2.9). Equating coefficients of equal powers of x in (4.15) we obtain (4.12). The proof of Theorem 2 is complete. h Remark 1. Setting ðr; ‘Þ ¼ ð2; 1Þ in (4.11) we have the full asymptotic expansion of Windschitl’s formula (4.4):

Cðx þ 1Þ 

x=2 pffiffiffiffiffiffiffiffiffixx  1 1 163 1019 x sinh þ  þ     ; 2px e x 810x6 170100x8 680400x10

ð4:16Þ

as x ! 1. Theorem 3. Let r – 0 be a given real number and ‘ P 0 be a given integer. The gamma function has the following asymptotic expansion:

!x‘ =r  pffiffiffiffiffiffiffi x 1 1 x 1 X dj  2p C xþ 2x tanh þ ; 2 e 2x j¼1 xj 

with the coefficients dj  dj ð‘; rÞ ðj 2 NÞ given by

x ! 1;

ð4:17Þ

C.-P. Chen, R.B. Paris / Applied Mathematics and Computation 250 (2015) 514–529

d2j1 ¼ c2j1 ð‘; rÞ;

d2j ¼ c2j ð‘; rÞ 

4ð22jþ2  1ÞB2jþ2 ; ð2j þ 2Þ!

523

ð4:18Þ

where cj ð‘; rÞ ðj 2 NÞ can be calculated using (2.11). Proof. The power series expansion of tanh t with t ¼ 1=ð2xÞ gives

2x tanh

1 X 1 4ð22j  1ÞB2j 1 ¼1þ ; 2x x2j2 ð2jÞ! j¼2

jxj >

1

p

:

ð4:19Þ

In view of (2.10) and (4.19), we can let

Cðx þ 12Þ

!r=x‘

pffiffiffiffiffiffiffi 2pðx=eÞx

 2x tanh

1 X 1 dj  ; 2x j¼1 xj

x ! 1;

ð4:20Þ

where dj ðj 2 NÞ are real numbers to be determined. It follows that 1 X cj j¼1

xj



1 1 X X 4ð22jþ2  1ÞB2jþ2 1 dj  ; 2j x ð2j þ 2Þ! xj j¼1 j¼1

x ! 1;

ð4:21Þ

where cj  cj ð‘; rÞ ðj 2 NÞ are given in (2.11). Equating coefficients of equal powers of x in (4.21) we obtain (4.18). The proof of Theorem 3 is complete. h Remark 2. Setting ðr; ‘Þ ¼ ð2; 1Þ in (4.17) we have the following full asymptotic expansion:



C xþ

 pffiffiffiffiffiffiffi x  x=2 1 x 1 31 6829  2p 2x tanh  þ     2 e 2x 25920x6 5443200x8

ð4:22Þ

as x ! 1. From (4.5) and (4.10), we derive



Cðx þ 1Þ pffiffiffi 1    x cosh 2x C x þ 12

x    1 ; 1þO 5 x

x ! 1:

This fact motivated us to observe the following. Theorem 4. Let r – 0 be a given real number and ‘ P 0 be a given integer. The following asymptotic expansion holds: 1 qj Cðx þ 1Þ pffiffiffi 1 X    x cosh þ 1 2x j¼1 xj C xþ2

!x‘ =r x ! 1;

;

ð4:23Þ

with the coefficients qj  qj ð‘; rÞ ðj 2 NÞ given by

q2j1 ¼ p2j1 ð‘; rÞ;

q2j ¼ p2j ð‘; rÞ 

1 22j ð2jÞ!

;

ð4:24Þ

where pj ð‘; rÞ ðj 2 NÞ can be calculated using (2.17). Proof. The Maclaurin expansion of cosh t with t ¼ 1=ð2xÞ gives

cosh

1 X 1 1 1 ¼1þ ; 2j 2j 2x x ð2jÞ! 2 j¼1

jxj – 0:

ð4:25Þ

In view of (2.16) and (4.25), we can let

Cðx þ 1Þ pffiffiffi xCðx þ 12Þ

!r=x‘  cosh

1 X qj 1  ; 2x j¼1 xj

x ! 1;

ð4:26Þ

where qj ðj 2 NÞ are real numbers to be determined. It follows that 1 X pj j¼1

xj



1 X

1 X qj 1  ; 2j xj 2 ð2jÞ! x j¼1

1

2j

j¼1

x ! 1;

ð4:27Þ

where pj  pj ð‘; rÞ ðj 2 NÞ are given in (2.17). Equating coefficients of equal powers of x in (4.27) we obtain (4.24). The proof of Theorem 3 is complete. h

524

C.-P. Chen, R.B. Paris / Applied Mathematics and Computation 250 (2015) 514–529

Remark 3. Setting ðr; ‘Þ ¼ ð1; 1Þ in (4.23) we have the following full asymptotic expansion:



Cðx þ 1Þ pffiffiffi 1 7 65    x cosh þ  þ  2x 5760x6 64512x8 C x þ 12

x x ! 1:

ð4:28Þ

 x    x   pffiffiffi 1 h1 Cðx þ 1Þ pffiffiffi 1 h2  < x cosh x cosh 1þ 5 <  1þ 5 ; 1 2x 2x x x C xþ2

ð5:1Þ

;

5. Inequalities Theorem 5. For x > 0,

with the best possible constants

h1 ¼ 0 and h2 ¼

7 : 5760

ð5:2Þ

7 Proof. We first prove the inequality (5.1) with h1 ¼ 0 and h2 ¼ 5760 . That is

 x  x   pffiffiffi 1 Cðx þ 1Þ pffiffiffi 1 7  < x cosh ; <  1þ x cosh 1 2x 2x 5760x5 C xþ2

x > 0:

ð5:3Þ

Lambert’s continued fraction [47, p. 349]

tanhðzÞ ¼

z 1þ

3þ

z2 z2 2 5þ z 7þ

is valid for all values of z. Hence for x > 0,

6x 1=ð2xÞ 1 1=ð2xÞ 60x2 þ 1 ¼ < tanh < ¼ : 2 2 2 12x þ 1 1 þ ð1=ð2xÞÞ 2x 1 þ ð1=ð2xÞÞ 12xð10x2 þ 1Þ 3

3þ

ð5:4Þ

ð1=ð2xÞÞ2 5

The proof of the inequality (5.3) make use of the inequality (5.4). The lower bound in (5.3) is obtained by considering the function f ðxÞ defined for x > 0 by

    1 1 1  ln x  x ln cosh : f ðxÞ ¼ ln Cðx þ 1Þ  ln C x þ 2 2 2x Differentiation yields

    1 1 1 1 1 0   ln cosh þ f ðxÞ ¼ wðx þ 1Þ  w x þ tanh ; 2 2x 2x 2x 2x

ð5:5Þ

and

   2    2 1 1 1 1 1 1 1 1 1 6x 00 þ 2  3 þ 3 tanh þ 2 3þ 3 f ðxÞ ¼ w0 ðx þ 1Þ  w0 x þ > w0 ðx þ 1Þ  w0 x þ 2 2x 4x 4x 2x 2 2x 4x 4x 12x2 þ 1 ¼: gðxÞ; by applying the left-hand inequality of (5.4). By using the recurrence formula

w0 ðx þ 1Þ ¼ w0 ðxÞ 

1 ; x2

ð5:6Þ

we find that

gðxÞ  gðx þ 1Þ ¼

hðxÞ 4x3 ð12x2

2

3

2

þ 1Þ ðx þ 1Þ ð2x þ 1Þ2 ð12x2 þ 24x þ 13Þ

;

with

hðxÞ ¼ 1540909 þ 8983929ðx  1Þ þ 22585735ðx  1Þ2 þ 32006952ðx  1Þ3 þ 27981444ðx  1Þ4 þ 15459984ðx  1Þ5 þ 5274000ðx  1Þ6 þ 1016064ðx  1Þ7 þ 84672ðx  1Þ8 :

C.-P. Chen, R.B. Paris / Applied Mathematics and Computation 250 (2015) 514–529

525

Hence, for x P 1,

gðxÞ > gðx þ 1Þ and gðxÞ > gðx þ nÞ: Therefore, for x P 1, 00

gðxÞ > lim gðx þ nÞ ¼ 0 and f ðxÞ > 0: n!1

We then obtain that 0

0

f ðxÞ < lim f ðtÞ ¼ 0 for x P 1: t!1

We now show that (5.7) is also valid for 0 < x  1. It follows from (5.5) that 0

f ðxÞ ¼ y1 ðxÞ þ y2 ðxÞ; with

  1 1 1  y1 ðxÞ ¼ ln cosh tanh ; 2x 2x 2x and

  1 1 y2 ðxÞ ¼ wðx þ 1Þ þ w x þ þ : 2 2x Differentiation yields

y01 ðxÞ ¼



4x3

1 1 cosh 2x

2 > 0:

Using the following representations:

Z

wðxÞ ¼



1

0

 et ext dt  t 1  et

in [1, p. 259, 6.3.21] and

Z

1 ¼ x

1

ext dt;

0

we conclude that

y2 ðxÞ ¼

Z

1



Z

1

0

 1 1 ext dt  t=2 2 e þ1

and

y02 ðxÞ ¼ 

0

  1 1 text dt < 0:  t=2 2 e þ1

Let 0 6 r 6 x 6 s 6 1. Since y1 ðxÞ is increasing and y2 ðxÞ is decreasing, we obtain 0

f ðxÞ P y1 ðrÞ þ y2 ðsÞ ¼: r1 ðr; sÞ: We divide the interval ½0; 1 into 100 subintervals:

½0; 1 ¼

99 [ k kþ1 > 0 for k ¼ 0; 1; 2; . . . ; 99: ; 100 100 k¼0

By direct computation we get

r1



 k kþ1 > 0 for k ¼ 0; 1; 2; . . . ; 99: ; 100 100

Hence, 0

f ðxÞ > 0 for x 2 0



k kþ1 and k ¼ 0; 1; 2; . . . ; 99: ; 100 100

This implies that f ðxÞ is negative on ð0; 1. We then obtain that for all x > 0,

f ðxÞ > lim f ðtÞ ¼ 0: t!1

ð5:7Þ

526

C.-P. Chen, R.B. Paris / Applied Mathematics and Computation 250 (2015) 514–529

This means that the first inequality in (5.3) holds for x > 0. The upper bound in (5.3) is obtained by considering the function uðxÞ defined for x > 0 by

      1 1 1 7  ln x  x ln cosh  ln 1 þ : uðxÞ ¼ ln Cðx þ 1Þ  ln C x þ 2 2 2x 5760x5 Differentiation yields

    1 1 1 1 1 35   ln cosh þ u0 ðxÞ ¼ wðx þ 1Þ  w x þ tanh þ ; 2 2x 2x 2x 2x xð5760x5 þ 7Þ and

ð5:8Þ

   2 1 1 1 1 1 35ð34560x5 þ 7Þ þ 2  3 þ 3 tanh u00 ðxÞ ¼ w0 ðx þ 1Þ  w0 x þ  2 2 2x 4x 4x 2x x2 ð5760x5 þ 7Þ    2 1 1 1 1 60x2 þ 1 35ð34560x5 þ 7Þ þ 2 3þ 3 < w0 ðx þ 1Þ  w0 x þ  ¼: v ðxÞ; 2 2 2 2x 4x 4x 12xð10x þ 1Þ x2 ð5760x5 þ 7Þ

by applying the right-hand inequality of (5.4). By using the recurrence formula (5.6), we find that

v ðxÞ  v ðx þ 1Þ ¼ 

w1 ðxÞ ; 576w2 ðxÞ

with

w1 ðxÞ ¼ 868270269104794924464036897 þ 18765499992969382192107571641ðx  1Þ þ    þ 14820540300656640000000ðx  1Þ30 has all coefficients positive, and 2

2

w2 ðxÞ ¼ x5 ð10x2 þ 1Þ ð5760x5 þ 7Þ ðx þ 1Þ5 ð2x þ 1Þ2 ð10x2 þ 20x þ 11Þ

2 2

 ð5760x5 þ 28800x4 þ 57600x3 þ 57600x2 þ 28800x þ 5767Þ : Hence, for x P 1,

v ðxÞ < v ðx þ 1Þ

and

v ðxÞ < v ðx þ nÞ:

Therefore, for x P 1,

v ðxÞ < n!1 lim v ðx þ nÞ ¼ 0

and u00 ðxÞ < 0:

We then obtain that

u0 ðxÞ > lim u0 ðtÞ ¼ 0 for x P 1:

ð5:9Þ

t!1

We now show that (5.9) is also valid for 0 < x  1. It follows from (5.8) that

u0 ðxÞ ¼ y3 ðxÞ þ y4 ðxÞ; where y3 ðxÞ ¼ y2 ðxÞ and

  1 1 1 35 þ y4 ðxÞ ¼  ln cosh tanh þ : 2x 2x 2x xð5760x5 þ 7Þ

Differentiation yields

y04 ðxÞ ¼ 

4x3



1

 1 2 cosh 2x



35ð34560x5 þ 7Þ x2 ð5760x5 þ 7Þ

2

< 0:

Let 0 6 r 6 x 6 s 6 1. Since y3 ðxÞ is increasing and y4 ðxÞ is decreasing, we obtain

u0 ðxÞ P y3 ðrÞ þ y4 ðsÞ ¼: r2 ðr; sÞ: The same as above, we divide the interval [0, 1] into 100 subintervals. By direct computation we get

r2



 k kþ1 > 0 for k ¼ 0; 1; 2; . . . ; 99: ; 100 100

Hence,

u0 ðxÞ > 0 for x 2



k kþ1 and k ¼ 0; 1; 2; . . . ; 99: ; 100 100

C.-P. Chen, R.B. Paris / Applied Mathematics and Computation 250 (2015) 514–529

527

This implies that u0 ðxÞ is positive on ð0; 1. We then obtain that for all x > 0,

uðxÞ < lim uðtÞ ¼ 0: t!1

This means that the second inequality in (5.3) holds for x > 0. The inequality (5.1) can be written as

h1 < x5



  x

1 1 xC x þ  1 < h2 ; cosh 2 2x

Cðx þ 1Þ pffiffiffi

x > 0:

We find that

" # Cðx þ 1Þ ¼ 0; h1 6 limþ x5 pffiffiffi   1    1 x x!0 xC x þ 12 cosh 2x and

"

#

7 lim x pffiffiffi  6 h2 :   1 ¼ 1 x 5760 xC x þ 12 cosh 2x 5

Cðx þ 1Þ

x!1

Hence, inequality (5.1) holds with the best possible constants given in Eq. (5.2). The proof of Theorem 5 is complete. h Remark 4. From (5.3), we derive the new inequality for the constant

!2

1 ð2nÞ!!    pffiffiffi 1 n 7 ð2n  1Þ!! 1 þ 5760n n cosh 2n 5


p:

1

ð2nÞ!!  pffiffiffi 1 n ð2n  1Þ!! n cosh 2n

!2 ;

ð5:10Þ

for n 2 N. Theorem 6. For x P 1,

x=2     pffiffiffiffiffiffiffi x  x=2 pffiffiffiffiffiffiffixx  1 31 1 x 1 < < 2x tanh 1 C x þ p 2x tanh : 2p 2 e 2x 51840x5 2 e 2x

ð5:11Þ

Proof. From the well-known continued fraction for w0 (see [47, p. 373])

  1 1 ¼ w0 x þ ; 2 x þ a1a2

x > 0;

xþxþ

where

ap ¼

p4 ; 4ð2p  1Þð2p þ 1Þ

p ¼ 1; 2; . . . ;

we find that for x > 0,

20xð84x2 þ 71Þ 1 ¼ 3ð560x4 þ 520x2 þ 27Þ x þ 121

4 xþ 1581 xþ140 x

  1 1 4ð15x2 þ 4Þ < < w0 x þ ¼ : 1 2 3xð20x2 þ 7Þ x þ 12 4 xþ15 x

The proof of the inequality (5.11) make use of the inequalities (2.22) and (5.12). The upper bound in (5.11) is obtained by considering the function UðxÞ defined for x P 1 by

    pffiffiffiffiffiffiffi 1 x x 1  lnð 2pÞ  x ln x þ x  lnð2xÞ  ln tanh : UðxÞ ¼ ln C x þ 2 2 2 2x Differentiation yields

    1 3 1 1 1 1 1  ln x   ln 2  ln tanh þ U 0 ðxÞ ¼ w x þ : 2 2 2 2 2 2x 2x sinh 1x Differentiating U 0 ðxÞ and applying the inequalities (2.22) and (5.12) we find

ð5:12Þ

528

C.-P. Chen, R.B. Paris / Applied Mathematics and Computation 250 (2015) 514–529

  cosh 1x 1 3 4ð15x2 þ 4Þ 1 1 7 31 3 þ 3 U 00 ðxÞ ¼ w0 x þ <  þ  þ þ  2 2 x ðcosh x  1Þ 2x 3xð20x2 þ 7Þ 2x 12x3 240x5 6048x7 2x ¼

3074x2  1085 < 0 for x P 1: 30240x7 ð20x2 þ 7Þ

We then obtain that for x P 1,

U 0 ðxÞ > lim U 0 ðtÞ ¼ 0 ) UðxÞ < lim UðtÞ ¼ 0: t!1

t!1

This means that the second inequality in (5.11) holds for x P 1. The lower bound in (5.11) is obtained by considering the function FðxÞ defined for x P 1 by

      pffiffiffiffiffiffiffi 1 x x 1 31  lnð 2pÞ  x ln x þ x  lnð2xÞ  ln tanh  ln 1  : FðxÞ ¼ ln C x þ 5 2 2 2 2x 51840x Differentiation yields

    1 3 1 1 1 1 1 155  ln x   ln 2  ln tanh þ F 0 ðxÞ ¼ w x þ  : 2 2 2 2 2 2x 2x sinh 1x xð51840x5  31Þ Differentiating F 0 ðxÞ and applying the inequalities (2.22) and (5.12) we find

  cosh 1x 1 8062156800x11  96422400x5  9642240x6 þ 2883x þ 9610 þ 3 F 00 ðxÞ ¼ w0 x þ  2 2 2 x ðcosh x  1Þ 2x2 ð51840x5  31Þ >

20xð84x2 þ 71Þ 1 1 7 31 127 þ  þ  þ 3ð560x4 þ 520x2 þ 27Þ 2x 12x3 240x5 6048x7 172800x9 

¼

8062156800x11  96422400x5  9642240x6 þ 2883x þ 9610 2x2 ð51840x5  31Þ GðxÞ

2

1209600x9 ð560x4 þ 520x2 þ 27Þð51840x5  31Þ

2

;

where

GðxÞ ¼ 35836321468874877 þ 500332411677183040ðx  1Þ þ 3243650443203650720ðx  1Þ2 þ 12942574029209436800ðx  1Þ3 þ 35510225676993688800ðx  1Þ4 þ 70870406789351495040ðx  1Þ5 þ 106102603246810676800ðx  1Þ6 þ 121054817642726534400ðx  1Þ7 þ 105767748316645632000ðx  1Þ8 þ 70425167657151744000ðx  1Þ9 þ 35177843816429875200ðx  1Þ10 þ 12782474258840064000ðx  1Þ11 þ 3194045448118272000ðx  1Þ12 þ 491271286947840000ðx  1Þ13 þ 35090806210560000ðx  1Þ14 : Hence, F 00 ðxÞ > 0 for x P 1. We then obtain that for x P 1,

F 0 ðxÞ < limF 0 ðtÞ ¼ 0 ) FðxÞ > limFðtÞ ¼ 0: t!1

t!1

This means that the first inequality in (5.11) holds for x P 1. The proof of Theorem 6 is complete. h Remark 5. Some computer experiments indicate that for x > 0,

x=2     pffiffiffiffiffiffiffi x  x=2   pffiffiffiffiffiffiffixx  1 #1 1 x 1 #2 < 2p 2x tanh 1 5
ð5:13Þ

31 with the best possible constants #1 ¼ 51840 and #2 ¼ 0.

  Although the double inequality (5.11) is given only for x P 1, its main utility is in the evaluation of C x þ 12 for large values of the argument.

References [1] M. Abramowitz, I.A. Stegun (Eds.), Handbook of mathematical functions with formulas, graphs, and mathematical tables, Appl. Math. Series 55, Ninth printing, National Bureau of Standards, Washington, D.C., 1972. [2] H. Alzer, On some inequalities for the gamma and psi functions, Math. Comp. 66 (1997) 373–389. [3] H. Alzer, Sharp upper and lower bounds for the gamma function, Proc. Royal Soc. Edinburgh 139A (2009) 709–718.

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