Two asymptotic expansions for gamma function developed by Windschitl’s formula

Lemma 3.1

The inequality

holds for allx > 0.

Proof

Let

Then we have

Hence, we conclude that

which proves (23), and the proof is done. □

Lemma 3.2

[43, [Lemma 7]. Letn ∈ ℕ andm ∈ ℕ ∪ {0} withn > m and letP_n(t) be anndegrees polynomial defined by

wherea_n, a_m > 0, a_i ≥ 0 for 0 ≤ i ≤ n − 1 withi ≠ m. Then, there is a unique numbert_m+1 ∈ (0, ∞) to satisfyP_n(t) = 0 such thatP_n(t) < 0 fort ∈ (0, t_m+1) andP_n(t) > 0 fort ∈ (t_m+1, ∞).

Consequently, for givent₀ > 0, ifP_n(t₀) > 0 thenP_n(t) > 0 fort ∈ (t₀, ∞) and ifP_n(t₀) < 0 thenP_n(t) < 0 fort ∈ (0, t₀).

Lemma 3.3

LetW₀₁(x), W01∗ (x), W₁(x), W_c1(x) and W_l1(x) be defined by (19), (20), (6), (21) and (22), respectively. Then we have

for allx ≥ 1.

Proof

1. The first inequality W₁(x) < W_c1(x) is equivalent to

   h1t=lnsinh⁡tt+1810t6−1+1135t4lnsinh⁡tt<0

   for t = 1/x ∈ (0, 1]. We have

   ddylny+1810t6−1+1135t4ln⁡y=−1135t4810y+135t2+t6y810y+t6<0

   for y > 1, which together with

   y=sinh⁡tt>1+16t2

   yields

   h1t<ln1+16t2+1810t6−1+1135t4ln1+16t2:=h11t.

   Differentiation leads us to

   1352t3h11′t=−2ln16t2+1−t2t6−135t2−1620t2+6t6+135t2+810:=h12t,h12′t=−4t3t12+9t10+540t8+8505t6+47385t4+328050t2+1312200t2+62t6+135t2+8102<0

   for t > 0. Therefore, we obtain h₁₂(t) < h₁₂(0) = 0, and so h₁₁(t) < h₁₁(0) = 0, which implies h₁(t) < 0 for t > 0.

2. The second inequality W_c1(x) < W01∗ (x) is equivalent to

   x21+1135x4lnxsinh⁡1x<x2lnxsinh⁡1x+ln1+11620x5,

   or equivalently,

   h2t=1270t3lnsinh⁡tt−ln1+11620t5<0

   for t = 1/x ∈ (0, 1]. Taking n = 2 in the inequalities (13) gives

   lnsinh⁡tt<16t2−1180t4+12835t6,

   which is applied to the expression of h₂(t):

   h2t<1270t316t2−1180t4+12835t6−ln1+11620t5:=h21t.

   Differentiation yields

   h21′t=t63402004t7−49t5+1050t3+6480t2−79380t5+1620<0

   for t ∈ (0, 1], which proves h₂(t) < 0 for t ∈ (0, 1].

3. The third inequality W01∗ (x) < W₀₁(x) is equivalent to

   1+11620x5<exp11620x5,

   which follows by a simple inequality 1 + y < e^y for y ≠ 0.

4. The fourth inequality W₀₁(x) < W_l1(x) is equivalent to

   x2lnxsinh1x+1810x7>x2lnxsinh⁡1x+11620x5,

   or equivalently,

   h3t=lnsinht+1810t7−lnsinh⁡t−1810t6>0

   for t = 1/x > 0. Denote h₃₀(t) = ln sinh t. Then by Taylor formula we have

   h3t=h30t+1810t7−h30t−1810t6 =t7810h30′t+12!t148102h30′′t+13!t218103h30′′′ξ−1810t6,

   where t < ξ < t + t⁷/810. Since h30′′′ (t) = 2(cosh t)/sinh³t > 0, we get

   h3t>1810t7cosh⁡tsinh⁡t−t142×81021sinh2⁡t−1810t6:=t6×h31t2×8102sinh2⁡t,

   where

   h31t=810tsinh⁡2t−810cosh⁡2t+810−t8.

   Due to

   h31t=540t4+144t6+1017t8+810∑n=5∞n−122n2n!t2n>0,

   we conclude that h₃(t) > 0 for t > 0, which completes the proof. □

Theorem 3.4

1. The function

   f1x=lnΓx+1−ln⁡2π−x+12ln⁡x+x−x2lnxsinh⁡1x+1810x6

   is strictly increasing and concave on [1, ∞).

2. Forx ≥ 1, we have

   β0xsinh⁡1x+1810x6x/2<Γx+12πxx/ex<xsinh⁡1x+1810x6x/2<xsinh⁡1xx21+1135x4<xsinh⁡1xx/21+11620x5<xsinh⁡1xx/2exp11620x5<2πxxexxsinh1x+1810x7x/2(25)

   with the best constant

   β0=e2πsinh⁡1+π/405≈0.99981.

Proof

1. Differentiation yields

   f1′x=ψx+1−ln⁡x−12x−12lnxsinh⁡1x+1810x6  +3135x6cosh⁡1x−135x7sinh⁡1x+1810x7sinh⁡1x+1,

   f1′′x=ψ′x+1−1x+12x2−3109350x14sinh2⁡1x+5940x7sinh⁡1x+135x5sinh⁡1x−1890x6cosh⁡1x−109350x12−1x810x7sinh⁡1x+12.

   Replacing x by x + 1/2 in inequality (23) yields

   ψ′x+1<1303780x4+7560x3+12705x2+8925x+30192x+163x4+126x3+217x2+154x+60,

   for x > − 1/2, and and applying the above inequality combined with the change of variable x = 1/t ∈ (0, 1] yield

   f1′′x<130t3019t4+8925t3+12705t2+7560t+3780t+260t4+154t3+217t2+126t+63−t+12t2−3t109350sinh2⁡t−1890t8cosh⁡t+5940t7sinh⁡t+135t9sinh⁡t−109350t2−t14810sinh⁡t+t72:=8102t×f11tt+2126t+217t2+154t3+60t4+63810sinh⁡t+t72,

   where

   f11t=p6tsinh2⁡t+p13tcosh⁡t−p14tsinh⁡t+p20t,  p6t=30t6+47t5−71815t4−210t3−259t2−3152t−63,  p13t=7810t8t+260t4+154t3+217t2+126t+63,p14t=t7127t7+77810t6+28571620t5+459736075t4+1547108t3+12341810t2+779t+15445, p20t=121870t20+257656100t19+136679841500t18+21787480t17+72700t16+74860t15+712150t14+30t7+137t6+5252t5+280t4+3152t3+63t2.

   To prove f₁₁(t) < 0 for t ∈ (0, 1], we use formula sinh²t = cosh²t − 1 to write f₁₁(t) as

   f31t=p6tcosh⁡t+p13tcosh⁡t−p14tsinh⁡t+p20t−p6t.

   Since the coefficients of polynomial − p₆(t) satisfy those conditions of Lemma 3.2, and − p₆(1) = 19 811/30 > 0, we see that − p₆(t) > 0 for t ∈ (0, 1]. It then follows from that

   p6tcosh⁡t+p13t<p6t+p13t=1427t13+959405t12+24554t11+39281t10+4918t9+4945t8+30t6+47t5−71815t4−210t3−259t2−3152t−63:=p13∗t.

   Application of Lemma 3.2 again with −p13∗ (1) = 173 959/270 > 0 yields −p13∗ (t) > 0 for t ∈ (0, 1], and so p₆(t) cosh t + p₁₃(t) < 0 for t ∈ (0, 1]. Since p₁₄(t) > 0 for t > 0, using the inequalities

   cosh⁡t>∑n=04t2n2n!=140320t8+1720t6+124t4+12t2+1,sinh⁡t>∑n=14t2n−12n−1!=15040t7+1120t5+16t3+t,

   we have

    f11t=p6tcosh⁡t+p13tcosh⁡t−p14tsinh⁡t+p20t−p6t<p6t∑n=04t2n2n!+p13t∑n=04t2n2n!−p14t∑n=14t2n−12n−1!+p20t−p6t=154190080t22+90071625702400t21+9615889109734912000t20+53514499405849600t19+739363013282175488000t18+173475972508226560t17+628751992090188800t16+252473483648t15−328873732480t14−232765193536t13−3620941870912t12−29209334560t11−29209386400t10:=t10p12t.

   From Lemma 3.2 and − p₁₂(1) = 67 766 507 802 179/3950 456 832 000 > 0 it follows that − p₁₂(t) > 0 for t ∈ (0, 1], and so f₁₁(t) < 0 for t ∈ (0, 1], which implies f1′′ (x) < 0 for x ≥ 1.

2. Using the increasing property of f₁ and noting that

   f11=ln⁡e2πsinh⁡1+π/405  and limx→∞f1x=0,

   we have

   ln⁡e2πsinh⁡1+π/405<ln⁡Γx+12πxxex−lnxsinh⁡1x+1810x6x/2<0,

   which imply the first and second inequalities of (25).

   The other ones of (25) follow from Lemma 3.3, which completes the proof. □