arXiv:1706.01971v1 [math.CA] 28 May 2017
Complete Monotonicity and Inequalities of Functions Involving Γ-function M. Al-Jararha∗ Department of Mathematics, Yarmouk University, Irbid, Jordan, 21163.
Keywords and Phrases: Gamma Function, Beta Function, completely monotonic functions, inequalities of gamma function. AMS (2000) Subject Classification: 33B15. Abstract In this paper, we investigate the complete monotonicity of some functions involving gamma function. Using the monotonic properties of these functions, we derived some inequalities involving gamma and beta functions. Such inequalities arising in probability theory.
1
Introduction
Completely monotonic functions play a major role in Probability and Mathematical Analysis due to their monotonic properties [9, 10, 26]. A function f (z) is called completely monotonic on an interval I ⊂ R if it has derivatives of any order f (n) (z), n = 0, 1, 2, 3, · · ·, and if (−1)n f (n) (z) ≥ 0 for all x ∈ I and all n ≥ 0. Recently, complete monotonicity of functions involving gamma and beta functions have been considered in many articles, for example [19, 29, 18, 8, 34, 20]. Also, many articles have appeared to provide various inequalities of functions that involving gamma and beta functions, see [13, 14, 22, 7, 31, 30, 24, 17, 2, 5, 6, 11, 15, 4, 28, 23, 12, 21, 35, 3]. In this paper, we investigate the complete monotonicity of some functions involving gamma function. Using the monotonic properties of these functions, we derive some inequalities involving gamma function, where gamma function is defined by Z ∞ e−t tx−1 dt, x > 0. Γ(x) = 0
∗
[email protected]
1
Due to the relation between gamma and beta function, we derived some inequalities that are involving beta function. Commonly, beta function is defined by Z 1 tx−1 (1 − t)y−1 dt, x and y > 0, B(x, y) = 0
and its relation with the gamma function is given by B(x, y) =
Γ(x + y) , x, y > 0. Γ(x)Γ(y)
The layout of the paper: In the first section, we prove our main results. In the second section, we apply some of our main results to prove the inequality Γ(x + 1)Γ(x − a − b + 1) Γ(y − a + 1)Γ(y − b + 1) ≥ 1, y ≥ x ≥ a + b > b ≥ a > 0. Γ(x − a + 1)Γ(x − b + 1) Γ(y + 1)Γ(y − a − b + 1) The last section is devoted for the concluding remarks.
2
The Main Results
In this section, we present and prove our main results. First, we present some useful definitions and theorems. Definition 2.1. A function f (z) is called completely monotonic on an interval I if it has derivatives of any order f (n) (z), n = 0, 1, 2, 3, · · ·, and if (−1)n f (n) (z) ≥ 0 for all x ∈ I and all n ≥ 0. If the above inequality is strict for all x ∈ I and all n ≥ 0, then f (z) is called strictly completely monotonic. Definition 2.2. A function f (z) is called logarithmically completely monotonic on an interval I if its logarithm has derivatives [ln f (z)](n) of orders n ≥ 1, and if (−1)n [ln f (z)](n) ≥ 0 for all x ∈ I and all n ≥ 1. If the above inequality is strict for all x ∈ I and all n ≥ 1, then f (z) is called strictly logarithmically completely monotonic. Theorem 2.1. [29]. Every (strict) logarithmically completely monotonic function is (strict) completely monotonic. Now, we turn to prove our main results. Theorem 2.2. Let a, b ≥ 0. Define f (z) =
Γ(z + 1)Γ(z − a − b + 1) , z > a + b − 1. Then f (z) Γ(z − a + 1)Γ(z − b + 1)
is completely monotonic function.
2
proof. Let f (z) =
Γ(z + 1)Γ(z − a − b + 1) . Then f (z) > 0, and Γ(z − a + 1)Γ(z − b + 1)
ln f (z) = ln Γ(z + 1) + ln Γ(z − a − b + 1) + ln Γ(z − a + 1) + ln Γ(z − b + 1).
(2.1)
By differentiating Eq. (2.1), we get d Γ′ (z + 1) Γ′ (z − a − b + 1) Γ′ (z − a + 1) Γ′ (z − b + 1) ln f (z) = + − − dz Γ(z + 1) Γ(z − a − b + 1) Γ(z − a + 1) Γ(z − b + 1) = Ψ(z + 1) + Ψ(z − a − b + 1) − Ψ(z − a + 1) − Ψ(z − b + 1),
(2.2)
where Ψ(z) is the Digamma function (the logarithmic differentiation of the Γ−function). By using the following integral representation of Digamma function Z ∞ −t (e − e−zt ) Ψ(z) = −γ + dt, Re(z) > 0, 1 − e−t 0 where γ = 0.577218... is Euler’s constant, we get d ln f (z) = Ψ(z + 1) + Ψ(z − a − b + 1) − Ψ(z − a + 1) − Ψ(z − b + 1) dz Z ∞ −t(z−a−b+1) e −(a+b)t −at −bt = − e − e − e + 1 dt 1 − e−t 0 Z ∞ −t(z−a−b+1) e −bt −at dt 1−e 1−e = − 1 − e−t 0 ≤ 0, ∀z > a + b − 1.
(2.3)
d Therefore, − dz ln f (z) ≥ 0, ∀z > a + b − 1. Inductively, we have
dn ln f (z) = (−1) dz n n
Z
0
∞ n−1 −t(z−a−b+1) t e
1 − e−t
1 − e−at
1 − e−bt dt ≥ 0, ∀z > a + b − 1. (2.4)
Hence, by Theorem 2.1, f (z) is completely monotonic function.
Remark 2.1. Since f (z) is completely monotonic. Then it is decreasing function. Hence, by using the asymptotic relation Γ(z + a) lim z (b−a) = 1, (2.5) z→∞ Γ(z + b) we get , Γ(z + 1)Γ(z − a − b + 1) ≥ 1, z > a + b − 1. Γ(z − a + 1)Γ(z − b + 1)
(2.6)
For a reference of the above asymptotic relation, see eq. 13 in [16] (see also, [1, 33]). Theorem 2.3. Let a1 , a2 , · · · , an ≥ 0, and let a ¯= z>a ¯ − 1 is completely monotonic function.
3
Pn
i=1 ai .
Then f (z) =
Γn−1 (z + 1)Γ(z − a ¯ + 1) Qn , i=1 Γ(z − ai + 1)
proof. Let f (z) =
Γn−1 (z + 1)Γ(z − a ¯ + 1) Qn , z>a ¯ − 1. Then i=1 Γ(z − ai + 1)
dn f (z) (−1)n dz n
=
Z
∞ (n−1) −t(z+1) t e
1 − e−t
0
≥ 0,
(n − 1) + ea¯t −
n X
eai t
i=1
z>a ¯ − 1.
!
dt
P This is correct since the function ξ(t) := (n−1)+ea¯ t − ni=1 eai t , t ≥ 0, is nonnegative function. In P P P P fact, ξ(0) = 0 and ξ ′ (t) = a ¯ea¯t − ni=1 ai eai t = ni=1 ai ea¯t − ni=1 ai eai t = ni=1 ai (ea¯t −eai t ) > 0, since the exponential function is increasing function. Hence, ξ ′ (0) = 0 and ξ ′ (t) > 0, ∀t > 0. Therefore, ξ(t) is nonnegative. Hence, by Theorem 2.1, f (z) is completely monotonic. P Remark 2.2. Let a1 , a2 , · · · , an ≥ 0, and let a ¯ = ni=1 ai . Since f (z) =
Γn−1 (z + 1)Γ(z − a ¯ + 1) Qn , ∀z > a ¯−1 i=1 Γ(z − ai + 1)
is completely monotonic function, and hence, it is decreasing function. Therefore, by using the asymptotic relation (2.5), we get f (z) =
Γn−1 (z + 1)Γ(z − a ¯ + 1) Qn ≥ 1, ∀z > a ¯ − 1. i=1 Γ(z − ai + 1)
Theorem 2.4. Let 0 ≤ a1 ≤ a2 ≤ · · · ≤ an , 0 ≤ b1 ≤ b2 ≤ · · · ≤ bn , and m = max{a1 , a2 , · · · , an , P P b1 , b2 , · · · , bn }. Moreover, assume that ki=1 bi ≤ ki=1 ai , k = 1, 2, · · · , n. Then Qn Γ(z − ai ) , z > m, f (z) = Qi=1 n i=1 Γ(z − bi ) is completely monotonic function. Qn Γ(z − ai ) proof. Let f (z) = Qi=1 , z > m. Then n i=1 Γ(z − bi ) dn f (z) (−1)n dz n
=
Z
0
∞ (n−1) −tz t e
1 − e−t
≥ 0, z > m.
n X i=1
eai t −
n X i=1
!
ebi t dt
Pk Pk This inequality holds since and the exponential function i=1 bi ≤ i=1 ai , k = 1, 2, · · · , n P P ex , x > 0 is convex and increasing function which implies that ni=1 eai t ≥ ni=1 ebi t (see page 12 in [25]). Hence, by Theorem 2.1, f (z) is completely monotonic function. Remark 2.3. Let 0 ≤ a1 ≤ a2 ≤ · · · ≤ an , 0 ≤ b1 ≤ b2 ≤ · · · ≤ bn , and m = max{a1 , a2 , · · · , an , Pk Pk Pn b1 , b2 , · · · , bn }. Moreover, assume that bi ≤ ai , k = 1, 2, · · · , n − 1 and i=1 i=1 i=1 ai = Pn f (z) is decreasing function as a result of the above theorem. Hence, by the limit i=1 bi . Then Qn i=1 Γ(z−ai ) Q (2.5), we get n Γ(z−bi ) ≥ 1, ∀z > m. i=1
4
Pn
i=1 ai Remark 2.4. Let 0 ≤ a1 ≤ a2 · · · ≤ an and let bi = a ¯ := , i = 1, 2, · · · , n. Then Pk Pn n Pn Pk k Pk a < a , k = 1, 2, · · · , n − 1, and a = b = k¯ a = i i i i i=1 i=1 i=1 i=1 i=1 n Qbni . Hence, f (z) = Qn Γ(z − ai ) i=1 Γ(z − ai ) is completely monotonic ∀z > a ¯. Moreover, the inequality i=1 ≥1 Γ(z − a ¯)n Γ(z −a ¯)n Qn Γ(z − ai + 1) holds ∀z > a ¯. By letting z = z + 1 in this inequality, we get that i=1 ≥ 1, ∀z > Γ(z − a ¯ + 1)n a ¯ − 1. For a particular case if we let n = 2, a1 = x > 0, a2 = y >, and z = x + y, then we Γ(x)Γ(y) (x + y)2 Γ(x + 1)Γ(y + 1) ≥ 1, ∀x, y ≥ 0. Particularly, we get ≥ , ∀x, y > 0. This get 2 2 4xy Γ( x+y Γ( x+y 2 + 1) 2 ) inequality has been proved in different method in [21].
Theorem 2.5. Let a ≥ 0 and define f (z) =
Γ(z + a)Γ(z − a) , z > a. Then f (z) is completely Γ(z)2
monotonic function. proof. Let f (z) =
Γ(x − a)Γ(x + a) , z > a. Then Γ(z)2 Z ∞ (n−1) −tz n t e n d f (z) (−1) = eat + e−at − 2 dt n −t dz 1−e 0 Z ∞ (n−1) −tz t e eat − 1 1 − e−at dt = −t 1−e 0 ≥ 0, z > a.
Hence, by Theorem 2.1, f (z) is completely monotonic function. Γ(z + a)Γ(z − a) , z > a is decreasing function. By using Γ(z)2 Γ(z + a) Γ(z + a)Γ(z − a) the asymptotic relation lim z (b−a) = 1, we get ≥ 1, ∀z > a. This z→∞ Γ(z + b) Γ(z)2 inequality has been proved in [15] by using classical integral inequalities. Remark 2.5. Let a ≥ 0. Then f (z) =
Remark 2.6. Using the same argument above, we can show that f (z) =
Γ(z + a + 1)Γ(z − a + 1) , z >a−1 Γ(z + 1)2
is completely monotonic, and so it is decreasing function. Consequently, the inequality Γ(z + a + 1)Γ(z − a + 1) ≥1 Γ(z + 1)2
(2.7)
holds ∀z > a − 1. Moreover, since f (z) is decreasing function, then f (z) ≤ f (0). Hence, we get 1≤
Γ(z + a + 1)Γ(z − a + 1) ≤ Γ(1 − a)Γ(1 + a), 0 ≤ a < 1, z ≥ 0. Γ(z + 1)2
Moreover, by using the formula Γ(z + 1) = zΓ(z), we get Γ(z + a)Γ(z − a) z2 z2 ≤ ≤ (Γ(1 − a)Γ(1 + a)), 0 ≤ a < 1, z > a. z 2 − a2 Γ(z)2 z 2 − a2 5
(2.8)
By using the formula Γ(1 − a)Γ(1 + a) = z2
πa sin πa ,
0 < a < 1 (see page 48,[32]), we get
Γ(z + a)Γ(z − a) πaz 2 z2 ≤ ≤ , 0 < a < 1, z > a. 2 2 −a Γ(z) sin πa(z 2 − a2 )
For the particular case a = 12 , we have Γ(z + 12 )Γ(z − 12 ) 4z 2 2πz 2 1 ≤ ≤ , z> . 2 2 2 4z − 1 Γ(z) 4z − 1 2 Let a =
1 2
in (2.8), then we have 1≤
Γ(z + 23 )Γ(z + 12 ) π ≤ , z ≥ 0. 2 Γ(z + 1) 2
Equivalently,
More precisely, we have
3
2 2z + 1
1
Γ(z + 12 ) ≤ ≤ Γ(z + 1)
π 2z + 1
1
, z ≥ 0.
2z 2 2z + 1
12
Γ(z + 12 ) ≤ ≤ Γ(z)
πz 2 2z + 1
12
, z > 0.
2
2
Applications to Probability
In this section, we prove the following inequality: Γ(x + 1)Γ(x − a − b + 1) Γ(y − a + 1)Γ(y − b + 1) ≥ 1, 0 < a ≤ b < a + b ≤ x ≤ y. Γ(x − a + 1)Γ(x − b + 1) Γ(y + 1)Γ(y − a − b + 1) This inequality arises in the Probability in the problems dealing with the general binomial coef Γ(α+1) α ficients β = Γ(β+1)Γ(α−β+1) . To prove this inequality, we present some useful remarks.
Remark 3.1. Let a, b > 0 and let M ≥ a + b. Then ∀z such that 0 < a ≤ b < a + b ≤ z < e−t(z−a−b+1) M , the function δ(z) = (1 − e−at )(1 − e−bt ) is nonnegative and is not identically 1 − e−t zero. Hence, as a consequence of the proof of Theorem 2.2, we have f ′ (z) < 0. Therefore, Γ(z + 1)Γ(z − a − b + 1) is strictly decreasing function on [a + b, M ). Hence, f (a + b) = f (z) = Γ(z − a + 1)Γ(z − b + 1) Γ(a + b + 1) Γ(z − a + 1)Γ(z − b + 1) > 1. Let g(z) = , z ≥ a + b > b ≥ a > 0. Then Γ(a + 1)Γ(b + 1) Γ(z + 1)Γ(z − a − b + 1) Γ(z − a + 1)Γ(z − b + 1) 1 , and so g(z) = ≤ 1, ∀z ≥ a + b > b ≥ a > 0. Also, for g(z) = f (z) Γ(z + 1)Γ(z − a − b + 1) z ∈ [a + b, M ), we have g′ (z) > 0. Remark 3.2. Let 0 < a ≤ b. Define h(x, y) = f (x)g(y), x, y ≥ a + b, where f (x) = Γ(y − a + 1)Γ(y − b + 1) Γ(x + 1)Γ(x − a − b + 1) and g(y) = . Then h(x, y) is positive and Γ(x − a + 1)Γ(x − b + 1), Γ(y + 1)Γ(y − a − b + 1) continuous on the rectangular domain D = [a + b, ∞) × [a + b, ∞). Using the asymptotic relation Γ(z + a) lim z (b−a) = 1, we have limk(x,y)k→∞ h(x, y) → 1. z→∞ Γ(z + b) 6
In the following theorem, we prove that h(x, y) ≥ 1 on the triangular domain Ω = {(x, y) | 0 < a ≤ b < a + b ≤ x ≤ y} ⊂ D. Theorem 3.1. Let h(x, y) =
Γ(x + 1)Γ(x − a − b + 1) Γ(y − a + 1)Γ(y − b + 1) , (x, y) ∈ Ω. Γ(x − a + 1)Γ(x − b + 1) Γ(y + 1)Γ(y − a − b + 1)
Then h(x, y) ≥ 1, ∀(x, y) ∈ Ω. proof. Since h(x, y) is positive, continuous, and limk(x,y)k→∞ h(x, y) → 1. Then, by using Remark 3.1 and Remark 3.2, there exists a sufficiently large constant M > 0, such that h(a + b, M ) =
Γ(a + b + 1) Γ(M − a + 1)Γ(M − b + 1) ≥ 1, Γ(a + 1)Γ(b + 1) Γ(M + 1)Γ(M − a − b + 1)
Now, define the closed and bounded rectangular region: R = {(x, y)| 0 < a ≤ b < a + b ≤ x ≤ y ≤ M } . Since h(x, y) is continuous function on R. Then it takes its absolute values on the boundaries of R, or at the points in R where ∇h(x, y) = 0 (∇h(x, y) is the gradient of h(x, y)). Clearly, ∇h(x, y) = f ′ (x)g(y)i + f (x)g′ (y)j, where f (x) and g(y) are defined in Remark 3.2. Moreover, by Remark 3.1, we have f (x) > 0, g(y) > 0, g ′ (y) > 0, and f ′ (x) < 0 in R. Hence, ∇h(x, y) 6= 0 in R. Hence, the absolute values of h(x, y) must occur at the boundaries of R. In fact, the boundaries of R are the line segments 1. l1 = {(x, M ) | a + b ≤ x ≤ M } , 2. l2 = {(a + b, y) | a + b ≤ x ≤ M } , and 3. l3 = {(x, x) | a + b ≤ x ≤ M } . Obviously, h(x, y) = h(x, x) = 1 on the line segment l3 . Moreover, on the line segment l1 , we have h(x, M ) =
Γ(x + 1)Γ(x − a − b + 1) Γ(M − a + 1)Γ(M − b + 1) , a + b ≤ x ≤ M, Γ(x − a + 1)Γ(x − b + 1) Γ(M + 1)Γ(M − a − b + 1)
which is decreasing function in x with a negative derivative. Similarly, on the line segment l1 , we have h(a + b, y) =
Γ(a + b + 1) Γ(y − a + 1)Γ(y − b + 1) , a + b ≤ y ≤ M, Γ(a + 1)Γ(b + 1) Γ(y + 1)Γ(y − a − b + 1)
which is increasing function in y with a positive derivative. Hence, h(x, y) takes its absolut values at the points (a + b, a + b), (M, M ), and (a + b, M ). Clearly, h(a + b, a + b) = h(M, M ) = 1. At the point (a + b, M ), we have h(a + b, M ) =
Γ(a + b + 1) Γ(M − a + 1)Γ(M − b + 1) ≥ 1. Γ(a + 1)Γ(b + 1) Γ(M + 1)Γ(M − a − b + 1)
This implies that h(x, y) ≥ 1, ∀(x, y) ∈ Ω. This completes the proof. Finally, we have the following remark: 7
Γ(x + 1)Γ(y − a + 1) , y > x > a > 0. Let x ∈ (a, y). Then there Γ(y + 1)Γ(x − a + 1) exists αx > 0, such that x = y − αx . Hence, we define Remark 3.3. Let ψ(x, y) =
ξ(y) = ψ(y − αx , y) =
Γ(y − αx + 1)Γ(y − a + 1) , y > αx + a > a > 0. Γ(y + 1)Γ(y − αx − a + 1)
Then, by using the above Remark 3.2, we have ψ(y − αx , y) =
Γ(y − αx + 1)Γ(y − a + 1) ≤ 1, ∀y > αx + a > a > 0. Γ(y + 1)Γ(y − αx − a + 1)
Therefore, ψ(x, y) =
Γ(x + 1)Γ(y − a + 1) ≤ 1, y > x > a > 0. Γ(y + 1)Γ(x − a + 1)
Consequently, we also have Γ(x + 1)Γ(y − a − b + 1) ≤ 1, y > x > a + b > b ≥ a > 0. Γ(y + 1)Γ(x − a − b + 1)
4
Concluding Remarks
We have seen in Remark 2.1 that Γ(z + 1)Γ(z − a − b + 1) ≥ 1, ∀z ≥ a + b − 1. Γ(z − a + 1)Γ(z − b + 1)
(4.1)
Let z = a + b, a, b ≥ 0 in (4.1), then we get Γ(a + b + 1) ≥ 1, ∀a, b ≥ 0. Γ(a + 1)Γ(b + 1)
(4.2)
By using the fact that Γ(z + 1) = zΓ(z), we have Γ(a + b + 1) (a + b) Γ(a + b) = ≥ 1, ∀a, b > 0. Γ(a + 1)Γ(b + 1) ab Γ(a)Γ(b)
(4.3)
ab Γ(a + b) ≥ , ∀a, b > 0. Γ(a)Γ(b) (a + b)
(4.4)
Hence,
Since B(a, b) =
Γ(a + b) , we get Γ(a)Γ(b)
(a + b) , ∀a, b > 0 ab y y B(x, y), we get Moreover, by using the fact B(x, y + 1) = B(x + 1, y) = x x+y B(a, b) ≤
B(a, b + 1) =
b b 1 B(a + 1, b) = B(a, b) ≤ , a, b > 0. a a+b a 8
(4.5)
1 1 Hence, B(a, b + 1) ≤ and B(a + 1, b) ≤ for a, b > 0. In [15], the authors proved the following a b inequality: 1 (4.6) B(a, b) ≤ , 0 < a, b ≤ 1. ab Clearly, if 0 < a, b ≤ 1 and a + b ≤ 1, the upper bound of ⌊(a, b) given in (4.5) is better than the one given in (4.6). Moreover, the inequality (4.5) is valid for a, b > 0 while the inequality (4.6) is restricted on a, b ∈ (0, 1]. Let a = b in Eq. (2.6) and Eq. (2.8). Then we get Γ(2a + 1) ≥ 1, a ≥ 0, Γ(a + 1)2
(4.7)
Γ(2a) a ≥ , a > 0, 2 Γ(a) 2
(4.8)
and
respectively. Assume that 0 < a ≤ 1. then we have the following inequality (see eq. 2.8 in [21]: Γ(a)2 2a − a2 ≥ , Γ(2a) a2
(4.9)
By combining (4.8) with (4.9), we get Γ(a)2 2 2a − a2 ≤ ≤ , 0 < a ≤ 1. 2 a Γ(2a) a Assume that a, b > 0 and 0 < z ≤ 1. Then inequality (4.1) implies that 1 Γ(z − a − b + 1) ≥ ≥ 1, 0 < a ≤ b < a + b ≤ z ≤ 1. Γ(z − a + 1)Γ(z − b + 1) Γ(z + 1)
(4.10)
This is correct since Γ(x + 1) ≤ 1, ∀z ∈ [0, 1]. Moreover, we have 1≤
Γ(a + b + 1) Γ(z + 1)Γ(z − a − b + 1) ≤ , z ≥ a + b > b ≥ a > 0. Γ(z − a + 1)Γ(z − b + 1) Γ(a + 1)Γ(b + 1)
(4.11)
Hence, for z ≥ a + b > b ≥ a > 0, we have 1 Γ(a + b + 1) Γ(z − a − b + 1) 1 ≤ ≤ . Γ(z + 1) Γ(z − a + 1)Γ(z − b + 1) Γ(z + 1) Γ(a + 1)Γ(b + 1)
(4.12)
Therefore, Γ(z − a − b + 1) = 0. z→∞ Γ(z − a + 1)Γ(z − b + 1) lim
Now, let a = b in (4.1), then we get Γ(z + 1)Γ(z − 2a + 1) ≥ 1, z > 2a − 1. Γ(z − a + 1)2 Equivalently,
Γ(z)Γ(z − 2a) (z − a)2 a2 ≥ = 1 + , z > 2a. Γ(z − a)2 z(z − 2a) z(z − 2a) 9
(4.13)
Clearly, Γ(z)Γ(z − 2a) ≥ 1, z > 2a. Γ(z − a)2 In fact, in [27] many lower bounds of given in the following inequalities:
and
Γ(z)Γ(z−2a) Γ(z−a)2
were presented. Some of these lower bounds are
Γ(z)Γ(z − 2a) a2 + z , z > 0 and z − 2a > 0, ≥ 2 Γ(z − a) z
(4.14)
a2 (z − 2) Γ(z)Γ(z − 2a) ≥ 1 + , z > 2, z − 2a > 0, and a 6= 0, −1. Γ(z − a)2 (z − a − 1)2
(4.15)
References [1] M. Abramowitz and I. A. Stegun, Handbook of Mathematical Functions with formulas, Graphs, and Mathematical Tables, Dover, new Yourk, 1965. [2] R. P. Agarwal, N. Elezovi´c, and J. Peˇcari´c, On Some Inequalities for Beta and Gamma Functions Via Some Classical Inequalities, Journal of Inequalities and Applications, 5(2005), 593–613. [3] H. Alzer, Some Gamma Function Inequalities, Mathematics of Computations, 60(1993), No. 201,337–346. [4] H. Alzer, On Some Inequalities for the Gamma and Psi functions, Mathematics of Computation, 66(1997), No. 217, 373–389. [5] H. Alzer, A Power Mean Inequality for the Gamma Function, Monatsh. Math., 131 (2000), 179-188. [6] H. Alzer, Sharp Inequalities for the beta function, Indag. Mathem., 12 (2001), No. 1, 15–21. [7] H. Alzer, Inequalities Involving Γ(x) and Γ( x1 ), Journal of Computational and Applied Mathematics, 192 (2006), 460–480. [8] H. Alzer and N. Batir, Monotonicity Properties of the Gamma Function, Applied Mathematical Letters, 20 (2007), 778–781. [9] H. Alzer and C. Berg, Some Classes of Completely Monotonic Functions, Annales Academiae Scientiarum Fennicae, 27(2002), 445–460. [10] H. Alzer and C. Berg, Some Classes of Completely Monotonic Functions, II, The Ramanujan Journal, 11(2006), No. 2, 225–248. [11] H. Alzer and A. Z. Grinshpan, Inequalities for the Gamma and q−gamma Functions, Journal of Approximation Theory, 144 (2007), 67–83.
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[12] N. Batir, Inequalities for the double Gamma Functions, J. Math. Anal. Appl., 351(2009), 182–185. [13] J. Bustoz and M. E. H. Ismail, On Gamma Function Inequality, Mathematics of Computation, 47(1986), No. 176, 659–667. [14] C. P. Chen and F. Qi, Inequalities Related to the gamma Function, The Australian Journal of Mathematical Analysis and Applications, 1 (2004), No. 1, 1–18. [15] S. S. Dragomir, R. P. Agarwal, and N. S. Barnett, Inequalities for Beta and Gamma Functions Via Some Classical and New Integral Inequalities, J. of Inequal. & Appl., 5 (2000), no. 2, pp. 103-165 [16] A. Erd´ely, W. Magnus, F. Oberhettinger, F. G. Tricomi, Higher Transcendental Functions, Volume I, McGraw Hill, New York, 1953. [17] M. Garg, A.Sharma, and P. Manohar , Some Inequalities Involving Ratios and Products of Gamma Function, Le Matematiche, LXVIII (2003), Fasc. II, pp. 219-225. [18] C. Giordano and A. Laforgia , Inequalities and Monotonicity Properties for the Gamma Function, Journal of Computational and Applied Mathematics, 133(2001), 387–396. [19] B. N. Guo and F.Qi, Inequalities and Monotonicity for the Ratio of Gamma Functions, 7(2003), No. 2, 239–247. [20] M. E. H. Ismail, L. Lorch, and M. E. Muldoon, Completely Monotonic Functions Associated with the Gamma Function and its q−Analogues, Journal of Mathematical Analysis and Applications, 116(1986), 1–9. [21] P. Iv´ ady, On the Beta Function Inequality, Journal of Mathematical Inequalities, 6(2012), No. 3, 333–341. [22] A. Laforgia , Further Inequalities for the Gamma Function, Mathematics of Computation, 42(1984), 597–600. [23] L. Maligranda, J. E. Pecaric, and L. E. Persson, Stolarsky’s Inequality with General Weights, Proceeding of the American Mathematical Society, 123, No. 7, 2113–2118. [24] M. Marcer, Some New Inequalities for the Gamma, Beta and Zeta Function, Journal of Inequalities in Pure and Applied Mathematics, 6 (2006), No. 1, Artical 29. [25] A. W. Marshall, I. Olkin, and B. C. Arnold, Inequalities: Theory of Majorization and Its Applications, Second Edition, Springer, New York, 2011. [26] K. S. Miller and S. G. Smako, Completely Monotonic Functions, Int. Transf. and Spec. Funct., 12(2001), No. 4, 389–402. [27] D. S. Mitrinovi´c and P. M. Vasi´c, Analytic Inequalities, Springer, New York, 1970.
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[28] K. Nantomah and M. M. Iddrisu, Extensions of Some Inequalities for the Gamma function, International Mathematical Forum, 9(2014), No. 5, 199–206. [29] F. Qi and C. P. Chen , A Complete Monotonicity Property of the Gamma Function, J. Math Anal. Appl., 296 (2004), pp. 603–607. [30] J. S´andor, A Note on Certain Inequalities for the Gamma Function, Journal of Inequalities in Pure and Applied Mathematics, 6 (2005), No. 3, Article 61. [31] A. S. Shabani, Some Inequalities of the Gamma Function, Journal of Inequalities in Pure and Applied Mathematics, 8 (2007), No. 2, Article 49. [32] N. M. Temme, Special Functions An Introduction to the Classical Functions of Mathematical Physics, John Wiley & Sons, New York, 1996. [33] F. Tricomi and A. Erd´elyi, The Asymptotic Expansion of a Ratio of Gamma Functions, Pacific Journal of Mathematics, 1(1951), No. 1, 133–142. [34] H. Vogt, A monotonicity Property of the Γ−Function, Journal of Inequalities in Pure and Applied Mathematics, 3 (2002), No. 5, Article 73. [35] S. Zhimin, D. Xiangkai, and Y. Li, On Some New Inequalities for the Gamma Function, Octagon Mathematical Magazine, 17(2009), No. 1, 14–18.
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