Analysis Mathematica, 40(2014), 1–11

DOI: 10.1007/s10476-014-0101-2

Inequalities for the beta function

HORST ALZER

Morsbacher Str. 10, 51545 Waldbrol, Germany; e-mail: H.Alzer@gmx.de

Received January 27, 2013.

Ab s t r a c t . We prove the following inequalities involving Euler’s beta function.(i) Let α and β be real numbers. The inequalities(

yz−x

xz−y zy−x

)α

≤ B(x, x)z−y B(z, z)y−x

B(y, y)z−x≤

(yz−x

xz−y zy−x

)β

hold for all x, y, z with 0 < x ≤ y ≤ z if and only if α ≤ 1/2 and β ≥ 1.(ii) Let a and b be non-negative real numbers. For all positive real numbers x and

y we have

δ(a, b) ≤ xaB(x+ b, y) + yaB(x, y + b)

(x+ y)aB(x, y)≤ Δ(a, b)

with the best possible bounds

δ(a, b) = min{2−a, 21−a−b} and Δ(a, b) = max{1, 21−a−b}.

1. Introduction and statement of results

The classical beta function, also known as Eulerian integral of the firstkind, is defined for positive real numbers x and y by

B(x, y) =∫ 1

0tx−1(1− t)y−1 dt.

The relationship between the beta function and the Eulerian integral of thesecond kind, the famous gamma function,

Γ(x) =∫ ∞

0e−ttx−1 dt,

0133–3852/$ 20.00c© 2014 Akademiai Kiado, Budapest

2 Horst Alzer

is given by the identity

B(x, y) =Γ(x)Γ(y)

Γ(x+ y).

Besides the integral formula we have the product and series representations

B(x, y) =( 1x+

1

y

) ∞∏n=1

1 + (x+ y)/n

(1 + x/n)(1 + y/n)=

=1

y

∞∑n=0

(−1)ny(y − 1) · · · (y − n)

n! (x+ n).

The most important facts on B(x, y) are collected, for example, in themonographs [6] and [21]. Interesting historical comments on the beta andgamma functions are given in [8].

The beta function has various applications not only in the theory ofspecial functions, but it also plays a role in other fields, like, for instance,statistics, mathematical physics, and graph theory; see [2], [10], [13], [19].Remarkable number-theoretic properties of B(x, y) are discussed in [20],and a characterization of the beta function via a functional equation can befound in [17].

In the past forty years, numerous inequalities for the beta function werepublished. We refer to [2], [3], [4], [5], [7], [9], [12], [14], [15], [16], and thereferences given therein. It is the aim of this paper to continue the researchon this subject and to present some new inequalities.

In [9] and [5] the authors proved that for all x, y > 0 the geometric andarithmetic means of B(x, x) and B(y, y) are separated by B(x, y), that is,√

B(x, x)B(y, y) ≤ B(x, y) ≤ B(x, x) +B(y, y)

2.

These inequalities have motivated the work on our first theorem which offersa Lyapunov-type inequality involving the three terms B(x, x), B(y, y), andB(z, z). We recall that inequalities of the form

f(y)z−x ≤ f(x)z−yf(z)y−x (x ≤ y ≤ z)

are named after A. Lyapunov, who presented in 1901 an inequality of thistype for finite sums; see [11, p. 27].

The following Lyapunov-type inequality for the beta function is valid:

(1.1) B(y, y)z−x ≤ B(x, x)z−yB(z, z)y−x (0 < x ≤ y ≤ z).

The proof of (1.1) is based upon the fact that Θ(x) = logB(x, x) is convexon (0,∞); see [9]. Applying Jensen’s inequality gives for all a, b > 0 andμ ∈ [0, 1]:

(1.2) Θ(μa+ (1− μ)b) ≤ μΘ(a) + (1− μ)Θ(b).

Inequalities for the beta function 3

Setting

a = x, b = z, and μ = (z − y)/(z − x) (with x < z)

in (1.2) yields (1.1).We note that the convexity of Θ leads to a companion of (1.1); see [22]:

1 ≤ B(x, x)(y−x)(z−x)B(y, y)(x−y)(z−y)B(z, z)(x−z)(y−z).

This inequality holds for all x, y, z ∈ (0, 1). The sign of equality is valid ifand only if x = y = z.

Our first theorem provides a refinement and a converse of (1.1).

Theor em 1. Let α and β be real numbers. The inequalities

(1.3)( yz−x

xz−y zy−x

)α ≤ B(x, x)z−y B(z, z)y−x

B(y, y)z−x≤

( yz−x

xz−y zy−x

)β

hold for all x, y, z with 0 < x ≤ y ≤ z if and only if α ≤ 1/2 and β ≥ 1.

The elegant identity

(1.4) B(x+ 1, y) +B(x, y + 1) = B(x, y)

which is given, for instance, in [18, p. 31] is a counterpart of the well-known recurrence formula for the gamma function, Γ(x + 1) = xΓ(x).Formula (1.4) leads to the question whether it is possible to compare thesum B(x+ b, y)+B(x, y+ b) (b > 0) with B(x, y). Our second theorem canbe regarded as an extension of (1.4). Indeed, setting a = 0, b = 1 in (1.5)and (1.6) below yields (1.4).

Theor em 2. Let a ≥ 0 and b > 0 be real numbers. For all positivereal numbers x and y we have

(1.5) δ(a, b) ≤ xaB(x+ b, y) + yaB(x, y + b)

(x+ y)aB(x, y)≤ Δ(a, b)

with the best possible bounds

(1.6) δ(a, b) = min{2−a, 21−a−b} and Δ(a, b) = max{1, 21−a−b}.In the next section we present a lemma which we need to prove our

theorems. In Section 3, we offer a proof of Theorem 1 and in Section 4 weestablish Theorem 2.

2. Lemma

We denote by ψ = Γ′/Γ the logarithmic derivative of the gammafunction. The properties given in the following lemma can be found in[1, Chapter 6].

4 Horst Alzer

Lemma. We have

(2.1) (−1)n+1ψ(n)(x) =

∫ ∞

0e−xt tn

1− e−tdt (x > 0; n ∈ N),

(2.2) ψ(n)(x+ 1) = ψ(n)(x) + (−1)nn!

xn+1(x > 0; 0 ≤ n ∈ Z),

(2.3) log Γ(x) ∼(x− 1

2

)log x− x+

1

2log(2π) +

1

12x− · · · (x→ ∞),

(2.4) (−1)n+1ψ(n)(x) ∼ (n− 1)!

xn+

n!

2xn+1+ · · · (x→ ∞; n ∈ N).

3. Proof of Theorem 1

First, we show: if α ≤ 1/2 and β ≥ 1, then (1.3) is valid for all numbersx, y, z with 0 < x ≤ y ≤ z. Let

Sλ(x) = logB(x, x) + λ log x (λ ∈ R).

Then, we have

S′′λ(x) = 2ψ′(x)− 4ψ′(2x) − λ

x2.

Using (2.1) with n = 1 gives

4ψ′(2x) =∫ ∞

0e−xt t

1− e−t/2dt =

=

∫ ∞

0e−xt

( t

1− e−t+

te−t/2

1− e−t

)dt = ψ′(x) + ψ′(x+

1

2

).

Next, we apply this duplication formula as well as the integral representa-tions (2.1) and

1

xn=

1

(n− 1)!

∫ ∞

0e−xttn−1 dt (x > 0; n ∈ N).

This yields

(3.1) S′′λ(x) = ψ′(x)− ψ′(x+ 1/2) − λ

x2=

∫ ∞

0e−xttS∗

λ(t) dt

with

S∗λ(t) =

et − et/2

et − 1− λ.

For α ≤ 1/2 we obtain

(3.2) S∗α(t) ≥

et − et/2

et − 1− 1

2=

(et/2 − 1)2

2(et − 1)> 0 for t > 0.

Inequalities for the beta function 5

If β ≥ 1, then we have

(3.3) S∗β(t) ≤

et − et/2

et − 1− 1 =

1− et/2

et − 1< 0 for t > 0.

From (3.1)–(3.3) we conclude that Sα and −Sβ are strictly convex on(0,∞). Thus, we get for a, b > 0, a �= b, and μ ∈ (0, 1):

(3.4) Sα(μa+ (1− μ)b) < μSα(a) + (1− μ)Sα(b)

and

(3.5) μSβ(a) + (1− μ)Sβ(b) < Sβ(μa+ (1− μ)b).

If x = y or y = z, then the sign of equality holds in (1.3). Let 0 < x < y <z. We set

a = x, b = z, and μ = (z − y)/(z − x)

in (3.4) and (3.5), respectively. This leads to (1.3) with “<” instead of “≤”.Next, we assume that (1.3) is valid for all x, y, z with 0 < x ≤ y ≤ z.

Let 0 < x < y < z. Then we obtain

(3.6) α ≤ T1(x, y, z)

T2(x, y, z)≤ β

with

T1(x, y, z) = (z − y) logB(x, x) + (y − x) logB(z, z) − (z − x) logB(y, y)

and

T2(x, y, z) = (z − x) log y − (z − y) log x− (y − x) log z > 0.

We have

limy→x

T1(x, y, z)

y − x= logB(z, z)− logB(x, x)− 2(z − x)[ψ(x) − ψ(2x)]

and

limy→x

T2(x, y, z)

y − x= log x− log z +

z

x− 1.

Hence it follows that

(3.7) limx→z

limy→x

T1(x, y, z)

T2(x, y, z)=

= limx→z

logB(z, z)− logB(x, x)− 2(z − x)[ψ(x) − ψ(2x)]

log x− log z + z/x− 1=

= 2z2[ψ′(z)− 2ψ′(2z)] = T3(z), say.

Applying (2.2) and (2.4) leads to the limit relations

(3.8) limz→∞T3(z) =

1

2, lim

z→0T3(z) = 1.

From (3.6)–(3.8) we get α ≤ 1/2 and β ≥ 1.

6 Horst Alzer

4. Proof of Theorem 2

In what follows, we asumme that 0 < x ≤ y. First, we prove theleft-hand side of (1.5) with δ(a, b) as given in (1.6). We distinguish twocases.

Case 1.1. b ≤ 1.Then, δ(a, b) = 2−a and we have to show that

(4.1)(x+ y

2

)aB(x, y) ≤ xaB(x+ b, y) + yaB(x, y + b).

Since∂

∂bB(x+ b, y) = B(x+ b, y)[ψ(x + b)− ψ(x+ y + b)] < 0,

we conclude that the expression on the right-hand side of (4.1) is strictlydecreasing with respect to b. Hence it follows that

(4.2) xaB(x+ 1, y) + yaB(x, y + 1) ≤ xaB(x+ b, y) + yaB(x, y + b).

Let

(4.3) Ω(a;x, y) =( 2x

x+ y

)aB(x+ 1, y) +

( 2y

x+ y

)aB(x, y + 1).

Since a → Ω(a;x, y) is convex on [0,∞), we get

∂

∂aΩ(a;x, y) ≥ ∂

∂aΩ(a;x, y)

∣∣∣a=0

=B(x+ 1, y)

x

∫ y

xlog

2t

x+ tdt ≥ 0

and

(4.4) Ω(a;x, y) ≥ Ω(0;x, y) = B(x+ 1, y) +B(x, y + 1) = B(x, y).

Combining (4.2)–(4.4) we conclude that (4.1) holds.

Case 1.2. b > 1.We have δ(a, b) = 21−a−b , so that the left-hand side of (1.5) can be

written as

21−bB(x, y) ≤( 2x

x+ y

)aB(x+ b, y) +

(4.5) +( 2y

x+ y

)aB(x, y + b) = Ω1(a, b;x, y), say.

The function a → Ω1(a, b;x, y) is convex on [0,∞). Thus, we have

(4.6)∂

∂aΩ1(a, b;x, y) ≥ ∂

∂aΩ1(a;x, y)

∣∣∣a=0 =

=Γ(x)Γ(x+ b)

Γ(x+ y + b)

[Γ(y)Γ(x)

log2x

x+ y+

Γ(y + b)

Γ(x+ b)log

2y

x+ y

].

Inequalities for the beta function 7

Using∂

∂b

Γ(y + b)

Γ(x+ b)=

Γ(y + b)

Γ(x+ b)[ψ(y + b)− ψ(x+ b)] ≥ 0

and b > 1 givesΓ(x+ y + b)

Γ(x)Γ(x+ b)

∂

∂aΩ1(a, b;x, y)

∣∣∣a=0 ≥

≥ Γ(y)

Γ(x)log

2x

x+ y+

Γ(y + 1)

Γ(x+ 1)log

2y

x+ y=

Γ(y)

Γ(x+ 1)

∫ y

xlog

2t

x+ tdt ≥ 0.

From (4.6) we find that

(4.7) Ω1(a, b;x, y) ≥ Ω1(0, b;x, y) = B(x+ b, y) +B(x, y + b).

We have

(4.8) B(x+ b, y) +B(x, y + b)− 21−bB(x, y) =

=

∫ 1

0tx−1(1− t)y−1[tb + (1− t)b − 21−b] dt ≥ 0,

so that (4.7) and (4.8) yield that (4.5) holds.Now, we prove the right-hand side of (1.5) with Δ(a, b) as defined in

(1.6). Again, we consider two cases.

Case 2.1. a+ b ≥ 1.Then, Δ(a, b) = 1. We have to show that

(4.9) xaB(x+ b, y) + yaB(x, y + b) ≤ (x+ y)aB(x, y).

Case 2.1.1. a ≥ 1.Since the sum on the left-hand side of (4.9) is decreasing with respect

to b, we obtain

xaB(x+ b, y) + yaB(x, y + b) ≤ (xa + ya)B(x, y) ≤ (x+ y)aB(x, y).

Case 2.1.2. a < 1.We have b ≥ 1− a > 0. This leads to the inequality

(4.10) xaB(x+b, y)+yaB(x, y+b) ≤ xaB(x+1−a, y)+yaB(x, y+1−a).We show that

(4.11) xaB(x+ 1− a, y) + yaB(x, y + 1− a) ≤ (x+ y)aB(x, y).

Let

(4.12) Ω2(a;x, y) =( x

x+ y

)aB(x+ 1− a, y).

Then we have

∂2

∂a2Ω2(a;x, y) =

( x

x+ y

)a ∫ 1

0tx−a(1− t)y−1

(log

x

x+ y− log t

)2dt ≥ 0.

8 Horst Alzer

This implies thata → Ω2(a;x, y) + Ω2(a; y, x)

is convex on [0, 1]. Hence it follows that

(4.13) Ω2(a;x, y) + Ω2(a; y, x) ≤≤ max{Ω2(0;x, y) + Ω2(0, y, x); Ω2(1;x, y) + Ω2(1; y, x)} = B(x, y).

From (4.12) and (4.13) we conclude that (4.11) is valid. Combining (4.10)and (4.11) gives (4.9).

Case 2.2. a+ b < 1.Since Δ(a, b) = 21−a−b, the second inequality in (1.5) is equivalent to

the following one:

(4.14)( 2x

x+ y

)aB(x+ b, y) +

( 2y

x+ y

)aB(x, y + b) ≤ 21−bB(x, y).

We denote the expression on the left-hand side of (4.14) by Ω3(a, b;x, y).The function a → Ω3(a, b;x, y) is convex, so that 0 ≤ a < 1− b gives

(4.15) Ω3(a, b;x, y) ≤ max{Ω3(0, b;x, y),Ω3(1− b, b;x, y)}.We have(4.16)

21−bB(x, y)− Ω3(0, b;x, y) =∫ 1

0tx−1(1− t)y−1[21−b − tb − (1− t)b] dt ≥ 0

and21−bB(x, y)− Ω3(1− b, b;x, y) =

= 21−b∫ 1

0tx−1(1− t)y−1

[1−

( x

x+ y

)1−btb −

( y

x+ y

)1−b(1− t)b

]dt.

Letλ = x/(x+ y) ∈ (0, 1).

Since b ∈ (0, 1), we obtain{( x

x+ y

)1−btb +

( y

x+ y

)1−b(1− t)b

}1/b=

={λ(t(x+ y)

x

)b+ (1− λ)

( (1− t)(x+ y)

y

)b}1/b ≤

≤ λt(x+ y)

x+ (1− λ)

(1− t)(x+ y)

y= 1,

see [11, p. 26]. It follows that the term in square brackets is non-negative.Thus we have

(4.17) 21−bB(x, y)− Ω3(1− b, b;x, y) ≥ 0.

Inequalities for the beta function 9

Combining (4.15)–(4.17) leads to

Ω3(a, b;x, y) ≤ 21−bB(x, y).

This implies that (4.14) is valid.It remains to prove that the upper and lower bounds given in (1.6) are

sharp. We define

Ω∗(a, b;x, y) =xaB(x+ b, y) + yaB(x, y + b)

(x+ y)aB(x, y).

Since

limx→0

Γ(2x)

Γ(x)= lim

x→0

Γ(2x+ 1)

2Γ(x+ 1)=

1

2,

we get

(4.18) limx→0

Ω∗(a, b;x, x) = limx→0

[21−a Γ(x+ b)

Γ(2x+ b)

Γ(2x)

Γ(x)

]= 2−a.

From (2.3) we obtain

(4.19) limx→∞ xv−uΓ(x+ u)

Γ(x+ v)= 1.

Applying (4.19) yields(4.20)

limx→∞Ω∗(a, b;x, x) = lim

x→∞

[21−a−b · (2x)b Γ(2x)

Γ(2x+ b)· x−bΓ(x+ b)

Γ(x)

]= 21−a−b.

Furthermore, we have

(4.21) limx→0

Ω∗(a, b;x, y) =

= limx→0

[ xa+1

Γ(x+ 1)· Γ(x+ y)

Γ(y)·B(x+ b, y)

(x+ y)a+( y

x+ y

)a · Γ(y + b)Γ(x+ y)

Γ(x+ y + b)Γ(y)

]= 1.

Let δ∗(a, b) and Δ∗(a, b) be the best possible lower and upper bounds forΩ∗(a, b;x, y). Then we have

(4.22) δ(a, b) ≤ δ∗(a, b) ≤ Ω∗(a, b;x, y) ≤ Δ∗(a, b) ≤ Δ(a, b).

The limit relations (4.18), (4.20), and (4.21) lead to

(4.23) δ∗(a, b) ≤ min{2−a, 1, 21−a−b} = δ(a, b)

and

(4.24) Δ(a, b) = max{2−a, 1, 21−a−b} ≤ Δ∗(a, b).

From (4.22)–(4.24) we conclude that

δ∗(a, b) = δ(a, b) and Δ∗(a, b) = Δ(a, b).

The proof of Theorem 2 is complete.

10 Horst Alzer

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Inequalities for the beta function 11

Neravenstva dl� beta-funkcii

HORST AL�CER

V rabote dokazyva�ts� sledu�wie neravenstva dl� beta-funkcii ��ilera.(i) Pust� α i β de�istvitel�nye qisla. Neravenstva(

yz−x

xz−y zy−x

)α

≤ B(x, x)z−y B(z, z)y−x

B(y, y)z−x≤

(yz−x

xz−y zy−x

)β

vypoln��ts� dl� vseh de�istvitel�nyh x, y, z, udovletvor��wih uslovi�m 0 <

x ≤ y ≤ z, togda i tol�ko togda, kogda α ≤ 1/2 i β ≥ 1.(ii) Pust� a i b de�istvitel�nye i neotricatel�nye qisla. Dl� vseh polo�i-

tel�nyh de�istvitel�nyh i neotricatel�nyh qisel x i y vypolneny neravenstva

δ(a, b) ≤ xaB(x+ b, y) + yaB(x, y + b)

(x+ y)aB(x, y)≤ Δ(a, b)

s toqnymi granicami

δ(a, b) = min{2−a, 21−a−b} i Δ(a, b) = max{1, 21−a−b}.