Beta and Gamma Functions: Definition, Properties, Solved Problems

The beta and gamma functions are one of the important improper integrals. There integrals converge for certain values. In this article, we will learn about beta and gamma functions with their definition of convergence, properties and some solved problems.

Beta Function

For integers m and n, let us consider the improper integral

$\int_0^1$ x^m-1 (1-x)^n-1.

This integral converges when m>0 and n>0.

Definition of Beta Function

The integral $\int_0^1$ x^m-1 (1-x)^n-1, m>0, n>0 is called the beta function, and it is denoted by the symbol B(m, n). So

B(m, n) = $\int_0^1$ x^m-1 (1-x)^n-1.

For example, the integral $\int_0^1$ x dx = $\int_0^1$ x^2-1 (1-x)^1-1 = B(2, 1)is a beta function.

Gamma Function

For an integer n, we consider the improper integral

$\int_0^\infty$ e^-x x^n-1.

This integral converges when n>0.

Definition of Gamma Function

The integral $\int_0^\infty$ e^-x x^n-1, n>0 is called the gamma function, and it is denoted by the symbol Γ(n). So

Γ(n) = $\int_0^\infty$ e^-x x^n-1.

For example, the integral $\int_0^1$ xe^-x dx = $\int_0^\infty$ e^-x x^2-1 = Γ(2) is a gamma function.

Properties of Beta and Gamma Functions

For m>0 and n>0, the properties of beta and gamma functions are listed below:

No.	Property
P1	B(m,n) = B(n,m)
P2	B(m,n) = 2 $\int_0^{\pi/2}$ sin2m-1θ cos2n-1θ dθ
P3	B(m, n) = $\int_0^{\infty} \dfrac{x^{m-1}}{(1+x)^{m+n}}dx$ (or) B(m, n) = $\int_0^{\infty} \dfrac{x^{n-1}}{(1+x)^{m+n}}dx$
P4	B(m, n) = $\int_0^1 \dfrac{x^{m-1}+x^{n-1}}{(1+x)^{m+n}}dx$
P5	Γ(n+1) = nΓ(n) Γ(n+1) = n!
P6	B(m, n) = $\dfrac{\Gamma(m) \Gamma(n)}{\Gamma(m+n)}$
P7	Γ(m)Γ(1-m) = $\dfrac{\pi}{\sin m\pi}$, 0<m<1.

Remarks: From the above table, we observe the following.

• We can interchange the roles of m and n in B(m, n).
• Beta and gamma functions are related by the formula: B(m, n) = {Γ(m) Γ(n)}/Γ(m+n).
• The formula Γ(n+1) = n! is valid for n=1, 2, 3, 4, ….

Solved Problems

Q1: Find the value of Γ(1/2).

Answer:

By the above property P2: B(m,n) = 2 $\int_0^{\pi/2}$ sin^2m-1θ cos^2n-1θ dθ we obtain that

B(1/2, 1/2) = 2 $\int_0^{\frac{\pi}{2}}$ sin2×1/2-1θ cos2×1/2-1θ dθ ⇒ B(1/2, 1/2) = 2 $\int_0^{\frac{\pi}{2}}$ dθ = 2 [θ]0π/2 = 2(π/2 – 0) = π.

Therefore, the value of B(1/2, 1/2) is equal to π.

Q2: Find the value of Γ(1/2).

Answer:

Putting m = $\frac{1}{2}$ and n = $\frac{1}{2}$ in the above property P6 (relation between beta and gamma functions), we have that

B(1/2, 1/2) = $\dfrac{\Gamma(\frac{1}{2}) \Gamma(\frac{1}{2})}{\Gamma(\frac{1}{2}+\frac{1}{2})}$

As Γ(1) =1, we deduce from above that

$\Gamma(\frac{1}{2}) = \sqrt{B(\frac{1}{2}, \frac{1}{2})}$

⇒ Γ(1/2) = √π, follows from Q1.

Q3: Find the integral $\int_0^{\pi/2}$ sin6θ dθ

Answer:

$\int_0^{\pi/2}$ sin⁶θ dθ

= $\dfrac{1}{2} \cdot 2$ $\int_0^{\pi/2}$ sin^{2⋅7/2 -1}θ cos^{2⋅1/2 -1}θ dθ

= $\dfrac{1}{2}$ B(7/2, 1/2), using the above property (P2).

= $\dfrac{1}{2} \dfrac{\Gamma(\frac{7}{2}) \Gamma(\frac{1}{2})}{\Gamma(\frac{7}{2}+\frac{1}{2})}$. This is obtained by using the relation between beta and gamma functions. Therefore,

$\int_0^{\pi/2}$ sin6θ dθ = $\dfrac{1}{2} \dfrac{\Gamma(\frac{7}{2}) \Gamma(\frac{1}{2})}{\Gamma(\frac{7}{2}+\frac{1}{2})}$ …(∗)

Now, using Γ(n+1) = nΓ(n) and the fact Γ(1/2) = √π, we deduce that

Γ(7/2) = Γ(5/2 +1) = 5/2 Γ(5/2) ⇒ Γ(7/2) = 5/2 ⋅ 3/2 Γ(3/2) ⇒ Γ(7/2) = 5/2 ⋅ 3/2 ⋅ 1/2 Γ(1/2) ⇒ Γ(7/2) = $\dfrac{15 \sqrt{\pi}}{8}$.

Noting Γ(4) = 3!, it follows from (∗) that

$\int_0^{\pi/2}$ sin⁶θ dθ = $\dfrac{1}{2} \dfrac{\frac{15 \sqrt{\pi}}{8} \times \sqrt{\pi}}{3!}$ = $\dfrac{5 \pi}{32}$.

Remark:

1. In a similar way, one can show that $\int_0^{\pi/2}$ cos⁶θ dθ = $\dfrac{5 \pi}{32}$.
2. Moreover, $\int_0^{\pi/2}$ sinⁿθ dθ = $\int_0^{\pi/2}$ cosⁿθ dθ