# 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

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

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

### Definition of Beta Function

The integral $\int_0^1$ xm-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) = $\int_0^1$ xm-1 (1-x)n-1.

For example, the integral $\int_0^1$ x dx = $\int_0^1$ x2-1 (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 xn-1.

This integral converges when n>0.

### Definition of Gamma Function

The integral $\int_0^\infty$ e-x xn-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 xn-1.

For example, the integral $\int_0^1$ xe-x dx = $\int_0^\infty$ e-x x2-1 = 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:

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

By the above property P2: B(m,n) = 2 $\int_0^{\pi/2}$ sin2m-1θ cos2n-1θ dθ we obtain that

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

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.

$\int_0^{\pi/2}$ sin6θ dθ

= $\dfrac{1}{2} \cdot 2$ $\int_0^{\pi/2}$ sin2⋅7/2 -1θ cos2⋅1/2 -1θ 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,

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

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
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