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.

Table of Contents

## 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 $\int_0^{\pi/2}$ sin^{2m-1}θ cos^{2n-1}θ 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-1}θ cos^{2n-1}θ dθ we obtain that

B(1/2, 1/2) = 2 $\int_0^{\frac{\pi}{2}}$ sin^{2×1/2-1}θ cos^{2×1/2-1}θ dθ⇒ B(1/2, 1/2) = 2 $\int_0^{\frac{\pi}{2}}$ dθ = 2 [θ] _{0}^{π/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 $\int_0^{\pi/2}$ sin^{6}θ dθ |

**Answer:**

$\int_0^{\pi/2}$ sin^{6}θ 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,

$\int_0^{\pi/2}$ sin^{6}θ 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

$\int_0^{\pi/2}$ sin^{6}θ dθ = $\dfrac{1}{2} \dfrac{\frac{15 \sqrt{\pi}}{8} \times \sqrt{\pi}}{3!}$ = $\dfrac{5 \pi}{32}$.

**Remark:**

- In a similar way, one can show that $\int_0^{\pi/2}$ cos
^{6}θ dθ = $\dfrac{5 \pi}{32}$. - Moreover, $\int_0^{\pi/2}$ sin
^{n}θ dθ = $\int_0^{\pi/2}$ cos^{n}θ dθ

This article is written by Dr. T. Mandal, Ph.D in Mathematics. On Mathstoon.com you will find Maths from very basic level to advanced level. Thanks for visiting.