 # Relation Between Beta And Gamma Function

Beta and gamma are the two most popular functions in mathematics. Gamma is a single variable function, whereas Beta is a two-variable function. The relation between beta and gamma function will help to solve many problems in physics and mathematics.

## Beta And Gamma Function

The relationship between beta and gamma function can be mathematically expressed as-

 $\beta (m,n)=\frac{\Gamma m\Gamma n}{\Gamma \left ( m+n \right )}$

Where,

• $\beta (m,n)$ is the beta function with two variables m and n.
• [latex\Gamma m\) is the gamma function with variable m.
• $\Gamma n$ is the gamma function with variable n.

### Relation between beta and gamma function derivation

Consider the general form of Gamma function is given by-

$\Gamma n=\int_{0}^{\infty }e^{-zx}x^{n-1}z^{n}dx$

Multiplying both the sides by $e^{-z}z^{m-1}$ and integrating with respect to z from 0 to 8 we get-

$\Rightarrow \Gamma n\int_{0}^{\infty }e^{-z}z^{m-1}dz=\int_{0}^{\infty }\int_{0}^{\infty }e^{-zx}x^{n-1}z^{n}e^{-z}z^{m-1}dzdx$ $=\int_{0}^{\infty }\int_{0}^{\infty }e^{-z\left ( x+1 \right )}x^{n-1}z^{\left ( m+n-1 \right )}dzdx$

Put z(x+1)=y

$\Rightarrow z=\frac{y}{x+1}\Rightarrow dz=\frac{dy}{x+1}$ $\Rightarrow \Gamma n\int_{0}^{\infty }e^{-z}z^{m-1}dz=\int_{0}^{\infty }x^{n-1}\int_{0}^{\infty }\left [ e^{-z\left ( x+1 \right )}z^{\left ( m+n-1 \right )}dz \right ]dx$ $=\int_{0}^{\infty }x^{n-1}\int_{0}^{\infty }\left [ e^{-y}\frac{y^{m+n-1}}{(1+x)^{m+n-1}}\left ( \frac{dy}{1+x} \right ) \right ]dx$ $=\int_{0}^{\infty }\frac{x^{n-1}}{(1+x)^{m+n}}\left [\int_{0}^{\infty } e^{-y}y^{m+n}dy \right ]dx$ $=\int_{0}^{\infty }\frac{x^{n-1}}{(1+x)^{m+n}}\Gamma \left ( m+n \right )dx$

Because $\Gamma \left ( m+n \right )=\int_{0}^{\infty } e^{-y}y^{m+n}dy$ $\Rightarrow \Gamma n\Gamma m=\int_{0}^{\infty }\frac{x^{n-1}}{(1+x)^{m+n}}\Gamma \left ( m+n \right )dx$

Because $\Gamma \left ( m \right )=\int_{0}^{\infty }e^{-z}z^{m-1}dz$ $\Rightarrow \frac{\Gamma n\Gamma m}{\Gamma \left ( m+n \right )}=\int_{0}^{\infty }\frac{x^{n-1}}{(1+x)^{m+n}}dx$

Thus, we arrive at-

$\Rightarrow \frac{\Gamma n\Gamma m}{\Gamma \left ( m+n \right )}=\beta (m,n)$

Because $\beta (m,n)=\int_{0}^{\infty }\frac{x^{n-1}}{(1+x)^{m+n}}dx$

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