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-

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


  • \(\beta (m,n)\) 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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