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-

(eta (m,n)=frac{Gamma mGamma n}{Gamma left ( m+n ight )})

Where,

  • (eta (m,n)) is the beta function with two variables m and n.
  • [latexGamma 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 nint_{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^{-zleft ( x+1 ight )}x^{n-1}z^{left ( m+n-1 ight )}dzdx)

Put z(x+1)=y

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

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

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

Thus, we arrive at-

(Rightarrow frac{Gamma nGamma m}{Gamma left ( m+n ight )}=eta (m,n))

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

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