Beta Function - Relationship Between Gamma Function and Beta Function | Relationship Gamma Function Beta Function

Relationship Between Gamma Function and Beta Function

To derive the integral representation of the beta function, write the product of two factorials as

$\Gamma(x)\Gamma(y) = \int_0^\infty\ e^{-u} u^{x-1}\,du \int_0^\infty\ e^{-v} v^{y-1}\,dv =\int_0^\infty\int_0^\infty\ e^{-u-v} u^{x-1}v^{y-1}\,du \,dv. \!$

Changing variables by putting u=zt, v=z(1-t) shows that this is

$\int_{z=0}^\infty\int_{t=0}^1\ e^{-z} (zt)^{x-1}(z(1-t))^{y-1}z\,dt \,dz =\int_{z=0}^\infty \ e^{-z}z^{x+y-1} \,dz\int_{t=0}^1t^{x-1}(1-t)^{y-1}\,dt. \!$

Hence

$\Gamma(x)\,\Gamma(y)=\Gamma(x+y)\Beta(x,y) .$

The stated identity may be seen as a particular case of the identity for the integral of a convolution. Taking

and, one has:

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