- The Beta function is defined for \(\mathrm { R e } ( \alpha ) > 0\) and \(\operatorname { R e } ( \beta ) > 0\) by
\[ B (\alpha , \beta) = \int_ {0} ^ {1} (1 - t) ^ {\alpha - 1} t ^ {\beta - 1} d t. \]
Prove that \(B ( \alpha , \beta ) = \frac { \Gamma ( \alpha ) \Gamma ( \beta ) } { \Gamma ( \alpha + \beta ) }\)
Show that \(B ( \alpha , \beta ) = \int _ { 0 } ^ { \infty } { \frac { u ^ { \alpha - 1 } } { ( 1 + u ) ^ { \alpha + \beta } } } d u .\)
[Hint: For part (a), note that
\[ \Gamma (\alpha) \Gamma (\beta) = \int_ {0} ^ {\infty} \int_ {0} ^ {\infty} t ^ {\alpha - 1} s ^ {\beta - 1} e ^ {- t - s} d t d s, \]
and make the change of variables \(s = u r , t = u ( 1 - r ) . ]\)
Solution
For \(\Re\alpha,\Re\beta>0\), \[ \Gamma(\alpha)\Gamma(\beta) =\int_0^\infty\int_0^\infty t^{\alpha-1}s^{\beta-1}e^{-t-s}\,dt\,ds. \] Set \[ t=u(1-r),\qquad s=ur, \] with \(u>0\) and \(0<r<1\). The Jacobian has absolute value \(u\), so \[ \Gamma(\alpha)\Gamma(\beta) =\int_0^\infty e^{-u}u^{\alpha+\beta-1}\,du \int_0^1(1-r)^{\alpha-1}r^{\beta-1}\,dr. \] Thus \[ \Gamma(\alpha)\Gamma(\beta) =\Gamma(\alpha+\beta)B(\alpha,\beta), \] and therefore \[ \boxed{B(\alpha,\beta)= \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}}. \]
For part (b), put \[ u=\frac{1-t}{t},\qquad t=\frac1{1+u},\qquad dt=-\frac{du}{(1+u)^2}. \] As \(t\) runs from \(0\) to \(1\), \(u\) runs from \(\infty\) to \(0\). Therefore \[ B(\alpha,\beta) =\int_0^\infty \frac{u^{\alpha-1}}{(1+u)^{\alpha+\beta}}\,du. \]