Question

Prove equation (6.5).

Short answer

It is proved that $B\left(p,q\right)=\underset{0}{\overset{\infty }{\int }}\frac{y^{p-1}}{{(1+y)}^{p+q}}dy$.

Step-by-step solution

Given information.

The given equation (6.5), is $B\left(p,q\right)=\underset{0}{\overset{\infty }{\int }}\frac{y^{p-1}}{{(1+y)}^{p+q}}dy$.

Definition of a Gamma function.

Beta function is defined as $\mathit{B}\mathbf{(}\mathit{p}\mathbf{,}\mathit{q}\mathbf{)}\mathbf{=}\underset{\mathbf{0}}{\overset{\mathbf{1}}{\mathbf{\int }}}{\mathit{x}}^{\mathbf{p}\mathbf{-}\mathbf{1}}{\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{x}\mathbf{)}}^{\mathbf{q}\mathbf{-}\mathbf{1}}\mathit{d}\mathit{x}$.

Perform necessary substitutions.

Make the following substitution and differentiate the terms.

$\begin{array}{c}y=\frac{x}{1-x}\\ dx=\frac{dy}{{(1+y)}^2}\end{array}$

Input these expressions in the integral.

Substitute these values in the integral.

$\begin{array}{c}B\left(p,q\right)=\underset{0}{\overset{\infty }{\int }}{\left(\frac{y}{1+y}\right)}^{p-1}{\left(1-\frac{y}{1+y}\right)}^{q-1}\frac{dy}{{(1+y)}^2}\\ =\underset{0}{\overset{\infty }{\int }}\frac{y^{p-1}}{{(1+y)}^{p-1}{(1+y)}^{q-1}{(1+y)}^2}dy\\ =\underset{0}{\overset{\infty }{\int }}\frac{y^{p-1}}{{(1+y)}^{p+q}}dy\end{array}$

Hence proved.