Question

Prove that $\mathit{B}\mathbf{(}\mathit{p}\mathbf{,}\mathit{q}\mathbf{)}\mathbf{=}\mathit{B}\mathbf{(}\mathit{q}\mathbf{,}\mathit{p}\mathbf{)}$. Hint:Put$\mathit{x}\mathbf{=}\mathbf{1}\mathbf{-}\mathit{y}$in Equation (6.1).

Short answer

The statement$B(p,q)=B(q,p)$ is proved.

Step-by-step solution

Given information

Beta function is given. A hint to substitute$x=1-y$ is given.

Definition of a Beta function

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

Begin with substitutions

Make the following substitution and differentiate the equation.

$\begin{array}{c}x=1-y\\ dy=-dx\end{array}$

Evaluate the resulting integral.

Simplify the resulting integral. Invert the boundary limits of the integral by negating the expression.

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

Since this results in the same integral as in the beginning, the statement$B(p,q)=B(q,p)$ is proved.