A Derivation of the Beta Function Representation \(\mathrm{B}(x,y) = 2^{1-x-y} \int_{0}^{1} (1+t)^{x-1}(1-t)^{y-1} + (1+t)^{y-1}(1-t)^{x-1} \mathrm{d} t\)

I used the beta function representation

\begin{equation}
\mathrm{B}(x,y) = 2^{1-x-y} \int\limits_{0}^{1} (1+t)^{x-1}(1-t)^{y-1} + (1+t)^{y-1}(1-t)^{x-1} \mathrm{d} t
\label{eq:1608051}
\tag{1}
\end{equation}

here to evaluate an integral. Now I will derive this result.

We begin with the basic integral definition of the beta function

\begin{equation}
\mathrm{B}(x,y) = \int\limits^{1}_{0} t^{x-1} (1-t)^{y-1} \mathrm{d} t
\label{eq:1608052}
\tag{2}
\end{equation}

then use the substitution \(t = \frac{z-a}{b-a}\), this yields

\begin{align}
\mathrm{B}(x,y) & = \frac{1}{b-a} \int\limits^{b}_{a} \left(\frac{z-a}{b-a}\right)^{x-1} \left(\frac{b-z}{b-a}\right)^{y-1} \mathrm{d} z \\
& = (b-a)^{-x-y+1} \int\limits^{b}_{a} (z-a)^{x-1} (b-z)^{y-1} \mathrm{d} z \\
& = 2^{-x-y+1} \int\limits_{-1}^{1} (z+1)^{x-1} (1-z)^{y-1} \mathrm{d} z \\
& = 2^{-x-y+1} \left[ \int\limits_{-1}^{0} (z+1)^{x-1} (1-z)^{y-1} \mathrm{d} z + \int\limits_{0}^{1} (z+1)^{x-1} (1-z)^{y-1} \mathrm{d} z \right]
\label{eq:1608053}
\tag{3}
\end{align}

Note that we let \(a=-1\) and \(b=1\).

For the rightmost integral, we let \(z=-w\) so that

\begin{equation}
\int\limits_{0}^{1} (z+1)^{x-1} (1-z)^{y-1} \mathrm{d} z = \int\limits_{0}^{1} (1-w)^{x-1} (1+w)^{y-1} \mathrm{d} w
\label{eq:1608054}
\tag{4}
\end{equation}

Substituting equation \eqref{eq:1608054} into equation \eqref{eq:1608053} yields equation \eqref{eq:1608051}.

The original substitution, \(t = \frac{z-a}{b-a}\) is a useful method of transforming the limits of integration from 0 to 1 into a to b.

Leave a Reply

Your email address will not be published. Required fields are marked *