# Derivation of Beta Function Expressions

In this post, I will derive some basic expressions of the beta function. I will follow the equation numbering of Higher Transcendental Functions (Bateman Manuscript), Volume 1, page 9 (print), page 35 (pdf).

We begin with the basic integral definition of the beta function

\mathrm{B}(x,y) = \int\limits^{1}_{0} t^{x-1} (1-t)^{y-1} \mathrm{d} t
\quad \mathrm{for} \,\, \mathrm{Re} \, x > 0 \,\, \mathrm{and} \,\, \mathrm{Re} \, y > 0
\label{eq:bf1}
\tag{1}

\mathrm{B}(x,y) = \int\limits_{0}^{\infty} \frac{v^{x-1}}{(1+v)^{x+y}} \mathrm{d} v
\quad \mathrm{for} \,\, \mathrm{Re} \, x > 0 \,\, \mathrm{and} \,\, \mathrm{Re} \, y > 0
\label{eq:bf2}
\tag{2}

To derive this equation, we begin with \eqref{eq:bf1} and make the substitution

t = \frac{v}{1+v}

to obtain

\mathrm{B}(x,y) = \int\limits_{0}^{\infty} \frac{v^{x-1}}{(1+v)^{x-1}} \Big(1 – \frac{v}{1+v}\Big)^{y-1} \frac{1}{(1+v)^{2}} \mathrm{d} v

simplification yields \eqref{eq:bf2}.

\mathrm{B}(x,y) = \int\limits_{0}^{1} \frac{(v^{x-1}+v^{y-1})}{(1+v)^{x+y}} \mathrm{d} v
\quad \mathrm{for} \,\, \mathrm{Re} \, x > 0 \,\, \mathrm{and} \,\, \mathrm{Re} \, y > 0
\label{eq:bf3}
\tag{3}

To obtain \eqref{eq:bf3} we begin with \eqref{eq:bf2} and break up the integral

\mathrm{B}(x,y) = \int\limits_{0}^{1} \frac{v^{x-1}}{(1+v)^{x+y}} \mathrm{d} v + \int\limits_{1}^{\infty} \frac{v^{x-1}}{(1+v)^{x+y}} \mathrm{d} v

and designate the last integral as I.

For I, we make the substitution $$w = v^{-1}$$

\mathrm{I} = \int\limits_{0}^{1} \frac{1}{w^{x-1}} \frac{w^{x+y}}{(1+w)^{x+y}} \frac{1}{w^{2}} \mathrm{d} w

Simplifying and then making the substitution for I yields \eqref{eq:bf3}.

\mathrm{B}(x,y) = \mathrm{B}(y,x)
\label{eq:bf4}
\tag{4}

Equation \eqref{eq:bf4} follows directly from equation \eqref{eq:bf3}.

\mathrm{B}(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x + y)}
\label{eq:bf5}
\tag{5}

To derive equation \eqref{eq:bf5}, let us start with the basic integral definition of the gamma function and then make the substitution $$x = at$$

\begin{align}
\Gamma(z) & = \int\limits_{0}^{\infty} x^{z-1} \mathrm{e}^{-x} \mathrm{d} x \\
& = a^{z} \int\limits_{0}^{\infty} t^{z-1} \mathrm{e}^{-at} \mathrm{d} t \\
\end{align}

Rearranging terms yields

\frac{1}{a^{z}} = \frac{1}{\Gamma(z)} \int\limits_{0}^{\infty} t^{z-1} \mathrm{e}^{-at} \mathrm{d} t

and making the substitutions $$z = \alpha + \beta$$ and $$a = 1 + v$$, we obtain

\frac{1}{(1+v)^{\alpha + \beta}} = \frac{1}{\Gamma(\alpha + \beta)} \int\limits_{0}^{\infty} t^{\alpha + \beta – 1} \mathrm{e}^{-(1+v)t} \mathrm{d} t

Combining this with equation \eqref{eq:bf2}, we have

\begin{align}
\mathrm{B}(\alpha,\beta) & = \frac{1}{\Gamma(\alpha + \beta)} \int\limits_{0}^{\infty} \int\limits_{0}^{\infty} v^{\beta – 1} t^{\alpha + \beta – 1}
\mathrm{e}^{-t} \mathrm{e}^{-vt} \mathrm{d} t \mathrm{d} v \\
& = \frac{1}{\Gamma(\alpha + \beta)} \int\limits_{0}^{\infty} \Big[\int\limits_{0}^{\infty} t^{\alpha – 1} \mathrm{e}^{-t} \mathrm{d} t \Big] t^{\beta} v^{\beta – 1} \mathrm{e}^{-vt} \mathrm{d} v \\
& = \frac{\Gamma(\alpha)}{\Gamma(\alpha + \beta)} \int\limits_{0}^{\infty} t^{\beta} v^{\beta – 1} \mathrm{e}^{-vt} \mathrm{d} v \\
& = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha + \beta)} \\
\end{align}

\mathrm{B}(x,y+1) = \frac{y}{x}\mathrm{B}(x+1,y) = \frac{y}{x+y}\mathrm{B}(x,y)
\label{eq:bf6}
\tag{6}

To derive equation \eqref{eq:bf6}, use \eqref{eq:bf5} to convert the beta functions into gamma functions and use

\Gamma(z+1) = z\Gamma(z)

\begin{align}
\mathrm{B}(x,y+1) & = \frac{\Gamma(x)\Gamma(y+1)}{\Gamma(x+y+1)} \\
& = \frac{y\Gamma(x)\Gamma(y)}{(x+y)\Gamma(x+y)} \\
& = \frac{y\mathrm{B}(x,y)}{x+y} \\
\end{align}

\begin{align}
\mathrm{B}(x+1,y) & = \frac{\Gamma(x+1)\Gamma(y)}{\Gamma(x+y+1)} \\
& = \frac{x\Gamma(x)\Gamma(y)}{(x+y)\Gamma(x+y)} \\
& = \frac{y\mathrm{B}(x,y)}{x+y} \\
\end{align}

\mathrm{B}(x,y)\mathrm{B}(x+y,z) = \mathrm{B}(y,z)\mathrm{B}(y+z,x) = \mathrm{B}(z,x)\mathrm{B}(x+z,y)
\label{eq:bf7}
\tag{7}

For equation \eqref{eq:bf7} we convert the beta functions to gamma functions.
\begin{align}
\mathrm{B}(x,y)\mathrm{B}(x+y,z) & = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x + y)} \times \frac{\Gamma(x + y)\Gamma(z)}{\Gamma(x + y + z)} \times
\frac{\Gamma(y + z)}{\Gamma(y + z)} \\
& = \frac{\Gamma(y)\Gamma(z)}{\Gamma(y + z)} \times \frac{\Gamma(y + z)\Gamma(x)}{\Gamma(x + y + z)} \\
& = \mathrm{B}(y,z)\mathrm{B}(y+z,x) \\
\end{align}

\begin{align}
\mathrm{B}(x,y)\mathrm{B}(x+y,z) & = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x + y)} \times \frac{\Gamma(x + y)\Gamma(z)}{\Gamma(x + y + z)} \times
\frac{\Gamma(x + z)}{\Gamma(x + z)} \\
& = \frac{\Gamma(z)\Gamma(x)}{\Gamma(z + x)} \times \frac{\Gamma(x + z)\Gamma(y)}{\Gamma(x + y + z)} \\
& = \mathrm{B}(z,x)\mathrm{B}(x+z,y) \\
\end{align}

\mathrm{B}(x,y)\mathrm{B}(x+y,z)\mathrm{B}(x+y+z,u) = \frac{\Gamma(x)\Gamma(y)\Gamma(z)\Gamma(u)}{\Gamma(x + y + z + u)}
\label{eq:bf8}
\tag{8}

Here, we use equation \eqref{eq:bf5}

\mathrm{B}(x,y)\mathrm{B}(x+y,z)\mathrm{B}(x+y+z,u) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x + y} \times \frac{\Gamma(x+y)\Gamma(z)}{\Gamma(x + y + z} \times \frac{\Gamma(x+y+z)\Gamma(u)}{\Gamma(x + y + z + u}

The generalization of equation \eqref{eq:bf8} is evident.

\frac{1}{\mathrm{B}(n,m)} = m \binom{n+m-1}{n-1} = n \binom{n+m-1}{m-1} \,\, \mathrm{for} \,\, n,m \in \mathbb{Z}^{+}
\label{eq:bf9}
\tag{9}

We use the following relationships and then invoke equation \eqref{eq:bf5}

n! = \Gamma(n+1), \quad n! = n(n-1)!, \quad \binom{n}{k} = \frac{n!}{k!(n-k)!}

\frac{1}{\mathrm{B}(n,m)} = \frac{\Gamma(n + m)}{\Gamma(n)\Gamma(m)} = \frac{(n + m – 1)!}{(n – 1)!(m – 1)!} =
m\frac{(n + m – 1)!}{m!(n – 1)!} = m \binom{n+m-1}{n-1}

We can do the same for the last part of equation \eqref{eq:bf9}.

Additional information about the beta function can be found at the following online references