# Beta Function as Integral of Power of t over Power of t plus 1

## Theorem

$\ds \map \Beta {x, y} = \int_{\mathop \to 0}^{\mathop \to \infty} \frac {t^{x - 1} } {\paren {1 + t}^{x + y} } \rd t$

where $\Beta$ denotes the Beta function.

## Proof

Consider the substitution $s = \dfrac t {1 + t}$.

We have the following:

$\dfrac 1 {1 + t} = 1 - s$
$t \to 0, s \to 0$
$t \to \infty, s \to 1$
$\rd s = \dfrac 1 {\paren {1 + t}^2} \rd t$

Then:

 $\ds \leadsto \ \$ $\ds \int_{\to 0}^{\to \infty} \frac {t^{x - 1} } {\paren {1 + t}^{x + y} } \rd t$ $=$ $\ds \int_{\to 0}^{\to \infty} \paren {\frac t {1 + t} }^{x - 1} \paren {\frac 1 {1 + t} }^{y - 1} \frac {\rd t} {\paren {1 + t}^2}$ $\ds$ $=$ $\ds \int_{\to 0}^{\to 1} s^{x - 1} \paren {1 - s}^{y - 1} \rd s$ Integration by Substitution $\ds$ $=$ $\ds \map \Beta {x, y}$ Definition of Beta Function

$\blacksquare$