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.

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 $

$=$

$\ds \int_{\to 0}^{\to 1} s^{x - 1} \paren {1 - s}^{y - 1} \rd s$

$\ds $

$=$

$\ds \map \Beta {x, y}$

Definition of Beta Function

$\blacksquare$