# Beta Function of x with y+1 by x+y over y

## Theorem

Let $\map \Beta {x, y}$ denote the Beta function.

Then:

$\map \Beta {x, y} = \dfrac {x + y} y \map \Beta {x, y + 1}$

## Proof

By definition of Beta function:

$\ds \map \Beta {x + 1, y} = \int_0^1 t^x \paren {1 - t}^{y - 1} \rd t$

With a view to expressing the primitive in the form:

$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

let:

 $\ds u$ $=$ $\ds t x$ $\ds \leadsto \ \$ $\ds \frac {\d u} {\d x}$ $=$ $\ds x t^{x - 1}$ Power Rule for Derivatives

and let:

 $\ds \frac {\d v} {\d x}$ $=$ $\ds \paren {1 - t}^{y - 1}$ $\ds \leadsto \ \$ $\ds v$ $=$ $\ds -\frac {\paren {1 - t}^y} y$ Primitive of Power

Then:

 $\ds$  $\ds \int_0^1 t^{x - 1} \paren {1 - t}^{y - 1} \rd t$ $\ds$ $=$ $\ds \sqbrk {t^x \frac {-\paren {1 - t}^y} y}_0^1 - \int_0^1 x t^{x - 1} \frac {-\paren {1 - t}^y} y \rd t$ Integration by Parts $\ds$ $=$ $\ds \frac x y \int_0^1 t^{x - 1} \paren {1 - t}^y \rd t$ as the left hand term vanishes $\ds$ $=$ $\ds \frac x y \map \Beta {x, y + 1}$ Definition of Beta Function

Hence:

 $\ds \map \Beta {x, y}$ $=$ $\ds \map \Beta {x + 1, y} + \map \Beta {x, y + 1}$ Beta Function of x+1 with y plus Beta Function of x with y+1 $\ds$ $=$ $\ds \frac x y \map \Beta {x, y + 1} + \map \Beta {x, y + 1}$ from above $\ds$ $=$ $\ds \frac {x + y} y \map \Beta {x, y + 1}$ simplifying

$\blacksquare$