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

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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

$\blacksquare$


Sources

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