Beta Function of Real Number with 1

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Theorem

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

Then:

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


Proof

By definition:

\(\ds \map \Beta {x, 1}\) \(=\) \(\ds \int_0^1 t^{x - 1} \paren {1 - t}^{1 - 1} \rd t\) Definition of Beta Function
\(\ds \) \(=\) \(\ds \int_0^1 t^{x - 1} \rd t\)
\(\ds \) \(=\) \(\ds \int_0^1 t^{x - 1} \rd t\)
\(\ds \) \(=\) \(\ds \intlimits {\frac {t^x} x} 0 1\)
\(\ds \) \(=\) \(\ds \frac 1 x - \frac 0 x\)
\(\ds \) \(=\) \(\ds \frac 1 x\)

$\Box$


\(\ds \map \Beta {1, x}\) \(=\) \(\ds \int_0^1 t^{1 - 1} \paren {1 - t}^{x - 1} \rd t\) Definition of Beta Function
\(\ds \) \(=\) \(\ds \int_0^1 \paren {1 - t}^{x - 1} \rd t\)
\(\ds \) \(=\) \(\ds \int_1^0 -\paren z^{x - 1} \rd z\) Integration by Substitution
\(\ds \) \(=\) \(\ds -\intlimits {\frac {z^x} x} 1 0\)
\(\ds \) \(=\) \(\ds -\frac 0 x + \frac 1 x\)
\(\ds \) \(=\) \(\ds \frac 1 x\)

$\blacksquare$


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