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 | |||||||||||
\(\ds \) | \(=\) | \(\ds -\intlimits {\frac {z^x} x} 1 0\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\frac 0 x + \frac 1 x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 x\) |
$\blacksquare$
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: Exercise $40 \ \text{(a)}$