Beta Function of Real Number with 1

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Theorem

Let $\Beta \left({x, y}\right)$ denote the Beta function.

Then:

$\Beta \left({x, 1}\right) = \Beta \left({1, x}\right) = \dfrac 1 x$


Proof

By definition:

\(\ds \Beta \left({x, 1}\right)\) \(=\) \(\ds \int_0^1 t^{x - 1} \left({1 - t}\right)^{1 - 1} \rd t\) Definition of Beta Function
\(\ds \) \(=\) \(\ds \int_0^1 t^{x - 1} \rd t\)
\(\ds \) \(=\) \(\ds \int_0^1 t^{x - 1} \rd t\)
\(\ds \) \(=\) \(\ds \left[{\frac {t^x} x}\right]_0^1\)
\(\ds \) \(=\) \(\ds \frac 1 x - \frac 0 x\)
\(\ds \) \(=\) \(\ds \frac 1 x\)

$\Box$


\(\ds \Beta \left({1, x}\right)\) \(=\) \(\ds \int_0^1 t^{1 - 1} \left({1 - t}\right)^{x - 1} \rd t\) Definition of Beta Function
\(\ds \) \(=\) \(\ds \int_0^1 \left({1 - t}\right)^{x - 1} \rd t\)
\(\ds \) \(=\) \(\ds \int_1^0 -\left({z}\right)^{x - 1} \rd z\) Integration by Substitution
\(\ds \) \(=\) \(\ds -\left[{\frac {z^x} x}\right]_1^0\)
\(\ds \) \(=\) \(\ds -\frac 0 x + \frac 1 x\)
\(\ds \) \(=\) \(\ds \frac 1 x\)

$\blacksquare$


Sources

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