A couple of results involving the beta function

In this note, I want to quickly record a couple of interesting results involving the beta function. First, consider the reciprocal beta function

$f(n) = \frac{\Gamma(\gamma)}{B(n, \gamma)}$

where

$B(n, \gamma) = \int_0^1 dx \ x^{n-1} (1 - x)^{\gamma - 1} = \frac{\Gamma(n) \Gamma(\gamma)}{\Gamma(n + \gamma)}$

It is easy to see by using Stirling’s formula

$\Gamma(p + 1) \sim p^p e^{-p} \sqrt{2 \pi p}$

in the numerator and denominator of $f(n)$ that

$f(n) = \frac{\Gamma(\gamma)}{B(n, \gamma)} = \frac{\Gamma(n + \gamma)}{\Gamma(n)} \sim n^{\gamma}$

for $n \gg \gamma$ with $\gamma$ fixed. Therefore, the reciprocal beta function $f(n)$ converges to a simple power law.

Second, we can easily show that the beta function converges when summed over $n$ by exchanging the summation and integration operations. We obtain the infinite sum of the beta function as

$\sum_{n=1}^{\infty} B(n, \gamma) = \sum_{n=1}^{\infty} \int_0^1 dx \ x^{n-1} (1 - x)^{\gamma - 1}$

$= \int_0^1 dx \ (1 - x)^{\gamma - 1} \bigg(\sum_{n=1}^{\infty} x^{n-1}\bigg) = \int_0^1 dx \ (1 - x)^{\gamma - 1} \cdot \frac{1}{(1 - x)} = \frac{1}{\gamma -1}$

A couple of results involving the beta function

Published by Dr Christian P. H. Salas

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