Limiting value of Beta function

Question

I would like to know the limiting value of the Beta function: $$ B(a,b) = \int_0^1 x^{a -1}(1-x)^{b-1}\,d x~ \tag 1. $$

For instance, what does (1) reduce to as

• $ a \to \infty$

• $ b\to \infty$

• $ a\to \infty, b\to \infty$?

Knowing the above would allow me to ( I hope) among others that

$$ \lim_{b\to\infty} \frac{\left(\frac{a}{b}\right)^{a} e^{wa}\left( 1 + \frac{a}{b} e^w \right)^{-(a+b)}}{B(a , b)} = \frac{a^a \exp(aw-ae^w)}{\Gamma(a)} \,,\,a,b \gt 0~,$$ where $$ \Gamma(a) = \int_0^\infty x^{a-1} e^{-x}~ dx~. $$

Answer 1

First note that when you write $a \to \infty$ what you are actually asking for is asymptotic behavior, i.e. $a >> b$ where $a$ is variable and $b$ is fixed

With that in mind, first note that $$B(a,b) = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$ We can thus get the cases $b \to \infty$ and $a \to \infty$ by using Stirling's Approximation. Since this is symmetric, we only really need to know one case. When you work it out, you should get $B(a,b) \stackrel{a \gg b}{\sim} \Gamma(b)a^{-b}$

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When both go to infinity we have to start worrying about size. Are $a$ and $b$ increasing at the same rate? Is one going to infinity at twice the rate of the other? There is no definitive answer unless you put some conditions on growth rate. Nevertheless, using Stirling again we find that $$B(a,b) \stackrel{a,b\gg 0}{\sim} \sqrt{2 \pi}\frac{a^{a-1/2}b^{b-1/2}}{(a+b)^{a+b-1/2}}$$ These formulas can be found here, on the Wikipedia page for the Beta Function

Answer 2

For $a$ or $b$ going to infinity, take the limit inside the integral and see that the integrand vanishes, so the integral term goes to zero. Same with when both go to infinity.

In more details:

Consider $$ B(a,b) = \int_0^1 x^{a -1}(1-x)^{b-1}\,d x~ \tag 1. $$

What happens when $a$ goes to infinity? This is equivalent to asking what happens in $\lim_{n\to\infty} \int_0^1 f_n(x)d x $ where, $f_n(x)=x^{a_n -1}(1-x)^{b-1}\,$ and $a_n\to\infty$.

Now, it would be really convenient if we could take the limit inside the integral, and it so happens that we can totally do that. We need to use a version of what is called the Monotone Convergence Theorem.

Once we take the limit inside, for any $x\in (0,1)$, for any $b$, $f_n(x)$ decreases monotonically to zero as $n\to\infty$. So, basically $f_n$ converges pointwise to the zero function. Now integration of the zero function of the interval is definitely zero.

Check this out if you want to know the technical details of the theorem we used to interchange limit and integration.

Monotone Convergence Theorem for non-negative decreasing sequence of measurable functions

Same happens for the other limits you want to know.