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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 $$ \Gamma(a) = \int_0^\infty x^{a-1} e^{-x}~ dx~. $$

L.bronze
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  • Stirling's Approximation is your friend here. Use it. – Brevan Ellefsen May 19 '17 at 23:49
  • For a or b going to infinity, take the limit inside and see that the integrand vanishes, so the term goes to zero. Same with when both go to infinity. – Juanito May 19 '17 at 23:51
  • @Juanito I would presume the OP is looking for asymptotic behavior, and not the limit itself – Brevan Ellefsen May 19 '17 at 23:53
  • As in closed form approximations of how fast things vanish, or something else? I would like to understand, why looking at the limit would not give that. (academic curiosity) – Juanito May 19 '17 at 23:55
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    @Juanito are you familiar with Stirling's approximation? Using it, we can get a simpler function that has roughly the same rate of decay as we go to infinity. In essence, you are right - we have an approximation to how fast it vanishes. However, we have far more information than that, because we can also get close approximations for any large values $\neq \infty$ – Brevan Ellefsen May 20 '17 at 00:24
  • Got it! Knew about Stirling's appr. – Juanito May 20 '17 at 01:40

2 Answers2

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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 $B(a,b) \stackrel{a \gg b}{\sim} \Gamma(b)a^{-b}$

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

Brevan Ellefsen
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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.

Juanito
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  • Please can expand on your answer a little bit? – L.bronze May 20 '17 at 00:34
  • I put in some of the details, let me know if you want any more. – Juanito May 20 '17 at 01:29
  • Full disclosure: MCT is a much stronger result, as it is defined for more general integrations than Riemann. And, I find it to be a very useful tool to think about a lot of problems. – Juanito May 20 '17 at 01:33
  • Also, "measurable function" is a generalization of continuous functions in this case, in case, you find that term ambiguous. The result holds for measurable functions, so also holds for continuous functions etc... – Juanito May 20 '17 at 01:39
  • Thanks for expanding on your answer. So basically, $B(a,b)\to 0$ as $b\to \infty$? – L.bronze May 20 '17 at 02:00
  • Yes, you can do similar steps. When both a and b go to infinity similarly. Do accept and upvote the answer if you found it useful. – Juanito May 20 '17 at 02:04