How can we prove the following asymptotic lower bound for the regularized Beta function when $n\rightarrow\infty$?
$$\int_0^{1} I_{2 t - t^2}\left(\frac{n - 1}{2}, \frac{1}{2}\right) dt=\Omega\left(\frac{1}{\sqrt{n}}\right)$$
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- $\begingroup$ The left side is not a function of $n$ once you've taken the limit. $\endgroup$Stopple– Stopple2020-12-21 21:25:03 +00:00Commented Dec 21, 2020 at 21:25
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1 Answer
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This integral can actually be evaluated in closed form, from which the large-$n$ asymptotics follows readily: $$\int_0^{1} dt\, I_{2 t - t^2}(a,b)= \frac{1}{B(a,b)}\int_0^1 dt\,\int_0^{2t-t^2} ds\,s^{a-1}(1-s)^{b-1}$$ $$=\frac{1}{B(a,b)}\int_0^1 ds\,\frac{s^{a-1} (1-s)^{b+\frac{1}{2}}}{1-s}=\frac{\Gamma (a) \Gamma \left(b+\frac{1}{2}\right)}{\Gamma \left(a+b+\frac{1}{2}\right) B(a,b)}$$ (for the first equality I substituted the integral expression for the Beta function and for the second equality I changed the order of integration).
So the desired integral is $$\int_0^{1} I_{2 t - t^2}\left(\tfrac{n - 1}{2}, \tfrac{1}{2}\right) dt = \frac{2}{(n-1) B\left(\frac{n-1}{2},\frac{1}{2}\right)}\rightarrow\sqrt{\frac{2}{\pi n}} +\frac{1}{(2n)^{3/2}\sqrt{\pi}}+\cdots$$
