Limit with regularised incomplete Beta function

Question

Let $\mathcal{I}_x(a,b)$ denote the regularised incomplete Beta function: $$ \mathcal{I}_x(a,b)=\frac{\Gamma(a+b)}{ \Gamma(a)\Gamma(b) }\int_0^x t^{a-1}(1-t)^{b-1}dt, $$ where $x \in (0,1)$ and $a,b>0$. Let $q>0$ and, for any $y>0$, denote $\log^qy=(\log y)^q$. What is the asymptotic behaviour of $$ \mathcal{I}_{\frac{\log^qn}{\sqrt{n}}}(a/\log n, b/ \log n) $$ as $n\to \infty$? I could not find any useful reference. I know that, as $n \to \infty$, $$ \frac{\Gamma((a+b)/\log n)}{ \Gamma(a/\log n)\Gamma(b /\log n) }\sim \frac{ab}{a+b}\frac{1}{\log n}; $$ as for the remaining factor, i.e. $$ \int_0^{\frac{\log^qn}{\sqrt{n}}} t^{a/\log n-1}(1-t)^{b/\log n-1}dt, $$ I've only managed to obtain the not so sharp lower bound $\left(\frac{\log^qn}{\sqrt{n}}\right)^{a/\log n} \to e^{-a/2}$ and upper bound $$ \left(1-\frac{\log^qn}{\sqrt{n}}\right)^{b/\log n-1} \frac{\log n}{a} \left( \frac{\log^qn}{\sqrt{n}} \right)^{a/\log n}\sim \frac{\log n}{a e^{a/2}}. $$ Thus I can conclude the limit of $\mathcal{I}_{\frac{\log^qn}{\sqrt{n}}}(a/\log n, b/ \log n)$ is finite: is it zero?

Answer 1

For $x\to 0+$, we can expand the factor $(1-t)^{\beta-1}$ about $t=0$ to find $$ I_x (\alpha ,\beta ) = \frac{{\Gamma (\alpha + \beta )}}{{\Gamma (\alpha )\Gamma (\beta )}}\frac{{x^\alpha }}{\alpha }\left( {1 + \mathcal{O}(x)} \right). $$ With your specific parameters, this gives $$ \frac{b}{{a + b}}\left( {\frac{{\log ^q n}}{{\sqrt n }}} \right)^{a/\log n} \left( {1 + \mathcal{O}\!\left( {\frac{{\log ^q n}}{{\sqrt n }}} \right)} \right) = \frac{{b\, \mathrm{e}^{ - a/2} }}{{a + b}}\left( {1 + \mathcal{O}\!\left( {\frac{{\log \log n}}{{\log n}}} \right)} \right), $$ as $n\to +\infty$.