I was trying to find the asymptotic expansion for $$\int_0^1 \sqrt{t(1-t)}(t+a)^{-x} \; \mathrm{d}t,$$ for $a>0$ as $x \rightarrow \infty$. I have already tried re-writing $$(t+a)^{-x}=\exp(- x\log(t+a)).$$ However, by using Laplace's method for asymptotic expansions I get that $f(t) = \sqrt{t(1-t)}$ vanishes everywhere.
Any ideas are appreciated/welcome. Thank you.
The minimum of the phase function occurs at the endpoint $t=0$. We have $$ \log (t + a) = \log a + \frac{t}{a} + \mathcal{O}(t^2 ) $$ and $$ \sqrt {t(1 - t)} = \sqrt t - \frac{1}{2}t^{3/2} + \mathcal{O}(t^{5/2} ) $$ as $t\to 0^+$. Thus, by Laplace's method, the leading order asymptotics is $$ \mathrm{e}^{ - x\log a} \Gamma\! \left( {\frac{3}{2}} \right)\frac{{a^{3/2} }}{{x^{3/2} }} = \frac{{\sqrt \pi }}{2}\frac{{a^{3/2 - x} }}{{x^{3/2} }}. $$
There ia an antiderivative which involves the Appell hypergeometric function of two variables.
Using the bounds, this leads to $$\int_0^1 \sqrt{t(1-t)}(t+a)^{-x} \,dt=\frac{\pi}{8} \, (a+1)^{-x} \,\, _2F_1\left(\frac{3}{2},x;3;\frac{1}{a+1}\right)$$ where appears the gaussian hypergeometric function. This is the exact solution.
I have been unable to find the asymptotics of the gaussian hypergeometric function when its second argument tends to infinity.
Since, in their answers, @gary and @Vajra proposed simpler approaches, I give up.
