I am currently dealing with a problem of bounding the expression $$ \frac{B(0.5,\beta)\max_{0\leq x<1} h(\alpha x)(1-x^2)^\beta }{\int_{-1}^{1}h(\alpha t)(1-t^2)^{\beta-1} dt} $$ for $\alpha,\beta >0$ and increasing $h()\geq 0$. I know that for some nice $h()$ this is bounded but not in general (e.g. indicator function can break it as $\beta \rightarrow \infty$). I tried to work out the asymptotic using Laplace's method:
For big $\beta$ (if I am not mistaken) this is trivial because $(1-x^2)$ has maximum at 0 and so $$ \int_{-1}^{1}h(\alpha t)(1-t^2)^{\beta-1} dt = \int_{-1}^{1}h(\alpha t)\exp((\beta-1)\log(1-t^2)) dt\approx h(0)\sqrt{\frac{\pi}{\beta-1}} $$ from which the above can be bounded easily by $h(\alpha x_0)/h(0)$ for $x_0$ being argmax of the numerator.
I would also like to investigate the case where $\alpha$ is big, under some form of $h()$ (e.g. exponential form $h(\alpha x) = c^{\alpha x}$), but since $h$ is increasing the maximum is not attained in the interior and the second function is 0 at $x=1$, is there any way to investigate this case under some restrictions of $h$? Or some other way how to bound the above expression uniformly under some form restrictions of $h()$ like positivity?