Let $ 0 < r < 1$, fix $x > 1$ and consider the integral
$ I_{r}(x) = \int_{1}^{\infty} \exp\left( - \frac{x^2}{2y^{2r}} - \frac{y^2}{2}\right) \frac{dy}{y^r}.$
In the investigation of iterating two Brownian motions the integral above arises in some form or another in the paper titled "Iterated Random Walk" L. Turban 2004 Europhys. Lett. 65 627. or a more recent paper "Fractional diffusion equations and processes with randomly varying time" Enzo Orsingher, Luisa Beghin http://arxiv.org/abs/1102.4729.
The result we have been trying to prove below can be thought of as an analysis of localizing around the maximum of the function $f(y) = \frac{x^2}{2y^{2r}} + \frac{y^2}{2}$ in the exponential of the integrand and then applying appropriate estimates along with integration by parts. I have always thought of this as a localization similar to what is used in Laplace's method but my advisor has explained to me that from a probabilistic perspective we are just applying a standard idea from large deviations and with that in mind much of our analysis has assumed that we are taking $x \gg 1$ much larger than one.
Question 1: Suppose that $p(x),q(x)>0$ are polynomials, $k(r)$ a constant depending on $r$ and that $c(r) = \frac{(r^{-\frac{r}{r+1}} +r^\frac{1}{r+1}) }{2}$ then is it true that $I_r (x) \leq k(r) \frac{p(x)}{q(x)} \exp(-c(r) x^{\frac{2}{1+r}}) $
We have a proof of this result under more strict hypothesis that is rather boring (in the sense it only uses calculus) and involves essentially integration by parts applied to the integral rewritten as I_r(x) = \int_{1}^{\infty} \frac{y^{-r}}{f'(y)} f'(y) \exp(-f(y)) dy. The draw back of our proof is it is not uniform in the sense that it only holds for all $x > x_0$ for some $x_0(r) \gg 1$ some constant much greater than $1$. We would like to relax this assumption and try to find a proof that hold for any $x>1$.