This is a follow up to Help with removing singularities involving $ \int_{1}^{\infty} \exp\left( - \frac{x^2}{2y^{2r}} - \frac{y^2}{2}\right) \frac{dy}{y^r}$
I am getting a little confused at the argument that was suggested to me which was based on replacing the limits in an integral using the fact that the singularity at the origin of our integrand in not a problem (the proof of removing the singularity is essentially contained in this post Why is $ \frac{\exp{\left( -\frac{x^2}{4y^{2r}} \right)}}{y^r} $ bounded on $[0,1]$?)
Let $ 0 < r < 1$, fix $x > 1$ and consider the integral
$ I_{1}(x) = \int_{0}^{\infty} \exp\left( - \frac{x^2}{2y^{2r}} - \frac{y^2}{2}\right) \frac{dy}{y^r}.$
Fix a constant $c^* = r^{\frac{1}{2r+2}} $ and let $x^* = x^{\frac{1}{1+r}}$.
Write $f(y) = \frac{x^2}{2y^{2r}} + \frac{y^2}{2}$ and note $c^* x^*$ is a local minimum of $f(y)$ so that it is a global max for $-f(y)$ on $[0, \infty)$.
(*) I am interested in bounds of the form $I_1 (x) \leq c_1(r) \exp( - f(c^* x^*))$ ($c_1(r)$ is constant depending only on $r$)
The argument presented to me said that "Without Loss of Generality" we can replace $\int_{0}^{\infty} \exp\left( - \frac{x^2}{2y^{2r}} - \frac{y^2}{2}\right) \frac{dy}{y^r}$ by $ \int_{\epsilon}^{\infty}\exp\left( - \frac{x^2}{2y^{2r}} - \frac{y^2}{2}\right) \frac{dy}{y^r} $ for some $\epsilon >0$ (the proof is based on the second post I cited) and then we just get the bound in (*) by observing that $I_1(x) \leq \exp( - f(c^* x^*)) \int_{\epsilon}^{\infty} y^{-r} dr$... This argument feels flawed to me is there any hopes of salvaging it?
Bonus Question: is there a way to get the upper bound in (*) than the one outlined in 2)?