Proving an asymptotic lower bound for the integral $\int_{0}^{\infty} \exp\left( - \frac{x^2}{2y^{2r}} - \frac{y^2}{2}\right) \frac{dy}{y^s}$ This is a follow up to the great answer posted to https://math.stackexchange.com/a/125991/7980
Let $ 0 < r < \infty, 0 < s < \infty$ , fix $x > 1$ and consider the integral 
$$ I_{1}(x) = \int_{0}^{\infty} \exp\left( - \frac{x^2}{2y^{2r}} - \frac{y^2}{2}\right) \frac{dy}{y^s}$$ 
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 $[0, \infty)$.
We are trying to determine if there exist upper and lower bounds of the same order for large x.  The coefficients in our bounds can be composed of rational functions in x or even more complicated as long as they do not have exponential growth. The Laplace expansion presented in the answer to the question cited above gives upper bounds.
In particular can we prove a specific lower bound:

Does there exist a positive constant $c_1(r,s)$ and such that for x>1 we have
  $$I_1 (x) > \frac{c_1(r,s)}{x} \exp( - f(c^* x^*))$$ (it is ok in the answer if the function $\frac{1}{x}$ in the upper bound is replaced by any rational function or power of $x$)


I think that if you make the change of variables $y = \lambda z$ with $\lambda = x^{\frac 2 {r+1} }\;i.e.  \frac {x^2} {\lambda^{2r}} = \lambda^2$ you convert it into $\lambda ^{s-1} \int e^{-\lambda^2 \frac 12(z^{-2r} + z^2)} \frac {dz}{z^s}$ which looks like a fairly normal laplace type expansion.
