Proving an asymptotic lower bound for the integral $\int_{0}^{\infty} \exp\left( - \frac{x^2}{2y^{2r}} - \frac{y^2}{2}\right) \frac{dy}{y^s}$

Question

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 $x$)

Answer 1

Let $$ \phi_{r,x}(y)=-\frac{x^2}{2y^{2r}}-\frac{y^2}{2}\tag{1} $$ Taking the first and second derivatives of $\phi_{r,x}(y)$ yields $$ \phi_{r,x}^\prime(y)=r\frac{x^2}{y^{2r+1}}-y\tag{2} $$ and $$ \phi_{r,x}^{\prime\prime}(y)=-(2r+1)r\frac{x^2}{y^{2r+2}}-1\tag{3} $$ Using $(2)$, $\phi_{r,x}(y)$ reaches a maximum at $y_0=(rx^2)^{\frac{1}{2r+2}}$. At that point, $$ \phi_{r,x}(y_0)=-\frac{r+1}{2r}(rx^2)^{\frac{1}{r+1}}\tag{4} $$ Furthermore, $(3)$ gives that $$ \frac12\phi_{r,x}^{\prime\prime}(y_0)=-(r+1)\tag{5} $$ Standard stationary phase methods yield $$ \begin{align} &\int_0^\infty\exp\left(-\frac{x^2}{2y^{2r}}-\frac{y^2}{2}\right)\frac{\mathrm{d}y}{y^s}\\ &\sim\exp\left(\phi_{r,x}(y_0)\right)\int_0^\infty\exp\left(-(r+1)(y-y_0)^2\right)\frac{\mathrm{d}y}{y^s}\\ &\sim y_0^{-s}\exp\left(\phi_{r,x}(y_0)\right)\int_{-\infty}^\infty\exp\left(-(r+1)y^2\right)\mathrm{d}y\\ &=(rx^2)^{\frac{-s}{2r+2}}\exp\left(-\frac{r+1}{2r}(rx^2)^{\frac{1}{r+1}}\right)\sqrt{\frac{\pi}{r+1}}\tag{6} \end{align} $$ Where $f(x)\sim g(x)$ means that $\lim\limits_{x\to\infty}f(x)/g(x)=1$.

Estimate $(5)$ says that the kind of estimate sought above can be achieved only when $s\le r+1$.

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Taking the derivative of $(2)$ yields $$ \phi_{r,x}^{\prime\prime\prime}(y)=(2r+1)(2r+2)r\frac{x^2}{y^{2r+4}}\tag{7} $$ which says that the second derivative of the exponent increases monotonically, whereas the second derivative of the quadratic approximation is constant. Since $\phi_{r,x}$ and its first and second derivatives match the quadratic approximation at $y_0$, we get that for $y\ge y_0$, $$ \phi_{r,x}(y)\ge\phi_{r,x}(y_0)-(r+1)(y-y_0)^2\tag{8} $$ Furthermore, since $(1+t)^{-s}\ge1-st$ for $t\ge0$, we get $$ \begin{align} &\int_0^\infty\exp\left(-\frac{x^2}{2y^{2r}}-\frac{y^2}{2}\right)\frac{\mathrm{d}y}{y^s}\\ &\ge\int_{y_0}^\infty\exp\left(\phi_{r,x}(y_0)-(r+1)(y-y_0)^2\right)y_0^{-s}\left(1-s\frac{y-y_0}{y_0}\right)\mathrm{d}y\\ &=y_0^{-s}\exp(\phi_{r,x}(y_0))\int_0^\infty\exp\left(-(r+1)t^2\right)\left(1-\frac{st}{y_0}\right)\mathrm{d}t\\ &=y_0^{-s}\exp(\phi_{r,x}(y_0))\left(\frac12\sqrt{\frac{\pi}{r+1}}-\frac{s}{2y_0\sqrt{r+1}}\right)\\ &=(rx^2)^{\frac{-s}{2r+2}}\exp\left(-\frac{r+1}{2r}(rx^2)^{\frac{1}{r+1}}\right)\left(\frac12\sqrt{\frac{\pi}{r+1}}-\frac{s}{2y_0\sqrt{r+1}}\right)\tag{9} \end{align} $$ For $x\ge x_0$, we get $(9)$ with $y_0=(rx_0^2)^\frac{1}{2r+2}$. This is the bound required as long as $s\le r+1$.

For example, if we set $\displaystyle x_0=\max\left(\frac{s^{r+1}}{\sqrt{r}},1\right)$, for $x\ge x_0$, we get $$ \begin{align} &\int_0^\infty\exp\left(-\frac{x^2}{2y^{2r}}-\frac{y^2}{2}\right)\frac{\mathrm{d}y}{y^s}\\ &\ge\left(\frac{\sqrt{\pi}-1}{2\sqrt{r+1}}r^{\frac{-s}{2r+2}}\right)x^{\frac{-s}{r+1}}\exp\left(-\frac{r+1}{2r}(rx^2)^{\frac{1}{r+1}}\right)\\ &\ge\left(\frac{\sqrt{\pi}-1}{2\sqrt{r+1}}r^{\frac{-s}{2r+2}}\right)\frac1x\exp\left(-\frac{r+1}{2r}(rx^2)^{\frac{1}{r+1}}\right)\tag{10} \end{align} $$ as long as $s\le r+1$.

Answer 2

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.