# 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 http://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 $x$)

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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 $s\le r+1$.

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 $s\le r+1$.

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Wow thank you for the nice answer. I know the first part of the asymptotic analysis is using the Taylor series expansion around the critical point but I am not quite sure how you go from the line above line(5) to line (5). Is there a reference you recommend? Finally why is it clear that the condition $s \leq r+1$ should hold? – user7980 Apr 7 '12 at 0:58
@user7980: as $x\to\infty$, the width of the range of $y$ for which the exponential is significant is fixed since the exponent is $-(r+1)(y-\alpha)^2$.Thus, the $y^{-s}$ is essentially $\alpha^{-s}$ and the integral of the gaussian is $\sqrt{\frac{\pi}{r+1}}$. So the factor in front of the exponential is $\alpha^{-s}$. In the factor out front, there is $x^{-\frac{s}{r+1}}$ and since you want something at least $x^{-1}$, we need $s\le r+1$. Does that make sense? – robjohn Apr 7 '12 at 1:15
Thanks for all your input - I think your answer is more or less correct but because I have posted a bounty I want a little bit more precise of a statment. In particular I was hoping to avoid the use of asympotic notation and prove something like $$I_1 (x) > \frac{c_1(r,s)}{x} \exp( - f(c^* x^*))$$ holds for all $x >1$ not just when $x$ tends to infinity – user7980 Apr 7 '12 at 8:57
I think the inequality $(9)$ is what you wanted. – robjohn Apr 8 '12 at 0:38
No, it is not always positive. You need to choose $x_0$ so that $s/y_0<\sqrt{\pi}$. For example, if we set $\displaystyle x_0=\frac{s^{r+1}}{\sqrt{r}}$, we get $s/y_0=1$. Then $\frac12\sqrt{\frac{\pi}{r+1}}-\frac{s}{2y_0\sqrt{r+1}}=\frac{\sqrt{\pi}-1}{2\sq‌​rt{r+1}}$ which is always positive. – robjohn Apr 8 '12 at 2:48

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.

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Thanks for the help. Could you elaborate a little on how to apply the Laplace expansion to get the desired bounds or list a reference that has a similar example worked? I am not so sure how to deal with the behavior of the singularity at the origin – user7980 Apr 2 '12 at 20:03