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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)?

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up vote 1 down vote accepted

Here are some nonasymptotic upper bounds, based on a quantitative Laplace expansion around the minimum.

Let $u=x^{1/(1+r)}$, $v=\frac{1+r}{1-r}$ and $w=r^{1/(2v)}$. The change of variables $y=uz^{1/(1-r)}$ yields $$ I_1(x)=\frac{u^{1-r}}{1-r}\int_0^{+\infty}\mathrm e^{-\frac12u^2g(z)}\mathrm dz, \quad\text{where}\quad g(z)=z^{1+v}+z^{1-v}. $$ One can check that $g'''(z)\leqslant0$ on $z\leqslant1$, $g'''(z)\geqslant0$ on $z\geqslant1$ and $g''(1)=2v^2$, hence $g''(z)\geqslant2v^2$ for every $z\gt0$. Furthermore, $g'(w)=0$ hence $g(z)\geqslant g(w)+v^2(z-w)^2$ for every $z\gt0$. This yields the upper bound $$ I_1(x)\leqslant\frac{u^{1-r}}{1-r}\mathrm e^{-\frac12u^2g(w)}\int_0^{+\infty}\mathrm e^{-\frac12u^2v^2(z-w)^2}\mathrm dz. $$ The last integral is a part of the integral of a non normalized gaussian density centered at $w$ and with variance the inverse of $u^2v^2$, hence $$ I_1(x)\leqslant\frac{u^{1-r}}{1-r}\mathrm e^{-\frac12u^2g(w)}\frac{\sqrt{2\pi}}{uv}, $$ that is, $$ I_1(x)\leqslant\frac{\sqrt{2\pi}}{1+r}x^{-r/(1+r)}\mathrm e^{-J(r)x^{2/(1+r)}},\qquad J(r)=\frac{1+r}{2r^{r/(1+r)}}. $$ One can replace $x^{-r/(1+r)}$ by a constant depending on $r$ only (but note that $I_1(x)\to+\infty$ when $x\to0$ and $r\to1$ hence this constant must be unbounded when $r\to1$).

To do so, use the fact that $I_1(x)\leqslant I_1(0)$ and consider separately the cases $x\leqslant1$ and $x\geqslant1$. This yields $$ I_1(x)\leqslant c_1(r)\mathrm e^{-J(r)x^{2/(1+r)}}, $$ for every $x\geqslant0$, as soon as $$ c_1(r)\geqslant\frac{\sqrt{2\pi}}{1+r}\quad\text{and}\quad c_1(r)\geqslant I_1(0)\mathrm e^{J(r)}=2^{-(1+r)/2}\Gamma\left(\frac{1-r}2\right)\mathrm e^{J(r)}. $$ Numerically, one can check that these conditions are fulfilled for $$ c_1(r)=2\Gamma\left(\frac{1-r}2\right). $$ Edit: No lower bound $I_1(x)\geqslant c_0(r)\mathrm e^{-J(r)x^{2/(1+r)}}$ can hold, with $c_0(r)$ independent on $x$, because of the negative power of $x$ in the upper bound. However, refining the estimates in the proof above, one can show that $$ \lim\limits_{x\to+\infty}x^{-2/(1+r)}\log I_1(x)=-J(r). $$ The same technique also provides upper bounds, at least for every $0\lt s\lt1$, of the integrals $$ I_{1}(x) = \int_{0}^{\infty} \exp\left( - \frac{x^2}{2y^{2r}} - \frac{y^2}{2}\right) \frac{\mathrm dy}{y^s}. $$ Following the very same steps, one gets $$ I_1(x)\leqslant\frac{\sqrt{2\pi}}{B(r,s)}x^{-s/(1+r)}\mathrm e^{-J(r)x^{2/(1+r)}}, $$ where $B(r,s)$ solves the equation $$ (1+r-s)B^2=(1+s)(1+r)\cdot C^{s}, $$ and $C=C(r,s)$ solves the equation $$ s(1+s)\cdot C^{1+r}=r(1+r-s)(1+2r-s). $$ One recovers $C(r,r)=1$ and $B(r,r)=1+r$, as in the upper bound above. This quantitative upper bound of $I_1(x)$ could even hold as long as the identities defining $B(r,s)$ involve only positive factors, that is, for every $0\leqslant s\lt1+r$, and one expects that the factor $1/y^s$ is irrelevant to the logarithmic asymptotics, that is, that, for every real $s$, $$ \lim\limits_{x\to+\infty}x^{-2/(1+r)}\log I_1(x)=-J(r). $$

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Thank you so much. I have posted multiple related questions with bounties in the past and have had a hard time getting any sort of feedback. I was about to post a bounty on this question as well and feel it is prudent to give you the award I just have to wait 23 more hours before I can award you the bounty. – user7980 Apr 1 '12 at 18:48
Is it possible to get a similar lower bound using this method? i.e something like $I_1(x)\geq c_0(r)\mathrm e^{-\frac12 A(r)x^{2/(1+r)}}$ – user7980 Apr 1 '12 at 18:49
And finally is it clear that this laplacle expansion can generalize to the case when the power in the denominator of integrand is different, that is for $s>0$ can apply the same proof technique to $\int_{0}^{\infty} \exp\left( - \frac{x^2}{2y^{2r}} - \frac{y^2}{2}\right) \frac{dy}{y^s}$? – user7980 Apr 1 '12 at 18:52
See Edit for some answers to both your comments. – Did Apr 1 '12 at 19:31
Thanks for your help. Our original guess was that something like the following estimate would hold asymptotically that is for $x>>1$ we where trying to prove something like $ \frac{c_0(r)}{x} \mathrm e^{-\frac12 A(r)x^{2/(1+r)}} \leq I_1(x)\leqslant \frac{c_1(r)}{x} \mathrm e^{-\frac12 \ A(r)x^{2/(1+r)}}$ and so thanks again for commenting on the lower bound – user7980 Apr 1 '12 at 22:49

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