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Prove that

$$\frac{1}{x+\sqrt{x^2+2}}

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    Just an idea... by making the change of variables $t^2=s^2+x^2$ we get $$e^{x^2}\int_x^{\infty}e^{-t^2} \ dt = \int_0^\infty \frac{s e^{-s^2}}{\sqrt{s^2+x^2}}\,ds,$$ which at least gets rid of the exponential outside and gives us a square root to work with.2012-12-15
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    As an addendum to my previous comment, it might be possible to get the lower bound by using the fact that the new integrand is unimodal and has a maximum when $$s^2 = \frac{x}{x+\sqrt{x^2+2}}.$$2012-12-15
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    @AntonioVargas: thanks for your idea.2012-12-16

2 Answers 2

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(The following argument is adapted from Dümbgen, ''Bounding Standard Gaussian Tail Probabilities.'')


Approximating $\displaystyle \int_x^{\infty} e^{-t^2} \, dt$

Suppose we want to approximate $\displaystyle \int_x^{\infty} e^{-t^2} \, dt$ with a function of the form $\dfrac{e^{-x^2}}{h(x)}.$ Let $$\Delta(x) = \frac{e^{-x^2}}{h(x)} - \int_x^{\infty} e^{-t^2} \, dt.$$

Then, if $h(x) \to \infty$ as $x \to \infty$, then $\Delta(x) \to 0$ as $x \to \infty$. Because of this, we have the following.

  • If $\Delta'(x) > 0$ for all $x \geq 0$ then $\Delta(x)$ increases to $0$. Therefore, $\dfrac{e^{-x^2}}{h(x)}$ is a lower bound on $\displaystyle \int_x^{\infty} e^{-t^2} \, dt$ for $x \geq 0$.
  • Similarly, if $\Delta'(x) < 0$ for all $x \geq 0$ then $\Delta(x)$ decreases to $0$. Therefore, $\dfrac{e^{-x^2}}{h(x)}$ is an upper bound on $\displaystyle \int_x^{\infty} e^{-t^2} \, dt$ for $x \geq 0$.

We have $$\Delta'(x) = \frac{e^{-x^2}}{h(x)^2} \left(h(x)^2 - 2xh(x) - h'(x) \right).$$ Thus the sign of $\Delta'(x)$ is determined by the sign of $f(x) = h(x)^2 - 2xh(x) - h'(x)$.

Given the bounds we're trying to show, let's consider functions of the form $h(x) = x + \sqrt{x^2 + c}$. Then $$f(x) = c - 1 - \frac{x}{\sqrt{x^2+c}}.$$ Thus $f(x)$ is decreasing on $[0, \infty)$.


The lower bound

To have $f(x) > 0$ for all $x \geq 0$, we need $$c > 1 + \frac{x}{\sqrt{x^2+c}}, \:\:\:\: x \geq 0.$$ The smallest value of $c$ for which this holds is $c = 2$. Therefore, $$\frac{1}{x + \sqrt{x^2+2}} < e^{x^2} \int_x^{\infty} e^{-t^2} \, dt, \:\:\:\: x \geq 0,$$ and $2$ is the smallest value of $c$ for which this bound holds for functions of the form $h(x) = x + \sqrt{x^2 + c}$.


The upper bound

To have $f(x) < 0$ for all $x \geq 0$ we can take $c = 1$. However, we can do better this because $f(x)$ is decreasing. If we find a larger value of $c$ such that $\Delta(0)= 0$, then we will have $f(x) > 0$ on $[0, x_0)$ for some $x_0$ and then $f(x) < 0$ on $(x_0, \infty)$. Thus $\Delta(x)$ will initially increase from $0$ and then decrease back to $0$, giving us a tighter upper bound. Since $$\Delta(0) = 0 \Longleftrightarrow \frac{1}{\sqrt{c}} = \int_0^{\infty} e^{-t^2}\, dt = \frac{\sqrt{\pi}}{2},$$ we have $c = \dfrac{4}{\pi}$ yielding a tighter upper bound than $c = 1$. Therefore, $$e^{x^2} \int_x^{\infty} e^{-t^2} \, dt \leq \frac{1}{x + \sqrt{x^2+\frac{\pi}{4}}}, \:\:\:\: x \geq 0,$$ and $\dfrac{\pi}{4}$ is the smallest value of $c$ for which this bound holds for functions of the form $h(x) = x + \sqrt{x^2 + c}$.

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    a simple and beautiful solution. Thanks!2012-12-16
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    @Chris'ssister: You're quite welcome. It was a fun problem to think about, and I learned some interesting things myself in the process.2012-12-16
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    wonderful answer!2012-12-16
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    (+1) Thank you very much for posting a link to this tech. report! I find it very interesting in that, several years ago (i.e., years before the publication date of that report), I wrote up a surprisingly similar report, taking a similar approach, finding similar bounds and producing similar plots. I distributed it to only a small handful of people and never attempted to publish it since I feared it would be of very limited interest to others and it was, by no means, a particularly "serious" work. I am pleased something so similar has been done and put out there by Dümbgen.2012-12-16
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    Incidentally, the *best* bounds of the form $$\frac{\alpha}{\gamma x + \sqrt{x^2 + \beta}}$$ (that are tight at both zero and infinity) were given in A. V. Boyd. Inequalities for Mills’ ratio. *Rep. Stat. Appl. Res., JUSE*, 6(2):1–3, 1959. They have an equally elegant and simple proof.2012-12-16
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    [Here](http://arxiv.org/abs/1012.2063) is a version of the Dümbgen technical report on the arXiv, just as an independent link that may be more stable.2012-12-16
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    @cardinal: Thanks for the comments and the Boyd reference! By the way, is the report that you wrote several years ago what you were referring to in your comment on [my blog](http://mikespivey.wordpress.com/2011/10/21/normaltails/) last year?2012-12-17
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    Yes, indeed. I had planned to post a short adaptation of a piece of that report as an answer to an old question of John D. Cook's. But, I fear that question is quite old (2009) and resurrecting it may unsettle some on MO; such "applied" questions are not so well received these days, it seems.2012-12-17
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    @cardinal: Is it [this one](http://mathoverflow.net/questions/1283/erfc-lower-bound)? John at least might appreciate your answer.2012-12-17
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    That's the one! I'll think about putting a little something together and posting it there. Thanks for the encouragement. It may just be the little push I need to get it done. :-) By the way, it's nice to have you back around and participating more regularly again.2012-12-17
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    @cardinal: Thanks. I'm enjoying being back. I was pretty busy this past year, and I only really started coming back to math.SE once the semester began winding down. We'll see how long my burst of activity on math.SE lasts into the spring term. :)2012-12-17
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The integral is equal to: $$\frac{\sqrt{\pi}}{2}\text{erfc}(x)$$ If you calculate the series expansion of the above function around infinity, you get: $$e^{-x^2}\left(\frac{1}{2x}-\frac{1}{4x^3}+\frac{3}{8x^5}+...\right)$$ I believe looking at the resultant fractional expression and the series expansion of the radicals should give you a clue, at least for $x\gg1$.

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    thanks for your suggestion. (+1)2012-12-16