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Prove that

$$\frac{1}{x+\sqrt{x^2+2}}<e^{x^2}\int\limits_x^{\infty}e^{-t^2} \, \text dt \le\frac{1}{x+\sqrt{x^2+\displaystyle\tfrac{4}{\pi}}}, \space (x\ge 0)$$

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    $\begingroup$ 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. $\endgroup$ Commented Dec 15, 2012 at 21:21
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    $\begingroup$ 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 $$s^2 = \frac{x}{x+\sqrt{x^2+2}}.$$ $\endgroup$ Commented Dec 15, 2012 at 21:39
  • $\begingroup$ @AntonioVargas: thanks for your idea. $\endgroup$ Commented Dec 16, 2012 at 22:12

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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  • $\begingroup$ a simple and beautiful solution. Thanks! $\endgroup$ Commented Dec 16, 2012 at 21:31
  • $\begingroup$ @Chris'ssister: You're quite welcome. It was a fun problem to think about, and I learned some interesting things myself in the process. $\endgroup$ Commented Dec 16, 2012 at 21:35
  • $\begingroup$ wonderful answer! $\endgroup$ Commented Dec 16, 2012 at 21:43
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    $\begingroup$ 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. $\endgroup$
    – cardinal
    Commented Dec 16, 2012 at 23:10
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    $\begingroup$ Here is a version of the Dümbgen technical report on the arXiv, just as an independent link that may be more stable. $\endgroup$
    – cardinal
    Commented Dec 16, 2012 at 23:26
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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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  • $\begingroup$ thanks for your suggestion. (+1) $\endgroup$ Commented Dec 16, 2012 at 22:13

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