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Prove that

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

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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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    @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