An inequality from the handbook of mathematical functions (by Abramowitz and Stegun) Prove that
$$\frac{1}{x+\sqrt{x^2+2}}<e^{x^2}\int\limits_x^{\infty}e^{-t^2} \, \text dt \le\frac{1}{x+\sqrt{x^2+\displaystyle\tfrac{4}{\pi}}}, \space (x\ge 0)$$
 A: 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 $x\gg1$.
A: (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 $h(x) = x + \sqrt{x^2 + c}$. 
