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For $x>0$, how to prove $$\frac{e^{-\frac{x^2}{2}}}{x}-\int_x^\infty e^{-\frac{t^2}{2}}dt\leq \frac{e^{-\frac{x^2}{2}}}{x^3}$$ I did the following let $f(x)=\frac{e^{-\frac{x^2}{2}}}{x^3}-\frac{e^{-\frac{x^2}{2}}}{x}+\int_x^\infty e^{-\frac{t^2}{2}}dt$. Observe that $f(0^+)=\infty$ and $f(\infty)=0$. I wish to show $f\geq 0$. Differentiating $f$ gives $$f'(x)=e^{-\frac{x^2}{2}}\left(2-\frac{3}{x^4}\right)$$ But $f'\geq 0$ for $x>(3/2)^{1/4}$, so $f$ increase up to 0 for large $x$. This means that $f(x)\leq 0$ for some $x$? That's contradictory to what I want to prove. Any one have a better idea?

nerd
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  • See this answer which shows that $$Q(x) > \phi(x) \left (\frac{1}{x} - \frac{1}{x^3}\right ) = \frac{\exp(-x^2/2)}{\sqrt{2\pi}}\left (\frac{1}{x} - \frac{1}{x^3}\right )~~ \text{for}~~ x > 0.$$ where $Q(x)$, the complementary standard normal distribution function, is a multiple of the integral whose value you want to bound. – Dilip Sarwate Sep 29 '15 at 02:49

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Start with $$ \int_{x}^{\infty} e^{-t^2/2} \, dt, $$ multiply the integrand by $1=t/t$, and integrate by parts: $$ \int_{x}^{\infty} te^{-t^2/2} \frac{1}{t} \, dt = \left[ -\frac{e^{-t^2/2}}{t} \right]_x^{\infty} - \int_x^{\infty} \frac{e^{-t^2/2}}{t^2} \, dt = \frac{e^{-x^2/2}}{x} - \int_x^{\infty} \frac{e^{-t^2/2}}{t^2} \, dt. $$ Rearranging, we have $$ \frac{e^{-x^2/2}}{x}-\int_{x}^{\infty} e^{-t^2/2} \, dt = \int_x^{\infty} \frac{e^{-t^2/2}}{t^2} \, dt $$ Now do the same trick again: $$ \int_x^{\infty} te^{-t^2/2}\frac{1}{t^3} \, dt = \left[ -\frac{e^{-t^2/2}}{t^3} \right]_x^{\infty} - 3\int_x^{\infty} \frac{e^{-t^2/2}}{t^4} \, dt = \frac{e^{-x^2/2}}{x^3} - 3\int_x^{\infty} \frac{e^{-t^2/2}}{t^4} \, dt $$ Hence $$ \frac{e^{-x^2/2}}{x}-\int_{x}^{\infty} e^{-t^2/2} \, dt = \int_x^{\infty} \frac{e^{-t^2/2}}{t^2} \, dt = \frac{e^{-x^2/2}}{x^3} - 3\int_x^{\infty} \frac{e^{-t^2/2}}{t^4} \, dt < \frac{e^{-x^2/2}}{x^3}, $$ since the integral $$ - 3\int_x^{\infty} \frac{e^{-t^2/2}}{t^4} \, dt $$ must be negative.

Chappers
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