Prove $\int_t^{\infty} e^{-x^2/2}\,dx > e^{-t^2/2}\left(\frac{1}{t} - \frac{1}{t^3}\right)$

Question

How to formally prove the following inequality -

$$\int_t^{\infty} e^{-x^2/2}\,dx > e^{-t^2/2}\left(\frac{1}{t} - \frac{1}{t^3}\right)$$

Answer 1

For a better lower-bound you may use the following proof by @robjohn:

$$x\int_x^\infty e^{-t^2/2}\,\mathrm{d}t \le \int_x^\infty e^{-t^2/2}\,t\,\mathrm{d}t =e^{-x^2/2}$$

Integrate both sides of the preceding:

$$ \begin{align} \int_s^\infty e^{-x^2/2}\,\mathrm{d}x &\ge\int_s^\infty x\int_x^\infty e^{-t^2/2}\,\mathrm{d}t\,\mathrm{d}x\\ &=\int_s^\infty\int_s^txe^{-t^2/2}\,\mathrm{d}x\,\mathrm{d}t\\ &=\int_s^\infty\frac12(t^2-s^2)e^{-t^2/2}\,\mathrm{d}t\\ \left(1+\frac12s^2\right)\int_s^\infty e^{-x^2/2}\,\mathrm{d}x &\ge\frac12\int_s^\infty t^2e^{-t^2/2}\,\mathrm{d}t\\ &=-\frac12\int_s^\infty t\,\mathrm{d}e^{-t^2/2}\\ &=\frac12se^{-s^2/2}+\frac12\int_s^\infty e^{-t^2/2}\,\mathrm{d}t\\ \left(s+\frac1s\right)\int_s^\infty e^{-x^2/2}\,\mathrm{d}x &\ge e^{-s^2/2} \end{align} $$

and note that $\displaystyle \left(s+\frac{1}{s}\right)^{-1} = \frac{s}{1+s^2} > \frac{1}{s} - \frac{1}{s^3}$ for $s > 0$,

which is equivalent to $\displaystyle s^4 > s^4 - 1$.

---

Expalanation for integartion by parts:

\begin{align*} &\int_x^{\infty} e^{-t^2/2} \mathrm dt\\ =& \int_x^{\infty} \color{red}{\frac{1}{t}} .\color{blue}{te^{-t^2/2}} \mathrm dt\\ =& \left[\color{red}{\frac{1}{t}} .\int\color{blue}{te^{-t^2/2}}\,dt\right]_x^{\infty} - \int_x^{\infty} \left( \color{red}{\frac{1}{t}} \right )' \left (\int\color{blue}{te^{-t^2/2}}\,dt\right)\mathrm dt\\ =& \frac{e^{-x^2/2}}{x} - \int_x^{\infty} \frac{e^{-t^2/2}}{t^2} \mathrm dt. \end{align*}

Now, for the second integral that we obtained in the previous line we employ similar idea as done above:

\begin{align*} &\int_x^{\infty} \frac{e^{-t^2/2}}{t^2} \mathrm dt\\ =& \int_x^{\infty} \color{red}{\frac{1}{t^3}} .\color{blue}{te^{-t^2/2}} \mathrm dt\\ =& \left[\color{red}{\frac{1}{t^3}} .\int\color{blue}{te^{-t^2/2}}\,dt\right]_x^{\infty} - \int_x^{\infty} \left( \color{red}{\frac{1}{t^3}} \right )' \left (\int\color{blue}{te^{-t^2/2}}\,dt\right)\mathrm dt\\ =& \frac{e^{-x^2/2}}{x^3} -3 \int_x^{\infty} \frac{e^{-t^2/2}}{t^4} \mathrm dt. \end{align*}

Combining these together we have $$\int_x^{\infty} e^{-t^2/2} = e^{-x^2/2}\left(\frac{1}{x} - \frac{1}{x^3}\right)+3 \int_x^{\infty} \frac{e^{-t^2/2}}{t^4} \mathrm dt$$

Since, the integrand $\dfrac{e^{-t^2/2}}{t^4}$ is always positive, we get the desired inequality:

$$\int_x^{\infty} e^{-t^2/2} > e^{-x^2/2}\left(\frac{1}{x} - \frac{1}{x^3}\right)$$