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To prove: $\displaystyle \int_x^\infty \exp \left(-\frac{t^2}{2}\right) \, dt < \frac{1}{x}\exp\left(-\frac{x^2}{2}\right)$, $\quad x>0$

We know

$$\int_x^\infty t^{-2} \exp\left(-\frac{t^2}{2}\right) \, dt \leq \exp\left(-\frac{x^2}{2}\right)\int_x ^\infty t^{-2} \, dt = \frac{1}{x}\exp\left(-\frac{x^2}{2}\right) $$

Does the above observation lead to solution? Please give a hint.

Sangchul Lee
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Vinay Deshpande
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1 Answers1

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You have obtained the right bound for a wrong integral!

$\int_x^{\infty} te^{-t^{2}/2}dt =-e^{-t^{2}/2}|_x^{\infty}=e^{-x^{2}/2}$. On $(x,\ \infty)$ note that $t>x$ so $\int_x^{\infty} te^{-t^{2}/2}dt >\int_x^{\infty} xe^{-t^{2}/2}dt$. Finish by pulling $x$ out of the integral.

Kavi Rama Murthy
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