Why does the following equality holds?
$\displaystyle \sqrt{\frac{2}{\pi}} \frac{e^{-y^2/2}}{y} - \sqrt{\frac{2}{\pi}} \int^{\infty} _{y} e^{-y^2/2} \cdot y^{-2} dy$
$\displaystyle = \sqrt{\frac{2}{\pi}} \frac{e^{-y^2/2}}{y} \cdot (1 + \mathcal{O}(\frac{1}{y^2}))$
It involves the Taylor series expansion, but I do not see what steps are taken to obtain the second line.
Integrate the integral by parts: $$ \int_y^{\infty} \frac{e^{-x^2/2}}{x^2} \, dx = \int_y^{\infty} \frac{xe^{-x^2/2}}{x^3} \, dx = \frac{e^{-y^2/2}}{y^3} + \int_y^{\infty} 3\frac{e^{-x^2/2}}{x^4} \, dx. $$ The former is obviously $e^{-y^2/2}O(y^{-3})$, for the latter, $e^{-x^2/2} \leq e^{-y^2/2}$, so the integral is bounded by $e^{-y^2/2}\int_y^{\infty} 3\frac{dy}{y^4} = y^{-3}e^{-y^2/2}$, and so the whole integral is $e^{-y^2/2}O(y^{-3})$, as required.