Proving estimate of tail of cumulative distribution function of standard normal distribution I tried to prove the following estimation, with no success:
$$\int_z^\infty e^{-t^2/2}dt= \frac{e^{-z^2/2}}{z}(1-z^{-1}+\mathcal{O}(z^{-2}))$$
Any advice on how to prove this?
 A: Hint: Integrate by parts the LHS twice, using $\displaystyle\int uv'=uv-\int u'v$ with $v(t)=\mathrm e^{-t^2/2}$ each time. This will show you that the formula in your post should read
$$
\int_z^\infty\mathrm e^{-t^2/2}\,\mathrm dt=\frac{\mathrm e^{-z^2/2}}z\,\left(1-\frac1{z^2}+O\left(\frac1{z^4}\right)\right).
$$
Edit: Here are the two integrations by parts mentioned above.


*

*First integration by parts: $u(t)=-1/t$, $v(t)=\mathrm e^{-t^2/2}$, results in
$$
\int_z^\infty\mathrm e^{-t^2/2}\,\mathrm dt=\frac{\mathrm e^{-z^2/2}}z-\int_z^\infty\frac1{t^2}\mathrm e^{-t^2/2}\,\mathrm dt.$$

*Second integration by parts: $u(t)=-1/t^3$, $v(t)=\mathrm e^{-t^2/2}$, results in
$$
\int_z^\infty\frac1{t^2}\mathrm e^{-t^2/2}\,\mathrm dt=\frac{\mathrm e^{-z^2/2}}{z^3}-3\int_z^\infty\frac1{t^4}\mathrm e^{-t^2/2}\,\mathrm dt.$$

*Estimate of the remainder term: $1/t^4\leqslant t/z^5$ on $(z,+\infty)$ hence
$$
0\leqslant\int_z^\infty\frac1{t^4}\mathrm e^{-t^2/2}\,\mathrm dt\leqslant\int_z^\infty\frac{t}{z^5}\mathrm e^{-t^2/2}\,\mathrm dt=\frac{\mathrm e^{-z^2/2}}{z^5}.$$

