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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?

Did
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Seneca
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1 Answers1

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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 $$ 0\leqslant\int_z^\infty\frac1{t^4}\mathrm e^{-t^2/2}\,\mathrm dt\leqslant\int_z^\infty\frac{t}{z^5}\mathrm e^{-t^2/2}\,\mathrm dt=\frac{\mathrm e^{-z^2/2}}{z^5}.$$
Did
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  • I dont get your hint. Two times integrating by parts gives: $\int_z^\infty e^{-t^2/2}dt= -ze^{-z^2/2}+\int_{z}^\infty t^2e^{-t^2/2} dt=-ze^{-z^2/2}-\frac{1}{3}z^3e^{-z^2/2}+\int_{z}^\infty\frac{ t^4}{3} e^{-t^2/2} dt$ – Seneca Jan 10 '14 at 07:34
  • Yes, that is because you did not read it: the hint is to consider $v(t)=\mathrm e^{-t^2/2}$, not $u(t)=\mathrm e^{-t^2/2}$. – Did Jan 10 '14 at 07:41
  • Ye I just recognized. Thank you very much I got it. – Seneca Jan 10 '14 at 07:50
  • I completed calculations. I think you have to integrate by parts three times. But anyway get your estimation. Thanks again. – Seneca Jan 10 '14 at 08:54
  • No, twice is enough. See Edit. – Did Jan 10 '14 at 09:05
  • Ah ok I did not manage your estimation and integrated a third time. So this was indeed not necessary. Thank you very much for your effort! – Seneca Jan 13 '14 at 12:23