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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.

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  • $\begingroup$ Put $y = \sqrt{t}$ and perform a partial integration where you differentiate the power of $t$ multiplying the exponential. You can do that repeatedly to derive the complete asymptotic expansion of the error function for large arguments. $\endgroup$ – Count Iblis Mar 26 '17 at 22:46
  • $\begingroup$ Please don't use the same letter for the bound variable inside an integral and the free one outside and in the limits: it's very easy to confuse the two, not to mention logically wrong. $\endgroup$ – Chappers Mar 26 '17 at 22:47
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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.

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