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$$I_1=\int_1^{\infty}\exp\left(-\left(\frac{x(2n-x)}{b}\right)^2\right)\mathrm dx,$$ I set $$t=\frac{x(2n-x)}{b},$$ and, solving for $x$ and $dt$ I got $$I_1=\frac{b}{2 n} \int_1^{\infty} e^{-t^2}\left(1-\frac{bt}{n^2}\right)^{\frac12}\mathrm dt.$$

I then expand $$\left(1-\frac{bt}{n^2}\right)^{\frac12} \approx 1+\frac{bt}{2n^2}$$ in the first two terms of Binomial series, and obtain something like $$I_1 \approx \frac{b}{2n}\left(\frac{\sqrt{\pi}(1-\mathrm{erf}(1))}{2}+\frac{b}{4n^2e}\right).$$

I am not 100% sure about this derivation (although the result is sensible), especially about the substitution, it seems I could misused integration of Gaussian function here.

  • 2
    Check your change of variables. The limits for the integral with respect to $t$ are obviously wrong and so your approximation, which uses the limits, is wrong. Also, your approach may have problems if $b$ is large.2011-06-26
  • 0
    @Steve: is this due to binomial expansion?2011-06-26

3 Answers 3