Evaluating $\int_{1}^{\infty}\exp(-(x(2n-x)/b)^2)\,\mathrm dx$

Question

$$I_1=\int_1^{\infty}\exp\left(-\left(\frac{x(2n-x)}{b}\right)^2\right)\mathrm dx,$$ 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.

Answer 1

The first two terms of the binomial expansion will only be a good approximation if $t$ is small, but you're integrating for $t$ out to infinity, so this seems like a losing proposition. You might be able to rescue something by splitting the range of integration into two parts, using that approximation for small $t$, and something else for large $t$ (where the integrand will be small anyway, so this might actually work).

Answer 2

Your substitution is right, except that it should have come out $$\int \frac{e^{-t^2}}{\sqrt{1-\frac{b t}{n^2}}} \, dt$$ rather than $$\int e^{-t^2} \sqrt{1-\frac{b t}{n^2}} \, dt$$