- #1
Alone
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I've got the next question which I just want to see if I got it right, and if not then do correct me.
we have [math]erf(x)=\frac{2}{\sqrt \pi} \int_{0}^{x} e^{-t^2} dt[/math]
and I want to find an asymptotic expansion of this function when [math]x\rightarrow \infty[/math].
So here's what I have done:
[math]erf(x)=\frac{2}{\sqrt \pi} \int_{0}^{\infty} e^{-t^2}dt - \int_{x}^{\infty} e^{-t^2} dt = 1 +\int_{\frac{1}{x}}^{0} e^{-\frac{1}{\xi ^2}}\frac{d\xi}{\xi ^2}[/math]
where in my last step I used the next change of variables: [math]\xi=\frac{1}{t}[/math].
Now, I am kind of stuck here, I mean [math]\xi[/math] is smaller than [math]\frac{1}{x}\rightarrow 0[/math], but I cannot expand [math]e^{-\frac{1}{\xi ^2}}[/math] with a power seris in [math]\frac{1}{\xi^2}[/math] so what to do now?
Thanks (the question appears in Murray's asymptotic anaysis page 27, question 2).
we have [math]erf(x)=\frac{2}{\sqrt \pi} \int_{0}^{x} e^{-t^2} dt[/math]
and I want to find an asymptotic expansion of this function when [math]x\rightarrow \infty[/math].
So here's what I have done:
[math]erf(x)=\frac{2}{\sqrt \pi} \int_{0}^{\infty} e^{-t^2}dt - \int_{x}^{\infty} e^{-t^2} dt = 1 +\int_{\frac{1}{x}}^{0} e^{-\frac{1}{\xi ^2}}\frac{d\xi}{\xi ^2}[/math]
where in my last step I used the next change of variables: [math]\xi=\frac{1}{t}[/math].
Now, I am kind of stuck here, I mean [math]\xi[/math] is smaller than [math]\frac{1}{x}\rightarrow 0[/math], but I cannot expand [math]e^{-\frac{1}{\xi ^2}}[/math] with a power seris in [math]\frac{1}{\xi^2}[/math] so what to do now?
Thanks (the question appears in Murray's asymptotic anaysis page 27, question 2).
