$$p(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$$ is the probability density function of the standard normal random variable. m-sigma quality control means that the probability of failure is less than $$F_m=2 \int_m^ \infty p(x)dx$$ Find a 3-term(i.e. 3 non-zero terms) asymptotic expansion of $F_m$, when $m$ is large.
I thought about first calculate the integration $2 \int_m^ \infty\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx$, then do the asymptotic expansion. The computer show the answer is on the last line of this following website. http://www.wolframalpha.com/input/?i=2+%5Cint_m%5E+%5Cinfty+%5Cfrac%7B1%7D%7B2%5Cpi%7De%5E%7B%5Cfrac%7B-x%5E2%7D%7B2%7D%7Ddx
I really cannot figure out how to derive this. Could anyone kindly help me. Thanks!
Try making the change of variables $x = m\sqrt{y+1}$ then using Watson's lemma.
Another method is to write
$$ \int_m^\infty e^{-x^2/2}\,dx = -\int_m^\infty x^{-1}\,d\left(e^{-x^2/2}\right) $$
then repeatedly integrate by parts:
$$ \begin{align} \int_m^\infty e^{-x^2/2}\,dx &= -\int_m^\infty x^{-1}\,d\left(e^{-x^2/2}\right) \\ &= \left. -x^{-1}e^{-x^2/2} \right|_m^\infty - \int_m^\infty x^{-2} e^{-x^2/2}\,dx \\ &= m^{-1} e^{-m^2/2} + \int_m^\infty x^{-3}\,d\left(e^{-x^2/2}\right) \\ &= \cdots \end{align} $$