|||
$1 \ {\rm mJy}=10^{-26} \ {\rm erg\ s^{-1}\ cm^{-2}\ Hz^{-1}}$
$1 \ {\rm Jy}=10^{-26} \ {\rm W\ m^{-2}\ Hz^{-1}}$
${\rm GAUSSINT}(x)$=$\frac{1}{\sqrt{2\pi}}\int^{x}_{-\infty}e^{-\frac{1}{2}t^2}{\rm d}t$
=$\frac{1}{\sqrt{\pi}}\int^{x}_{-\infty}e^{-\left(\frac{t}{\sqrt{2}}\right)^2}{\rm d}\left(\frac{t}{\sqrt{2}}\right)$
=$\frac{1}{\sqrt{\pi}}\int^{\frac{x}{\sqrt{2}}}_{-\infty}e^{-s^2}{\rm d}s$
=$\frac{1}{\sqrt{\pi}}\int^{\frac{x}{\sqrt{2}}}_{0}e^{-s^2}{\rm d}s$+$\frac{1}{\sqrt{\pi}}\int^{0}_{-\infty}e^{-s^2}{\rm d}s$
=$\frac{1}{2}{\rm erf}(\frac{x}{\sqrt{2}})$+$\frac{1}{2}$
Archiver|手机版|科学网 ( 京ICP备07017567号-12 )
GMT+8, 2024-4-26 00:40
Powered by ScienceNet.cn
Copyright © 2007- 中国科学报社