证明拉马努金等式：∫(z,+∞)e^(-x^2/2)dx=e^(-z^2/2)/(z+1/(z+2/(z+3/(z+…))))

永远

\(\displaystyle\int_z^\infty  {{e^{ - \frac{{{x^2}}}{2}}}} dx = \frac{{{e^{ - \frac{{{z^2}}}{2}}}}}{{z + \frac{1}{{z + \frac{2}{{z + \frac{3}{{z + \frac{4}{{z +  \ddots }}}}}}}}}}\)

wilsony

挺吓人的