test jhj
erfc ( x ) = 2 π ∫ x ∞ e − t 2 d t = e − x 2 x π ∑ n = 0 ∞ ( − 1 ) n ( 2 n ) ! n ! ( 2 x ) 2 n {\displaystyle \operatorname {erfc} (x)={\frac {2}{\sqrt {\pi }}}\int _{x}^{\infty }e^{-t^{2}}\,dt={\frac {e^{-x^{2}}}{x{\sqrt {\pi }}}}\sum _{n=0}^{\infty }(-1)^{n}{\frac {(2n)!}{n!(2x)^{2n}}}}
