1.1Bearbeiten

{\displaystyle \int _{0}^{\infty }{\text{erf}}^{\;2}\left({\sqrt {x}}\,\right)\,e^{-ax}\,dx={\frac {4}{a\pi }}\cdot {\frac {\operatorname {arccot} {\sqrt {1+a}}}{\sqrt {1+a}}}\qquad {\text{Re}}(a)>0}

Beweis

{\displaystyle I=\int _{0}^{\infty }{\text{erf}}^{\;2}\left({\sqrt {x}}\,\right)\,e^{-ax}\,dx=\int _{0}^{\infty }{\text{erf}}^{\;2}(x)\,e^{-ax^{2}}\,2x\,dx} ist nach partieller Integration

{\displaystyle \left[-{\frac {1}{a}}\,e^{-ax^{2}}\,{\text{erf}}^{\;2}(x)\right]_{0}^{\infty }+\int _{0}^{\infty }{\frac {1}{a}}\,e^{-ax^{2}}\,2\,{\text{erf}}(x)\,{\frac {2}{\sqrt {\pi }}}\,e^{-x^{2}}\,dx={\frac {2}{a{\sqrt {\pi }}}}\int _{0}^{\infty }2\,{\text{erf}}(x)\,e^{-(a+1)x^{2}}\,dx}.

Nach der Ersetzung {\displaystyle {\text{erf}}(x)={\frac {2}{\sqrt {\pi }}}\int _{0}^{x}e^{-t^{2}}\,dt={\frac {2}{\sqrt {\pi }}}\int _{0}^{1}e^{-x^{2}t^{2}}\,x\,dt} ist

{\displaystyle I={\frac {4}{a\pi }}\int _{0}^{\infty }\int _{0}^{1}2\,e^{-x^{2}t^{2}}\,x\,e^{-(a+1)x^{2}}\,dt\,dx={\frac {4}{a\pi }}\int _{0}^{1}\int _{0}^{\infty }2x\,e^{-(t^{2}+a+1)x^{2}}\,dx\,dt}

{\displaystyle ={\frac {4}{a\pi }}\int _{0}^{1}{\frac {1}{t^{2}+a+1}}\,dt={\frac {4}{a\pi }}\cdot {\frac {\operatorname {arccot} {\sqrt {1+a}}}{\sqrt {1+a}}}}.