1.1Bearbeiten
- {displaystyle int _{0}^{infty }{ ext{erf}}^{;2}left({sqrt {x}}\,
ight)\,e^{-ax}\,dx={frac {4}{api }}cdot {frac {operatorname {arccot} {sqrt {1+a}}}{sqrt {1+a}}}qquad { ext{Re}}(a)>0}
{displaystyle I=int _{0}^{infty }{ ext{erf}}^{;2}left({sqrt {x}}\,
ight)\,e^{-ax}\,dx=int _{0}^{infty }{ ext{erf}}^{;2}(x)\,e^{-ax^{2}}\,2x\,dx}
{displaystyle left[-{frac {1}{a}}\,e^{-ax^{2}}\,{ ext{erf}}^{;2}(x)
ight]_{0}^{infty }+int _{0}^{infty }{frac {1}{a}}\,e^{-ax^{2}}\,2\,{ ext{erf}}(x)\,{frac {2}{sqrt {pi }}}\,e^{-x^{2}}\,dx={frac {2}{a{sqrt {pi }}}}int _{0}^{infty }2\,{ ext{erf}}(x)\,e^{-(a+1)x^{2}}\,dx}![{displaystyle left[-{frac {1}{a}}\,e^{-ax^{2}}\,{ ext{erf}}^{;2}(x)
ight]_{0}^{infty }+int _{0}^{infty }{frac {1}{a}}\,e^{-ax^{2}}\,2\,{ ext{erf}}(x)\,{frac {2}{sqrt {pi }}}\,e^{-x^{2}}\,dx={frac {2}{a{sqrt {pi }}}}int _{0}^{infty }2\,{ ext{erf}}(x)\,e^{-(a+1)x^{2}}\,dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b89c076acda51ae2c7a43cedff2bc4bf9bf19320)
Nach der Ersetzung {displaystyle { ext{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}
{displaystyle I={frac {4}{api }}int _{0}^{infty }int _{0}^{1}2\,e^{-x^{2}t^{2}}\,x\,e^{-(a+1)x^{2}}\,dt\,dx={frac {4}{api }}int _{0}^{1}int _{0}^{infty }2x\,e^{-(t^{2}+a+1)x^{2}}\,dx\,dt}
{displaystyle ={frac {4}{api }}int _{0}^{1}{frac {1}{t^{2}+a+1}}\,dt={frac {4}{api }}cdot {frac {operatorname {arccot} {sqrt {1+a}}}{sqrt {1+a}}}}
