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# Formelsammlung Mathematik: Bestimmte Integrale: Form R(x,exp,erf)

1周前 (05-04) 7次浏览

##### 1.1Bearbeiten
{displaystyle int _{0}^{infty }{text{erf}}^{;2}left({sqrt {x}},right),e^{-ax},dx={frac {4}{api }}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}{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}}}}.