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

1周前 (05-04) 4次浏览

##### 0.1Bearbeiten
{displaystyle int _{0}^{infty }Wleft({frac {1}{x^{2}}}right)dx={sqrt {2pi }}}
Beweis

In der Formel {displaystyle int _{0}^{infty }left[Wleft({frac {1}{x^{2}}}right)right]^{alpha }dx=alpha cdot 2^{alpha -1/2}cdot Gamma left(alpha -{frac {1}{2}}right)} setze {displaystyle alpha =1},

dann ist {displaystyle int _{0}^{infty }Wleft({frac {1}{x^{2}}}right),dx={sqrt {2}}cdot Gamma left({frac {1}{2}}right)={sqrt {2pi }}}.

##### 0.2Bearbeiten
{displaystyle int _{0}^{infty }{frac {W(x)}{x,{sqrt {x}}}},dx=2cdot {sqrt {2pi }}}
Beweis

{displaystyle int _{0}^{infty }Wleft({frac {1}{x^{2}}}right)dx} ist nach Substitution {displaystyle xmapsto x^{-1/2}} gleich {displaystyle {frac {1}{2}},int _{0}^{infty }{frac {W(x)}{x,{sqrt {x}}}},dx}.

##### 1.1Bearbeiten
{displaystyle int _{0}^{infty }left[Wleft({frac {1}{x^{2}}}right)right]^{alpha }dx=alpha cdot 2^{alpha -1/2}cdot Gamma left(alpha -{frac {1}{2}}right)qquad {text{Re}}(alpha )>{frac {1}{2}}}
Beweis

Die Funktion {displaystyle f(x)=left[Wleft({frac {1}{x^{2}}}right)right]^{alpha }} besitzt die Umkehrfunktion {displaystyle g(x)=x^{-{frac {1}{2alpha }}}cdot e^{-1/2cdot x^{1/alpha }}}.

Nun ist {displaystyle int _{0}^{infty }f(x),dx=int _{0}^{infty }g(x),dx=int _{0}^{infty }g(x^{alpha }),alpha ,x^{alpha -1},dx}

{displaystyle =alpha int _{0}^{infty }x^{-1/2+alpha -1},e^{-1/2cdot x},dx=alpha cdot 2^{alpha -1/2}cdot Gamma left(alpha -{frac {1}{2}}right)}.