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

0.1Bearbeiten

{displaystyle int _{0}^{infty }Wleft({frac {1}{x^{2}}} ight)dx={sqrt {2pi }}}

Beweis

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

dann ist {displaystyle int _{0}^{infty }Wleft({frac {1}{x^{2}}} ight)\,dx={sqrt {2}}cdot Gamma left({frac {1}{2}} ight)={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}}} ight)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}}} ight) ight]^{alpha }dx=alpha cdot 2^{alpha -1/2}cdot Gamma left(alpha -{frac {1}{2}} ight)qquad { ext{Re}}(alpha )>{frac {1}{2}}}

Beweis

Die Funktion {displaystyle f(x)=left[Wleft({frac {1}{x^{2}}} ight) ight]^{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}} ight)}.