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

 

0.1Bearbeiten

{\displaystyle \int _{0}^{\infty }W\left({\frac {1}{x^{2}}}\right)dx={\sqrt {2\pi }}}

Beweis

In der Formel {\displaystyle \int _{0}^{\infty }\left[W\left({\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 }W\left({\frac {1}{x^{2}}}\right)\,dx={\sqrt {2}}\cdot \Gamma \left({\frac {1}{2}}\right)={\sqrt {2\pi }}}.

 

 

 

0.2Bearbeiten

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

Beweis

{\displaystyle \int _{0}^{\infty }W\left({\frac {1}{x^{2}}}\right)dx} ist nach Substitution {\displaystyle x\mapsto 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[W\left({\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[W\left({\frac {1}{x^{2}}}\right)\right]^{\alpha }} besitzt die Umkehrfunktion {\displaystyle g(x)=x^{-{\frac {1}{2\alpha }}}\cdot e^{-1/2\cdot 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^{-1/2\cdot x}\,dx=\alpha \cdot 2^{\alpha -1/2}\cdot \Gamma \left(\alpha -{\frac {1}{2}}\right)}.