0.1Bearbeiten
- {\displaystyle \int _{0}^{\infty }W\left({\frac {1}{x^{2}}}\right)dx={\sqrt {2\pi }}}
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}
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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 }}}
{\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}
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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}}}
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 }}}
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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)}.