0.1Bearbeiten

{\displaystyle \int _{0}^{1}\log \left(\tan {\frac {\pi x}{2}}\right)\,dx=0}

ohne Beweis

 

 

0.2Bearbeiten

{\displaystyle \int _{0}^{\pi }\log ^{2}\left(\tan {\frac {x}{2}}\right)dx={\frac {\pi ^{3}}{4}}}

Beweis

Die Funktion {\displaystyle f(x)=-{\frac {1}{2}}\log \left(\tan {\frac {x}{2}}\right)} besitzt die Fourierreihenentwicklung {\displaystyle \sum _{k=0}^{\infty }{\frac {\cos(2k+1)x}{2k+1}}}.

Nach der Parsevalschen Gleichung {\displaystyle {\frac {1}{\pi }}\int _{-\pi }^{\pi }|f(x)|^{2}\,dx={\frac {a_{0}^{2}}{2}}+\sum _{k=1}^{\infty }{\Big (}a_{k}^{2}+b_{k}^{2}{\Big )}}

gilt dann {\displaystyle {\frac {1}{4\pi }}\int _{-\pi }^{\pi }\log ^{2}\left(\tan {\frac {x}{2}}\right)dx=\sum _{k=0}^{\infty }{\frac {1}{(2k+1)^{2}}}={\frac {\pi ^{2}}{8}}}.

 

0.3Bearbeiten

{\displaystyle \int _{0}^{\pi }\log ^{2}\left(\tan {\frac {x}{4}}\right)dx={\frac {\pi ^{3}}{4}}}

Beweis

Die Funktion {\displaystyle f(x)=\log ^{2}\left(\tan {\frac {x}{2}}\right)} besitzt die Symmetrie {\displaystyle f(\pi -x)=f(x)\,}.

{\displaystyle \int _{0}^{\pi }f\left({\frac {x}{2}}\right)dx=2\int _{0}^{\frac {\pi }{2}}f(x)\,dx} ist daher {\displaystyle \int _{0}^{\pi }f(x)\,dx={\frac {\pi ^{3}}{4}}}.

 

0.4Bearbeiten

{\displaystyle \int _{\pi /4}^{\pi /2}\log \log \tan x\,dx={\frac {\pi }{2}}\,\log \left({\sqrt {2\pi }}\,\,{\frac {\Gamma \left({\frac {3}{4}}\right)}{\Gamma \left({\frac {1}{4}}\right)}}\right)}

1. Beweis (Vardisches Integral)

{\displaystyle I:=\int _{\pi /4}^{\pi /2}\log \log \tan x\,dx} ist nach Substitution {\displaystyle x\mapsto \arctan e^{x}} gleich {\displaystyle {\frac {1}{2}}\int _{0}^{\infty }{\frac {\log x}{\cosh x}}\,dx}.

Und das ist {\displaystyle {\frac {\pi }{2}}\,\int _{0}^{\infty }{\frac {\log \pi x}{\cosh \pi x}}\,dx={\frac {\pi }{2}}\log {\sqrt {\pi }}\int _{-\infty }^{\infty }{\frac {dx}{\cosh \pi x}}+{\frac {\pi }{8}}\int _{-\infty }^{\infty }{\frac {\log x^{2}}{\cosh \pi x}}\,dx}.

Dabei ist {\displaystyle \int _{-\infty }^{\infty }{\frac {dx}{\cosh \pi x}}=1} und nach der Formel {\displaystyle \int _{-\infty }^{\infty }{\frac {\log(\alpha ^{2}+x^{2})}{\cosh \pi x}}\,dx=4\,\log \left({\sqrt {2}}\;{\frac {\Gamma \left({\frac {3}{4}}+{\frac {\alpha }{2}}\right)}{\Gamma \left({\frac {1}{4}}+{\frac {\alpha }{2}}\right)}}\right)} für {\displaystyle \alpha \geq 0}

ist {\displaystyle {\frac {\pi }{8}}\,\int _{-\infty }^{\infty }{\frac {\log x^{2}}{\cosh \pi x}}\,dx={\frac {\pi }{2}}\,\log \left({\sqrt {2}}\,\,{\frac {\Gamma \left({\frac {3}{4}}\right)}{\Gamma \left({\frac {1}{4}}\right)}}\right)}. Also ist {\displaystyle I={\frac {\pi }{2}}\,\log \left({\sqrt {2\pi }}\;{\frac {\Gamma \left({\frac {3}{4}}\right)}{\Gamma \left({\frac {1}{4}}\right)}}\right)}.

2. Beweis

In der Formel {\displaystyle \int _{0}^{1}{\frac {\log \log \left({\frac {1}{x}}\right)}{1+2\cos \alpha \pi \cdot x+x^{2}}}\,dx={\frac {\pi }{2\sin \alpha \pi }}\left(\alpha \log 2\pi +\log {\frac {\Gamma \left({\frac {1}{2}}+{\frac {\alpha }{2}}\right)}{\Gamma \left({\frac {1}{2}}-{\frac {\alpha }{2}}\right)}}\right)}

setze {\displaystyle \alpha ={\frac {1}{2}}\,:\quad \int _{0}^{1}{\frac {\log \log \left({\frac {1}{x}}\right)}{1+x^{2}}}\,dx={\frac {\pi }{2}}\left(\log {\sqrt {2\pi }}+\log {\frac {\Gamma \left({\frac {3}{4}}\right)}{\Gamma \left({\frac {1}{4}}\right)}}\right)}

Durch die Substitution {\displaystyle x\mapsto \cot x} ergibt sich die besagte Gleichung.