Zurück zu Bestimmte Integrale
Die Funktion f(x)=−12log(tanx2){\displaystyle f(x)=-{\frac {1}{2}}\log \left(\tan {\frac {x}{2}}\right)} besitzt die Fourierreihenentwicklung ∑k=0∞cos(2k+1)x2k+1{\displaystyle \sum _{k=0}^{\infty }{\frac {\cos(2k+1)x}{2k+1}}} . Nach der Parsevalschen Gleichung 1π∫−ππ|f(x)|2dx=a022+∑k=1∞(ak2+bk2){\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 14π∫−ππlog2(tanx2)dx=∑k=0∞1(2k+1)2=π28{\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}}} .
Die Funktion f(x)=log2(tanx2){\displaystyle f(x)=\log ^{2}\left(\tan {\frac {x}{2}}\right)} besitzt die Symmetrie f(π−x)=f(x){\displaystyle f(\pi -x)=f(x)\,} . ∫0πf(x2)dx=2∫0π2f(x)dx{\displaystyle \int _{0}^{\pi }f\left({\frac {x}{2}}\right)dx=2\int _{0}^{\frac {\pi }{2}}f(x)\,dx} ist daher ∫0πf(x)dx=π34{\displaystyle \int _{0}^{\pi }f(x)\,dx={\frac {\pi ^{3}}{4}}} .
I:=∫π/4π/2loglogtanxdx{\displaystyle I:=\int _{\pi /4}^{\pi /2}\log \log \tan x\,dx} ist nach Substitution x↦arctanex{\displaystyle x\mapsto \arctan e^{x}} gleich 12∫0∞logxcoshxdx{\displaystyle {\frac {1}{2}}\int _{0}^{\infty }{\frac {\log x}{\cosh x}}\,dx} . Und das ist π2∫0∞logπxcoshπxdx=π2logπ∫−∞∞dxcoshπx+π8∫−∞∞logx2coshπxdx{\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 ∫−∞∞dxcoshπx=1{\displaystyle \int _{-\infty }^{\infty }{\frac {dx}{\cosh \pi x}}=1} und nach der Formel ∫−∞∞log(α2+x2)coshπxdx=4log(2Γ(34+α2)Γ(14+α2)){\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 α≥0{\displaystyle \alpha \geq 0} ist π8∫−∞∞logx2coshπxdx=π2log(2Γ(34)Γ(14)){\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 I=π2log(2πΓ(34)Γ(14)){\displaystyle I={\frac {\pi }{2}}\,\log \left({\sqrt {2\pi }}\;{\frac {\Gamma \left({\frac {3}{4}}\right)}{\Gamma \left({\frac {1}{4}}\right)}}\right)} .
In der Formel ∫01loglog(1x)1+2cosαπ⋅x+x2dx=π2sinαπ(αlog2π+logΓ(12+α2)Γ(12−α2)){\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 α=12:∫01loglog(1x)1+x2dx=π2(log2π+logΓ(34)Γ(14)){\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 x↦cotx{\displaystyle x\mapsto \cot x} ergibt sich die besagte Gleichung.