0.1Bearbeiten

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

Beweis

Aus der Fourierreihendarstellung {\displaystyle \log \left(2\cos {\frac {x}{2}}\right)=\sum _{k=1}^{\infty }{\frac {(-1)^{k-1}}{k}}\,\cos kx}

folgt {\displaystyle \int _{0}^{\pi }\log \left(2\cos {\frac {x}{2}}\right)dx=\sum _{k=1}^{\infty }{\frac {(-1)^{k-1}}{k}}\,\int _{0}^{\pi }\cos kx\,dx=0}.

 

 

 

0.2Bearbeiten

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

Beweis

Aus der Fourierreihendarstellung {\displaystyle \log \left(2\cos {\frac {x}{2}}\right)=\sum _{k=1}^{\infty }{\frac {(-1)^{k-1}}{k}}\,\cos kx}

folgt {\displaystyle \int _{0}^{\frac {\pi }{2}}\log \left(2\cos {\frac {x}{2}}\right)dx=\sum _{k=1}^{\infty }{\frac {(-1)^{k-1}}{k}}\,\int _{0}^{\frac {\pi }{2}}\cos kx\,dx=\sum _{k=1}^{\infty }{\frac {(-1)^{k-1}\,\sin {\frac {k\pi }{2}}}{k^{2}}}=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)^{2}}}=G}.

 

 

 

0.3Bearbeiten

{\displaystyle \int _{0}^{\pi }x^{2}\,\log ^{2}\left(2\cos {\frac {x}{2}}\right)\,dx={\frac {11\pi ^{5}}{180}}}

Beweis

Nach der Cauchyschen Cosinus-Integralformel ist {\displaystyle f(y):={\frac {1}{\pi }}\int _{0}^{\pi }\left(2\cos {\frac {\theta }{2}}\right)^{x}\cos y\theta \,d\theta ={\frac {\Gamma (x+1)}{\Gamma \left({\frac {x}{2}}+y+1\right)\,\Gamma \left({\frac {x}{2}}-y+1\right)}}}.

Durch logarithmisches Differenzieren ergibt sich:

{\displaystyle \log f(y)=\log \Gamma (x+1)-\log \Gamma \left({\frac {x}{2}}+y+1\right)-\log \Gamma \left({\frac {x}{2}}-y+1\right)}

{\displaystyle \Rightarrow \,{\frac {f'(y)}{f(y)}}=-\psi \left({\frac {x}{2}}+y+1\right)+\psi \left({\frac {x}{2}}-y+1\right)\,\Rightarrow \,f'(y)=f(y)\cdot \left(\psi \left({\frac {x}{2}}-y+1\right)-\psi \left({\frac {x}{2}}+y+1\right)\right)\,\Rightarrow \,f'(0)=0}

Nach der Produktregel ist dann

{\displaystyle f''(y)=f'(y)\cdot \left(\psi \left({\frac {x}{2}}-y+1\right)-\psi \left({\frac {x}{2}}+y+1\right)\right)+f(y)\cdot \left(-\psi '\left({\frac {x}{2}}-y+1\right)-\psi '\left({\frac {x}{2}}+y+1\right)\right)}

{\displaystyle \Rightarrow \,f''(0)=-2\cdot f(0)\cdot \psi '\left({\frac {x}{2}}+1\right)=-2\cdot {\frac {\Gamma (x+1)}{\Gamma ^{2}\left({\frac {x}{2}}+1\right)}}\cdot \psi '\left({\frac {x}{2}}+1\right)}.

Also ist {\displaystyle g(x):={\frac {1}{\pi }}\int _{0}^{\pi }\left(2\cos {\frac {\theta }{2}}\right)^{x}\,\theta ^{2}\,d\theta =2\cdot {\frac {\Gamma (x+1)}{\Gamma ^{2}\left({\frac {x}{2}}+1\right)}}\cdot \psi '\left({\frac {x}{2}}+1\right)}.

Setzt man {\displaystyle h(x):={\frac {\Gamma (x+1)}{\Gamma ^{2}\left({\frac {x}{2}}+1\right)}}}, so ist nach logarithmischer Differenzation {\displaystyle \log h(x)=\log \Gamma (x+1)-2\log \Gamma \left({\frac {x}{2}}+1\right)}

{\displaystyle \Rightarrow \,{\frac {h'(x)}{h(x)}}=\psi (x+1)-2\cdot \psi \left({\frac {x}{2}}+1\right)\cdot {\frac {1}{2}}\,\Rightarrow \,h'(x)=h(x)\cdot \left(\psi (x+1)-\psi \left({\frac {x}{2}}+1\right)\right)\,\Rightarrow \,h'(0)=0}.

Nach der Produktregel ist {\displaystyle h''(x)=h'(x)\cdot \left(\psi (x+1)-\psi \left({\frac {x}{2}}+1\right)\right)+h(x)\cdot \left(\psi '(x+1)-\psi '\left({\frac {x}{2}}+1\right)\cdot {\frac {1}{2}}\right)}

{\displaystyle \Rightarrow \,h''(0)=\psi '(1)-{\frac {1}{2}}\psi '(1)={\frac {1}{2}}\psi '(1)={\frac {\pi ^{2}}{12}}}.

Die Taylorreihenentwicklung von {\displaystyle h\,} beginnt daher wie folgt {\displaystyle h(x)=1+{\frac {\pi ^{2}}{24}}\cdot x^{2}+...}

Und wegen {\displaystyle \psi (x+1)=-\gamma +\zeta (2)\cdot x-\zeta (3)\cdot x^{2}+\zeta (4)\cdot x^{3}-...} ist {\displaystyle \psi '(x+1)=\zeta (2)-2\zeta (3)\cdot x+3\zeta (4)\cdot x^{2}-...}

und somit ist {\displaystyle \psi '\left({\frac {x}{2}}+1\right)=\zeta (2)-\zeta (3)\cdot x+{\frac {3}{4}}\zeta (4)\cdot x^{2}-...}

{\displaystyle h(x)\cdot \psi '\left({\frac {x}{2}}+1\right)=\left(1+{\frac {\pi ^{2}}{24}}\cdot x^{2}+...\right)\cdot \left(\zeta (2)-\zeta (3)\cdot x+{\frac {3}{4}}\zeta (4)\cdot x^{2}-...\right)=\zeta (2)-\zeta (3)\cdot x+\left({\frac {\pi ^{2}}{24}}\zeta (2)+{\frac {3}{4}}\zeta (4)\right)\cdot x^{2}+...}

{\displaystyle g''(0)={\frac {1}{\pi }}\int _{0}^{\pi }\log ^{2}\left(2\cos {\frac {\theta }{2}}\right)\cdot \theta ^{2}\,d\theta =2\cdot 2\cdot \left({\frac {\pi ^{2}}{24}}\zeta (2)+{\frac {3}{4}}\zeta (4)\right)={\frac {11\,\pi ^{4}}{180}}}

 

 

 

0.4Bearbeiten

{\displaystyle \int _{0}^{\frac {\pi }{2}}{\frac {x^{2}}{x^{2}+\log ^{2}(2\cos x)}}\,dx={\frac {\pi }{8}}\left(1-\gamma +\log 2\pi \right)}

ohne Beweis    

 

 

1.1Bearbeiten

{\displaystyle \int _{0}^{\pi }\log \left(1-2\alpha \cos x+\alpha ^{2}\right)dx=\left\{{\begin{matrix}0&|\alpha |\leq 1\\\\2\pi \log |\alpha |&|\alpha |>1\end{matrix}}\right.\qquad ,\qquad \alpha \in \mathbb {R} }

Beweis

Für {\displaystyle |\alpha |\leq 1} und {\displaystyle 0<x<\pi } ist

{\displaystyle -{\frac {1}{2}}\log \left(1-2\alpha \cos x+\alpha ^{2}\right)=\sum _{k=1}^{\infty }{\frac {\alpha ^{k}\,\cos kx}{k}}}.

{\displaystyle \Rightarrow \,I(\alpha ):=\int _{0}^{\pi }\log \left(1-2\alpha \cos x+\alpha ^{2}\right)dx=-2\sum _{k=1}^{\infty }{\frac {\alpha ^{k}}{k}}\,\int _{0}^{\pi }\cos kx\,dx=0}

Und für {\displaystyle |\alpha |>1} ist

{\displaystyle I(\alpha )=\int _{0}^{\pi }\left[\log(\alpha ^{2})+\log \left({\frac {1}{\alpha ^{2}}}+{\frac {2}{\alpha }}\cos x+1\right)\right]dx=2\pi \log |\alpha |+\underbrace {I\left({\frac {1}{\alpha }}\right)} _{=0}}.