0.1Bearbeiten

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

Beweis

Verwende die Fourierreihe {\displaystyle -\log \left(2\sin {\frac {x}{2}}\right)=\sum _{n=1}^{\infty }{\frac {\cos nx}{n}}}.

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

 

0.2Bearbeiten

{\displaystyle \int _{0}^{\frac {\pi }{3}}\log ^{2}\left(2\sin {\frac {x}{2}}\right)\,dx={\frac {7\pi ^{3}}{108}}}

Beweis

Es sei {\displaystyle \mathbb {H} =\mathbb {C} \setminus \{z\in \mathbb {R} \,|\,z\leq 0\}} und {\displaystyle n\geq 2\,} eine natürliche Zahl.

Die Funktion {\displaystyle f_{n}:\mathbb {H} \to \mathbb {C} \,,\,z\mapsto {\frac {(-\log z)^{n-1}}{1-z}}} ist auf ganz {\displaystyle \mathbb {H} \,} holomorph,

wenn man sie an ihrer hebbaren Definitionslücke {\displaystyle z=1\,} stetig fortsetzt.


{\displaystyle F_{n}:\,\,]-\pi ,\pi [\to \mathbb {C} \,,\,\theta \mapsto \int _{1}^{e^{-i\theta }}f_{n}(z)\,dz} ist nach der Substitution {\displaystyle z\to {\frac {1}{z}}}

gleich {\displaystyle \int _{1}^{e^{i\theta }}{\frac {(\log z)^{n-1}}{1-{\frac {1}{z}}}}\,{\frac {-dz}{z^{2}}}=\int _{1}^{e^{i\theta }}{\frac {(\log z)^{n-1}}{z\,(1-z)}}\,dz}.

Und das ist nach der Partialbruchzerlegung {\displaystyle {\frac {1}{z\,(1-z)}}={\frac {1}{z}}+{\frac {1}{1-z}}}

gleich {\displaystyle \int _{1}^{e^{i\theta }}{\frac {(\log z)^{n-1}}{z}}\,dz+\int _{1}^{e^{i\theta }}{\frac {(\log z)^{n-1}}{1-z}}\,dz=\left[{\frac {1}{n}}\,(\log z)^{n}\right]_{1}^{e^{i\theta }}+(-1)^{n-1}\int _{1}^{e^{i\theta }}f_{n}(z)\,dz}.

Also ist {\displaystyle F_{n}(\theta )={\frac {(i\theta )^{n}}{n}}+{\overline {F_{n}(\theta )}}\qquad (1)}


{\displaystyle G_{n}:\,\,]0,2\pi [\to \mathbb {C} \,,\,\theta \mapsto \int _{0}^{1-e^{i\theta }}f_{n}(z)\,dz} ist nach der Substitution {\displaystyle z\to 1-z\,} gleich

{\displaystyle -\int _{1}^{e^{i\theta }}{\frac {\left[-\log(1-z)\right]^{n-1}}{z}}\,dz}. Und das ist nach der Substitution {\displaystyle z\to e^{ix}\,} gleich

{\displaystyle -i\int _{0}^{\theta }\left[-\log(1-e^{ix})\right]^{n-1}\,dx}, wobei {\displaystyle -\log(1-e^{ix})=-\log \left(2\sin {\frac {x}{2}}\right)+i\,{\frac {\pi -x}{2}}} ist.

Also ist {\displaystyle G_{n}(\theta )=-i\int _{0}^{\theta }\left[-\log \left(2\sin {\frac {x}{2}}\right)+i\,{\frac {\pi -x}{2}}\right]^{n-1}\,dx\qquad (2)}


{\displaystyle G_{n}(\theta )=\int _{0}^{1-e^{i\theta }}f_{n}(z)\,dz} lässt sich aufspalten in {\displaystyle \int _{0}^{1}f_{n}(z)\,dz+\int _{1}^{1-e^{i\theta }}f_{n}(z)\,dz},

wobei {\displaystyle \int _{0}^{1}f_{n}(z)\,dz=\Gamma (n)\,\zeta (n)} ist. Setzt man {\displaystyle \theta ={\frac {\pi }{3}}\,}, so ist {\displaystyle 1-e^{i\theta }=e^{-i\theta }\,}.

Daher gilt {\displaystyle G_{n}\left({\frac {\pi }{3}}\right)=\Gamma (n)\zeta (n)+F_{n}\left({\frac {\pi }{3}}\right)\qquad (3)}



Betrachte nun den Fall {\displaystyle \theta ={\frac {\pi }{3}}} und {\displaystyle n=3\,:}

Aus {\displaystyle (1)\,\,\,F_{3}\left({\frac {\pi }{3}}\right)={\frac {\left(i{\frac {\pi }{3}}\right)^{3}}{3}}+{\overline {F_{3}\left({\frac {\pi }{3}}\right)}}}

folgt {\displaystyle {\text{Im}}\left[F_{3}\left({\frac {\pi }{3}}\right)\right]={\frac {1}{2i}}\,\left(F_{3}\left({\frac {\pi }{3}}\right)-{\overline {F_{3}\left({\frac {\pi }{3}}\right)}}\right)={\frac {1}{2i}}\,{\frac {i^{3}\pi ^{3}}{3^{4}}}=-{\frac {\pi ^{3}}{162}}}.

Aus {\displaystyle (2)\,\,\,G_{3}\left({\frac {\pi }{3}}\right)=-i\int _{0}^{\frac {\pi }{3}}\left[-\log \left(2\sin {\frac {x}{2}}\right)+i\,{\frac {\pi -x}{2}}\right]^{2}\,dx}

{\displaystyle =\int _{0}^{\frac {\pi }{3}}(\pi -x)\,\log \left(2\sin {\frac {x}{2}}\right)\,dx-i\int _{0}^{\frac {\pi }{3}}\left[\log ^{2}\left(2\sin {\frac {x}{2}}\right)-\left({\frac {\pi -x}{2}}\right)^{2}\right]dx}

folgt {\displaystyle {\text{Im}}\left[G_{3}\left({\frac {\pi }{3}}\right)\right]=\int _{0}^{\frac {\pi }{3}}\left({\frac {\pi -x}{2}}\right)^{2}\,dx-\int _{0}^{\frac {\pi }{3}}\log ^{2}\left(2\sin {\frac {x}{2}}\right)\,dx}.

Und aus {\displaystyle (3)\,\,\,G_{3}\left({\frac {\pi }{3}}\right)=2\zeta (3)+F_{3}\left({\frac {\pi }{3}}\right)} folgt {\displaystyle {\text{Im}}\left[G_{3}\left({\frac {\pi }{3}}\right)\right]={\text{Im}}\left[F_{3}\left({\frac {\pi }{3}}\right)\right]}.

Also ist {\displaystyle \int _{0}^{\frac {\pi }{3}}\log ^{2}\left(2\sin {\frac {x}{2}}\right)\,dx-\int _{0}^{\frac {\pi }{3}}{\frac {(\pi -x)^{2}}{4}}\,dx={\frac {\pi ^{3}}{162}}}.

Und somit ist {\displaystyle \int _{0}^{\frac {\pi }{3}}\log ^{2}\left(2\sin {\frac {x}{2}}\right)\,dx={\frac {7\pi ^{3}}{108}}}.

 

0.3Bearbeiten

{\displaystyle \int _{0}^{\pi }\log ^{2}\left(2\sin {\frac {x}{2}}\right)\,dx={\frac {\pi ^{3}}{12}}}

ohne Beweis

 

 

0.4Bearbeiten

{\displaystyle \int _{0}^{\frac {\pi }{3}}x\log ^{2}\left(2\sin {\frac {x}{2}}\right)\,dx={\frac {17\pi ^{4}}{6480}}}

Beweis

Betrachte nun den Fall {\displaystyle \theta ={\frac {\pi }{3}}} und {\displaystyle n=4\,:}

Aus {\displaystyle (1)\,\,\,F_{4}\left({\frac {\pi }{3}}\right)={\frac {\left(i{\frac {\pi }{3}}\right)^{4}}{4}}-{\overline {F_{4}\left({\frac {\pi }{3}}\right)}}}

folgt {\displaystyle {\text{Re}}\left[F_{4}\left({\frac {\pi }{3}}\right)\right]={\frac {1}{2}}\,\left(F_{4}\left({\frac {\pi }{3}}\right)+{\overline {F_{4}\left({\frac {\pi }{3}}\right)}}\right)={\frac {1}{2}}\,{\frac {\pi ^{4}}{4\cdot 3^{4}}}={\frac {\pi ^{4}}{648}}}.

Aus {\displaystyle (2)\,\,\,G_{4}\left({\frac {\pi }{3}}\right)=-i\int _{0}^{\frac {\pi }{3}}\left[-\log \left(2\sin {\frac {x}{2}}\right)+i\,{\frac {\pi -x}{2}}\right]^{3}\,dx}

{\displaystyle =\int _{0}^{\frac {\pi }{3}}\left[3\log ^{2}\left(2\sin {\frac {x}{2}}\right){\frac {\pi -x}{2}}-\left({\frac {\pi -x}{2}}\right)^{3}\right]\,dx+i\int _{0}^{\frac {\pi }{3}}\left[\log ^{3}\left(2\sin {\frac {x}{2}}\right)-3\log \left(2\sin {\frac {x}{2}}\right)\,\left({\frac {\pi -x}{2}}\right)^{2}\right]\,dx}

folgt {\displaystyle {\text{Re}}\left[G_{4}\left({\frac {\pi }{3}}\right)\right]={\frac {3\pi }{2}}\int _{0}^{\frac {\pi }{3}}\log ^{2}\left(2\sin {\frac {x}{2}}\right)\,dx-{\frac {3}{2}}\int _{0}^{\frac {\pi }{3}}x\log ^{2}\left(2\sin {\frac {x}{2}}\right)\,dx-\int _{0}^{\frac {\pi }{3}}{\frac {(\pi -x)^{3}}{8}}dx}.

Aus dem Fall {\displaystyle n=3\,} ist bereits bekannt, dass {\displaystyle \int _{0}^{\frac {\pi }{3}}\log ^{2}\left(2\sin {\frac {x}{2}}\right)\,dx={\frac {7\pi ^{3}}{108}}} ist.

Also ist {\displaystyle {\text{Re}}\left[G_{4}\left({\frac {\pi }{3}}\right)\right]=-{\frac {3}{2}}\int _{0}^{\frac {\pi }{3}}x\log ^{2}\left(2\sin {\frac {x}{2}}\right)\,dx+{\frac {187\pi ^{4}}{2592}}}.

Und aus {\displaystyle (3)\,\,\,G_{4}\left({\frac {\pi }{3}}\right)=\Gamma (4)\zeta (4)+F_{4}\left({\frac {\pi }{3}}\right)} folgt {\displaystyle {\text{Re}}\left[G_{4}\left({\frac {\pi }{3}}\right)\right]-6\cdot {\frac {\pi ^{4}}{90}}={\text{Re}}\left[F_{4}\left({\frac {\pi }{3}}\right)\right]}.

Also ist {\displaystyle -{\frac {3}{2}}\int _{0}^{\frac {\pi }{3}}x\log ^{2}\left(2\sin {\frac {x}{2}}\right)\,dx+{\frac {187\pi ^{4}}{2592}}-6\cdot {\frac {\pi ^{4}}{90}}={\frac {\pi ^{4}}{648}}}.

Und somit ist {\displaystyle \int _{0}^{\frac {\pi }{3}}x\log ^{2}\left(2\sin {\frac {x}{2}}\right)\,dx={\frac {17\pi ^{4}}{6480}}}.