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Formelsammlung Mathematik: Bestimmte Integrale: Form R(x,log,sin)

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0.1
${\displaystyle \int _{0}^{\frac {\pi }{2}}\log \left(2\sin {\frac {x}{2}}\right)dx=-G}$
Beweis

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

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

0.2
${\displaystyle \int _{0}^{\frac {\pi }{3}}\log ^{2}\left(2\sin {\frac {x}{2}}\right)\,dx={\frac {7\pi ^{3}}{108}}}$
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 ${\displaystyle \int _{0}^{\frac {\pi }{3}}\log ^{2}\left(2\sin {\frac {x}{2}}\right)\,dx={\frac {7\pi ^{3}}{108}}}$.

0.3
${\displaystyle \int _{0}^{\pi }\log ^{2}\left(2\sin {\frac {x}{2}}\right)\,dx={\frac {\pi ^{3}}{12}}}$
ohne Beweis

0.4
${\displaystyle \int _{0}^{\frac {\pi }{3}}x\log ^{2}\left(2\sin {\frac {x}{2}}\right)\,dx={\frac {17\pi ^{4}}{6480}}}$
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 ${\displaystyle \int _{0}^{\frac {\pi }{3}}x\log ^{2}\left(2\sin {\frac {x}{2}}\right)\,dx={\frac {17\pi ^{4}}{6480}}}$.