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

Zurück zu Bestimmte Integrale

0.1

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

Beweis

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

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

0.2

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

Beweis

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

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

wenn man sie an ihrer hebbaren Definitionslücke $z=1\,$ stetig fortsetzt.


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

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

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

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

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


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

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

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

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


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

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

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



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

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

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

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

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

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

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

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

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

0.3

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

ohne Beweis

0.4

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

Beweis

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

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

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

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

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

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

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

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

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

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

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