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

2021-05-05 03:01:39 作者：互联网

0.1Bearbeiten

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

\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}}} $-\log\left(2 \sin \frac{x}{2}\right)=\sum_{n=1}^\infty \frac{\cos nx}{n}$ .

{\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} $-\int_0^{\frac{\pi}{2}} \log\left(2 \sin \frac{x}{2}\right) dx=\sum_{n=1}^\infty \frac{1}{n} \int_0^{\frac{\pi}{2}} \cos nx \, dx=\sum_{n=1}^\infty \frac{\sin \frac{n\pi}{2}}{n^2}=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2}=G$

0.2Bearbeiten

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

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

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

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

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

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} $\int_1^{e^{i\theta}} \frac{(\log z)^{n-1}}{1-\frac{1}{z}}\, \frac{-dz}{z^2}=\int_1^{e^{i\theta}} \frac{(\log z)^{n-1}}{z\, (1-z)}\, dz$ .

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

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} $\int_1^{e^{i\theta}} \frac{(\log z)^{n-1}}{z} \, dz+\int_1^{e^{i\theta}} \frac{(\log z)^{n-1}}{1-z} \, dz=\left[\frac{1}{n}\, (\log z)^n \right]_1^{e^{i\theta}}+(-1)^{n-1} \int_1^{e^{i\theta}} f_n(z)\, dz$ .

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

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

{\displaystyle -\int _{1}^{e^{i\theta }}{\frac {\left[-\log(1-z)\right]^{n-1}}{z}}\,dz} $-\int_1^{e^{i\theta}} \frac{\left[-\log(1-z)\right]^{n-1}}{z}\, dz$ . Und das ist nach der Substitution {\displaystyle z\to e^{ix}\,} $z\to e^{ix}\,$ gleich

{\displaystyle -i\int _{0}^{\theta }\left[-\log(1-e^{ix})\right]^{n-1}\,dx} $-i \int_0^\theta \left[-\log(1-e^{ix})\right]^{n-1}\, dx$ , wobei {\displaystyle -\log(1-e^{ix})=-\log \left(2\sin {\frac {x}{2}}\right)+i\,{\frac {\pi -x}{2}}} $-\log(1-e^{ix})=-\log\left(2\sin\frac{x}{2}\right)+i\,\frac{\pi-x}{2}$ 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)} $G_n(\theta)=-i\int_0^\theta \left[-\log\left(2\sin\frac{x}{2}\right)+i\,\frac{\pi-x}{2}\right]^{n-1}\, dx \qquad (2)$

{\displaystyle G_{n}(\theta )=\int _{0}^{1-e^{i\theta }}f_{n}(z)\,dz} $G_n(\theta)=\int_0^{1-e^{i\theta}} f_n(z)\, dz$ lässt sich aufspalten in {\displaystyle \int _{0}^{1}f_{n}(z)\,dz+\int _{1}^{1-e^{i\theta }}f_{n}(z)\,dz} $\int_0^1 f_n(z)\, dz+\int_1^{1-e^{i\theta}} f_n(z)\, dz$ ,

wobei {\displaystyle \int _{0}^{1}f_{n}(z)\,dz=\Gamma (n)\,\zeta (n)} $\int_0^1 f_n(z)\, dz=\Gamma(n)\, \zeta(n)$ ist. Setzt man {\displaystyle \theta ={\frac {\pi }{3}}\,} $\theta=\frac{\pi}{3}\,$ , so ist {\displaystyle 1-e^{i\theta }=e^{-i\theta }\,} $1-e^{i\theta}=e^{-i\theta}\,$ .

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

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

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)}}} $(1) \,\,\, F_3\left(\frac{\pi}{3}\right)=\frac{\left(i\frac{\pi}{3}\right)^3}{3}+\overline{F_3\left(\frac{\pi}{3}\right)}$

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}}} $\text{Im}\left[F_3\left(\frac{\pi}{3}\right)\right]=\frac{1}{2i}\, \left(F_3\left(\frac{\pi}{3}\right)-\overline{F_3\left(\frac{\pi}{3}\right)}\right)=\frac{1}{2i}\, \frac{i^3 \pi^3}{3^4}=-\frac{\pi^3}{162}$ .

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} $(2) \,\,\, G_3\left(\frac{\pi}{3}\right)=-i\int_0^{\frac{\pi}{3}} \left[-\log\left(2\sin\frac{x}{2}\right)+i\,\frac{\pi-x}{2}\right]^2 \, dx$

{\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} $=\int_0^{\frac{\pi}{3}} (\pi-x)\, \log\left(2\sin\frac{x}{2}\right)\, dx-i\int_0^{\frac{\pi}{3}} \left[\log^2\left(2\sin\frac{x}{2}\right)-\left(\frac{\pi-x}{2}\right)^2\right] dx$

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} $\text{Im}\left[G_3\left(\frac{\pi}{3}\right)\right]=\int_0^{\frac{\pi}{3}} \left(\frac{\pi-x}{2}\right)^2 \, dx-\int_0^{\frac{\pi}{3}} \log^2\left(2\sin\frac{x}{2}\right) \, dx$ .

Und aus {\displaystyle (3)\,\,\,G_{3}\left({\frac {\pi }{3}}\right)=2\zeta (3)+F_{3}\left({\frac {\pi }{3}}\right)} $(3) \,\,\, G_3\left(\frac{\pi}{3}\right)=2\zeta(3)+F_3\left(\frac{\pi}{3}\right)$ folgt {\displaystyle {\text{Im}}\left[G_{3}\left({\frac {\pi }{3}}\right)\right]={\text{Im}}\left[F_{3}\left({\frac {\pi }{3}}\right)\right]} $\text{Im}\left[G_3\left(\frac{\pi}{3}\right)\right]=\text{Im}\left[F_3\left(\frac{\pi}{3}\right)\right]$ .

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}}} $\int_0^{\frac{\pi}{3}} \log^2\left(2\sin\frac{x}{2}\right) \, dx-\int_0^{\frac{\pi}{3}} \frac{(\pi-x)^2}{4} \, dx=\frac{\pi^3}{162}$ .

Und somit ist {\displaystyle \int _{0}^{\frac {\pi }{3}}\log ^{2}\left(2\sin {\frac {x}{2}}\right)\,dx={\frac {7\pi ^{3}}{108}}} $\int_0^{\frac{\pi}{3}} \log^2\left(2\sin\frac{x}{2}\right) \, dx=\frac{7\pi^3}{108}$ .

0.3Bearbeiten

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

\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}}}

\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}}} $\theta=\frac{\pi}{3}$ und {\displaystyle n=4\,:} $n=4\, :$

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)}}} $(1) \,\,\, F_4\left(\frac{\pi}{3}\right)=\frac{\left(i\frac{\pi}{3}\right)^4}{4}-\overline{F_4\left(\frac{\pi}{3}\right)}$

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}}} $\text{Re}\left[F_4\left(\frac{\pi}{3}\right)\right]=\frac12 \, \left(F_4\left(\frac{\pi}{3}\right)+\overline{F_4\left(\frac{\pi}{3}\right)}\right)=\frac{1}{2}\, \frac{\pi^4}{4\cdot 3^4}=\frac{\pi^4}{648}$ .

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} $(2) \,\,\, G_4\left(\frac{\pi}{3}\right)=-i\int_0^{\frac{\pi}{3}} \left[-\log\left(2\sin\frac{x}{2}\right)+i\,\frac{\pi-x}{2}\right]^3 \, dx$

{\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} $=\int_0^{\frac{\pi}{3}} \left[3\log^2\left(2\sin\frac{x}{2}\right)\frac{\pi-x}{2}-\left(\frac{\pi-x}{2}\right)^3\right]\, dx+i \int_0^{\frac{\pi}{3}} \left[\log^3\left(2\sin\frac{x}{2}\right)-3\log\left(2\sin\frac{x}{2}\right) \, \left(\frac{\pi-x}{2}\right)^2 \right]\, dx$

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} $\text{Re}\left[G_4\left(\frac{\pi}{3}\right)\right]=\frac{3\pi}{2}\int_0^{\frac{\pi}{3}} \log^2\left(2\sin\frac{x}{2}\right) \, dx-\frac32 \int_0^{\frac{\pi}{3}} x\log^2\left(2\sin\frac{x}{2}\right) \, dx-\int_0^{\frac{\pi}{3}} \frac{(\pi-x)^3}{8} dx$ .

Aus dem Fall {\displaystyle n=3\,} $n=3\,$ 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}}} $\int_0^{\frac{\pi}{3}} \log^2\left(2\sin\frac{x}{2}\right) \, dx=\frac{7\pi^3}{108}$ 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}}} $\text{Re}\left[G_4\left(\frac{\pi}{3}\right)\right]=-\frac32\int_0^{\frac{\pi}{3}} x\log^2\left(2\sin\frac{x}{2}\right) \, dx+\frac{187\pi^4}{2592}$ .

Und aus {\displaystyle (3)\,\,\,G_{4}\left({\frac {\pi }{3}}\right)=\Gamma (4)\zeta (4)+F_{4}\left({\frac {\pi }{3}}\right)} $(3) \,\,\, G_4\left(\frac{\pi}{3}\right)=\Gamma(4)\zeta(4)+F_4\left(\frac{\pi}{3}\right)$ 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]} $\text{Re}\left[G_4\left(\frac{\pi}{3}\right)\right]-6\cdot\frac{\pi^4}{90}=\text{Re}\left[F_4\left(\frac{\pi}{3}\right)\right]$ .

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}}} $-\frac32\int_0^{\frac{\pi}{3}} x\log^2\left(2\sin\frac{x}{2}\right) \, dx+\frac{187\pi^4}{2592}-6\cdot\frac{\pi^4}{90}=\frac{\pi^4}{648}$ .

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}}} $\int_0^{\frac{\pi}{3}} x\log^2\left(2\sin\frac{x}{2}\right) \, dx=\frac{17\pi^4}{6480}$ .

标签：Integrale,right,frac,log,Form,int,pi,displaystyle,left
来源： https://www.cnblogs.com/Eufisky/p/14730815.html