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Formelsammlung Mathematik: Bestimmte Integrale: Form R(x,arctan)

2021-05-05 02:32:14 作者：互联网

0.1Bearbeiten

{\displaystyle \int _{0}^{1}{\frac {\arctan x}{x}}\,dx=G}

\int _{0}^{1}{\frac {\arctan x}{x}}\,dx=G

Beweis

Benutze die Reihenentwicklung {\displaystyle \arctan x=\sum _{k=0}^{\infty }(-1)^{k}\,{\frac {x^{2k+1}}{2k+1}}} $\arctan x=\sum _{k=0}^{\infty }(-1)^{k}\,{\frac {x^{2k+1}}{2k+1}}$ .

{\displaystyle \int _{0}^{1}{\frac {\arctan x}{x}}\,dx=\sum _{k=0}^{\infty }(-1)^{k}\int _{0}^{1}{\frac {x^{2k}}{2k+1}}\,dx=\sum _{k=0}^{\infty }(-1)^{k}\,{\frac {1}{(2k+1)^{2}}}=G} $\int _{0}^{1}{\frac {\arctan x}{x}}\,dx=\sum _{k=0}^{\infty }(-1)^{k}\int _{0}^{1}{\frac {x^{2k}}{2k+1}}\,dx=\sum _{k=0}^{\infty }(-1)^{k}\,{\frac {1}{(2k+1)^{2}}}=G$

0.2Bearbeiten

{\displaystyle \int _{0}^{\infty }{\frac {\arctan x}{1+x^{2}}}\,dx={\frac {\pi ^{2}}{8}}}

\int _{0}^{\infty }{\frac {\arctan x}{1+x^{2}}}\,dx={\frac {\pi ^{2}}{8}}

Beweis

{\displaystyle \int _{0}^{\infty }{\frac {\arctan x}{1+x^{2}}}\,dx=\left[{\frac {1}{2}}\arctan ^{2}x\right]_{0}^{\infty }={\frac {1}{2}}\,\left({\frac {\pi }{2}}\right)^{2}={\frac {\pi ^{2}}{8}}} $\int _{0}^{\infty }{\frac {\arctan x}{1+x^{2}}}\,dx=\left[{\frac {1}{2}}\arctan ^{2}x\right]_{0}^{\infty }={\frac {1}{2}}\,\left({\frac {\pi }{2}}\right)^{2}={\frac {\pi ^{2}}{8}}$

0.3Bearbeiten

{\displaystyle \int _{0}^{\infty }{\frac {\arctan x}{1-x^{2}}}\,dx=-G}

\int _{0}^{\infty }{\frac {\arctan x}{1-x^{2}}}\,dx=-G

ohne Beweis

0.4Bearbeiten

{\displaystyle \int _{0}^{\infty }{\frac {x\,\arctan x}{1+x^{4}}}\,dx={\frac {\pi ^{2}}{16}}}

\int _{0}^{\infty }{\frac {x\,\arctan x}{1+x^{4}}}\,dx={\frac {\pi ^{2}}{16}}

ohne Beweis

0.5Bearbeiten

{\displaystyle \int _{0}^{\infty }{\frac {x\,\arctan x}{1-x^{4}}}\,dx=-{\frac {\pi }{8}}\,\log 2}

\int _{0}^{\infty }{\frac {x\,\arctan x}{1-x^{4}}}\,dx=-{\frac {\pi }{8}}\,\log 2

ohne Beweis

0.6Bearbeiten

{\displaystyle \int _{0}^{1}{\frac {\arctan \left(x^{3+{\sqrt {8}}\,}\right)}{1+x^{2}}}\,dx={\frac {1}{8}}\,\log(2)\cdot \log \left(1+{\sqrt {2}}\,\right)}

\int _{0}^{1}{\frac {\arctan \left(x^{3+{\sqrt {8}}\,}\right)}{1+x^{2}}}\,dx={\frac {1}{8}}\,\log(2)\cdot \log \left(1+{\sqrt {2}}\,\right)

ohne Beweis

1.1Bearbeiten

{\displaystyle \int _{-\infty }^{\infty }{\frac {\arctan ax}{x\,(1+x^{2})}}\,dx=\pi \log(1+a)\qquad a\geq 0}

\int _{-\infty }^{\infty }{\frac {\arctan ax}{x\,(1+x^{2})}}\,dx=\pi \log(1+a)\qquad a\geq 0

Beweis

Integriere die Formel {\displaystyle \int _{-\infty }^{\infty }{\frac {dx}{(1+t^{2}x^{2})(1+x^{2})}}={\frac {\pi }{1+t}}} $\int _{-\infty }^{\infty }{\frac {dx}{(1+t^{2}x^{2})(1+x^{2})}}={\frac {\pi }{1+t}}$ nach {\displaystyle t\,} $t\,$ von {\displaystyle 0\,} $0\,$ bis {\displaystyle a\,} $a\,$ .

1.2Bearbeiten

{\displaystyle \int _{0}^{\infty }{\frac {\arctan ax}{x\,(1-x^{2})}}\,dx={\frac {\pi }{4}}\log(1+a^{2})\qquad a\geq 0}

\int _{0}^{\infty }{\frac {\arctan ax}{x\,(1-x^{2})}}\,dx={\frac {\pi }{4}}\log(1+a^{2})\qquad a\geq 0

ohne Beweis

1.3Bearbeiten

{\displaystyle \int _{0}^{\infty }{\frac {\arctan \alpha x}{x\,{\sqrt {1-x^{2}}}}}\,dx={\frac {\pi }{2}}\,{\text{arsinh}}\,\alpha }

\int _{0}^{\infty }{\frac {\arctan \alpha x}{x\,{\sqrt {1-x^{2}}}}}\,dx={\frac {\pi }{2}}\,{\text{arsinh}}\,\alpha

Beweis

In der Formel {\displaystyle \int _{0}^{\frac {\pi }{2}}{\frac {1}{a^{2}\,\cos ^{2}x+b^{2}\,\sin ^{2}x}}\,dx={\frac {\pi }{2ab}}} $\int _{0}^{\frac {\pi }{2}}{\frac {1}{a^{2}\,\cos ^{2}x+b^{2}\,\sin ^{2}x}}\,dx={\frac {\pi }{2ab}}$ für {\displaystyle a,b>0\,} $a,b>0\,$

setze {\displaystyle a=1\,} $a=1\,$ und {\displaystyle b={\sqrt {1+t^{2}}}} $b={\sqrt {1+t^{2}}}$ .

{\displaystyle \int _{0}^{\frac {\pi }{2}}{\frac {1}{\underbrace {\cos ^{2}x+\sin ^{2}x} _{=1}+t^{2}\,\sin ^{2}x}}\,dx={\frac {\pi }{2}}\,{\frac {1}{\sqrt {1+t^{2}}}}} $\int _{0}^{\frac {\pi }{2}}{\frac {1}{\underbrace {\cos ^{2}x+\sin ^{2}x} _{=1}+t^{2}\,\sin ^{2}x}}\,dx={\frac {\pi }{2}}\,{\frac {1}{\sqrt {1+t^{2}}}}$

Nach Substitution {\displaystyle x\to \arcsin x\,} $x\to \arcsin x\,$ ist {\displaystyle \int _{0}^{\frac {\pi }{2}}{\frac {1}{1+t^{2}x^{2}}}\,{\frac {dx}{\sqrt {1-x^{2}}}}={\frac {\pi }{2}}\,{\frac {1}{\sqrt {1+t^{2}}}}} $\int _{0}^{\frac {\pi }{2}}{\frac {1}{1+t^{2}x^{2}}}\,{\frac {dx}{\sqrt {1-x^{2}}}}={\frac {\pi }{2}}\,{\frac {1}{\sqrt {1+t^{2}}}}$ .

Integriere nun nach {\displaystyle t\,} $t\,$ von {\displaystyle 0\,} $0\,$ bis {\displaystyle \alpha \,:\,\,\int _{0}^{\frac {\pi }{2}}\left[{\frac {\arctan tx}{x}}\right]_{0}^{\alpha }\,{\frac {dx}{\sqrt {1-x^{2}}}}={\frac {\pi }{2}}\,{\text{arsinh}}\,\alpha } $\alpha \,:\,\,\int _{0}^{\frac {\pi }{2}}\left[{\frac {\arctan tx}{x}}\right]_{0}^{\alpha }\,{\frac {dx}{\sqrt {1-x^{2}}}}={\frac {\pi }{2}}\,{\text{arsinh}}\,\alpha$ .

1.4Bearbeiten

{\displaystyle \int _{0}^{\infty }{\frac {\arctan x}{1+2\cos \alpha \,\,x+x^{2}}}\,dx={\frac {\pi }{4}}\,{\frac {\alpha }{\sin \alpha }}\qquad -{\frac {\pi }{2}}<{\text{Re}}(\alpha )<{\frac {\pi }{2}}}

\int _{0}^{\infty }{\frac {\arctan x}{1+2\cos \alpha \,\,x+x^{2}}}\,dx={\frac {\pi }{4}}\,{\frac {\alpha }{\sin \alpha }}\qquad -{\frac {\pi }{2}}<{\text{Re}}(\alpha )<{\frac {\pi }{2}}

Beweis

Nach Substitution {\displaystyle x\mapsto {\frac {1}{x}}} $x\mapsto {\frac {1}{x}}$ lässt sich das Integral auch schreiben als {\displaystyle \int _{0}^{\infty }{\frac {\arctan {\frac {1}{x}}}{1+2\cos \alpha \,\,x+x^{2}}}\,dx} $\int _{0}^{\infty }{\frac {\arctan {\frac {1}{x}}}{1+2\cos \alpha \,\,x+x^{2}}}\,dx$ .

Addiert man beide Darstellungen, so ist {\displaystyle 2I=\int _{0}^{\infty }{\frac {\arctan x+\arctan {\frac {1}{x}}}{1+2\cos \alpha \,\,x+x^{2}}}\,dx} $2I=\int _{0}^{\infty }{\frac {\arctan x+\arctan {\frac {1}{x}}}{1+2\cos \alpha \,\,x+x^{2}}}\,dx$ . Der Zähler ist konstant {\displaystyle {\frac {\pi }{2}}} ${\frac {\pi }{2}}$ .

Somit ist {\displaystyle I={\frac {\pi }{4}}\int _{0}^{\infty }{\frac {1}{1+2\cos \alpha \,\,x+x^{2}}}\,dx={\frac {\pi }{4}}\left[{\frac {1}{\sin \alpha }}\arctan {\frac {x+\cos \alpha }{\sin \alpha }}\right]_{0}^{\infty }={\frac {\pi }{4}}\,{\frac {\alpha }{\sin \alpha }}} $I={\frac {\pi }{4}}\int _{0}^{\infty }{\frac {1}{1+2\cos \alpha \,\,x+x^{2}}}\,dx={\frac {\pi }{4}}\left[{\frac {1}{\sin \alpha }}\arctan {\frac {x+\cos \alpha }{\sin \alpha }}\right]_{0}^{\infty }={\frac {\pi }{4}}\,{\frac {\alpha }{\sin \alpha }}$ .

1.5Bearbeiten

{\displaystyle \int _{0}^{1}{\frac {2a^{2}}{a^{2}+x^{2}}}\,{\frac {\arctan {\sqrt {2a^{2}+x^{2}}}}{\sqrt {2a^{2}+x^{2}}}}\,dx=\pi \,\arctan {\frac {1}{\sqrt {2a^{2}+1}}}-\left(\arctan {\frac {1}{a}}\right)^{2}}

\int _{0}^{1}{\frac {2a^{2}}{a^{2}+x^{2}}}\,{\frac {\arctan {\sqrt {2a^{2}+x^{2}}}}{\sqrt {2a^{2}+x^{2}}}}\,dx=\pi \,\arctan {\frac {1}{\sqrt {2a^{2}+1}}}-\left(\arctan {\frac {1}{a}}\right)^{2}

Beweis (Ahmedsches Integral)

Es ist {\displaystyle {\frac {1}{p}}+{\frac {1}{q}}={\frac {p+q}{p\,q}}\Rightarrow {\frac {1}{p\,(p+q)}}+{\frac {1}{q\,(p+q)}}={\frac {1}{p\,q}}} ${\frac {1}{p}}+{\frac {1}{q}}={\frac {p+q}{p\,q}}\Rightarrow {\frac {1}{p\,(p+q)}}+{\frac {1}{q\,(p+q)}}={\frac {1}{p\,q}}$ .

Setze {\displaystyle p=a^{2}+x^{2}\,} $p=a^{2}+x^{2}\,$ und {\displaystyle q=a^{2}+y^{2}\,} $q=a^{2}+y^{2}\,$ und integriere nach {\displaystyle x\,} $x\,$ und {\displaystyle y\,} $y\,$ jeweils von {\displaystyle 0\,} $0\,$ bis {\displaystyle 1\,} $1\,$ .

{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {dx\,dy}{(a^{2}+x^{2})(2a^{2}+x^{2}+y^{2})}}+\int _{0}^{1}\int _{0}^{1}{\frac {dy\,dx}{(a^{2}+y^{2})(2a^{2}+x^{2}+y^{2})}}=\int _{0}^{1}{\frac {dx}{a^{2}+x^{2}}}\cdot \int _{0}^{1}{\frac {dy}{a^{2}+y^{2}}}} $\int _{0}^{1}\int _{0}^{1}{\frac {dx\,dy}{(a^{2}+x^{2})(2a^{2}+x^{2}+y^{2})}}+\int _{0}^{1}\int _{0}^{1}{\frac {dy\,dx}{(a^{2}+y^{2})(2a^{2}+x^{2}+y^{2})}}=\int _{0}^{1}{\frac {dx}{a^{2}+x^{2}}}\cdot \int _{0}^{1}{\frac {dy}{a^{2}+y^{2}}}$

Vertauscht man die Rollen von {\displaystyle x\,} $x\,$ und {\displaystyle y\,} $y\,$ , so erkennt man, dass beide Integrale auf der linken Seite gleich sind und dass beide Integrale auf der rechten Seite gleich sind.

Also ist {\displaystyle 2\int _{0}^{1}\int _{0}^{1}{\frac {dx}{a^{2}+x^{2}}}\,{\frac {dy}{{\sqrt {2a^{2}+x^{2}}}^{2}+y^{2}}}=\left({\frac {1}{a}}\arctan {\frac {1}{a}}\right)^{2}} $2\int _{0}^{1}\int _{0}^{1}{\frac {dx}{a^{2}+x^{2}}}\,{\frac {dy}{{\sqrt {2a^{2}+x^{2}}}^{2}+y^{2}}}=\left({\frac {1}{a}}\arctan {\frac {1}{a}}\right)^{2}$

{\displaystyle \Rightarrow \int _{0}^{1}{\frac {2a^{2}}{a^{2}+x^{2}}}\,\left.{\frac {\arctan {\frac {y}{\sqrt {2a^{2}+x^{2}}}}}{\sqrt {2a^{2}+x^{2}}}}\right|_{0}^{1}\,dx=\left(\arctan {\frac {1}{a}}\right)^{2}} $\Rightarrow \int _{0}^{1}{\frac {2a^{2}}{a^{2}+x^{2}}}\,\left.{\frac {\arctan {\frac {y}{\sqrt {2a^{2}+x^{2}}}}}{\sqrt {2a^{2}+x^{2}}}}\right|_{0}^{1}\,dx=\left(\arctan {\frac {1}{a}}\right)^{2}$ .

Schreibe nun {\displaystyle \arctan {\frac {1}{\sqrt {2a^{2}+x^{2}}}}} $\arctan {\frac {1}{\sqrt {2a^{2}+x^{2}}}}$ als {\displaystyle {\frac {\pi }{2}}-\arctan {\sqrt {2a^{2}+x^{2}}}} ${\frac {\pi }{2}}-\arctan {\sqrt {2a^{2}+x^{2}}}$ .

{\displaystyle \underbrace {\int _{0}^{1}{\frac {2a^{2}}{a^{2}+x^{2}}}\,{\frac {\frac {\pi }{2}}{\sqrt {2a^{2}+x^{2}}}}\,dx} _{\pi \left.\arctan {\frac {x}{\sqrt {2a^{2}+x^{2}}}}\right|_{0}^{1}}-\int _{0}^{1}{\frac {2a^{2}}{a^{2}+x^{2}}}\,{\frac {\arctan {\sqrt {2a^{2}+x^{2}}}}{\sqrt {2a^{2}+x^{2}}}}\,dx=\left(\arctan {\frac {1}{a}}\right)^{2}} $\underbrace {\int _{0}^{1}{\frac {2a^{2}}{a^{2}+x^{2}}}\,{\frac {\frac {\pi }{2}}{\sqrt {2a^{2}+x^{2}}}}\,dx} _{\pi \left.\arctan {\frac {x}{\sqrt {2a^{2}+x^{2}}}}\right|_{0}^{1}}-\int _{0}^{1}{\frac {2a^{2}}{a^{2}+x^{2}}}\,{\frac {\arctan {\sqrt {2a^{2}+x^{2}}}}{\sqrt {2a^{2}+x^{2}}}}\,dx=\left(\arctan {\frac {1}{a}}\right)^{2}$

标签：Integrale,frac,Form,int,Mathematik,arctan,dx,pi,displaystyle
来源： https://www.cnblogs.com/Eufisky/p/14730801.html