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

 

0.1Bearbeiten
{\displaystyle \int _{0}^{1}{\frac {\arctan x}{x}}\,dx=G}{\displaystyle \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}}}{\displaystyle \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}{\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}

 
0.2Bearbeiten
{\displaystyle \int _{0}^{\infty }{\frac {\arctan x}{1+x^{2}}}\,dx={\frac {\pi ^{2}}{8}}}{\displaystyle \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}}}{\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}}}

 
0.3Bearbeiten
{\displaystyle \int _{0}^{\infty }{\frac {\arctan x}{1-x^{2}}}\,dx=-G}{\displaystyle \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}}}{\displaystyle \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}{\displaystyle \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)}{\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)}
ohne Beweis

 

 
1.1Bearbeiten
{\displaystyle \int _{-\infty }^{\infty }{\frac {\arctan ax}{x\,(1+x^{2})}}\,dx=\pi \log(1+a)\qquad a\geq 0}{\displaystyle \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}}}{\displaystyle \int _{-\infty }^{\infty }{\frac {dx}{(1+t^{2}x^{2})(1+x^{2})}}={\frac {\pi }{1+t}}} nach {\displaystyle t\,}{\displaystyle 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}{\displaystyle \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 }{\displaystyle \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}}}{\displaystyle \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\,}{\displaystyle a,b>0\,}

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

Nach Substitution {\displaystyle x\to \arcsin x\,}{\displaystyle 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}}}}}{\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}}}}}.

Integriere nun nach {\displaystyle t\,}{\displaystyle 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 }{\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 }.

 
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}}}{\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}}}
Beweis

Nach Substitution {\displaystyle x\mapsto {\frac {1}{x}}}{\displaystyle 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}{\displaystyle \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}{\displaystyle 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 }}}{\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 }}}.

 
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}}{\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}}
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}}}{\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}}}.

Setze {\displaystyle p=a^{2}+x^{2}\,}{\displaystyle p=a^{2}+x^{2}\,} und {\displaystyle q=a^{2}+y^{2}\,}{\displaystyle 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}}}}{\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}}}}

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}}{\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}}

{\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}}{\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}}.

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

posted on 2021-05-05 02:29  Eufisky  阅读(46)  评论(0)  编辑  收藏  举报

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