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 ight]_{0}^{infty }={frac {1}{2}}\,left({frac {pi }{2}} ight)^{2}={frac {pi ^{2}}{8}}}{displaystyle int _{0}^{infty }{frac {arctan x}{1+x^{2}}}\,dx=left[{frac {1}{2}}arctan ^{2}x ight]_{0}^{infty }={frac {1}{2}}\,left({frac {pi }{2}} ight)^{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}}\,} ight)}{1+x^{2}}}\,dx={frac {1}{8}}\,log(2)cdot log left(1+{sqrt {2}}\, ight)}{displaystyle int _{0}^{1}{frac {arctan left(x^{3+{sqrt {8}}\,} ight)}{1+x^{2}}}\,dx={frac {1}{8}}\,log(2)cdot log left(1+{sqrt {2}}\, ight)}
ohne Beweis
 
1.1Bearbeiten
{displaystyle int _{-infty }^{infty }{frac {arctan ax}{x\,(1+x^{2})}}\,dx=pi log(1+a)qquad ageq 0}{displaystyle int _{-infty }^{infty }{frac {arctan ax}{x\,(1+x^{2})}}\,dx=pi log(1+a)qquad ageq 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 ageq 0}{displaystyle int _{0}^{infty }{frac {arctan ax}{x\,(1-x^{2})}}\,dx={frac {pi }{4}}log(1+a^{2})qquad ageq 0}
ohne Beweis
 
1.3Bearbeiten
{displaystyle int _{0}^{infty }{frac {arctan alpha x}{x\,{sqrt {1-x^{2}}}}}\,dx={frac {pi }{2}}\,{ ext{arsinh}}\,alpha }{displaystyle int _{0}^{infty }{frac {arctan alpha x}{x\,{sqrt {1-x^{2}}}}}\,dx={frac {pi }{2}}\,{ ext{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 o arcsin x\,}{displaystyle x o 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}} ight]_{0}^{alpha }\,{frac {dx}{sqrt {1-x^{2}}}}={frac {pi }{2}}\,{ ext{arsinh}}\,alpha }{displaystyle alpha \,:\,\,int _{0}^{frac {pi }{2}}left[{frac {arctan tx}{x}} ight]_{0}^{alpha }\,{frac {dx}{sqrt {1-x^{2}}}}={frac {pi }{2}}\,{ ext{arsinh}}\,alpha }.

 
1.4Bearbeiten
{displaystyle int _{0}^{infty }{frac {arctan x}{1+2cos alpha \,\,x+x^{2}}}\,dx={frac {pi }{4}}\,{frac {alpha }{sin alpha }}qquad -{frac {pi }{2}}<{ ext{Re}}(alpha )<{frac {pi }{2}}}{displaystyle int _{0}^{infty }{frac {arctan x}{1+2cos alpha \,\,x+x^{2}}}\,dx={frac {pi }{4}}\,{frac {alpha }{sin alpha }}qquad -{frac {pi }{2}}<{ ext{Re}}(alpha )<{frac {pi }{2}}}
Beweis

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

Addiert man beide Darstellungen, so ist {displaystyle 2I=int _{0}^{infty }{frac {arctan x+arctan {frac {1}{x}}}{1+2cos alpha \,\,x+x^{2}}}\,dx}{displaystyle 2I=int _{0}^{infty }{frac {arctan x+arctan {frac {1}{x}}}{1+2cos 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+2cos alpha \,\,x+x^{2}}}\,dx={frac {pi }{4}}left[{frac {1}{sin alpha }}arctan {frac {x+cos alpha }{sin alpha }} ight]_{0}^{infty }={frac {pi }{4}}\,{frac {alpha }{sin alpha }}}{displaystyle I={frac {pi }{4}}int _{0}^{infty }{frac {1}{1+2cos alpha \,\,x+x^{2}}}\,dx={frac {pi }{4}}left[{frac {1}{sin alpha }}arctan {frac {x+cos alpha }{sin alpha }} ight]_{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}} ight)^{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}} ight)^{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 2int _{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}} ight)^{2}}{displaystyle 2int _{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}} ight)^{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}}}} ight|_{0}^{1}\,dx=left(arctan {frac {1}{a}} ight)^{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}}}} ight|_{0}^{1}\,dx=left(arctan {frac {1}{a}} ight)^{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}}}} ight|_{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}} ight)^{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}}}} ight|_{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}} ight)^{2}}

【推广】 免费学中医，健康全家人
原文地址：https://www.cnblogs.com/Eufisky/p/14730801.html