• 微信公众号：美女很有趣。 工作之余，放松一下，关注即送10G+美女照片！

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

1周前 (05-04) 6次浏览

0.1Bearbeiten
{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 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}}}
Beweis

{displaystyle int _{0}^{infty }{frac {arctan x}{1+x^{2}}},dx=left[{frac {1}{2}}arctan ^{2}xright]_{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}
ohne Beweis

0.4Bearbeiten
{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}
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)}
ohne Beweis

1.1Bearbeiten
{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}}} nach {displaystyle t,} von {displaystyle 0,} bis {displaystyle 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}
ohne Beweis

1.3Bearbeiten
{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}}} für {displaystyle a,b>0,}

setze {displaystyle a=1,} und {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}}}}}

Nach Substitution {displaystyle xto 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}}}}}.

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

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}}<{text{Re}}(alpha )<{frac {pi }{2}}}
Beweis

Nach Substitution {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}.

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}. Der Zähler ist konstant {displaystyle {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 }}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}}
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}}}.

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

Vertauscht man die Rollen von {displaystyle x,} und {displaystyle 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}}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}}}}} als {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}}