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

0.1Bearbeiten

{displaystyle int _{0}^{1}log left( an {frac {pi x}{2}} ight)\,dx=0}

ohne Beweis

 

0.2Bearbeiten

{displaystyle int _{0}^{pi }log ^{2}left( an {frac {x}{2}} ight)dx={frac {pi ^{3}}{4}}}

Beweis

Die Funktion {displaystyle f(x)=-{frac {1}{2}}log left( an {frac {x}{2}} ight)} besitzt die Fourierreihenentwicklung {displaystyle sum _{k=0}^{infty }{frac {cos(2k+1)x}{2k+1}}}.

Nach der Parsevalschen Gleichung {displaystyle {frac {1}{pi }}int _{-pi }^{pi }|f(x)|^{2}\,dx={frac {a_{0}^{2}}{2}}+sum _{k=1}^{infty }{Big (}a_{k}^{2}+b_{k}^{2}{Big )}}

gilt dann {displaystyle {frac {1}{4pi }}int _{-pi }^{pi }log ^{2}left( an {frac {x}{2}} ight)dx=sum _{k=0}^{infty }{frac {1}{(2k+1)^{2}}}={frac {pi ^{2}}{8}}}.

 

0.3Bearbeiten

{displaystyle int _{0}^{pi }log ^{2}left( an {frac {x}{4}} ight)dx={frac {pi ^{3}}{4}}}

Beweis

Die Funktion {displaystyle f(x)=log ^{2}left( an {frac {x}{2}} ight)} besitzt die Symmetrie {displaystyle f(pi -x)=f(x)\,}.

{displaystyle int _{0}^{pi }fleft({frac {x}{2}} ight)dx=2int _{0}^{frac {pi }{2}}f(x)\,dx} ist daher {displaystyle int _{0}^{pi }f(x)\,dx={frac {pi ^{3}}{4}}}.

 

0.4Bearbeiten

{displaystyle int _{pi /4}^{pi /2}log log an x\,dx={frac {pi }{2}}\,log left({sqrt {2pi }}\,\,{frac {Gamma left({frac {3}{4}} ight)}{Gamma left({frac {1}{4}} ight)}} ight)}

1. Beweis (Vardisches Integral)

{displaystyle I:=int _{pi /4}^{pi /2}log log an x\,dx} ist nach Substitution {displaystyle xmapsto arctan e^{x}} gleich {displaystyle {frac {1}{2}}int _{0}^{infty }{frac {log x}{cosh x}}\,dx}.

Und das ist {displaystyle {frac {pi }{2}}\,int _{0}^{infty }{frac {log pi x}{cosh pi x}}\,dx={frac {pi }{2}}log {sqrt {pi }}int _{-infty }^{infty }{frac {dx}{cosh pi x}}+{frac {pi }{8}}int _{-infty }^{infty }{frac {log x^{2}}{cosh pi x}}\,dx}.

Dabei ist {displaystyle int _{-infty }^{infty }{frac {dx}{cosh pi x}}=1} und nach der Formel {displaystyle int _{-infty }^{infty }{frac {log(alpha ^{2}+x^{2})}{cosh pi x}}\,dx=4\,log left({sqrt {2}};{frac {Gamma left({frac {3}{4}}+{frac {alpha }{2}} ight)}{Gamma left({frac {1}{4}}+{frac {alpha }{2}} ight)}} ight)} für {displaystyle alpha geq 0}

ist {displaystyle {frac {pi }{8}}\,int _{-infty }^{infty }{frac {log x^{2}}{cosh pi x}}\,dx={frac {pi }{2}}\,log left({sqrt {2}}\,\,{frac {Gamma left({frac {3}{4}} ight)}{Gamma left({frac {1}{4}} ight)}} ight)}. Also ist {displaystyle I={frac {pi }{2}}\,log left({sqrt {2pi }};{frac {Gamma left({frac {3}{4}} ight)}{Gamma left({frac {1}{4}} ight)}} ight)}.

2. Beweis

In der Formel {displaystyle int _{0}^{1}{frac {log log left({frac {1}{x}} ight)}{1+2cos alpha pi cdot x+x^{2}}}\,dx={frac {pi }{2sin alpha pi }}left(alpha log 2pi +log {frac {Gamma left({frac {1}{2}}+{frac {alpha }{2}} ight)}{Gamma left({frac {1}{2}}-{frac {alpha }{2}} ight)}} ight)}

setze {displaystyle alpha ={frac {1}{2}}\,:quad int _{0}^{1}{frac {log log left({frac {1}{x}} ight)}{1+x^{2}}}\,dx={frac {pi }{2}}left(log {sqrt {2pi }}+log {frac {Gamma left({frac {3}{4}} ight)}{Gamma left({frac {1}{4}} ight)}} ight)}

Durch die Substitution {displaystyle xmapsto cot x} ergibt sich die besagte Gleichung.