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

1.1Bearbeiten

{displaystyle int _{-infty }^{infty }{frac {log(alpha ^{2}+x^{2})}{cosh pi x}}\,dx=4log left({sqrt {2}}\,\,{frac {Gamma left({frac {3}{4}}+{frac {alpha }{2}} ight)}{Gamma left({frac {1}{4}}+{frac {alpha }{2}} ight)}} ight)qquad { ext{Re}}(alpha )>0}

Beweis

Setzt man {displaystyle f(z)={frac {2alpha }{(alpha ^{2}+z^{2})cosh pi z}}}, so ist

{displaystyle 2pi i\,;{ ext{res}}(f,ialpha )={frac {2pi }{cos alpha pi }}=pi cot left[left({frac {1}{4}}+{frac {alpha }{2}} ight)pi ight]-pi cot left[left({frac {3}{4}}+{frac {alpha }{2}} ight)pi ight]}.

Und {displaystyle 2pi isum _{k=0}^{infty }{ ext{res}}left(f,i\,{frac {2k+1}{2}} ight)=sum _{k=0}^{infty }(-1)^{k}left[{frac {4}{2k+1+2alpha }}-{frac {4}{2k+1-2alpha }} ight]}

{displaystyle =left[psi left({frac {3}{4}}+{frac {alpha }{2}} ight)-psi left({frac {1}{4}}+{frac {alpha }{2}} ight) ight]-left[psi left({frac {3}{4}}-{frac {alpha }{2}} ight)-psi left({frac {1}{4}}-{frac {alpha }{2}} ight) ight]},

wobei {displaystyle psi left({frac {3}{4}}-{frac {alpha }{2}} ight)=psi left(1-left({frac {1}{4}}+{frac {alpha }{2}} ight) ight)=psi left({frac {1}{4}}+{frac {alpha }{2}} ight)+pi cot left[left({frac {1}{4}}+{frac {alpha }{2}} ight)pi ight]}

und {displaystyle psi left({frac {1}{4}}-{frac {alpha }{2}} ight)=psi left(1-left({frac {3}{4}}+{frac {alpha }{2}} ight) ight)=psi left({frac {3}{4}}+{frac {alpha }{2}} ight)+pi cot left[left({frac {3}{4}}+{frac {alpha }{2}} ight)pi ight]} ist.



{displaystyle int _{-infty }^{infty }f(x)\,dx=lim _{N o infty }oint _{gamma _{N}}f\,dz=2pi i\,sum _{{ ext{Im}}>0}{ ext{res}}f=2left[psi left({frac {3}{4}}+{frac {alpha }{2}} ight)-psi left({frac {1}{4}}+{frac {alpha }{2}} ight) ight]}.

Integriere nun beide Seiten nach {displaystyle alpha \,}:

{displaystyle underbrace {int _{-infty }^{infty }{frac {log(alpha ^{2}+x^{2})}{cosh pi x}}\,dx} _{=:U(alpha )}=underbrace {4left[log Gamma left({frac {3}{4}}+{frac {alpha }{2}} ight)-log Gamma left({frac {1}{4}}+{frac {alpha }{2}} ight) ight]} _{=:V(alpha )}+C}

Wegen {displaystyle U(alpha )-log(alpha ^{2})=U(alpha )-int _{-infty }^{infty }{frac {log(alpha ^{2})}{cosh pi x}}\,dx=int _{-infty }^{infty }{frac {log left(1+{frac {x^{2}}{alpha ^{2}}} ight)}{cosh pi x}}\,dx o 0} für {displaystyle alpha o infty \,}

und {displaystyle V(alpha )-log(alpha ^{2}) o -4log {sqrt {2}}} muss {displaystyle C=4log {sqrt {2}}} sein.

Daher lässt sich die rechte Seite auch schreiben als {displaystyle 4log left({sqrt {2}}\,;{frac {Gamma left({frac {3}{4}}+{frac {alpha }{2}} ight)}{Gamma left({frac {1}{4}}+{frac {alpha }{2}} ight)}} ight)}.

 

1.2Bearbeiten

{displaystyle int _{0}^{infty }{frac {log x}{cosh x+cos pi alpha }}\,dx={frac {pi }{sin pi alpha }}log left((2pi )^{alpha }\,\,{frac {Gamma left({frac {1+alpha }{2}} ight)}{Gamma left({frac {1-alpha }{2}} ight)}} ight)qquad 0<alpha <1}

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)qquad 0<alpha <1}

substituiere {displaystyle xmapsto e^{-x}}.