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

1周前 (05-04) 8次浏览

##### 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}}right)}{Gamma left({frac {1}{4}}+{frac {alpha }{2}}right)}}right)qquad {text{Re}}(alpha )>0}
Beweis

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

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

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

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

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

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

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

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}}right)-log Gamma left({frac {1}{4}}+{frac {alpha }{2}}right)right]} _{=: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}}}right)}{cosh pi x}},dxto 0} für {displaystyle alpha to infty ,}

und {displaystyle V(alpha )-log(alpha ^{2})to -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}}right)}{Gamma left({frac {1}{4}}+{frac {alpha }{2}}right)}}right)}.

##### 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}}right)}{Gamma left({frac {1-alpha }{2}}right)}}right)qquad 0<alpha <1}
Beweis

In der Formel

{displaystyle int _{0}^{1}{frac {log log left({frac {1}{x}}right)}{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}}right)}{Gamma left({frac {1}{2}}-{frac {alpha }{2}}right)}}right)qquad 0<alpha <1}

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