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=4\log \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 {2\alpha }{(\alpha ^{2}+z^{2})\cosh \pi z}}}, so ist

{\displaystyle 2\pi i\,\;{\text{res}}(f,i\alpha )={\frac {2\pi }{\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 2\pi i\sum _{k=0}^{\infty }{\text{res}}\left(f,i\,{\frac {2k+1}{2}}\right)=\sum _{k=0}^{\infty }(-1)^{k}\left[{\frac {4}{2k+1+2\alpha }}-{\frac {4}{2k+1-2\alpha }}\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 _{N\to \infty }\oint _{\gamma _{N}}f\,dz=2\pi i\,\sum _{{\text{Im}}>0}{\text{res}}f=2\left[\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 {4\left[\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}}\,dx\to 0} für {\displaystyle \alpha \to \infty \,}

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

Daher lässt sich die rechte Seite auch schreiben als {\displaystyle 4\log \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((2\pi )^{\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+2\cos \alpha \pi \cdot x+x^{2}}}\,dx={\frac {\pi }{2\sin \alpha \pi }}\left(\alpha \log 2\pi +\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 x\mapsto e^{-x}}.