0.1Bearbeiten

{\displaystyle \int _{-\infty }^{\infty }{\frac {1}{1+x^{2}}}\,{\frac {x}{\sinh \pi x}}\,dx=2\log 2-1}

ohne Beweis (Abels Integral)

 

 

1.1Bearbeiten

{\displaystyle \int _{-\infty }^{\infty }{\frac {x^{n-1}}{\sinh x}}\,dx={\frac {2^{n}(2^{n}-1)|B_{n}|}{n}}\;{\frac {\pi ^{n}}{2^{n-1}}}\qquad n\in \mathbb {Z} ^{\geq 2}}

ohne Beweis

 

 

1.2Bearbeiten

{\displaystyle \int _{0}^{\infty }{\frac {x^{\alpha -1}}{\sinh x}}\,dx=2\,\Gamma (\alpha )\,\lambda (\alpha )\qquad {\text{Re}}(\alpha )>1}

Beweis

Für {\displaystyle x>0\,} ist {\displaystyle {\frac {1}{\sinh x}}={\frac {2}{e^{x}-e^{-x}}}=2e^{-x}{\frac {1}{1-e^{-2x}}}=2e^{-x}\sum _{k=0}^{\infty }e^{-2kx}=2\sum _{k=0}^{\infty }e^{-(2k+1)x}}.

Also ist {\displaystyle {\frac {x^{\alpha -1}}{\sinh x}}=2\sum _{k=0}^{\infty }x^{\alpha -1}\,e^{-(2k+1)x}} und somit ist

{\displaystyle \int _{0}^{\infty }{\frac {x^{\alpha -1}}{\sinh x}}\,dx=2\sum _{k=0}^{\infty }\int _{0}^{\infty }x^{\alpha -1}\,e^{-(2k+1)x}\,dx=2\sum _{k=0}^{\infty }{\frac {\Gamma (\alpha )}{(2k+1)^{\alpha }}}=2\,\Gamma (\alpha )\,\lambda (\alpha )}.

 

1.3Bearbeiten

{\displaystyle \int _{0}^{\infty }{\frac {x^{\alpha -1}}{\sinh ^{2}x}}\,dx={\frac {\Gamma (\alpha )\,\zeta (\alpha -1)}{2^{\alpha -2}}}\qquad {\text{Re}}(\alpha )>2}

Beweis

Für {\displaystyle x>0\,} ist {\displaystyle {\frac {1}{\sinh ^{2}x}}={\frac {4}{(e^{x}-e^{-x})^{2}}}={\frac {4\,e^{-2x}}{(1-e^{-2x})^{2}}}=4\sum _{k=1}^{\infty }k\,e^{-2kx}}.

Also ist {\displaystyle \int _{0}^{\infty }{\frac {x^{\alpha -1}}{\sinh ^{2}x}}\,dx=4\sum _{k=1}^{\infty }k\int _{0}^{\infty }x^{\alpha -1}\,e^{-2kx}\,dx=4\sum _{k=1}^{\infty }k\,{\frac {\Gamma (\alpha )}{(2k)^{\alpha }}}={\frac {\Gamma (\alpha )\,\zeta (\alpha -1)}{2^{\alpha -2}}}}.

 

1.4Bearbeiten

{\displaystyle \int _{-\infty }^{\infty }{\frac {\sinh \alpha x}{\sinh x}}\,dx=\pi \,\tan \left({\frac {\alpha \pi }{2}}\right)\qquad -1<\mathrm {Re} (\alpha )<1}

Beweis

{\displaystyle \int _{-\infty }^{\infty }{\frac {\sinh \alpha x}{\sinh x}}\,dx}   {\displaystyle =2\int _{0}^{\infty }{\frac {e^{\alpha x}-e^{-\alpha x}}{e^{x}-e^{-x}}}\,dx=\psi \left({\frac {1+\alpha }{2}}\right)-\psi \left({\frac {1-\alpha }{2}}\right)}   {\displaystyle =\pi \,\tan \left({\frac {\alpha \pi }{2}}\right)}

 

1.5Bearbeiten

{\displaystyle \int _{-\infty }^{\infty }{\frac {\sinh \alpha x}{\sinh \pi x}}\,{\frac {1}{1+x^{2}}}\,dx=-\alpha \,\cos \alpha +2\sin \alpha \,\log \left(2\,\cos {\frac {\alpha }{2}}\right)\qquad -\pi <\mathrm {Re} (\alpha )<\pi }

Beweis

{\displaystyle y(\alpha ):=\int _{-\infty }^{\infty }{\frac {\sinh \alpha x}{\sinh \pi x}}\,{\frac {1}{1+x^{2}}}\,dx}

{\displaystyle \Rightarrow y''(\alpha )+y(\alpha )=\int _{-\infty }^{\infty }{\frac {\sinh \alpha x}{\sinh \pi x}}\,dx=\tan {\frac {\alpha }{2}}}

mit {\displaystyle y(0)=0\,} und {\displaystyle y'(0)=\int _{-\infty }^{\infty }{\frac {1}{\sinh \pi x}}\,{\frac {x}{1+x^{2}}}\,dx=-1+2\log 2}.


Ansatz (Variation der Konstante):

{\displaystyle y(x)=c(x)\sin x+d(x)\cos x\,}

{\displaystyle y'(x)=c(x)\cos x-d(x)\sin x+\underbrace {c'(x)\sin x+d'(x)\cos x} _{=0\,{\text{(Forderung)}}}}

{\displaystyle y''(x)=-c(x)\sin x-d(x)\cos x+c'(x)\cos x-d'(x)\sin x\,}

Also ist {\displaystyle y''(x)+y(x)=c'(x)\cos x-d'(x)\sin x=\tan {\frac {x}{2}}} und {\displaystyle c'(x)\sin x+d'(x)\cos x=0\,}.

{\displaystyle {\begin{pmatrix}\cos x&-\sin x\\\sin x&\cos x\end{pmatrix}}{\begin{pmatrix}c'(x)\\d'(x)\end{pmatrix}}={\begin{pmatrix}\tan {\frac {x}{2}}\\0\end{pmatrix}}}

{\displaystyle \Rightarrow {\begin{pmatrix}c'(x)\\d'(x)\end{pmatrix}}={\begin{pmatrix}\cos x&\sin x\\-\sin x&\cos x\end{pmatrix}}{\begin{pmatrix}\tan {\frac {x}{2}}\\0\end{pmatrix}}={\begin{pmatrix}\cos x\,\,\tan {\frac {x}{2}}\\-\sin x\,\,\tan {\frac {x}{2}}\end{pmatrix}}}

{\displaystyle \Rightarrow {\begin{pmatrix}c(x)\\d(x)\end{pmatrix}}={\begin{pmatrix}-\cos x+2\log \left(2\cos {\frac {x}{2}}\right)\\\sin x-x\end{pmatrix}}}, wegen {\displaystyle {\begin{pmatrix}c(0)\\d(0)\end{pmatrix}}={\begin{pmatrix}y'(0)\\y(0)\end{pmatrix}}={\begin{pmatrix}-1+2\log 2\\0\end{pmatrix}}}.

Somit ist {\displaystyle y(x)=\left[-\cos x+2\log \left(2\cos {\frac {x}{2}}\right)\right]\sin x+(\sin x-x)\cos x=-x\cos x+2\sin x\,\log \left(2\cos {\frac {x}{2}}\right)}.

 

2.1Bearbeiten

{\displaystyle \int _{-\infty }^{\infty }{\frac {x}{\sinh \alpha \pi x}}\,{\frac {1}{x^{2}+\beta ^{2}}}\,dx=-{\frac {1}{\alpha \beta }}+\psi \left({\frac {\alpha \beta }{2}}+{\frac {1}{2}}\right)-\psi \left({\frac {\alpha \beta }{2}}\right)\qquad {\text{Re}}(\alpha )\,,\,{\text{Re}}(\beta )>0}

ohne Beweis