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

 

0.1Bearbeiten
{\displaystyle \int _{-\infty }^{\infty }{\frac {1}{1+x^{2}}}\,{\frac {1}{\cosh {\frac {\pi x}{2}}}}\,dx=2\log 2}{\displaystyle \int _{-\infty }^{\infty }{\frac {1}{1+x^{2}}}\,{\frac {1}{\cosh {\frac {\pi x}{2}}}}\,dx=2\log 2}
ohne Beweis

 

 
1.1Bearbeiten
{\displaystyle \int _{-\infty }^{\infty }{\frac {x^{n-1}}{\cosh x}}\,dx=|E_{n-1}|\;{\frac {\pi ^{n}}{2^{n-1}}}\qquad n\in \mathbb {Z} ^{\geq 1}}{\displaystyle \int _{-\infty }^{\infty }{\frac {x^{n-1}}{\cosh x}}\,dx=|E_{n-1}|\;{\frac {\pi ^{n}}{2^{n-1}}}\qquad n\in \mathbb {Z} ^{\geq 1}}
ohne Beweis

 

 
1.2Bearbeiten
{\displaystyle \int _{0}^{\infty }{\frac {x^{2n}}{\cosh 2\pi x+1}}\,dx=(-1)^{n}\,{\frac {B_{2n}\left({\frac {1}{2}}\right)}{2\pi }}\qquad n\in \mathbb {N} }{\displaystyle \int _{0}^{\infty }{\frac {x^{2n}}{\cosh 2\pi x+1}}\,dx=(-1)^{n}\,{\frac {B_{2n}\left({\frac {1}{2}}\right)}{2\pi }}\qquad n\in \mathbb {N} }
1. Beweis

Aus der Formel {\displaystyle \int _{0}^{\infty }{\frac {x^{2n}}{\cosh 2\pi x-\cos 2\pi \alpha }}\,dx=(-1)^{n+1}\,{\frac {\frac {B_{2n+1}(\alpha )}{2n+1}}{\sin 2\pi \alpha }}}{\displaystyle \int _{0}^{\infty }{\frac {x^{2n}}{\cosh 2\pi x-\cos 2\pi \alpha }}\,dx=(-1)^{n+1}\,{\frac {\frac {B_{2n+1}(\alpha )}{2n+1}}{\sin 2\pi \alpha }}} folgt

{\displaystyle \int _{0}^{\infty }{\frac {x^{2n}}{\cosh 2\pi x+1}}\,dx=(-1)^{n+1}\,\lim _{\alpha \to 1/2}{\frac {\frac {B_{2n+1}(\alpha )}{2n+1}}{\sin 2\pi \alpha }}{\stackrel {\text{L.H.}}{\,\,\,\,=\,\,\,\,}}(-1)^{n+1}\lim _{\alpha \to 1/2}{\frac {B_{2n}(\alpha )}{2\pi \cos 2\pi \alpha }}=(-1)^{n}\,{\frac {B_{2n}\left({\frac {1}{2}}\right)}{2\pi }}}{\displaystyle \int _{0}^{\infty }{\frac {x^{2n}}{\cosh 2\pi x+1}}\,dx=(-1)^{n+1}\,\lim _{\alpha \to 1/2}{\frac {\frac {B_{2n+1}(\alpha )}{2n+1}}{\sin 2\pi \alpha }}{\stackrel {\text{L.H.}}{\,\,\,\,=\,\,\,\,}}(-1)^{n+1}\lim _{\alpha \to 1/2}{\frac {B_{2n}(\alpha )}{2\pi \cos 2\pi \alpha }}=(-1)^{n}\,{\frac {B_{2n}\left({\frac {1}{2}}\right)}{2\pi }}}.

2. Beweis

Betrachte die Formel {\displaystyle \int _{-\infty }^{\infty }{\frac {\cosh ax}{\cosh 2\pi x+\cos 2\pi b}}\,dx={\frac {\sin ab}{\sin \left({\frac {a}{2}}\right)\,\sin 2\pi b}}}{\displaystyle \int _{-\infty }^{\infty }{\frac {\cosh ax}{\cosh 2\pi x+\cos 2\pi b}}\,dx={\frac {\sin ab}{\sin \left({\frac {a}{2}}\right)\,\sin 2\pi b}}} im Grenzfall {\displaystyle b\to 0}{\displaystyle b\to 0}:

{\displaystyle f(a):=\int _{-\infty }^{\infty }{\frac {\cosh ax}{\cosh 2\pi x+1}}\,dx={\frac {1}{\pi }}\,{\frac {\frac {a}{2}}{\sin {\frac {a}{2}}}}}{\displaystyle f(a):=\int _{-\infty }^{\infty }{\frac {\cosh ax}{\cosh 2\pi x+1}}\,dx={\frac {1}{\pi }}\,{\frac {\frac {a}{2}}{\sin {\frac {a}{2}}}}}

Auf der einen Seite ist nun {\displaystyle f^{(2n)}(a)=\int _{-\infty }^{\infty }{\frac {\cosh ax\cdot x^{2n}}{\cosh 2\pi x+1}}\,dx\,\,\Rightarrow \,\,f^{(2n)}(0)=\int _{-\infty }^{\infty }{\frac {x^{2n}}{\cosh 2\pi x+1}}\,dx}{\displaystyle f^{(2n)}(a)=\int _{-\infty }^{\infty }{\frac {\cosh ax\cdot x^{2n}}{\cosh 2\pi x+1}}\,dx\,\,\Rightarrow \,\,f^{(2n)}(0)=\int _{-\infty }^{\infty }{\frac {x^{2n}}{\cosh 2\pi x+1}}\,dx}.

Auf der anderen Seite hat die Taylorreihenentwicklung von {\displaystyle f\,}f\, folgende Form:

{\displaystyle f(a)={\frac {1}{\pi }}\sum _{n=0}^{\infty }(-1)^{n}\left({\frac {1}{2^{2n-1}}}-1\right)B_{2n}\,{\frac {a^{2n}}{(2n)!}}\,\,\Rightarrow \,\,f^{(2n)}(0)={\frac {1}{\pi }}\,(-1)^{n}\,\left({\frac {1}{2^{2n-1}}}-1\right)B_{2n}}{\displaystyle f(a)={\frac {1}{\pi }}\sum _{n=0}^{\infty }(-1)^{n}\left({\frac {1}{2^{2n-1}}}-1\right)B_{2n}\,{\frac {a^{2n}}{(2n)!}}\,\,\Rightarrow \,\,f^{(2n)}(0)={\frac {1}{\pi }}\,(-1)^{n}\,\left({\frac {1}{2^{2n-1}}}-1\right)B_{2n}}

Also ist {\displaystyle \int _{-\infty }^{\infty }{\frac {x^{2n}}{\cosh 2\pi x+1}}\,dx={\frac {1}{\pi }}\,(-1)^{n}\,\left({\frac {1}{2^{2n-1}}}-1\right)B_{2n}=(-1)^{n}\,{\frac {B_{2n}\left({\frac {1}{2}}\right)}{\pi }}}{\displaystyle \int _{-\infty }^{\infty }{\frac {x^{2n}}{\cosh 2\pi x+1}}\,dx={\frac {1}{\pi }}\,(-1)^{n}\,\left({\frac {1}{2^{2n-1}}}-1\right)B_{2n}=(-1)^{n}\,{\frac {B_{2n}\left({\frac {1}{2}}\right)}{\pi }}}.

 
1.3Bearbeiten
{\displaystyle \int _{0}^{\infty }{\frac {x^{\alpha -1}}{\cosh x}}\,dx=2\,\Gamma (\alpha )\,\beta (\alpha )\qquad {\text{Re}}(\alpha )>0}{\displaystyle \int _{0}^{\infty }{\frac {x^{\alpha -1}}{\cosh x}}\,dx=2\,\Gamma (\alpha )\,\beta (\alpha )\qquad {\text{Re}}(\alpha )>0}
Beweis

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

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

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

 
1.4Bearbeiten
{\displaystyle \int _{0}^{\infty }{\frac {x^{\alpha }}{\cosh ^{2}x}}\,dx={\frac {2\alpha }{2^{\alpha }}}\,\Gamma (\alpha )\,\eta (\alpha )\qquad {\text{Re}}(\alpha )>-1}{\displaystyle \int _{0}^{\infty }{\frac {x^{\alpha }}{\cosh ^{2}x}}\,dx={\frac {2\alpha }{2^{\alpha }}}\,\Gamma (\alpha )\,\eta (\alpha )\qquad {\text{Re}}(\alpha )>-1}
ohne Beweis

 

 
1.5Bearbeiten
{\displaystyle \int _{-\infty }^{\infty }{\frac {\cosh \alpha x}{\cosh x}}\,dx=\pi \,\sec \left({\frac {\alpha \pi }{2}}\right)\qquad -1<\mathrm {Re} (\alpha )<1}{\displaystyle \int _{-\infty }^{\infty }{\frac {\cosh \alpha x}{\cosh x}}\,dx=\pi \,\sec \left({\frac {\alpha \pi }{2}}\right)\qquad -1<\mathrm {Re} (\alpha )<1}
ohne Beweis

 

 
1.6Bearbeiten
{\displaystyle \int _{-\infty }^{\infty }{\frac {\cosh \alpha x}{\cosh \pi x}}\,{\frac {1}{1+x^{2}}}\,dx=4\cos \left({\frac {\alpha }{2}}\right)-\,\pi \cos \alpha -\,2\sin \alpha \,\,\log \tan \left({\frac {\pi }{4}}+{\frac {\alpha }{4}}\right)\qquad -\pi <\mathrm {Re} (\alpha )<\pi }{\displaystyle \int _{-\infty }^{\infty }{\frac {\cosh \alpha x}{\cosh \pi x}}\,{\frac {1}{1+x^{2}}}\,dx=4\cos \left({\frac {\alpha }{2}}\right)-\,\pi \cos \alpha -\,2\sin \alpha \,\,\log \tan \left({\frac {\pi }{4}}+{\frac {\alpha }{4}}\right)\qquad -\pi <\mathrm {Re} (\alpha )<\pi }
Beweis

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

{\displaystyle \Rightarrow y''(\alpha )+y(\alpha )=\int _{-\infty }^{\infty }{\frac {\cosh \alpha x}{\cosh \pi x}}\,dx=\sec \left({\frac {\alpha }{2}}\right)}{\displaystyle \Rightarrow y''(\alpha )+y(\alpha )=\int _{-\infty }^{\infty }{\frac {\cosh \alpha x}{\cosh \pi x}}\,dx=\sec \left({\frac {\alpha }{2}}\right)} mit {\displaystyle y(\pi )=\int _{-\infty }^{\infty }{\frac {dx}{1+x^{2}}}=\pi }{\displaystyle y(\pi )=\int _{-\infty }^{\infty }{\frac {dx}{1+x^{2}}}=\pi } und {\displaystyle y'(0)=0}{\displaystyle y'(0)=0}.


Ansatz (Variation der Konstante):

{\displaystyle y(x)=c(x)\sin x+d(x)\cos x\,}{\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)\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\,}{\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=\sec {\frac {x}{2}}}{\displaystyle y''(x)+y(x)=c'(x)\cos x-d'(x)\sin x=\sec {\frac {x}{2}}}.

{\displaystyle {\begin{pmatrix}\cos x&-\sin x\\\sin x&\cos x\end{pmatrix}}{\begin{pmatrix}c'(x)\\d'(x)\end{pmatrix}}={\begin{pmatrix}\sec {\frac {x}{2}}\\0\end{pmatrix}}}{\displaystyle {\begin{pmatrix}\cos x&-\sin x\\\sin x&\cos x\end{pmatrix}}{\begin{pmatrix}c'(x)\\d'(x)\end{pmatrix}}={\begin{pmatrix}\sec {\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}\sec {\frac {x}{2}}\\0\end{pmatrix}}={\begin{pmatrix}\cos x\,\,\sec {\frac {x}{2}}\\-\sin x\,\,\sec {\frac {x}{2}}\end{pmatrix}}={\begin{pmatrix}\left(2\cos ^{2}{\frac {x}{2}}-1\right)\sec {\frac {x}{2}}\\-2\sin {\frac {x}{2}}\cos {\frac {x}{2}}\,\sec {\frac {x}{2}}\end{pmatrix}}={\begin{pmatrix}2\cos {\frac {x}{2}}-\sec {\frac {x}{2}}\\-2\sin {\frac {x}{2}}\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}\sec {\frac {x}{2}}\\0\end{pmatrix}}={\begin{pmatrix}\cos x\,\,\sec {\frac {x}{2}}\\-\sin x\,\,\sec {\frac {x}{2}}\end{pmatrix}}={\begin{pmatrix}\left(2\cos ^{2}{\frac {x}{2}}-1\right)\sec {\frac {x}{2}}\\-2\sin {\frac {x}{2}}\cos {\frac {x}{2}}\,\sec {\frac {x}{2}}\end{pmatrix}}={\begin{pmatrix}2\cos {\frac {x}{2}}-\sec {\frac {x}{2}}\\-2\sin {\frac {x}{2}}\end{pmatrix}}}

{\displaystyle \Rightarrow {\begin{pmatrix}c(x)\\d(x)\end{pmatrix}}={\begin{pmatrix}4\sin {\frac {x}{2}}-2\log \tan \left({\frac {\pi }{4}}+{\frac {x}{4}}\right)\\4\cos {\frac {x}{2}}-\pi \end{pmatrix}}}{\displaystyle \Rightarrow {\begin{pmatrix}c(x)\\d(x)\end{pmatrix}}={\begin{pmatrix}4\sin {\frac {x}{2}}-2\log \tan \left({\frac {\pi }{4}}+{\frac {x}{4}}\right)\\4\cos {\frac {x}{2}}-\pi \end{pmatrix}}}, wegen {\displaystyle c(0)=y'(0)=0}{\displaystyle c(0)=y'(0)=0} und {\displaystyle -d(\pi )=y(\pi )=\pi }{\displaystyle -d(\pi )=y(\pi )=\pi }.

Somit ist {\displaystyle y(x)=\left[4\sin {\frac {x}{2}}-2\log \tan \left({\frac {\pi }{4}}+{\frac {x}{4}}\right)\right]\sin x+\left[4\cos {\frac {x}{2}}-\pi \right]\cos x}{\displaystyle y(x)=\left[4\sin {\frac {x}{2}}-2\log \tan \left({\frac {\pi }{4}}+{\frac {x}{4}}\right)\right]\sin x+\left[4\cos {\frac {x}{2}}-\pi \right]\cos x}

{\displaystyle =4\sin {\frac {x}{2}}\,\sin x-2\sin x\,\log \tan \left({\frac {\pi }{4}}+{\frac {x}{4}}\right)+4\cos {\frac {x}{2}}\,\cos x-\pi \,\cos x}{\displaystyle =4\sin {\frac {x}{2}}\,\sin x-2\sin x\,\log \tan \left({\frac {\pi }{4}}+{\frac {x}{4}}\right)+4\cos {\frac {x}{2}}\,\cos x-\pi \,\cos x}

{\displaystyle =4\cos \left({\frac {x}{2}}\right)-\,\pi \cos x-\,2\sin x\,\,\log \tan \left({\frac {\pi }{4}}+{\frac {x}{4}}\right)}{\displaystyle =4\cos \left({\frac {x}{2}}\right)-\,\pi \cos x-\,2\sin x\,\,\log \tan \left({\frac {\pi }{4}}+{\frac {x}{4}}\right)}.

 
2.1Bearbeiten
{\displaystyle \int _{-\infty }^{\infty }{\frac {1}{\cosh \alpha \pi x}}\,{\frac {\beta }{x^{2}+\beta ^{2}}}\,dx=\psi \left({\frac {\alpha \beta }{2}}+{\frac {3}{4}}\right)-\psi \left({\frac {\alpha \beta }{2}}+{\frac {1}{4}}\right)\qquad {\text{Re}}(\alpha )\,,\,{\text{Re}}(\beta )>0}{\displaystyle \int _{-\infty }^{\infty }{\frac {1}{\cosh \alpha \pi x}}\,{\frac {\beta }{x^{2}+\beta ^{2}}}\,dx=\psi \left({\frac {\alpha \beta }{2}}+{\frac {3}{4}}\right)-\psi \left({\frac {\alpha \beta }{2}}+{\frac {1}{4}}\right)\qquad {\text{Re}}(\alpha )\,,\,{\text{Re}}(\beta )>0}
ohne Beweis

 

 
2.2Bearbeiten
{\displaystyle \int _{-\infty }^{\infty }{\frac {\cosh \beta x}{(\cosh x)^{\alpha }}}dx=2^{\alpha -1}\,{\frac {\Gamma \left({\frac {\alpha +\beta }{2}}\right)\,\Gamma \left({\frac {\alpha -\beta }{2}}\right)}{\Gamma (\alpha )}}\qquad {\text{Re}}(\alpha )>{\text{Re}}(\beta )\geq 0}{\displaystyle \int _{-\infty }^{\infty }{\frac {\cosh \beta x}{(\cosh x)^{\alpha }}}dx=2^{\alpha -1}\,{\frac {\Gamma \left({\frac {\alpha +\beta }{2}}\right)\,\Gamma \left({\frac {\alpha -\beta }{2}}\right)}{\Gamma (\alpha )}}\qquad {\text{Re}}(\alpha )>{\text{Re}}(\beta )\geq 0}
Beweis

{\displaystyle I:=\int _{-\infty }^{\infty }{\frac {2\cosh \beta x}{(2\cosh x)^{\alpha }}}\,dx=\int _{-\infty }^{\infty }{\frac {e^{\beta x}+e^{-\beta x}}{(e^{x}+e^{-x})^{\alpha }}}\,dx=\int _{0}^{\infty }{\frac {x^{\beta }+x^{-\beta }}{(x+x^{-1})^{\alpha }}}\,{\frac {dx}{x}}}{\displaystyle I:=\int _{-\infty }^{\infty }{\frac {2\cosh \beta x}{(2\cosh x)^{\alpha }}}\,dx=\int _{-\infty }^{\infty }{\frac {e^{\beta x}+e^{-\beta x}}{(e^{x}+e^{-x})^{\alpha }}}\,dx=\int _{0}^{\infty }{\frac {x^{\beta }+x^{-\beta }}{(x+x^{-1})^{\alpha }}}\,{\frac {dx}{x}}}

ist nach Substitution {\displaystyle x\mapsto {\frac {\sin u}{\cos u}}}{\displaystyle x\mapsto {\frac {\sin u}{\cos u}}} gleich {\displaystyle \int _{0}^{\frac {\pi }{2}}{\frac {\left({\frac {\sin u}{\cos u}}\right)^{\beta }+\left({\frac {\cos u}{\sin u}}\right)^{\beta }}{\left({\frac {\sin u}{\cos u}}+{\frac {\cos u}{\sin u}}\right)^{\alpha }}}\,{\frac {du}{\sin u\cdot \cos u}}}{\displaystyle \int _{0}^{\frac {\pi }{2}}{\frac {\left({\frac {\sin u}{\cos u}}\right)^{\beta }+\left({\frac {\cos u}{\sin u}}\right)^{\beta }}{\left({\frac {\sin u}{\cos u}}+{\frac {\cos u}{\sin u}}\right)^{\alpha }}}\,{\frac {du}{\sin u\cdot \cos u}}}.

Dabei ist {\displaystyle {\frac {\sin u}{\cos u}}+{\frac {\cos u}{\sin u}}={\frac {\sin ^{2}u+\cos ^{2}u}{\sin u\cdot \cos u}}={\frac {1}{\sin u\cdot \cos u}}}{\displaystyle {\frac {\sin u}{\cos u}}+{\frac {\cos u}{\sin u}}={\frac {\sin ^{2}u+\cos ^{2}u}{\sin u\cdot \cos u}}={\frac {1}{\sin u\cdot \cos u}}}.

Also ist {\displaystyle I=\int _{0}^{\frac {\pi }{2}}\left(\sin u\cdot \cos u\right)^{\alpha -1}\left(\left({\frac {\sin u}{\cos u}}\right)^{\beta }+\left({\frac {\cos u}{\sin u}}\right)^{\beta }\right)du}{\displaystyle I=\int _{0}^{\frac {\pi }{2}}\left(\sin u\cdot \cos u\right)^{\alpha -1}\left(\left({\frac {\sin u}{\cos u}}\right)^{\beta }+\left({\frac {\cos u}{\sin u}}\right)^{\beta }\right)du}

{\displaystyle \int _{0}^{\frac {\pi }{2}}(\sin u)^{\alpha +\beta -1}\,(\cos u)^{\alpha -\beta -1}\,du+\int _{0}^{\frac {\pi }{2}}(\sin u)^{\alpha -\beta -1}\,(\cos u)^{\alpha +\beta -1}\,du}{\displaystyle \int _{0}^{\frac {\pi }{2}}(\sin u)^{\alpha +\beta -1}\,(\cos u)^{\alpha -\beta -1}\,du+\int _{0}^{\frac {\pi }{2}}(\sin u)^{\alpha -\beta -1}\,(\cos u)^{\alpha +\beta -1}\,du}

{\displaystyle =B\left({\frac {\alpha +\beta }{2}},{\frac {\alpha -\beta }{2}}\right)={\frac {\Gamma \left({\frac {\alpha +\beta }{2}}\right)\,\Gamma \left({\frac {\alpha -\beta }{2}}\right)}{\Gamma (\alpha )}}}{\displaystyle =B\left({\frac {\alpha +\beta }{2}},{\frac {\alpha -\beta }{2}}\right)={\frac {\Gamma \left({\frac {\alpha +\beta }{2}}\right)\,\Gamma \left({\frac {\alpha -\beta }{2}}\right)}{\Gamma (\alpha )}}}.

 
2.3Bearbeiten
{\displaystyle \int _{0}^{\infty }{\frac {x^{2n}\,\sin 2\pi \alpha }{\cosh 2\pi x-\cos 2\pi \alpha }}\,dx=(-1)^{n+1}\,{\frac {B_{2n+1}(\alpha )}{2n+1}}\qquad n\in \mathbb {N} \,,\,0<{\text{Re}}(\alpha )<1}{\displaystyle \int _{0}^{\infty }{\frac {x^{2n}\,\sin 2\pi \alpha }{\cosh 2\pi x-\cos 2\pi \alpha }}\,dx=(-1)^{n+1}\,{\frac {B_{2n+1}(\alpha )}{2n+1}}\qquad n\in \mathbb {N} \,,\,0<{\text{Re}}(\alpha )<1}
Beweis

Setze {\displaystyle F_{n}(x):={\frac {1}{i^{n}}}\,{\frac {\pi }{2}}\,\int _{-\infty }^{\infty }y^{n}\,{\frac {1}{\sin ^{2}\pi (x+iy)}}\,dy}{\displaystyle F_{n}(x):={\frac {1}{i^{n}}}\,{\frac {\pi }{2}}\,\int _{-\infty }^{\infty }y^{n}\,{\frac {1}{\sin ^{2}\pi (x+iy)}}\,dy}.

{\displaystyle F_{0}(x)={\frac {\pi }{2}}\int _{-\infty }^{\infty }{\frac {1}{\sin ^{2}\pi (x+iy)}}\,dy=\left[{\frac {i}{2}}\,\cot \pi (x+iy)\right]_{y=-\infty }^{y=\infty }={\frac {1}{2}}+{\frac {1}{2}}=1}{\displaystyle F_{0}(x)={\frac {\pi }{2}}\int _{-\infty }^{\infty }{\frac {1}{\sin ^{2}\pi (x+iy)}}\,dy=\left[{\frac {i}{2}}\,\cot \pi (x+iy)\right]_{y=-\infty }^{y=\infty }={\frac {1}{2}}+{\frac {1}{2}}=1}

{\displaystyle F_{n}(0)=-{\frac {1}{i^{n}}}\,{\frac {\pi }{2}}\,\int _{-\infty }^{\infty }{\frac {y^{n}}{\sinh ^{2}\pi y}}=B_{n}\qquad n\geq 2}{\displaystyle F_{n}(0)=-{\frac {1}{i^{n}}}\,{\frac {\pi }{2}}\,\int _{-\infty }^{\infty }{\frac {y^{n}}{\sinh ^{2}\pi y}}=B_{n}\qquad n\geq 2}

{\displaystyle F_{n}\left({\frac {1}{2}}\right)={\frac {1}{i^{n}}}\,{\frac {\pi }{2}}\,\int _{-\infty }^{\infty }{\frac {y^{n}}{\cosh ^{2}\pi y}}\,dy=\left({\frac {1}{2^{n-1}}}-1\right)B_{n}\qquad n\geq 0}{\displaystyle F_{n}\left({\frac {1}{2}}\right)={\frac {1}{i^{n}}}\,{\frac {\pi }{2}}\,\int _{-\infty }^{\infty }{\frac {y^{n}}{\cosh ^{2}\pi y}}\,dy=\left({\frac {1}{2^{n-1}}}-1\right)B_{n}\qquad n\geq 0}

{\displaystyle F_{n}'(x)=-{\frac {1}{i^{n}}}\,\pi ^{2}\,\int _{-\infty }^{\infty }y^{n}\,{\frac {\cos \pi (x+iy)}{\sin ^{3}\pi (x+iy)}}\,dy}{\displaystyle F_{n}'(x)=-{\frac {1}{i^{n}}}\,\pi ^{2}\,\int _{-\infty }^{\infty }y^{n}\,{\frac {\cos \pi (x+iy)}{\sin ^{3}\pi (x+iy)}}\,dy}

{\displaystyle =\underbrace {\left[-{\frac {1}{i^{n-1}}}\,{\frac {\pi }{2}}\,y^{n}\,{\frac {1}{\sin ^{2}\pi (x+iy)}}\right]_{y=-\infty }^{y=\infty }} _{=0}+{\frac {1}{i^{n-1}}}\,{\frac {\pi }{2}}\,\int _{-\infty }^{\infty }n\,y^{n-1}\,{\frac {1}{\sin ^{2}\pi (x+iy)}}\,dy=n\,F_{n-1}(x)}{\displaystyle =\underbrace {\left[-{\frac {1}{i^{n-1}}}\,{\frac {\pi }{2}}\,y^{n}\,{\frac {1}{\sin ^{2}\pi (x+iy)}}\right]_{y=-\infty }^{y=\infty }} _{=0}+{\frac {1}{i^{n-1}}}\,{\frac {\pi }{2}}\,\int _{-\infty }^{\infty }n\,y^{n-1}\,{\frac {1}{\sin ^{2}\pi (x+iy)}}\,dy=n\,F_{n-1}(x)}

Daher muss {\displaystyle F_{n}(x)}{\displaystyle F_{n}(x)} das Bernoulli-Polynom {\displaystyle B_{n}(x)}{\displaystyle B_{n}(x)} sein.

{\displaystyle B_{2n+1}(x)={\frac {1}{i^{2n+1}}}\,{\frac {\pi }{2}}\,\int _{-\infty }^{\infty }y^{2n+1}\,{\frac {1}{\sin ^{2}\pi (x+iy)}}\,dy}{\displaystyle B_{2n+1}(x)={\frac {1}{i^{2n+1}}}\,{\frac {\pi }{2}}\,\int _{-\infty }^{\infty }y^{2n+1}\,{\frac {1}{\sin ^{2}\pi (x+iy)}}\,dy}

{\displaystyle =\lim _{M\to \infty }\left(\left[{\frac {1}{i^{2n}}}\,{\frac {1}{2}}\,y^{2n+1}\,\cot \pi (x+iy)\right]_{y=-M}^{y=M}-{\frac {1}{i^{2n}}}\,{\frac {1}{2}}\,\int _{-M}^{M}(2n+1)\,y^{2n}\,\cot \pi (x+iy)\,dy\right)}{\displaystyle =\lim _{M\to \infty }\left(\left[{\frac {1}{i^{2n}}}\,{\frac {1}{2}}\,y^{2n+1}\,\cot \pi (x+iy)\right]_{y=-M}^{y=M}-{\frac {1}{i^{2n}}}\,{\frac {1}{2}}\,\int _{-M}^{M}(2n+1)\,y^{2n}\,\cot \pi (x+iy)\,dy\right)}

Wegen {\displaystyle \cot \pi (x+iy)+\cot \pi (x-iy)=2\,{\frac {\sin 2\pi x}{\cosh 2\pi y-\cos 2\pi x}}}{\displaystyle \cot \pi (x+iy)+\cot \pi (x-iy)=2\,{\frac {\sin 2\pi x}{\cosh 2\pi y-\cos 2\pi x}}} ist

{\displaystyle \lim _{M\to \infty }M^{2n+1}{\Big (}\cot \pi (x+iM)+\cot \pi (x-iM){\Big )}=0}{\displaystyle \lim _{M\to \infty }M^{2n+1}{\Big (}\cot \pi (x+iM)+\cot \pi (x-iM){\Big )}=0} und daher ist

{\displaystyle 2\,(-1)^{n+1}\,{\frac {B_{2n+1}(x)}{2n+1}}={\text{p.V.}}\int _{-\infty }^{\infty }y^{2n}\,\cot \pi (x+iy)\,dy={\text{p.V.}}\int _{-\infty }^{\infty }y^{2n}\,\cot \pi (x-iy)\,dy}{\displaystyle 2\,(-1)^{n+1}\,{\frac {B_{2n+1}(x)}{2n+1}}={\text{p.V.}}\int _{-\infty }^{\infty }y^{2n}\,\cot \pi (x+iy)\,dy={\text{p.V.}}\int _{-\infty }^{\infty }y^{2n}\,\cot \pi (x-iy)\,dy}.

Also ist {\displaystyle \int _{-\infty }^{\infty }y^{2n}\,{\frac {\sin 2\pi x}{\cosh 2\pi y-\cos 2\pi x}}\,dy}{\displaystyle \int _{-\infty }^{\infty }y^{2n}\,{\frac {\sin 2\pi x}{\cosh 2\pi y-\cos 2\pi x}}\,dy}

{\displaystyle ={\frac {1}{2}}\int _{-\infty }^{\infty }y^{2n}{\Big (}\cot \pi (x+iy)+\cot \pi (x-iy){\Big )}dy=2\,(-1)^{n+1}\,{\frac {B_{2n+1}(x)}{2n+1}}}{\displaystyle ={\frac {1}{2}}\int _{-\infty }^{\infty }y^{2n}{\Big (}\cot \pi (x+iy)+\cot \pi (x-iy){\Big )}dy=2\,(-1)^{n+1}\,{\frac {B_{2n+1}(x)}{2n+1}}}.

 
2.4Bearbeiten
{\displaystyle \int _{-\infty }^{\infty }{\frac {\cosh ax}{\cosh 2\pi x+\cos 2\pi b}}\,dx={\frac {\sin ab}{\sin \left({\frac {a}{2}}\right)\,\sin 2\pi b}}\qquad 0<\left|{\text{Re}}(a)\right|<2\pi \;\;,\;\;0<\left|{\text{Re}}(b)\right|<{\frac {1}{2}}}{\displaystyle \int _{-\infty }^{\infty }{\frac {\cosh ax}{\cosh 2\pi x+\cos 2\pi b}}\,dx={\frac {\sin ab}{\sin \left({\frac {a}{2}}\right)\,\sin 2\pi b}}\qquad 0<\left|{\text{Re}}(a)\right|<2\pi \;\;,\;\;0<\left|{\text{Re}}(b)\right|<{\frac {1}{2}}}
Beweis

Lobatschewskiintegral2.PNG
Setzt man {\displaystyle f(z)={\frac {\cosh az}{\cosh 2\pi z+\cos 2\pi b}}}{\displaystyle f(z)={\frac {\cosh az}{\cosh 2\pi z+\cos 2\pi b}}}, so ist {\displaystyle f(z+i)={\frac {\cosh az\,\cos a+i\sinh az\,\sin a}{\cosh 2\pi z+\cos 2\pi b}}}{\displaystyle f(z+i)={\frac {\cosh az\,\cos a+i\sinh az\,\sin a}{\cosh 2\pi z+\cos 2\pi b}}}.

Da der Imaginärteil eine ungerade Funktion ist, gilt {\displaystyle \int _{-\infty }^{\infty }f(x+i)\,dx=\cos a\,\int _{-\infty }^{\infty }f(x)\,dx}{\displaystyle \int _{-\infty }^{\infty }f(x+i)\,dx=\cos a\,\int _{-\infty }^{\infty }f(x)\,dx}.

Für {\displaystyle R\to \infty \,}R\to \infty \, verschwinden die Integrale über den vertikalen Strecken, daher ist

{\displaystyle \lim _{R\to \infty }\oint f\,dz=\int _{-\infty }^{\infty }f(x)\,dx-\int _{-\infty }^{\infty }f(x+i)\,dx=(1-\cos a)\int _{-\infty }^{\infty }f\,dx=2\,\sin ^{2}\left({\frac {a}{2}}\right)\,\int _{-\infty }^{\infty }f\,dx}{\displaystyle \lim _{R\to \infty }\oint f\,dz=\int _{-\infty }^{\infty }f(x)\,dx-\int _{-\infty }^{\infty }f(x+i)\,dx=(1-\cos a)\int _{-\infty }^{\infty }f\,dx=2\,\sin ^{2}\left({\frac {a}{2}}\right)\,\int _{-\infty }^{\infty }f\,dx}.

Auf der anderen Seite ist nach dem Residuensatz {\displaystyle \lim _{R\to \infty }\oint f\,dz=2\pi i\,{\text{res}}\left(f,i\left({\frac {1}{2}}-b\right)\right)+2\pi i\,{\text{res}}\left(f,i\left({\frac {1}{2}}+b\right)\right)}{\displaystyle \lim _{R\to \infty }\oint f\,dz=2\pi i\,{\text{res}}\left(f,i\left({\frac {1}{2}}-b\right)\right)+2\pi i\,{\text{res}}\left(f,i\left({\frac {1}{2}}+b\right)\right)}

{\displaystyle ={\frac {\cos \left(a\left({\frac {1}{2}}-b\right)\right)}{\sin 2\pi b}}-{\frac {\cos \left(a\left({\frac {1}{2}}+b\right)\right)}{\sin 2\pi b}}={\frac {2\,\sin \left({\frac {a}{2}}\right)\,\sin ab}{\sin 2\pi b}}}{\displaystyle ={\frac {\cos \left(a\left({\frac {1}{2}}-b\right)\right)}{\sin 2\pi b}}-{\frac {\cos \left(a\left({\frac {1}{2}}+b\right)\right)}{\sin 2\pi b}}={\frac {2\,\sin \left({\frac {a}{2}}\right)\,\sin ab}{\sin 2\pi b}}}.

Daraus folgt {\displaystyle \int _{-\infty }^{\infty }f(x)\,dx={\frac {\sin ab}{\sin \left({\frac {a}{2}}\right)\,\sin 2\pi b}}}{\displaystyle \int _{-\infty }^{\infty }f(x)\,dx={\frac {\sin ab}{\sin \left({\frac {a}{2}}\right)\,\sin 2\pi b}}}.

 
2.5Bearbeiten
{\displaystyle \int _{-\infty }^{\infty }{\frac {\cosh \pi x\,\cosh ax}{\cosh 2\pi x+\cos 2\pi b}}\,dx={\frac {\cos ab}{2\,\cos \left({\frac {a}{2}}\right)\,\cos \pi b}}\qquad 0<\left|{\text{Re}}(a)\right|<\pi \;\;,\;\;0<\left|{\text{Re}}(b)\right|<{\frac {1}{2}}}{\displaystyle \int _{-\infty }^{\infty }{\frac {\cosh \pi x\,\cosh ax}{\cosh 2\pi x+\cos 2\pi b}}\,dx={\frac {\cos ab}{2\,\cos \left({\frac {a}{2}}\right)\,\cos \pi b}}\qquad 0<\left|{\text{Re}}(a)\right|<\pi \;\;,\;\;0<\left|{\text{Re}}(b)\right|<{\frac {1}{2}}}
Beweis

Lobatschewskiintegral2.PNG
Setzt man {\displaystyle f(z)={\frac {\cosh \pi z\,\cosh az}{\cosh 2\pi z+\cos 2\pi b}}}{\displaystyle f(z)={\frac {\cosh \pi z\,\cosh az}{\cosh 2\pi z+\cos 2\pi b}}}, so ist {\displaystyle f(z+i)=-{\frac {\cosh \pi z\,(\cosh az\,\cos a+i\sinh az\,\sin a)}{\cosh 2\pi z+\cos 2\pi b}}}{\displaystyle f(z+i)=-{\frac {\cosh \pi z\,(\cosh az\,\cos a+i\sinh az\,\sin a)}{\cosh 2\pi z+\cos 2\pi b}}}.

Da der Imaginärteil eine ungerade Funktion ist, gilt {\displaystyle \int _{-\infty }^{\infty }f(x+i)\,dx=-\cos a\,\int _{-\infty }^{\infty }f(x)\,dx}{\displaystyle \int _{-\infty }^{\infty }f(x+i)\,dx=-\cos a\,\int _{-\infty }^{\infty }f(x)\,dx}.

Für {\displaystyle R\to \infty \,}R\to \infty \, verschwinden die Integrale über den vertikalen Strecken, daher ist

{\displaystyle \lim _{R\to \infty }\oint f\,dz=\int _{-\infty }^{\infty }f(x)\,dx-\int _{-\infty }^{\infty }f(x+i)\,dx=(1+\cos a)\int _{-\infty }^{\infty }f\,dx=2\,\cos ^{2}\left({\frac {a}{2}}\right)\,\int _{-\infty }^{\infty }f\,dx}{\displaystyle \lim _{R\to \infty }\oint f\,dz=\int _{-\infty }^{\infty }f(x)\,dx-\int _{-\infty }^{\infty }f(x+i)\,dx=(1+\cos a)\int _{-\infty }^{\infty }f\,dx=2\,\cos ^{2}\left({\frac {a}{2}}\right)\,\int _{-\infty }^{\infty }f\,dx}.

Auf der anderen Seite ist nach dem Residuensatz {\displaystyle \lim _{R\to \infty }\oint f\,dz=2\pi i\,{\text{res}}\left(f,i\left({\frac {1}{2}}-b\right)\right)+2\pi i\,{\text{res}}\left(f,i\left({\frac {1}{2}}+b\right)\right)}{\displaystyle \lim _{R\to \infty }\oint f\,dz=2\pi i\,{\text{res}}\left(f,i\left({\frac {1}{2}}-b\right)\right)+2\pi i\,{\text{res}}\left(f,i\left({\frac {1}{2}}+b\right)\right)}

{\displaystyle ={\frac {\cos \left(a\left({\frac {1}{2}}-b\right)\right)\sin b\pi }{\sin 2\pi b}}-{\frac {\cos \left(a\left({\frac {1}{2}}+b\right)\right)\sin b\pi }{\sin 2\pi b}}=2\cos \left({\frac {a}{2}}\right)\cos ab\,{\frac {\sin b\pi }{2\sin b\pi \,\cos b\pi }}={\frac {\cos \left({\frac {a}{2}}\right)\,\cos ab}{\cos \pi b}}}{\displaystyle ={\frac {\cos \left(a\left({\frac {1}{2}}-b\right)\right)\sin b\pi }{\sin 2\pi b}}-{\frac {\cos \left(a\left({\frac {1}{2}}+b\right)\right)\sin b\pi }{\sin 2\pi b}}=2\cos \left({\frac {a}{2}}\right)\cos ab\,{\frac {\sin b\pi }{2\sin b\pi \,\cos b\pi }}={\frac {\cos \left({\frac {a}{2}}\right)\,\cos ab}{\cos \pi b}}}.

Daraus folgt {\displaystyle \int _{-\infty }^{\infty }f(x)\,dx={\frac {\cos ab}{2\,\cos \left({\frac {a}{2}}\right)\,\cos \pi b}}}{\displaystyle \int _{-\infty }^{\infty }f(x)\,dx={\frac {\cos ab}{2\,\cos \left({\frac {a}{2}}\right)\,\cos \pi b}}}.

posted on 2021-05-05 02:26  Eufisky  阅读(39)  评论(0)  编辑  收藏  举报

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