• 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=2log 2}{displaystyle int _{-infty }^{infty }{frac {1}{1+x^{2}}}\,{frac {1}{cosh {frac {pi x}{2}}}}\,dx=2log 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 nin 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 nin mathbb {Z} ^{geq 1}}
    ohne Beweis
     
    1.2Bearbeiten
    {displaystyle int _{0}^{infty }{frac {x^{2n}}{cosh 2pi x+1}}\,dx=(-1)^{n}\,{frac {B_{2n}left({frac {1}{2}} ight)}{2pi }}qquad nin mathbb {N} }{displaystyle int _{0}^{infty }{frac {x^{2n}}{cosh 2pi x+1}}\,dx=(-1)^{n}\,{frac {B_{2n}left({frac {1}{2}} ight)}{2pi }}qquad nin mathbb {N} }
    1. Beweis

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

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

    2. Beweis

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

    {displaystyle f(a):=int _{-infty }^{infty }{frac {cosh ax}{cosh 2pi x+1}}\,dx={frac {1}{pi }}\,{frac {frac {a}{2}}{sin {frac {a}{2}}}}}{displaystyle f(a):=int _{-infty }^{infty }{frac {cosh ax}{cosh 2pi 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 axcdot x^{2n}}{cosh 2pi x+1}}\,dx\,\,Rightarrow \,\,f^{(2n)}(0)=int _{-infty }^{infty }{frac {x^{2n}}{cosh 2pi x+1}}\,dx}{displaystyle f^{(2n)}(a)=int _{-infty }^{infty }{frac {cosh axcdot x^{2n}}{cosh 2pi x+1}}\,dx\,\,Rightarrow \,\,f^{(2n)}(0)=int _{-infty }^{infty }{frac {x^{2n}}{cosh 2pi 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 ight)B_{2n}\,{frac {a^{2n}}{(2n)!}}\,\,Rightarrow \,\,f^{(2n)}(0)={frac {1}{pi }}\,(-1)^{n}\,left({frac {1}{2^{2n-1}}}-1 ight)B_{2n}}{displaystyle f(a)={frac {1}{pi }}sum _{n=0}^{infty }(-1)^{n}left({frac {1}{2^{2n-1}}}-1 ight)B_{2n}\,{frac {a^{2n}}{(2n)!}}\,\,Rightarrow \,\,f^{(2n)}(0)={frac {1}{pi }}\,(-1)^{n}\,left({frac {1}{2^{2n-1}}}-1 ight)B_{2n}}

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

     
    1.3Bearbeiten
    {displaystyle int _{0}^{infty }{frac {x^{alpha -1}}{cosh x}}\,dx=2\,Gamma (alpha )\,eta (alpha )qquad { ext{Re}}(alpha )>0}{displaystyle int _{0}^{infty }{frac {x^{alpha -1}}{cosh x}}\,dx=2\,Gamma (alpha )\,eta (alpha )qquad { ext{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}=2sum _{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}=2sum _{k=0}^{infty }(-1)^{k}\,e^{-(2k+1)x}}.

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

     
    1.4Bearbeiten
    {displaystyle int _{0}^{infty }{frac {x^{alpha }}{cosh ^{2}x}}\,dx={frac {2alpha }{2^{alpha }}}\,Gamma (alpha )\,eta (alpha )qquad { ext{Re}}(alpha )>-1}{displaystyle int _{0}^{infty }{frac {x^{alpha }}{cosh ^{2}x}}\,dx={frac {2alpha }{2^{alpha }}}\,Gamma (alpha )\,eta (alpha )qquad { ext{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}} ight)qquad -1<mathrm {Re} (alpha )<1}{displaystyle int _{-infty }^{infty }{frac {cosh alpha x}{cosh x}}\,dx=pi \,sec left({frac {alpha pi }{2}} ight)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=4cos left({frac {alpha }{2}} ight)-\,pi cos alpha -\,2sin alpha \,\,log an left({frac {pi }{4}}+{frac {alpha }{4}} ight)qquad -pi <mathrm {Re} (alpha )<pi }{displaystyle int _{-infty }^{infty }{frac {cosh alpha x}{cosh pi x}}\,{frac {1}{1+x^{2}}}\,dx=4cos left({frac {alpha }{2}} ight)-\,pi cos alpha -\,2sin alpha \,\,log an left({frac {pi }{4}}+{frac {alpha }{4}} ight)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}} ight)}{displaystyle Rightarrow y''(alpha )+y(alpha )=int _{-infty }^{infty }{frac {cosh alpha x}{cosh pi x}}\,dx=sec left({frac {alpha }{2}} ight)} 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\,{ ext{(Forderung)}}}}{displaystyle y'(x)=c(x)cos x-d(x)sin x+underbrace {c'(x)sin x+d'(x)cos x} _{=0\,{ ext{(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 {egin{pmatrix}cos x&-sin x\sin x&cos xend{pmatrix}}{egin{pmatrix}c'(x)\d'(x)end{pmatrix}}={egin{pmatrix}sec {frac {x}{2}}\0end{pmatrix}}}{displaystyle {egin{pmatrix}cos x&-sin x\sin x&cos xend{pmatrix}}{egin{pmatrix}c'(x)\d'(x)end{pmatrix}}={egin{pmatrix}sec {frac {x}{2}}\0end{pmatrix}}}

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

    {displaystyle Rightarrow {egin{pmatrix}c(x)\d(x)end{pmatrix}}={egin{pmatrix}4sin {frac {x}{2}}-2log an left({frac {pi }{4}}+{frac {x}{4}} ight)\4cos {frac {x}{2}}-pi end{pmatrix}}}{displaystyle Rightarrow {egin{pmatrix}c(x)\d(x)end{pmatrix}}={egin{pmatrix}4sin {frac {x}{2}}-2log an left({frac {pi }{4}}+{frac {x}{4}} ight)\4cos {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[4sin {frac {x}{2}}-2log an left({frac {pi }{4}}+{frac {x}{4}} ight) ight]sin x+left[4cos {frac {x}{2}}-pi ight]cos x}{displaystyle y(x)=left[4sin {frac {x}{2}}-2log an left({frac {pi }{4}}+{frac {x}{4}} ight) ight]sin x+left[4cos {frac {x}{2}}-pi ight]cos x}

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

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

     
    2.1Bearbeiten
    {displaystyle int _{-infty }^{infty }{frac {1}{cosh alpha pi x}}\,{frac {eta }{x^{2}+eta ^{2}}}\,dx=psi left({frac {alpha eta }{2}}+{frac {3}{4}} ight)-psi left({frac {alpha eta }{2}}+{frac {1}{4}} ight)qquad { ext{Re}}(alpha )\,,\,{ ext{Re}}(eta )>0}{displaystyle int _{-infty }^{infty }{frac {1}{cosh alpha pi x}}\,{frac {eta }{x^{2}+eta ^{2}}}\,dx=psi left({frac {alpha eta }{2}}+{frac {3}{4}} ight)-psi left({frac {alpha eta }{2}}+{frac {1}{4}} ight)qquad { ext{Re}}(alpha )\,,\,{ ext{Re}}(eta )>0}
    ohne Beweis
     
    2.2Bearbeiten
    {displaystyle int _{-infty }^{infty }{frac {cosh eta x}{(cosh x)^{alpha }}}dx=2^{alpha -1}\,{frac {Gamma left({frac {alpha +eta }{2}} ight)\,Gamma left({frac {alpha -eta }{2}} ight)}{Gamma (alpha )}}qquad { ext{Re}}(alpha )>{ ext{Re}}(eta )geq 0}{displaystyle int _{-infty }^{infty }{frac {cosh eta x}{(cosh x)^{alpha }}}dx=2^{alpha -1}\,{frac {Gamma left({frac {alpha +eta }{2}} ight)\,Gamma left({frac {alpha -eta }{2}} ight)}{Gamma (alpha )}}qquad { ext{Re}}(alpha )>{ ext{Re}}(eta )geq 0}
    Beweis

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

    ist nach Substitution {displaystyle xmapsto {frac {sin u}{cos u}}}{displaystyle xmapsto {frac {sin u}{cos u}}} gleich {displaystyle int _{0}^{frac {pi }{2}}{frac {left({frac {sin u}{cos u}} ight)^{eta }+left({frac {cos u}{sin u}} ight)^{eta }}{left({frac {sin u}{cos u}}+{frac {cos u}{sin u}} ight)^{alpha }}}\,{frac {du}{sin ucdot cos u}}}{displaystyle int _{0}^{frac {pi }{2}}{frac {left({frac {sin u}{cos u}} ight)^{eta }+left({frac {cos u}{sin u}} ight)^{eta }}{left({frac {sin u}{cos u}}+{frac {cos u}{sin u}} ight)^{alpha }}}\,{frac {du}{sin ucdot cos u}}}.

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

    Also ist {displaystyle I=int _{0}^{frac {pi }{2}}left(sin ucdot cos u ight)^{alpha -1}left(left({frac {sin u}{cos u}} ight)^{eta }+left({frac {cos u}{sin u}} ight)^{eta } ight)du}{displaystyle I=int _{0}^{frac {pi }{2}}left(sin ucdot cos u ight)^{alpha -1}left(left({frac {sin u}{cos u}} ight)^{eta }+left({frac {cos u}{sin u}} ight)^{eta } ight)du}

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

    {displaystyle =Bleft({frac {alpha +eta }{2}},{frac {alpha -eta }{2}} ight)={frac {Gamma left({frac {alpha +eta }{2}} ight)\,Gamma left({frac {alpha -eta }{2}} ight)}{Gamma (alpha )}}}{displaystyle =Bleft({frac {alpha +eta }{2}},{frac {alpha -eta }{2}} ight)={frac {Gamma left({frac {alpha +eta }{2}} ight)\,Gamma left({frac {alpha -eta }{2}} ight)}{Gamma (alpha )}}}.

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

    {displaystyle F_{n}left({frac {1}{2}} ight)={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 ight)B_{n}qquad ngeq 0}{displaystyle F_{n}left({frac {1}{2}} ight)={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 ight)B_{n}qquad ngeq 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)}} ight]_{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)}} ight]_{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 o infty }left(left[{frac {1}{i^{2n}}}\,{frac {1}{2}}\,y^{2n+1}\,cot pi (x+iy) ight]_{y=-M}^{y=M}-{frac {1}{i^{2n}}}\,{frac {1}{2}}\,int _{-M}^{M}(2n+1)\,y^{2n}\,cot pi (x+iy)\,dy ight)}{displaystyle =lim _{M o infty }left(left[{frac {1}{i^{2n}}}\,{frac {1}{2}}\,y^{2n+1}\,cot pi (x+iy) ight]_{y=-M}^{y=M}-{frac {1}{i^{2n}}}\,{frac {1}{2}}\,int _{-M}^{M}(2n+1)\,y^{2n}\,cot pi (x+iy)\,dy ight)}

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

    {displaystyle lim _{M o infty }M^{2n+1}{Big (}cot pi (x+iM)+cot pi (x-iM){Big )}=0}{displaystyle lim _{M o 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}}={ ext{p.V.}}int _{-infty }^{infty }y^{2n}\,cot pi (x+iy)\,dy={ ext{p.V.}}int _{-infty }^{infty }y^{2n}\,cot pi (x-iy)\,dy}{displaystyle 2\,(-1)^{n+1}\,{frac {B_{2n+1}(x)}{2n+1}}={ ext{p.V.}}int _{-infty }^{infty }y^{2n}\,cot pi (x+iy)\,dy={ ext{p.V.}}int _{-infty }^{infty }y^{2n}\,cot pi (x-iy)\,dy}.

    Also ist {displaystyle int _{-infty }^{infty }y^{2n}\,{frac {sin 2pi x}{cosh 2pi y-cos 2pi x}}\,dy}{displaystyle int _{-infty }^{infty }y^{2n}\,{frac {sin 2pi x}{cosh 2pi y-cos 2pi 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 2pi x+cos 2pi b}}\,dx={frac {sin ab}{sin left({frac {a}{2}} ight)\,sin 2pi b}}qquad 0<left|{ ext{Re}}(a) ight|<2pi ;;,;;0<left|{ ext{Re}}(b) ight|<{frac {1}{2}}}{displaystyle int _{-infty }^{infty }{frac {cosh ax}{cosh 2pi x+cos 2pi b}}\,dx={frac {sin ab}{sin left({frac {a}{2}} ight)\,sin 2pi b}}qquad 0<left|{ ext{Re}}(a) ight|<2pi ;;,;;0<left|{ ext{Re}}(b) ight|<{frac {1}{2}}}
    Beweis

    Lobatschewskiintegral2.PNG
    Setzt man {displaystyle f(z)={frac {cosh az}{cosh 2pi z+cos 2pi b}}}{displaystyle f(z)={frac {cosh az}{cosh 2pi z+cos 2pi b}}}, so ist {displaystyle f(z+i)={frac {cosh az\,cos a+isinh az\,sin a}{cosh 2pi z+cos 2pi b}}}{displaystyle f(z+i)={frac {cosh az\,cos a+isinh az\,sin a}{cosh 2pi z+cos 2pi 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 o infty \,}R o infty \, verschwinden die Integrale über den vertikalen Strecken, daher ist

    {displaystyle lim _{R o 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}} ight)\,int _{-infty }^{infty }f\,dx}{displaystyle lim _{R o 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}} ight)\,int _{-infty }^{infty }f\,dx}.

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

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

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

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

    Lobatschewskiintegral2.PNG
    Setzt man {displaystyle f(z)={frac {cosh pi z\,cosh az}{cosh 2pi z+cos 2pi b}}}{displaystyle f(z)={frac {cosh pi z\,cosh az}{cosh 2pi z+cos 2pi b}}}, so ist {displaystyle f(z+i)=-{frac {cosh pi z\,(cosh az\,cos a+isinh az\,sin a)}{cosh 2pi z+cos 2pi b}}}{displaystyle f(z+i)=-{frac {cosh pi z\,(cosh az\,cos a+isinh az\,sin a)}{cosh 2pi z+cos 2pi 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 o infty \,}R o infty \, verschwinden die Integrale über den vertikalen Strecken, daher ist

    {displaystyle lim _{R o 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}} ight)\,int _{-infty }^{infty }f\,dx}{displaystyle lim _{R o 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}} ight)\,int _{-infty }^{infty }f\,dx}.

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

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

    Daraus folgt {displaystyle int _{-infty }^{infty }f(x)\,dx={frac {cos ab}{2\,cos left({frac {a}{2}} ight)\,cos pi b}}}{displaystyle int _{-infty }^{infty }f(x)\,dx={frac {cos ab}{2\,cos left({frac {a}{2}} ight)\,cos pi b}}}.
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  • 原文地址：https://www.cnblogs.com/Eufisky/p/14730799.html
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