zoukankan      html  css  js  c++  java

  • Formelsammlung Mathematik: Bestimmte Integrale: Form R(x)

    1.1Bearbeiten
    {displaystyle int _{0}^{1}{frac {1-x^{z-1}}{1-x}}\,dx=gamma +psi (z)qquad { ext{Re}}(z)>0}{displaystyle int _{0}^{1}{frac {1-x^{z-1}}{1-x}}\,dx=gamma +psi (z)qquad { ext{Re}}(z)>0}
    Beweis (Formel nach Gauß)

    {displaystyle {frac {1-x^{z-1}}{1-x}}=sum _{k=1}^{infty }left(x^{k-1}-x^{k+z-2} ight)}{displaystyle {frac {1-x^{z-1}}{1-x}}=sum _{k=1}^{infty }left(x^{k-1}-x^{k+z-2} ight)}

    {displaystyle Rightarrow int _{0}^{1}{frac {1-x^{z-1}}{1-x}}\,dx=sum _{k=1}^{infty }int _{0}^{1}left(x^{k-1}-x^{k+z-2} ight)dx=-{frac {1}{z}}+sum _{k=1}^{infty }left({frac {1}{k}}-{frac {1}{k+z}} ight)=gamma +psi (z+1)-{frac {1}{z}}=gamma +psi (z)}{displaystyle Rightarrow int _{0}^{1}{frac {1-x^{z-1}}{1-x}}\,dx=sum _{k=1}^{infty }int _{0}^{1}left(x^{k-1}-x^{k+z-2} ight)dx=-{frac {1}{z}}+sum _{k=1}^{infty }left({frac {1}{k}}-{frac {1}{k+z}} ight)=gamma +psi (z+1)-{frac {1}{z}}=gamma +psi (z)}

     
    1.2Bearbeiten
    {displaystyle int _{0}^{1}{frac {1-x^{alpha -1}}{{sqrt {1-x^{2}}}^{\,3}}}dx=2^{alpha -2}{frac {Gamma ^{2}left({frac {alpha }{2}} ight)}{Gamma (alpha -1)}}}{displaystyle int _{0}^{1}{frac {1-x^{alpha -1}}{{sqrt {1-x^{2}}}^{\,3}}}dx=2^{alpha -2}{frac {Gamma ^{2}left({frac {alpha }{2}} ight)}{Gamma (alpha -1)}}}
    ohne Beweis
     
    1.3Bearbeiten
    {displaystyle int _{0}^{1}xcdot {sqrt {frac {1-alpha ^{2}x^{2}}{1-x^{2}}}}\,dx={frac {1}{2}}+{frac {1-alpha ^{2}}{2alpha }}\,{ ext{artanh}}\,alpha }{displaystyle int _{0}^{1}xcdot {sqrt {frac {1-alpha ^{2}x^{2}}{1-x^{2}}}}\,dx={frac {1}{2}}+{frac {1-alpha ^{2}}{2alpha }}\,{ ext{artanh}}\,alpha }
    Beweis

    {displaystyle I:=int _{0}^{1}xcdot {sqrt {frac {1-alpha ^{2}x^{2}}{1-x^{2}}}}\,dx}{displaystyle I:=int _{0}^{1}xcdot {sqrt {frac {1-alpha ^{2}x^{2}}{1-x^{2}}}}\,dx}

    Nach der Substitution {displaystyle x={sqrt {frac {1-u^{2}}{1-alpha ^{2}u^{2}}}}}{displaystyle x={sqrt {frac {1-u^{2}}{1-alpha ^{2}u^{2}}}}} bzw. {displaystyle u={sqrt {frac {1-x^{2}}{1-alpha ^{2}x^{2}}}}}{displaystyle u={sqrt {frac {1-x^{2}}{1-alpha ^{2}x^{2}}}}} ist

    {displaystyle {frac {dx}{du}}={frac {{sqrt {1-alpha ^{2}u^{2}}}cdot {frac {-u}{sqrt {1-u^{2}}}}-{sqrt {1-u^{2}}}cdot {frac {-alpha ^{2}u}{sqrt {1-alpha ^{2}u^{2}}}}}{1-alpha ^{2}u^{2}}}=-{frac {(1-alpha ^{2}u^{2})cdot u-(1-u^{2})cdot alpha ^{2}u}{{sqrt {1-u^{2}}}cdot {sqrt {1-alpha ^{2}u^{2}}}^{3}}}=-{frac {(1-alpha ^{2})cdot u}{{sqrt {1-u^{2}}}cdot {sqrt {1-alpha ^{2}u^{2}}}^{3}}}}{displaystyle {frac {dx}{du}}={frac {{sqrt {1-alpha ^{2}u^{2}}}cdot {frac {-u}{sqrt {1-u^{2}}}}-{sqrt {1-u^{2}}}cdot {frac {-alpha ^{2}u}{sqrt {1-alpha ^{2}u^{2}}}}}{1-alpha ^{2}u^{2}}}=-{frac {(1-alpha ^{2}u^{2})cdot u-(1-u^{2})cdot alpha ^{2}u}{{sqrt {1-u^{2}}}cdot {sqrt {1-alpha ^{2}u^{2}}}^{3}}}=-{frac {(1-alpha ^{2})cdot u}{{sqrt {1-u^{2}}}cdot {sqrt {1-alpha ^{2}u^{2}}}^{3}}}}.

    {displaystyle I=int _{0}^{1}{sqrt {frac {1-u^{2}}{1-alpha ^{2}u^{2}}}}cdot {frac {1}{u}}cdot {frac {(1-alpha ^{2})cdot u}{{sqrt {1-u^{2}}}cdot {sqrt {1-alpha ^{2}u^{2}}}^{3}}}cdot du=(1-alpha ^{2})cdot int _{0}^{1}{frac {du}{(1-alpha ^{2}u^{2})^{2}}}}{displaystyle I=int _{0}^{1}{sqrt {frac {1-u^{2}}{1-alpha ^{2}u^{2}}}}cdot {frac {1}{u}}cdot {frac {(1-alpha ^{2})cdot u}{{sqrt {1-u^{2}}}cdot {sqrt {1-alpha ^{2}u^{2}}}^{3}}}cdot du=(1-alpha ^{2})cdot int _{0}^{1}{frac {du}{(1-alpha ^{2}u^{2})^{2}}}}

    {displaystyle (1-alpha ^{2})cdot {frac {1}{2}}cdot left[{frac {u}{1-alpha ^{2}u^{2}}}+{frac {{ ext{artanh}}(alpha u)}{alpha }} ight]_{0}^{1}=(1-alpha ^{2})cdot {frac {1}{2}}cdot left({frac {1}{1-alpha ^{2}}}+{frac {{ ext{artanh}}\,alpha }{alpha }} ight)={frac {1}{2}}+{frac {1-alpha ^{2}}{2alpha }}\,{ ext{artanh}}\,alpha }{displaystyle (1-alpha ^{2})cdot {frac {1}{2}}cdot left[{frac {u}{1-alpha ^{2}u^{2}}}+{frac {{ ext{artanh}}(alpha u)}{alpha }} ight]_{0}^{1}=(1-alpha ^{2})cdot {frac {1}{2}}cdot left({frac {1}{1-alpha ^{2}}}+{frac {{ ext{artanh}}\,alpha }{alpha }} ight)={frac {1}{2}}+{frac {1-alpha ^{2}}{2alpha }}\,{ ext{artanh}}\,alpha }

     
    1.4Bearbeiten
    {displaystyle int _{0}^{infty }left{{frac {1}{x}} ight}x^{s-1}\,dx=-{frac {zeta (s)}{s}}qquad 0<{ ext{Re}}(s)<1}{displaystyle int _{0}^{infty }left{{frac {1}{x}} ight}x^{s-1}\,dx=-{frac {zeta (s)}{s}}qquad 0<{ ext{Re}}(s)<1}
    Beweis

    {displaystyle I_{N}:=-sint _{1/N}^{infty }left{{frac {1}{x}} ight}x^{s-1}\,dx}{displaystyle I_{N}:=-sint _{1/N}^{infty }left{{frac {1}{x}} ight}x^{s-1}\,dx} ist nach Substitution {displaystyle xmapsto {frac {1}{x}}}{displaystyle xmapsto {frac {1}{x}}} gleich {displaystyle -sint _{0}^{N}{x}\,{frac {1}{x^{s-1}}}\,{frac {dx}{x^{2}}}=int _{0}^{N}{x}\,{frac {d}{dx}}{frac {1}{x^{s}}}\,dx}{displaystyle -sint _{0}^{N}{x}\,{frac {1}{x^{s-1}}}\,{frac {dx}{x^{2}}}=int _{0}^{N}{x}\,{frac {d}{dx}}{frac {1}{x^{s}}}\,dx}.

    Dies ist nach Eulerscher Summenformel {displaystyle sum _{n=1}^{N}{frac {1}{n^{s}}}-int _{0}^{N}{frac {1}{x^{s}}}\,dx=sum _{n=1}^{N}{frac {1}{n^{s}}}-{frac {N^{1-s}}{1-s}}}{displaystyle sum _{n=1}^{N}{frac {1}{n^{s}}}-int _{0}^{N}{frac {1}{x^{s}}}\,dx=sum _{n=1}^{N}{frac {1}{n^{s}}}-{frac {N^{1-s}}{1-s}}}, woraus {displaystyle lim _{N o infty }I_{N}=zeta (s)\,}{displaystyle lim _{N o infty }I_{N}=zeta (s)\,} folgt.

     
    2.1Bearbeiten
    {displaystyle int _{0}^{infty }{frac {x^{m-1}}{1+x+...+x^{n-1}}}\,dx={frac {pi }{n}}left[cot left({frac {mpi }{n}} ight)-cot left({frac {(m+1)pi }{n}} ight) ight]qquad 0<m<n-1}{displaystyle int _{0}^{infty }{frac {x^{m-1}}{1+x+...+x^{n-1}}}\,dx={frac {pi }{n}}left[cot left({frac {mpi }{n}} ight)-cot left({frac {(m+1)pi }{n}} ight) ight]qquad 0<m<n-1}
    Beweis

    Integriere {displaystyle f(z)={frac {z^{m-1}}{1+...+z^{n-1}}}}{displaystyle f(z)={frac {z^{m-1}}{1+...+z^{n-1}}}} entlang dem Kreissektor, der durch den Ursprung, {displaystyle Rin mathbb {R} _{+}}{displaystyle Rin mathbb {R} _{+}} und {displaystyle R\,e^{ipi /n}}{displaystyle R\,e^{ipi /n}} als Eckpunkte beschränkt wird.

    Das Integral über dem Kreisbogen geht gegen null für {displaystyle R o infty \,}R o infty \,. Also ist {displaystyle I=int _{mathbb {R} _{+}}f\,dz=int _{e^{ipi /n}\,mathbb {R} _{+}}f\,dz=int _{0}^{infty }fleft(e^{ipi /n}z ight)\,e^{ipi /n}\,dz}{displaystyle I=int _{mathbb {R} _{+}}f\,dz=int _{e^{ipi /n}\,mathbb {R} _{+}}f\,dz=int _{0}^{infty }fleft(e^{ipi /n}z ight)\,e^{ipi /n}\,dz}.

    Nachdem sich {displaystyle f(z)\,}{displaystyle f(z)\,} auch als {displaystyle {frac {1-z}{1-z^{n}}}\,z^{m-1}}{displaystyle {frac {1-z}{1-z^{n}}}\,z^{m-1}} schreiben lässt, ist {displaystyle I=int _{0}^{infty }{frac {1-e^{ipi /n}z}{1+z^{n}}}\,e^{i(m-1)pi /n}\,z^{m-1}\,e^{ipi /n}\,dz}{displaystyle I=int _{0}^{infty }{frac {1-e^{ipi /n}z}{1+z^{n}}}\,e^{i(m-1)pi /n}\,z^{m-1}\,e^{ipi /n}\,dz}

    {displaystyle =e^{impi /n}\,int _{0}^{infty }{frac {z^{m-1}}{1+z^{n}}}\,dz+e^{i(m+1)pi /n}\,int _{0}^{infty }{frac {z^{m}}{1+z^{n}}}\,dz}{displaystyle =e^{impi /n}\,int _{0}^{infty }{frac {z^{m-1}}{1+z^{n}}}\,dz+e^{i(m+1)pi /n}\,int _{0}^{infty }{frac {z^{m}}{1+z^{n}}}\,dz}

    {displaystyle =left[cos left({frac {mpi }{n}} ight)+isin left({frac {mpi }{n}} ight) ight]\,{frac {pi }{n}}\,{frac {1}{sin left({frac {mpi }{n}} ight)}}-left[cos left({frac {(m+1)pi }{n}} ight)+isin left({frac {(m+1)pi }{n}} ight) ight]\,{frac {pi }{n}}\,{frac {1}{sin left({frac {(m+1)pi }{n}} ight)}}}{displaystyle =left[cos left({frac {mpi }{n}} ight)+isin left({frac {mpi }{n}} ight) ight]\,{frac {pi }{n}}\,{frac {1}{sin left({frac {mpi }{n}} ight)}}-left[cos left({frac {(m+1)pi }{n}} ight)+isin left({frac {(m+1)pi }{n}} ight) ight]\,{frac {pi }{n}}\,{frac {1}{sin left({frac {(m+1)pi }{n}} ight)}}}.

    Der Imaginärteil hebt sich auf und übrig bleibt {displaystyle I={frac {pi }{n}}left[cot left({frac {mpi }{n}} ight)-cot left({frac {(m+1)pi }{n}} ight) ight]}{displaystyle I={frac {pi }{n}}left[cot left({frac {mpi }{n}} ight)-cot left({frac {(m+1)pi }{n}} ight) ight]}.

     
    2.2Bearbeiten
    {displaystyle B(alpha ,eta )=int _{0}^{1}x^{alpha -1}(1-x)^{eta -1}\,dxqquad { ext{Re}}(alpha )\,,\,{ ext{Re}}(eta )>0}{displaystyle B(alpha ,eta )=int _{0}^{1}x^{alpha -1}(1-x)^{eta -1}\,dxqquad { ext{Re}}(alpha )\,,\,{ ext{Re}}(eta )>0}
    ohne Beweis (Definition der Betafunktion)
     
    2.3Bearbeiten
    {displaystyle int _{0}^{infty }{frac {x^{alpha -1}}{(1+x)^{alpha +eta }}}\,dx=B(alpha ,eta )qquad { ext{Re}}(alpha )\,,\,{ ext{Re}}(eta )>0}{displaystyle int _{0}^{infty }{frac {x^{alpha -1}}{(1+x)^{alpha +eta }}}\,dx=B(alpha ,eta )qquad { ext{Re}}(alpha )\,,\,{ ext{Re}}(eta )>0}
    Beweis

    {displaystyle B(alpha ,eta )=int _{0}^{1}x^{alpha -1}(1-x)^{eta -1}\,dx}{displaystyle B(alpha ,eta )=int _{0}^{1}x^{alpha -1}(1-x)^{eta -1}\,dx} ist nach der Substitution {displaystyle x o {frac {1}{1+x}}}{displaystyle x o {frac {1}{1+x}}} gleich {displaystyle int _{0}^{infty }{frac {x^{eta -1}}{(1+x)^{alpha +eta }}}\,dx}{displaystyle int _{0}^{infty }{frac {x^{eta -1}}{(1+x)^{alpha +eta }}}\,dx}.

    Und auf Grund der Symmetrie {displaystyle B(alpha ,eta )=B(eta ,alpha )\,}{displaystyle B(alpha ,eta )=B(eta ,alpha )\,} ist das das selbe wie {displaystyle int _{0}^{infty }{frac {x^{alpha -1}}{(1+x)^{alpha +eta }}}\,dx}{displaystyle int _{0}^{infty }{frac {x^{alpha -1}}{(1+x)^{alpha +eta }}}\,dx}.

     
    2.4Bearbeiten
    {displaystyle int _{0}^{infty }{frac {x^{alpha -1}}{1+x^{eta }}}\,dx={frac {pi }{eta }};csc left({frac {alpha pi }{eta }} ight)qquad 0<operatorname {Re} (alpha )<operatorname {Re} (eta )}{displaystyle int _{0}^{infty }{frac {x^{alpha -1}}{1+x^{eta }}}\,dx={frac {pi }{eta }};csc left({frac {alpha pi }{eta }} ight)qquad 0<operatorname {Re} (alpha )<operatorname {Re} (eta )}
    1. Beweis

    Es sei {displaystyle { ext{Log}}:mathbb {C} setminus {0} o mathbb {C} }{displaystyle { ext{Log}}:mathbb {C} setminus {0} o mathbb {C} } definiert durch {displaystyle re^{ivarphi }mapsto log r+ivarphi }{displaystyle re^{ivarphi }mapsto log r+ivarphi } für {displaystyle r>0\,}{displaystyle r>0\,} und {displaystyle 0leq varphi <2pi }{displaystyle 0leq varphi <2pi }.

    Für {displaystyle 0<m<n\,}{displaystyle 0<m<n\,} gilt die Partialbruchzerlegung {displaystyle {frac {x^{m-1}}{1+x^{n}}}={frac {1}{n}}\,sum _{k=0}^{n-1}{frac {left(e^{frac {impi }{n}} ight)^{2k+1}}{left(e^{frac {ipi }{n}} ight)^{2k+1}-x}}}{displaystyle {frac {x^{m-1}}{1+x^{n}}}={frac {1}{n}}\,sum _{k=0}^{n-1}{frac {left(e^{frac {impi }{n}} ight)^{2k+1}}{left(e^{frac {ipi }{n}} ight)^{2k+1}-x}}}.

    Also ist {displaystyle int _{0}^{R}{frac {x^{m-1}}{1+x^{n}}}\,dx={frac {1}{n}}sum _{k=0}^{n-1}left[-{ ext{Log}}left(e^{{frac {ipi }{n}}(2k+1)}-x ight) ight]_{0}^{R}\,left(e^{frac {impi }{n}} ight)^{2k+1}}{displaystyle int _{0}^{R}{frac {x^{m-1}}{1+x^{n}}}\,dx={frac {1}{n}}sum _{k=0}^{n-1}left[-{ ext{Log}}left(e^{{frac {ipi }{n}}(2k+1)}-x ight) ight]_{0}^{R}\,left(e^{frac {impi }{n}} ight)^{2k+1}}

    {displaystyle =underbrace {{frac {1}{n}}sum _{k=0}^{n-1}-{ ext{Log}}left(e^{{frac {ipi }{n}}(2k+1)}-R ight)\,left(e^{frac {impi }{n}} ight)^{2k+1}} _{ ext{1.Summe}}+underbrace {{frac {1}{n}}sum _{k=0}^{n-1}{frac {ipi }{n}}(2k+1)left(e^{frac {impi }{n}} ight)^{2k+1}} _{ ext{2.Summe}}}{displaystyle =underbrace {{frac {1}{n}}sum _{k=0}^{n-1}-{ ext{Log}}left(e^{{frac {ipi }{n}}(2k+1)}-R ight)\,left(e^{frac {impi }{n}} ight)^{2k+1}} _{ ext{1.Summe}}+underbrace {{frac {1}{n}}sum _{k=0}^{n-1}{frac {ipi }{n}}(2k+1)left(e^{frac {impi }{n}} ight)^{2k+1}} _{ ext{2.Summe}}}.

    Nun soll gezeigt werden, dass die 1.Summe für {displaystyle R o infty \,}R o infty \, gegen null konvergiert und die 2.Summe gleich {displaystyle {frac {pi }{n}}\,{frac {1}{sin {frac {mpi }{n}}}}}{displaystyle {frac {pi }{n}}\,{frac {1}{sin {frac {mpi }{n}}}}} ist.

    Für {displaystyle x eq pm 1\,}{displaystyle x eq pm 1\,} gilt {displaystyle sum _{k=0}^{n-1}x^{2k+1}={frac {x^{2n}-1}{x-{frac {1}{x}}}}}{displaystyle sum _{k=0}^{n-1}x^{2k+1}={frac {x^{2n}-1}{x-{frac {1}{x}}}}} und {displaystyle sum _{k=0}^{n-1}(2k+1)\,x^{2k+1}={frac {n\,x^{2n}}{{frac {1}{2}}left(x-{frac {1}{x}} ight)}}-(1-x^{2n})\,{frac {x+{frac {1}{x}}}{left(x-{frac {1}{x}} ight)^{2}}}}{displaystyle sum _{k=0}^{n-1}(2k+1)\,x^{2k+1}={frac {n\,x^{2n}}{{frac {1}{2}}left(x-{frac {1}{x}} ight)}}-(1-x^{2n})\,{frac {x+{frac {1}{x}}}{left(x-{frac {1}{x}} ight)^{2}}}}.

    Setzt man {displaystyle x=e^{frac {impi }{n}}}{displaystyle x=e^{frac {impi }{n}}}, so ist {displaystyle sum _{k=0}^{n-1}(2k+1)left(e^{frac {impi }{n}} ight)^{2k+1}={frac {n}{i\,sin {frac {mpi }{n}}}}}{displaystyle sum _{k=0}^{n-1}(2k+1)left(e^{frac {impi }{n}} ight)^{2k+1}={frac {n}{i\,sin {frac {mpi }{n}}}}}. Also ist die 2.Summe gleich {displaystyle {frac {pi }{n}}\,{frac {1}{sin {frac {mpi }{n}}}}}{displaystyle {frac {pi }{n}}\,{frac {1}{sin {frac {mpi }{n}}}}}.

    Und wegen {displaystyle sum _{k=0}^{n-1}left(e^{frac {impi }{n}} ight)^{2k+1}=0}{displaystyle sum _{k=0}^{n-1}left(e^{frac {impi }{n}} ight)^{2k+1}=0} lässt sich die 1.Summe schreiben als

    {displaystyle {frac {1}{n}}sum _{k=0}^{n-1}underbrace {left[{ ext{Log}}(-R)-{ ext{Log}}left(e^{{frac {ipi }{n}}(2k+1)}-R ight) ight]} _{ o 0\,{ ext{wenn}}\,R o infty }\,left(e^{frac {impi }{n}} ight)^{2k+1}}{displaystyle {frac {1}{n}}sum _{k=0}^{n-1}underbrace {left[{ ext{Log}}(-R)-{ ext{Log}}left(e^{{frac {ipi }{n}}(2k+1)}-R ight) ight]} _{ o 0\,{ ext{wenn}}\,R o infty }\,left(e^{frac {impi }{n}} ight)^{2k+1}}.

    Damit ist {displaystyle int _{0}^{infty }{frac {x^{m-1}}{1+x^{n}}}\,dx={frac {pi }{n}}\,{frac {1}{sin {frac {mpi }{n}}}}}{displaystyle int _{0}^{infty }{frac {x^{m-1}}{1+x^{n}}}\,dx={frac {pi }{n}}\,{frac {1}{sin {frac {mpi }{n}}}}} gezeigt. Substituiert man {displaystyle x\,}x\, durch {displaystyle x^{frac {eta }{n}}}{displaystyle x^{frac {eta }{n}}}, so ist {displaystyle int _{0}^{infty }{frac {x^{eta {frac {m}{n}}-1}}{1+x^{eta }}}\,dx={frac {pi }{eta }}\,{frac {1}{sin {frac {mpi }{n}}}}}{displaystyle int _{0}^{infty }{frac {x^{eta {frac {m}{n}}-1}}{1+x^{eta }}}\,dx={frac {pi }{eta }}\,{frac {1}{sin {frac {mpi }{n}}}}}.

    Für reelle {displaystyle alpha ,eta \,}{displaystyle alpha ,eta \,} folgt die Behauptung, wenn man eine Folge rationaler Zahlen {displaystyle {frac {m}{n}}}{displaystyle {frac {m}{n}}} konstruiert, die gegen {displaystyle {frac {alpha }{eta }}}frac{alpha}{eta} konvergiert.

    2. Beweis

    Spalte auf in {displaystyle int _{0}^{1}{frac {x^{alpha -1}}{1+x^{eta }}}\,dx+int _{1}^{infty }{frac {x^{alpha -1}}{1+x^{eta }}}\,dx}{displaystyle int _{0}^{1}{frac {x^{alpha -1}}{1+x^{eta }}}\,dx+int _{1}^{infty }{frac {x^{alpha -1}}{1+x^{eta }}}\,dx}. Das erste Integral ist {displaystyle {frac {1}{2eta }}left[psi left({frac {1}{2}}+{frac {alpha }{2eta }} ight)-psi left({frac {alpha }{2eta }} ight) ight]}{displaystyle {frac {1}{2eta }}left[psi left({frac {1}{2}}+{frac {alpha }{2eta }} ight)-psi left({frac {alpha }{2eta }} ight) ight]}.

    Und das zweite Integral ist nach Substitution {displaystyle xmapsto {frac {1}{x}}}{displaystyle xmapsto {frac {1}{x}}} gleich {displaystyle int _{0}^{1}{frac {x^{eta -alpha -1}}{1+x^{eta }}}\,dx}{displaystyle int _{0}^{1}{frac {x^{eta -alpha -1}}{1+x^{eta }}}\,dx}, und somit gleich {displaystyle {frac {1}{2eta }}left[underbrace {psi left({frac {1}{2}}+{frac {eta -alpha }{2eta }} ight)} _{psi left(1-{frac {alpha }{2eta }} ight)}-underbrace {psi left({frac {eta -alpha }{2eta }} ight)} _{psi left({frac {1}{2}}-{frac {alpha }{2eta }} ight)} ight]}{displaystyle {frac {1}{2eta }}left[underbrace {psi left({frac {1}{2}}+{frac {eta -alpha }{2eta }} ight)} _{psi left(1-{frac {alpha }{2eta }} ight)}-underbrace {psi left({frac {eta -alpha }{2eta }} ight)} _{psi left({frac {1}{2}}-{frac {alpha }{2eta }} ight)} ight]}.

    Also ist {displaystyle int _{0}^{infty }{frac {x^{alpha -1}}{1+x^{eta }}}\,dx={frac {1}{2eta }}left[underbrace {psi left(1-{frac {alpha }{2eta }} ight)-psi left({frac {alpha }{2eta }} ight)} _{pi \,cot {frac {alpha pi }{2eta }}}+underbrace {psi left({frac {1}{2}}+{frac {alpha }{2eta }} ight)-psi left({frac {1}{2}}-{frac {alpha }{2eta }} ight)} _{_{pi \, an {frac {alpha pi }{2eta }}}} ight]={frac {pi }{eta }}\,{frac {1}{sin {frac {alpha pi }{eta }}}}}{displaystyle int _{0}^{infty }{frac {x^{alpha -1}}{1+x^{eta }}}\,dx={frac {1}{2eta }}left[underbrace {psi left(1-{frac {alpha }{2eta }} ight)-psi left({frac {alpha }{2eta }} ight)} _{pi \,cot {frac {alpha pi }{2eta }}}+underbrace {psi left({frac {1}{2}}+{frac {alpha }{2eta }} ight)-psi left({frac {1}{2}}-{frac {alpha }{2eta }} ight)} _{_{pi \, an {frac {alpha pi }{2eta }}}} ight]={frac {pi }{eta }}\,{frac {1}{sin {frac {alpha pi }{eta }}}}}.


    Anders formuliert kann das erste Integral {displaystyle int _{0}^{1}{frac {x^{alpha -1}}{1+x^{eta }}}\,dx}{displaystyle int _{0}^{1}{frac {x^{alpha -1}}{1+x^{eta }}}\,dx} geschrieben werden als {displaystyle sum _{k=0}^{infty }{frac {(-1)^{k}}{alpha +eta k}}}{displaystyle sum _{k=0}^{infty }{frac {(-1)^{k}}{alpha +eta k}}}

    und das zweite Integral {displaystyle int _{0}^{1}{frac {x^{eta -alpha -1}}{1+x^{eta }}}\,dx}{displaystyle int _{0}^{1}{frac {x^{eta -alpha -1}}{1+x^{eta }}}\,dx} geschrieben werden als {displaystyle sum _{k=0}^{infty }{frac {(-1)^{k}}{eta -alpha +eta k}}=sum _{k=-infty }^{-1}{frac {(-1)^{k}}{alpha +eta k}}}{displaystyle sum _{k=0}^{infty }{frac {(-1)^{k}}{eta -alpha +eta k}}=sum _{k=-infty }^{-1}{frac {(-1)^{k}}{alpha +eta k}}}.

    Also ist {displaystyle int _{0}^{infty }{frac {x^{alpha -1}}{1+x^{eta }}}\,dx=sum _{kin mathbb {Z} }{frac {(-1)^{k}}{alpha +eta \,k}}}{displaystyle int _{0}^{infty }{frac {x^{alpha -1}}{1+x^{eta }}}\,dx=sum _{kin mathbb {Z} }{frac {(-1)^{k}}{alpha +eta \,k}}}, was gerade die Partialbruchentwicklung von {displaystyle {frac {pi }{eta }}\,csc left({frac {alpha pi }{eta }} ight)}{displaystyle {frac {pi }{eta }}\,csc left({frac {alpha pi }{eta }} ight)} ist.

    3. Beweis

    Für {displaystyle 0<{ ext{Re}}(alpha )<eta in mathbb {R} }{displaystyle 0<{ ext{Re}}(alpha )<eta in mathbb {R} } und ein {displaystyle nin mathbb {Z} ^{>2}}{displaystyle nin mathbb {Z} ^{>2}} definiere {displaystyle f(z)={frac {z^{n{frac {alpha }{eta }}-1}}{1+z^{n}}}}{displaystyle f(z)={frac {z^{n{frac {alpha }{eta }}-1}}{1+z^{n}}}}.

    Integriere {displaystyle f\,}f\, entlang dem Kreissektor {displaystyle gamma _{R}\,}{displaystyle gamma _{R}\,}, der durch den Ursprung, {displaystyle R>0\,}R>0\, und {displaystyle R\,e^{frac {2pi i}{n}}}{displaystyle R\,e^{frac {2pi i}{n}}} als Eckpunkte beschränkt wird.

    Das Integral über dem Kreisbogen geht gegen null für {displaystyle R o infty \,}R o infty \,.

    Also ist {displaystyle I:=lim _{R o infty }oint _{gamma _{R}}f\,dz=int _{mathbb {R} _{+}}f\,dz-int _{e^{frac {2pi i}{n}}mathbb {R} _{+}}f\,dz}{displaystyle I:=lim _{R o infty }oint _{gamma _{R}}f\,dz=int _{mathbb {R} _{+}}f\,dz-int _{e^{frac {2pi i}{n}}mathbb {R} _{+}}f\,dz}, wobei letztes Integral

    {displaystyle int _{mathbb {R} _{+}}fleft(e^{frac {2pi i}{n}}z ight)\,e^{frac {2pi i}{n}}\,dz=e^{2pi i{frac {alpha }{eta }}}\,int _{mathbb {R} _{+}}f\,dz}{displaystyle int _{mathbb {R} _{+}}fleft(e^{frac {2pi i}{n}}z ight)\,e^{frac {2pi i}{n}}\,dz=e^{2pi i{frac {alpha }{eta }}}\,int _{mathbb {R} _{+}}f\,dz} ist. Und somit ist {displaystyle I=left(1-e^{2pi i{frac {alpha }{eta }}} ight)int _{mathbb {R} _{+}}f\,dz}{displaystyle I=left(1-e^{2pi i{frac {alpha }{eta }}} ight)int _{mathbb {R} _{+}}f\,dz}.

    Nach dem Residuensatz ist {displaystyle I=2pi i\,{ ext{res}}left(f,z=e^{frac {ipi }{n}} ight)=-{frac {2pi i}{n}}e^{ipi {frac {alpha }{eta }}}}{displaystyle I=2pi i\,{ ext{res}}left(f,z=e^{frac {ipi }{n}} ight)=-{frac {2pi i}{n}}e^{ipi {frac {alpha }{eta }}}}.

    Daher muss gelten {displaystyle int _{mathbb {R} _{+}}f\,dz={frac {2pi i}{n}}\,{frac {e^{ipi {frac {alpha }{eta }}}}{e^{2pi i{frac {alpha }{eta }}}-1}}={frac {pi }{n}}\,{frac {1}{sin {frac {alpha pi }{eta }}}}}{displaystyle int _{mathbb {R} _{+}}f\,dz={frac {2pi i}{n}}\,{frac {e^{ipi {frac {alpha }{eta }}}}{e^{2pi i{frac {alpha }{eta }}}-1}}={frac {pi }{n}}\,{frac {1}{sin {frac {alpha pi }{eta }}}}}.

    Nach Substitution {displaystyle zmapsto z^{frac {eta }{n}}}{displaystyle zmapsto z^{frac {eta }{n}}} ergibt sich die Behauptung zumindest für reelles {displaystyle eta \,}{displaystyle eta \,}.

    4. Beweis

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

    Nach Substitution {displaystyle u=tcdot x^{eta }quad left[x=left({frac {u}{t}} ight)^{frac {1}{eta }} ight]}{displaystyle u=tcdot x^{eta }quad left[x=left({frac {u}{t}} ight)^{frac {1}{eta }} ight]} ist

    {displaystyle I=int _{0}^{infty }int _{0}^{infty }left({frac {u}{t}} ight)^{{frac {alpha }{eta }}-{frac {1}{eta }}}\,e^{-t}\,e^{-u}\,{frac {1}{eta }}\,left({frac {u}{t}} ight)^{{frac {1}{eta }}-1}\,{frac {1}{t}}\,du\,dt={frac {1}{eta }}int _{0}^{infty }int _{0}^{infty }left({frac {u}{t}} ight)^{{frac {alpha }{eta }}-1}\,e^{-u}\,e^{-t}\,{frac {1}{t}}\,du\,dt}{displaystyle I=int _{0}^{infty }int _{0}^{infty }left({frac {u}{t}} ight)^{{frac {alpha }{eta }}-{frac {1}{eta }}}\,e^{-t}\,e^{-u}\,{frac {1}{eta }}\,left({frac {u}{t}} ight)^{{frac {1}{eta }}-1}\,{frac {1}{t}}\,du\,dt={frac {1}{eta }}int _{0}^{infty }int _{0}^{infty }left({frac {u}{t}} ight)^{{frac {alpha }{eta }}-1}\,e^{-u}\,e^{-t}\,{frac {1}{t}}\,du\,dt}

    {displaystyle ={frac {1}{eta }}int _{0}^{infty }u^{{frac {alpha }{eta }}-1}\,e^{-u}\,du\,\,int _{0}^{infty }t^{-{frac {alpha }{eta }}}\,e^{-t}\,dt={frac {1}{eta }}\,Gamma left({frac {alpha }{eta }} ight)\,Gamma left(1-{frac {alpha }{eta }} ight)={frac {1}{eta }}\,{frac {pi }{sin left({frac {alpha pi }{eta }} ight)}}}{displaystyle ={frac {1}{eta }}int _{0}^{infty }u^{{frac {alpha }{eta }}-1}\,e^{-u}\,du\,\,int _{0}^{infty }t^{-{frac {alpha }{eta }}}\,e^{-t}\,dt={frac {1}{eta }}\,Gamma left({frac {alpha }{eta }} ight)\,Gamma left(1-{frac {alpha }{eta }} ight)={frac {1}{eta }}\,{frac {pi }{sin left({frac {alpha pi }{eta }} ight)}}}.

     
    2.5Bearbeiten
    {displaystyle int _{0}^{infty }{frac {x^{alpha -1}}{1-x^{eta }}}\,dx={frac {pi }{eta }};cot left({frac {alpha pi }{eta }} ight)qquad 0<{ ext{Re}}(alpha )<{ ext{Re}}(eta )}{displaystyle int _{0}^{infty }{frac {x^{alpha -1}}{1-x^{eta }}}\,dx={frac {pi }{eta }};cot left({frac {alpha pi }{eta }} ight)qquad 0<{ ext{Re}}(alpha )<{ ext{Re}}(eta )}
    ohne Beweis
     
    2.6Bearbeiten
    {displaystyle int _{0}^{1}{frac {x^{alpha -1}}{1+x^{eta }}}\,dx={frac {1}{2eta }}left[psi left({frac {1}{2}}+{frac {alpha }{2eta }} ight)-psi left({frac {alpha }{2eta }} ight) ight]}{displaystyle int _{0}^{1}{frac {x^{alpha -1}}{1+x^{eta }}}\,dx={frac {1}{2eta }}left[psi left({frac {1}{2}}+{frac {alpha }{2eta }} ight)-psi left({frac {alpha }{2eta }} ight) ight]}
    Beweis

    {displaystyle int _{0}^{1}{frac {x^{alpha -1}}{1+x^{eta }}}\,dx=int _{0}^{1}x^{alpha -1}\,sum _{k=0}^{infty }(-x^{eta })^{k}\,dx=sum _{k=0}^{infty }(-1)^{k}\,int _{0}^{1}x^{alpha -1+eta \,k}\,dx=sum _{k=0}^{infty }{frac {(-1)^{k}}{alpha +eta \,k}}}{displaystyle int _{0}^{1}{frac {x^{alpha -1}}{1+x^{eta }}}\,dx=int _{0}^{1}x^{alpha -1}\,sum _{k=0}^{infty }(-x^{eta })^{k}\,dx=sum _{k=0}^{infty }(-1)^{k}\,int _{0}^{1}x^{alpha -1+eta \,k}\,dx=sum _{k=0}^{infty }{frac {(-1)^{k}}{alpha +eta \,k}}}

    Und diese Reihe konvergiert gegen {displaystyle {frac {1}{2eta }}left[psi left({frac {1}{2}}+{frac {alpha }{2eta }} ight)-psi left({frac {alpha }{2eta }} ight) ight]}{displaystyle {frac {1}{2eta }}left[psi left({frac {1}{2}}+{frac {alpha }{2eta }} ight)-psi left({frac {alpha }{2eta }} ight) ight]}.

     
    2.7Bearbeiten
    {displaystyle int _{0}^{infty }{frac {x^{alpha }}{x^{2}+2cos heta cdot x+1}}\,dx={frac {pi }{sin alpha pi }}\,{frac {sin alpha heta }{sin heta }}qquad -1<{ ext{Re}}(alpha )<1\,,\,-pi <{ ext{Re}}( heta )<pi }{displaystyle int _{0}^{infty }{frac {x^{alpha }}{x^{2}+2cos heta cdot x+1}}\,dx={frac {pi }{sin alpha pi }}\,{frac {sin alpha heta }{sin heta }}qquad -1<{ ext{Re}}(alpha )<1\,,\,-pi <{ ext{Re}}( heta )<pi }
    Beweis

    Verwende die Formel {displaystyle int _{0}^{infty }f(x)\,x^{alpha }\,dx={frac {pi }{sin alpha pi }}\,sum { ext{res}}{Big (}f(-z)\,z^{alpha }{Big )}}{displaystyle int _{0}^{infty }f(x)\,x^{alpha }\,dx={frac {pi }{sin alpha pi }}\,sum { ext{res}}{Big (}f(-z)\,z^{alpha }{Big )}}.

    {displaystyle f(z)={frac {1}{z^{2}+2cos heta cdot z+1}}={frac {1}{2isin heta }}left({frac {1}{z+e^{-i heta }}}-{frac {1}{z+e^{i heta }}} ight)}{displaystyle f(z)={frac {1}{z^{2}+2cos heta cdot z+1}}={frac {1}{2isin heta }}left({frac {1}{z+e^{-i heta }}}-{frac {1}{z+e^{i heta }}} ight)}

    {displaystyle Rightarrow \,f(-z)\,z^{alpha }={frac {1}{2isin heta }}left({frac {z^{alpha }}{z-e^{i heta }}}-{frac {z^{alpha }}{z-e^{-i heta }}} ight)}{displaystyle Rightarrow \,f(-z)\,z^{alpha }={frac {1}{2isin heta }}left({frac {z^{alpha }}{z-e^{i heta }}}-{frac {z^{alpha }}{z-e^{-i heta }}} ight)}

    {displaystyle Rightarrow \,sum { ext{res}}{Big (}f(-z)\,z^{alpha }{Big )}={frac {1}{2isin heta }}\,{Big (}left(e^{i heta } ight)^{alpha }-left(e^{-i heta } ight)^{alpha }{Big )}={frac {sin alpha heta }{sin heta }}}{displaystyle Rightarrow \,sum { ext{res}}{Big (}f(-z)\,z^{alpha }{Big )}={frac {1}{2isin heta }}\,{Big (}left(e^{i heta } ight)^{alpha }-left(e^{-i heta } ight)^{alpha }{Big )}={frac {sin alpha heta }{sin heta }}}

     
    2.8Bearbeiten
    {displaystyle int _{0}^{infty }{frac {1;;;{ ext{oder}};;;x^{2}}{(x^{2}+2xcos alpha +1)(x^{2}+2xcos eta +1)}}\,dx={frac {1}{2}}\,{frac {alpha cot alpha -eta cot eta }{cos alpha -cos eta }}}{displaystyle int _{0}^{infty }{frac {1;;;{ ext{oder}};;;x^{2}}{(x^{2}+2xcos alpha +1)(x^{2}+2xcos eta +1)}}\,dx={frac {1}{2}}\,{frac {alpha cot alpha -eta cot eta }{cos alpha -cos eta }}}
    ohne Beweis
     
    2.9Bearbeiten
    {displaystyle int _{0}^{infty }{frac {x}{(x^{2}+2xcos alpha +1)(x^{2}+2xcos eta +1)}}\,dx=-{frac {1}{2}}\,{frac {alpha csc alpha -eta csc eta }{cos alpha -cos eta }}}{displaystyle int _{0}^{infty }{frac {x}{(x^{2}+2xcos alpha +1)(x^{2}+2xcos eta +1)}}\,dx=-{frac {1}{2}}\,{frac {alpha csc alpha -eta csc eta }{cos alpha -cos eta }}}
    Beweis

    {displaystyle {frac {x}{(x^{2}+2xcos alpha +1)(x^{2}+2xcos eta +1)}}}{displaystyle {frac {x}{(x^{2}+2xcos alpha +1)(x^{2}+2xcos eta +1)}}}

    {displaystyle ={frac {1}{2\,(cos eta -cos alpha )}}\,left({frac {1}{x^{2}+2xcos alpha +1}}-{frac {1}{x^{2}+2xcos eta +1}} ight)}{displaystyle ={frac {1}{2\,(cos eta -cos alpha )}}\,left({frac {1}{x^{2}+2xcos alpha +1}}-{frac {1}{x^{2}+2xcos eta +1}} ight)}

    {displaystyle ={frac {1}{2\,(cos eta -cos alpha )}}\,left({frac {1}{(x+cos alpha )^{2}+sin ^{2}alpha }}-{frac {1}{(x+cos eta )^{2}+sin ^{2}eta }} ight)}{displaystyle ={frac {1}{2\,(cos eta -cos alpha )}}\,left({frac {1}{(x+cos alpha )^{2}+sin ^{2}alpha }}-{frac {1}{(x+cos eta )^{2}+sin ^{2}eta }} ight)}

    Also ist

    {displaystyle int _{0}^{infty }{frac {x}{(x^{2}+2xcos alpha +1)(x^{2}+2xcos eta +1)}}\,dx}{displaystyle int _{0}^{infty }{frac {x}{(x^{2}+2xcos alpha +1)(x^{2}+2xcos eta +1)}}\,dx}

    {displaystyle ={frac {1}{2\,(cos eta -cos alpha )}}\,left(int _{0}^{infty }{frac {dx}{(x+cos alpha )^{2}+sin ^{2}alpha }}-int _{0}^{infty }{frac {dx}{(x+cos eta )^{2}+sin ^{2}eta }} ight)}{displaystyle ={frac {1}{2\,(cos eta -cos alpha )}}\,left(int _{0}^{infty }{frac {dx}{(x+cos alpha )^{2}+sin ^{2}alpha }}-int _{0}^{infty }{frac {dx}{(x+cos eta )^{2}+sin ^{2}eta }} ight)}

    {displaystyle ={frac {1}{2\,(cos eta -cos alpha )}}\,left(left[{frac {1}{sin alpha }}arctan left({frac {x+cos alpha }{sin alpha }} ight) ight]_{0}^{infty }-left[{frac {1}{sin eta }}arctan left({frac {x+cos eta }{sin eta }} ight) ight]_{0}^{infty } ight)}{displaystyle ={frac {1}{2\,(cos eta -cos alpha )}}\,left(left[{frac {1}{sin alpha }}arctan left({frac {x+cos alpha }{sin alpha }} ight) ight]_{0}^{infty }-left[{frac {1}{sin eta }}arctan left({frac {x+cos eta }{sin eta }} ight) ight]_{0}^{infty } ight)}

    {displaystyle ={frac {1}{2\,(cos eta -cos alpha )}}\,left({frac {alpha }{sin alpha }}-{frac {eta }{sin eta }} ight)}{displaystyle ={frac {1}{2\,(cos eta -cos alpha )}}\,left({frac {alpha }{sin alpha }}-{frac {eta }{sin eta }} ight)}.

     
    2.10Bearbeiten
    {displaystyle int _{-infty }^{infty }{frac {1}{1+{frac {x^{2}}{a^{2}}}}}\,prod _{k=1}^{infty }{frac {1+{frac {x^{2}}{(b+k)^{2}}}}{1+{frac {x^{2}}{(a+k)^{2}}}}}\,dx={sqrt {pi }}\,\,{frac {Gamma left(a+{frac {1}{2}} ight)cdot Gamma (b+1)cdot Gamma left(b-a+{frac {1}{2}} ight)}{Gamma (a)cdot Gamma left(b+{frac {1}{2}} ight)cdot Gamma left(b-a+1 ight)}}qquad 0<a<b+{frac {1}{2}}}{displaystyle int _{-infty }^{infty }{frac {1}{1+{frac {x^{2}}{a^{2}}}}}\,prod _{k=1}^{infty }{frac {1+{frac {x^{2}}{(b+k)^{2}}}}{1+{frac {x^{2}}{(a+k)^{2}}}}}\,dx={sqrt {pi }}\,\,{frac {Gamma left(a+{frac {1}{2}} ight)cdot Gamma (b+1)cdot Gamma left(b-a+{frac {1}{2}} ight)}{Gamma (a)cdot Gamma left(b+{frac {1}{2}} ight)cdot Gamma left(b-a+1 ight)}}qquad 0<a<b+{frac {1}{2}}}
    Beweis (Formel nach Ramanujan)

    Es sei {displaystyle f(z)=prod _{ell =1}^{n}left((b+ell )^{2}+z^{2} ight)\,,\,g(z)=prod _{ell =0}^{n}left((a+ell )^{2}+z^{2} ight)}{displaystyle f(z)=prod _{ell =1}^{n}left((b+ell )^{2}+z^{2} ight)\,,\,g(z)=prod _{ell =0}^{n}left((a+ell )^{2}+z^{2} ight)} und {displaystyle gamma _{R}\,}{displaystyle gamma _{R}\,} der Halbmond in der oberen komplexen Halbebene.

    {displaystyle f{Big (}i(a+k){Big )}=prod _{ell =1}^{n}left((b+ell )^{2}-(a+k)^{2} ight)=prod _{ell =1}^{n}{Big (}(b+a+k+ell )(b-a-k+ell ){Big )}={frac {(b+a+k+n)!}{(b+a+k)!}}\,{frac {(b-a-k+n)!}{(b-a-k)!}}}{displaystyle f{Big (}i(a+k){Big )}=prod _{ell =1}^{n}left((b+ell )^{2}-(a+k)^{2} ight)=prod _{ell =1}^{n}{Big (}(b+a+k+ell )(b-a-k+ell ){Big )}={frac {(b+a+k+n)!}{(b+a+k)!}}\,{frac {(b-a-k+n)!}{(b-a-k)!}}}

    {displaystyle g'{Big (}i(a+k){Big )}=prod _{ell =0}^{k-1}underbrace {left((a+ell )^{2}-(a+k)^{2} ight)} _{(ell -k)(2a+k+ell )}cdot 2i(a+k)cdot prod _{ell =k+1}^{n}underbrace {left((a+ell )^{2}-(a+k)^{2} ight)} _{(ell -k)(2a+k+ell )}}{displaystyle g'{Big (}i(a+k){Big )}=prod _{ell =0}^{k-1}underbrace {left((a+ell )^{2}-(a+k)^{2} ight)} _{(ell -k)(2a+k+ell )}cdot 2i(a+k)cdot prod _{ell =k+1}^{n}underbrace {left((a+ell )^{2}-(a+k)^{2} ight)} _{(ell -k)(2a+k+ell )}}

    {displaystyle =(-1)^{k}\,k!\,{frac {(2a+2k-1)!}{(2a+k-1)!}}cdot 2i(a+k)cdot (n-k)!\,{frac {(2a+k+n)!}{(2a+2k)!}}=i\,(-1)^{k}\,{frac {(2a+k+n)!}{(2a+k-1)!}}\,k!\,(n-k)!}{displaystyle =(-1)^{k}\,k!\,{frac {(2a+2k-1)!}{(2a+k-1)!}}cdot 2i(a+k)cdot (n-k)!\,{frac {(2a+k+n)!}{(2a+2k)!}}=i\,(-1)^{k}\,{frac {(2a+k+n)!}{(2a+k-1)!}}\,k!\,(n-k)!}

    {displaystyle int _{-infty }^{infty }{frac {1}{a^{2}+x^{2}}}\,{frac {(b+1)^{2}+x^{2}}{(a+1)^{2}+x^{2}}}cdots {frac {(b+n)^{2}+x^{2}}{(a+n)^{2}+x^{2}}}\,dx=lim _{R o infty }oint _{gamma _{R}}{frac {f(z)}{g(z)}}\,dz=2pi isum _{k=0}^{n}{ ext{res}}left({frac {f}{g}},i(a+k) ight)}{displaystyle int _{-infty }^{infty }{frac {1}{a^{2}+x^{2}}}\,{frac {(b+1)^{2}+x^{2}}{(a+1)^{2}+x^{2}}}cdots {frac {(b+n)^{2}+x^{2}}{(a+n)^{2}+x^{2}}}\,dx=lim _{R o infty }oint _{gamma _{R}}{frac {f(z)}{g(z)}}\,dz=2pi isum _{k=0}^{n}{ ext{res}}left({frac {f}{g}},i(a+k) ight)}

    {displaystyle =2pi i\,sum _{k=0}^{n}{frac {f{Big (}i(a+k){Big )}}{g'{Big (}i(a+k){Big )}}}=2pi sum _{k=0}^{n}(-1)^{k}\,{frac {(2a+k-1)!}{(2a+k+n)!}}\,{frac {1}{k!\,(n-k)!}}\,{frac {(b+a+k+n)!\,(b-a-k+n)!}{(b+a+k)!\,(b-a-k)!}}}{displaystyle =2pi i\,sum _{k=0}^{n}{frac {f{Big (}i(a+k){Big )}}{g'{Big (}i(a+k){Big )}}}=2pi sum _{k=0}^{n}(-1)^{k}\,{frac {(2a+k-1)!}{(2a+k+n)!}}\,{frac {1}{k!\,(n-k)!}}\,{frac {(b+a+k+n)!\,(b-a-k+n)!}{(b+a+k)!\,(b-a-k)!}}}

    Wenn man beide Seiten mit {displaystyle a^{2}\,(a+1)^{2}cdots (a+n)^{2}={frac {(a+n)!^{2}}{(a-1)!^{2}}}}{displaystyle a^{2}\,(a+1)^{2}cdots (a+n)^{2}={frac {(a+n)!^{2}}{(a-1)!^{2}}}} durchmultipliziert

    und mit {displaystyle (b+1)^{2}\,(b+2)^{2}cdots (b+n)^{2}={frac {(b+n)!^{2}}{b!^{2}}}}{displaystyle (b+1)^{2}\,(b+2)^{2}cdots (b+n)^{2}={frac {(b+n)!^{2}}{b!^{2}}}} durchdividiert, so ist {displaystyle I_{n}:=int _{-infty }^{infty }{frac {1}{1+{frac {x^{2}}{a^{2}}}}}cdot {frac {1+{frac {x^{2}}{(b+1)^{2}}}}{1+{frac {x^{2}}{(a+1)^{2}}}}}cdots {frac {1+{frac {x^{2}}{(b+n)^{2}}}}{1+{frac {x^{2}}{(a+n)^{2}}}}}\,dx}{displaystyle I_{n}:=int _{-infty }^{infty }{frac {1}{1+{frac {x^{2}}{a^{2}}}}}cdot {frac {1+{frac {x^{2}}{(b+1)^{2}}}}{1+{frac {x^{2}}{(a+1)^{2}}}}}cdots {frac {1+{frac {x^{2}}{(b+n)^{2}}}}{1+{frac {x^{2}}{(a+n)^{2}}}}}\,dx}

    {displaystyle =2pi \,{frac {b!^{2}}{(a-1)!^{2}}}sum _{k=0}^{n}underbrace {(-1)^{k}\,{frac {(2a+k-1)!}{k!}}} _{={frac {(-2a)!\,(2a-1)!}{k!\,(-2a-k)!}}}\,{frac {1}{(b+a+k)!\,(b-a-k)!}}\,underbrace {{frac {(b+a+k+n)!\,(b-a-k+n)!}{(2a+k+n)!\,(n-k)!}}\,{frac {(a+n)!^{2}}{(b+n)!^{2}}}} _{A_{n,k}}}{displaystyle =2pi \,{frac {b!^{2}}{(a-1)!^{2}}}sum _{k=0}^{n}underbrace {(-1)^{k}\,{frac {(2a+k-1)!}{k!}}} _{={frac {(-2a)!\,(2a-1)!}{k!\,(-2a-k)!}}}\,{frac {1}{(b+a+k)!\,(b-a-k)!}}\,underbrace {{frac {(b+a+k+n)!\,(b-a-k+n)!}{(2a+k+n)!\,(n-k)!}}\,{frac {(a+n)!^{2}}{(b+n)!^{2}}}} _{A_{n,k}}}.

    Mit einem {displaystyle 0leq k<M\,}{displaystyle 0leq k<M\,} lässt sich letzte Summe folgendermaßen aufspalten:

    {displaystyle 2pi \,{frac {b!^{2}\,(-2a)!\,(2a-1)!}{(a-1)!^{2}}}\,left(sum _{k=0}^{M-1}{frac {A_{n,k}}{k!\,(b+a+k)!\,(-2a-k)!\,(b-a-k)!}}+sum _{k=M}^{n}{frac {A_{n,k}}{k!\,(b+a+k)!\,(-2a-k)!\,(b-a-k)!}} ight)}{displaystyle 2pi \,{frac {b!^{2}\,(-2a)!\,(2a-1)!}{(a-1)!^{2}}}\,left(sum _{k=0}^{M-1}{frac {A_{n,k}}{k!\,(b+a+k)!\,(-2a-k)!\,(b-a-k)!}}+sum _{k=M}^{n}{frac {A_{n,k}}{k!\,(b+a+k)!\,(-2a-k)!\,(b-a-k)!}} ight)}

    Für alle {displaystyle 0leq k<M}{displaystyle 0leq k<M} gilt {displaystyle lim _{n o infty }A_{n,k}=1\,}{displaystyle lim _{n o infty }A_{n,k}=1\,}.

    Also ist {displaystyle I:=lim _{n o infty }I_{n}=2pi \,{frac {b!^{2}\,(-2a)!\,(2a-1)!}{(a-1)!^{2}}}\,left(sum _{k=0}^{M-1}{frac {1}{k!\,(b+a+k)!\,(-2a-k)!\,(b-a-k)!}}+varepsilon _{M} ight)}{displaystyle I:=lim _{n o infty }I_{n}=2pi \,{frac {b!^{2}\,(-2a)!\,(2a-1)!}{(a-1)!^{2}}}\,left(sum _{k=0}^{M-1}{frac {1}{k!\,(b+a+k)!\,(-2a-k)!\,(b-a-k)!}}+varepsilon _{M} ight)}.

    Für {displaystyle M o infty \,}{displaystyle M o infty \,} geht {displaystyle varepsilon _{M}\,}{displaystyle varepsilon _{M}\,} gegen null und die hypergeometrische Reihe {displaystyle sum _{k=0}^{infty }{frac {1}{k!\,(b+a+k)!\,(-2a-k)!\,(b-a-k)!}}}{displaystyle sum _{k=0}^{infty }{frac {1}{k!\,(b+a+k)!\,(-2a-k)!\,(b-a-k)!}}} lässt sich

    nach der Formel {displaystyle sum _{k=0}^{infty }{frac {1}{k!\,(alpha +k)!\,(eta -k)!\,(gamma -k)!}}={frac {(alpha +eta +gamma )!}{eta !\,gamma !\,(alpha +eta )!\,(alpha +gamma )!}}}{displaystyle sum _{k=0}^{infty }{frac {1}{k!\,(alpha +k)!\,(eta -k)!\,(gamma -k)!}}={frac {(alpha +eta +gamma )!}{eta !\,gamma !\,(alpha +eta )!\,(alpha +gamma )!}}} für {displaystyle alpha +eta +gamma >-1\,}{displaystyle alpha +eta +gamma >-1\,},

    schreiben als {displaystyle {frac {(2b-2a)!}{(-2a)!\,(b-a)!\,(b-a)!\,(2b)!}}}{displaystyle {frac {(2b-2a)!}{(-2a)!\,(b-a)!\,(b-a)!\,(2b)!}}} wenn {displaystyle 2b-2a>-1\,}{displaystyle 2b-2a>-1\,} bzw. {displaystyle a<b+{frac {1}{2}}}{displaystyle a<b+{frac {1}{2}}} ist. Also ist {displaystyle I=2pi \,{frac {b!^{2}\,(2a-1)!\,(2b-2a)!}{(a-1)!^{2}\,(b-a)!^{2}\,(2b)!}}}{displaystyle I=2pi \,{frac {b!^{2}\,(2a-1)!\,(2b-2a)!}{(a-1)!^{2}\,(b-a)!^{2}\,(2b)!}}}.

    Und das lässt sich unter Verwendung der Legendreschen Verdopplungsformel schreiben als {displaystyle {sqrt {pi }}\,{frac {Gamma left(a+{frac {1}{2}} ight)cdot Gamma (b+1)cdot Gamma left(b-a+{frac {1}{2}} ight)}{Gamma (a)cdot Gamma left(b+{frac {1}{2}} ight)cdot Gamma left(b-a+1 ight)}}}{displaystyle {sqrt {pi }}\,{frac {Gamma left(a+{frac {1}{2}} ight)cdot Gamma (b+1)cdot Gamma left(b-a+{frac {1}{2}} ight)}{Gamma (a)cdot Gamma left(b+{frac {1}{2}} ight)cdot Gamma left(b-a+1 ight)}}}.

     
    2.11Bearbeiten
    {displaystyle int _{0}^{infty }{frac {x^{2m-1}}{prod limits _{k=1}^{n}(x^{2}+k^{2})}}\,dx=2cdot (-1)^{m-1}sum _{k=1}^{n}{frac {(-1)^{k}\,k^{2m}\,log k}{(n+k)!\,(n-k)!}}qquad m<n}{displaystyle int _{0}^{infty }{frac {x^{2m-1}}{prod limits _{k=1}^{n}(x^{2}+k^{2})}}\,dx=2cdot (-1)^{m-1}sum _{k=1}^{n}{frac {(-1)^{k}\,k^{2m}\,log k}{(n+k)!\,(n-k)!}}qquad m<n}
    Beweis

    Aus der Partialbruchzerlegung {displaystyle {frac {x^{2m-1}}{prod limits _{k=1}^{n}(x^{2}+k^{2})}}=sum _{k=1}^{n}{frac {(-1)^{m+k}\,k^{2m}}{(n+k)!\,(n-k)!}}cdot {frac {2x}{x^{2}+k^{2}}}}{displaystyle {frac {x^{2m-1}}{prod limits _{k=1}^{n}(x^{2}+k^{2})}}=sum _{k=1}^{n}{frac {(-1)^{m+k}\,k^{2m}}{(n+k)!\,(n-k)!}}cdot {frac {2x}{x^{2}+k^{2}}}}

    folgt {displaystyle sum _{k=1}^{n}{frac {(-1)^{m+k}\,k^{2m}cdot 2}{(n+k)!\,(n-k)!}}=lim _{x o infty }sum _{k=1}^{n}{frac {(-1)^{m+k}\,k^{2m}}{(n+k)!\,(n-k)!}}cdot {frac {2x^{2}}{x^{2}+k^{2}}}=lim _{x o infty }{frac {x^{2m}}{prod limits _{k=1}^{n}(x^{2}+k^{2})}}=0}{displaystyle sum _{k=1}^{n}{frac {(-1)^{m+k}\,k^{2m}cdot 2}{(n+k)!\,(n-k)!}}=lim _{x o infty }sum _{k=1}^{n}{frac {(-1)^{m+k}\,k^{2m}}{(n+k)!\,(n-k)!}}cdot {frac {2x^{2}}{x^{2}+k^{2}}}=lim _{x o infty }{frac {x^{2m}}{prod limits _{k=1}^{n}(x^{2}+k^{2})}}=0}.

    Also verschwindet {displaystyle sum _{k=1}^{n}{frac {(-1)^{m+k}\,k^{2m}}{(n+k)!\,(n-k)!}}cdot log(R^{2})}{displaystyle sum _{k=1}^{n}{frac {(-1)^{m+k}\,k^{2m}}{(n+k)!\,(n-k)!}}cdot log(R^{2})} für alle {displaystyle R>0}R>0.

    Nun ist {displaystyle int _{0}^{infty }{frac {x^{2m-1}}{prod limits _{k=1}^{n}(x^{2}+k^{2})}}\,dx=lim _{R o infty }sum _{k=1}^{n}{frac {(-1)^{m+k}\,k^{2m}}{(n+k)!\,(n-k)!}}cdot int _{0}^{R}{frac {2x}{x^{2}+k^{2}}}\,dx}{displaystyle int _{0}^{infty }{frac {x^{2m-1}}{prod limits _{k=1}^{n}(x^{2}+k^{2})}}\,dx=lim _{R o infty }sum _{k=1}^{n}{frac {(-1)^{m+k}\,k^{2m}}{(n+k)!\,(n-k)!}}cdot int _{0}^{R}{frac {2x}{x^{2}+k^{2}}}\,dx}

    {displaystyle =lim _{R o infty }sum _{k=1}^{n}{frac {(-1)^{m+k}\,k^{2m}}{(n+k)!\,(n-k)!}}cdot left[log(R^{2}+k^{2})-log(k^{2}) ight]}{displaystyle =lim _{R o infty }sum _{k=1}^{n}{frac {(-1)^{m+k}\,k^{2m}}{(n+k)!\,(n-k)!}}cdot left[log(R^{2}+k^{2})-log(k^{2}) ight]}

    {displaystyle =lim _{R o infty }sum _{k=1}^{n}{frac {(-1)^{m+k}\,k^{2m}}{(n+k)!\,(n-k)!}}cdot left[log left(1+{frac {k^{2}}{R^{2}}} ight)-log(k^{2}) ight]=sum _{k=1}^{n}{frac {(-1)^{m+k}\,k^{2m}}{(n+k)!\,(n-k)!}}\,(-2cdot log k)}{displaystyle =lim _{R o infty }sum _{k=1}^{n}{frac {(-1)^{m+k}\,k^{2m}}{(n+k)!\,(n-k)!}}cdot left[log left(1+{frac {k^{2}}{R^{2}}} ight)-log(k^{2}) ight]=sum _{k=1}^{n}{frac {(-1)^{m+k}\,k^{2m}}{(n+k)!\,(n-k)!}}\,(-2cdot log k)}.

     
    2.12Bearbeiten
    {displaystyle int _{0}^{infty }{frac {x^{2m-1}}{prod limits _{k=0}^{n}{Big (}x^{2}+(2k+1)^{2}{Big )}}}\,dx=sum _{k=0}^{n}{frac {(-1)^{m+k}\,(2k+1)^{2m-1}\,log(2k+1)}{2^{2n}\,(n+k+1)!\,(n-k)!}}qquad mleq n}{displaystyle int _{0}^{infty }{frac {x^{2m-1}}{prod limits _{k=0}^{n}{Big (}x^{2}+(2k+1)^{2}{Big )}}}\,dx=sum _{k=0}^{n}{frac {(-1)^{m+k}\,(2k+1)^{2m-1}\,log(2k+1)}{2^{2n}\,(n+k+1)!\,(n-k)!}}qquad mleq n}
    Beweis

    Aus der Partialbruchzerlegung {displaystyle {frac {x^{2m-1}}{prod limits _{k=0}^{n}{Big (}x^{2}+(2k+1)^{2}{Big )}}}=sum _{k=0}^{n}{frac {(-1)^{m-1+k}\,(2k+1)^{2m-1}}{2^{2n}\,(n+k+1)!\,(n-k)!}}cdot {frac {x}{x^{2}+(2k+1)^{2}}}}{displaystyle {frac {x^{2m-1}}{prod limits _{k=0}^{n}{Big (}x^{2}+(2k+1)^{2}{Big )}}}=sum _{k=0}^{n}{frac {(-1)^{m-1+k}\,(2k+1)^{2m-1}}{2^{2n}\,(n+k+1)!\,(n-k)!}}cdot {frac {x}{x^{2}+(2k+1)^{2}}}}

    folgt {displaystyle sum _{k=0}^{n}{frac {(-1)^{m-1+k}\,(2k+1)^{2m-1}}{2^{2n}\,(n+k+1)!\,(n-k)!}}=lim _{x o infty }sum _{k=0}^{n}{frac {(-1)^{m-1+k}\,(2k+1)^{2m-1}}{2^{2n}\,(n+k+1)!\,(n-k)!}}cdot {frac {x^{2}}{x^{2}+(2k+1)^{2}}}=lim _{x o infty }{frac {x^{2m}}{prod limits _{k=0}^{n}{Big (}x^{2}+(2k+1)^{2}{Big )}}}=0}{displaystyle sum _{k=0}^{n}{frac {(-1)^{m-1+k}\,(2k+1)^{2m-1}}{2^{2n}\,(n+k+1)!\,(n-k)!}}=lim _{x o infty }sum _{k=0}^{n}{frac {(-1)^{m-1+k}\,(2k+1)^{2m-1}}{2^{2n}\,(n+k+1)!\,(n-k)!}}cdot {frac {x^{2}}{x^{2}+(2k+1)^{2}}}=lim _{x o infty }{frac {x^{2m}}{prod limits _{k=0}^{n}{Big (}x^{2}+(2k+1)^{2}{Big )}}}=0}.

    Also verschwindet {displaystyle sum _{k=0}^{n}{frac {(-1)^{m-1+k}\,(2k+1)^{2m-1}}{2^{2n}\,(n+k+1)!\,(n-k)!}}\,log(R^{2})}{displaystyle sum _{k=0}^{n}{frac {(-1)^{m-1+k}\,(2k+1)^{2m-1}}{2^{2n}\,(n+k+1)!\,(n-k)!}}\,log(R^{2})} für alle {displaystyle R>0}R>0.

    Nun ist {displaystyle int _{0}^{infty }{frac {x^{2m-1}}{prod limits _{k=0}^{n}{Big (}x^{2}+(2k+1)^{2}{Big )}}}\,dx=lim _{R o infty }sum _{k=0}^{n}{frac {(-1)^{m-1+k}\,(2k+1)^{2m-1}}{2^{2n+1}\,(n+k+1)!\,(n-k)!}}\,int _{0}^{R}{frac {2x}{x^{2}+(2k+1)^{2}}}\,dx}{displaystyle int _{0}^{infty }{frac {x^{2m-1}}{prod limits _{k=0}^{n}{Big (}x^{2}+(2k+1)^{2}{Big )}}}\,dx=lim _{R o infty }sum _{k=0}^{n}{frac {(-1)^{m-1+k}\,(2k+1)^{2m-1}}{2^{2n+1}\,(n+k+1)!\,(n-k)!}}\,int _{0}^{R}{frac {2x}{x^{2}+(2k+1)^{2}}}\,dx}

    {displaystyle =lim _{R o infty }sum _{k=0}^{n}{frac {(-1)^{m-1+k}\,(2k+1)^{2m-1}}{2^{2n+1}\,(n+k+1)!\,(n-k)!}}left[log {Big (}R^{2}+(2k+1)^{2}{Big )}-log {Big (}(2k+1)^{2}{Big )} ight]}{displaystyle =lim _{R o infty }sum _{k=0}^{n}{frac {(-1)^{m-1+k}\,(2k+1)^{2m-1}}{2^{2n+1}\,(n+k+1)!\,(n-k)!}}left[log {Big (}R^{2}+(2k+1)^{2}{Big )}-log {Big (}(2k+1)^{2}{Big )} ight]}

    {displaystyle =lim _{R o infty }sum _{k=0}^{n}{frac {(-1)^{m-1+k}\,(2k+1)^{2m-1}}{2^{2n+1}\,(n+k+1)!\,(n-k)!}}left[log left(1+{frac {(2k+1)^{2}}{R^{2}}} ight)-log {Big (}(2k+1)^{2}{Big )} ight]}{displaystyle =lim _{R o infty }sum _{k=0}^{n}{frac {(-1)^{m-1+k}\,(2k+1)^{2m-1}}{2^{2n+1}\,(n+k+1)!\,(n-k)!}}left[log left(1+{frac {(2k+1)^{2}}{R^{2}}} ight)-log {Big (}(2k+1)^{2}{Big )} ight]}

    {displaystyle =sum _{k=0}^{n}{frac {(-1)^{m+k}\,(2k+1)^{2m-1}\,log(2k+1)}{2^{2n}\,(n+k+1)!\,(n-k)!}}}{displaystyle =sum _{k=0}^{n}{frac {(-1)^{m+k}\,(2k+1)^{2m-1}\,log(2k+1)}{2^{2n}\,(n+k+1)!\,(n-k)!}}}.

     
    2.13Bearbeiten
    {displaystyle int _{0}^{infty }{frac {x^{2m}}{prod limits _{k=1}^{n}(x^{2}+k^{2})}}\,dx=pi cdot sum _{k=1}^{n}{frac {(-1)^{m-1+k}\,k^{2m+1}}{(n+k)!\,(n-k)!}}qquad m<n}{displaystyle int _{0}^{infty }{frac {x^{2m}}{prod limits _{k=1}^{n}(x^{2}+k^{2})}}\,dx=pi cdot sum _{k=1}^{n}{frac {(-1)^{m-1+k}\,k^{2m+1}}{(n+k)!\,(n-k)!}}qquad m<n}
    Beweis

    Aus der Partialbruchzerlegung {displaystyle {frac {x^{2m}}{prod limits _{k=1}^{n}(x^{2}+k^{2})}}=sum _{k=1}^{n}{frac {(-1)^{m-1+k}\,k^{2m+2}}{(n+k)!\,(n-k)!}}cdot {frac {2}{x^{2}+k^{2}}}}{displaystyle {frac {x^{2m}}{prod limits _{k=1}^{n}(x^{2}+k^{2})}}=sum _{k=1}^{n}{frac {(-1)^{m-1+k}\,k^{2m+2}}{(n+k)!\,(n-k)!}}cdot {frac {2}{x^{2}+k^{2}}}}

    folgt unmittelbar {displaystyle int _{0}^{infty }{frac {x^{2m}}{prod limits _{k=1}^{n}(x^{2}+k^{2})}}\,dx=sum _{k=1}^{n}{frac {(-1)^{m-1+k}\,k^{2m+2}}{(n+k)!\,(n-k)!}}cdot underbrace {int _{0}^{infty }{frac {2}{x^{2}+k^{2}}}\,dx} _{={frac {pi }{k}}}}{displaystyle int _{0}^{infty }{frac {x^{2m}}{prod limits _{k=1}^{n}(x^{2}+k^{2})}}\,dx=sum _{k=1}^{n}{frac {(-1)^{m-1+k}\,k^{2m+2}}{(n+k)!\,(n-k)!}}cdot underbrace {int _{0}^{infty }{frac {2}{x^{2}+k^{2}}}\,dx} _{={frac {pi }{k}}}}.

     
    2.14Bearbeiten
    {displaystyle int _{0}^{infty }{frac {x^{2m}}{prod limits _{k=0}^{n}{Big (}x^{2}+(2k+1)^{2}{Big )}}}\,dx=pi cdot sum _{k=0}^{n}{frac {(-1)^{m+k}\,(2k+1)^{2m}}{2^{2n+1}\,(n+k+1)!\,(n-k)!}}qquad mleq n}{displaystyle int _{0}^{infty }{frac {x^{2m}}{prod limits _{k=0}^{n}{Big (}x^{2}+(2k+1)^{2}{Big )}}}\,dx=pi cdot sum _{k=0}^{n}{frac {(-1)^{m+k}\,(2k+1)^{2m}}{2^{2n+1}\,(n+k+1)!\,(n-k)!}}qquad mleq n}
    Beweis

    Aus der Partialbruchzerlegung {displaystyle {frac {x^{2m}}{prod limits _{k=0}^{n}{Big (}x^{2}+(2k+1)^{2}{Big )}}}=sum _{k=0}^{n}{frac {(-1)^{m+k}\,(2k+1)^{2m+1}}{2^{2n}\,(n+k+1)!\,(n-k)!}}cdot {frac {1}{x^{2}+(2k+1)^{2}}}}{displaystyle {frac {x^{2m}}{prod limits _{k=0}^{n}{Big (}x^{2}+(2k+1)^{2}{Big )}}}=sum _{k=0}^{n}{frac {(-1)^{m+k}\,(2k+1)^{2m+1}}{2^{2n}\,(n+k+1)!\,(n-k)!}}cdot {frac {1}{x^{2}+(2k+1)^{2}}}}

    folgt unmittelbar {displaystyle int _{0}^{infty }{frac {x^{2m}}{prod limits _{k=0}^{n}{Big (}x^{2}+(2k+1)^{2}{Big )}}}\,dx=sum _{k=0}^{n}{frac {(-1)^{m+k}\,(2k+1)^{2m+1}}{2^{2n}\,(n+k+1)!\,(n-k)!}}cdot underbrace {int _{0}^{infty }{frac {dx}{x^{2}+(2k+1)^{2}}}} _{={frac {1}{2}}cdot {frac {pi }{2k+1}}}}{displaystyle int _{0}^{infty }{frac {x^{2m}}{prod limits _{k=0}^{n}{Big (}x^{2}+(2k+1)^{2}{Big )}}}\,dx=sum _{k=0}^{n}{frac {(-1)^{m+k}\,(2k+1)^{2m+1}}{2^{2n}\,(n+k+1)!\,(n-k)!}}cdot underbrace {int _{0}^{infty }{frac {dx}{x^{2}+(2k+1)^{2}}}} _{={frac {1}{2}}cdot {frac {pi }{2k+1}}}}.

     
    2.15Bearbeiten
    {displaystyle int _{0}^{infty }left({frac {2}{1+{sqrt {1+4x}}}} ight)^{p}cdot x^{s-1}\,dx={frac {pcdot Gamma (s)cdot Gamma (p-2s)}{Gamma (p-s+1)}}qquad 0<{ ext{Re}}(s)<{frac {1}{2}}cdot { ext{Re}}(p)}{displaystyle int _{0}^{infty }left({frac {2}{1+{sqrt {1+4x}}}} ight)^{p}cdot x^{s-1}\,dx={frac {pcdot Gamma (s)cdot Gamma (p-2s)}{Gamma (p-s+1)}}qquad 0<{ ext{Re}}(s)<{frac {1}{2}}cdot { ext{Re}}(p)}
    Beweis

    {displaystyle I:=int _{0}^{infty }left({frac {2}{1+{sqrt {1+4x}}}} ight)^{p}cdot x^{s-1}\,dx}{displaystyle I:=int _{0}^{infty }left({frac {2}{1+{sqrt {1+4x}}}} ight)^{p}cdot x^{s-1}\,dx}

    Nach Substitution {displaystyle x=y+y^{2}}{displaystyle x=y+y^{2}} ist

    {displaystyle I=int _{0}^{infty }left({frac {2}{1+(1+2y)}} ight)^{p}cdot y^{s-1}\,(1+y)^{s-1}cdot (1+2y)\,dy=int _{0}^{infty }y^{s-1}cdot (1+y)^{s-p-1}cdot (1+2y)\,dy}{displaystyle I=int _{0}^{infty }left({frac {2}{1+(1+2y)}} ight)^{p}cdot y^{s-1}\,(1+y)^{s-1}cdot (1+2y)\,dy=int _{0}^{infty }y^{s-1}cdot (1+y)^{s-p-1}cdot (1+2y)\,dy}.

    Nach Substitution {displaystyle y={frac {z}{1-z}}}{displaystyle y={frac {z}{1-z}}} ist

    {displaystyle I=int _{0}^{1}{frac {z^{s-1}}{(1-z)^{s-1}}}cdot {frac {1}{(1-z)^{s-p-1}}}cdot {frac {1+z}{1-z}}cdot {frac {dz}{(1-z)^{2}}}=int _{0}^{1}z^{s-1}cdot (1-z)^{p-2s-1}\,(1+z)\,dz}{displaystyle I=int _{0}^{1}{frac {z^{s-1}}{(1-z)^{s-1}}}cdot {frac {1}{(1-z)^{s-p-1}}}cdot {frac {1+z}{1-z}}cdot {frac {dz}{(1-z)^{2}}}=int _{0}^{1}z^{s-1}cdot (1-z)^{p-2s-1}\,(1+z)\,dz}

    {displaystyle ={frac {Gamma (s)cdot Gamma (p-2s)}{Gamma (p-s)}}+{frac {Gamma (s+1)cdot Gamma (p-2s)}{Gamma (p-s+1)}}={frac {(p-s)cdot Gamma (s)cdot Gamma (p-2s)}{(p-s)cdot Gamma (p-s)}}+{frac {scdot Gamma (s)cdot Gamma (p-2s)}{Gamma (p-s+1)}}={frac {pcdot Gamma (s)cdot Gamma (p-2s)}{Gamma (p-s+1)}}}{displaystyle ={frac {Gamma (s)cdot Gamma (p-2s)}{Gamma (p-s)}}+{frac {Gamma (s+1)cdot Gamma (p-2s)}{Gamma (p-s+1)}}={frac {(p-s)cdot Gamma (s)cdot Gamma (p-2s)}{(p-s)cdot Gamma (p-s)}}+{frac {scdot Gamma (s)cdot Gamma (p-2s)}{Gamma (p-s+1)}}={frac {pcdot Gamma (s)cdot Gamma (p-2s)}{Gamma (p-s+1)}}}.

     
    3.1Bearbeiten
    {displaystyle int _{0}^{infty }{frac {x^{alpha -1}}{(w+x)^{eta }}}\,dx={frac {B(alpha ,eta -alpha )}{w^{eta -alpha }}}qquad 0<{ ext{Re}}(alpha )<{ ext{Re}}(eta )\,,;{ ext{Re}}(w)>0}{displaystyle int _{0}^{infty }{frac {x^{alpha -1}}{(w+x)^{eta }}}\,dx={frac {B(alpha ,eta -alpha )}{w^{eta -alpha }}}qquad 0<{ ext{Re}}(alpha )<{ ext{Re}}(eta )\,,;{ ext{Re}}(w)>0}
    Beweis

    Setzt man {displaystyle f(z)={frac {z^{alpha -1}}{(1+z)^{alpha +eta }}}}{displaystyle f(z)={frac {z^{alpha -1}}{(1+z)^{alpha +eta }}}}, so ist {displaystyle f\,}f\, auf {displaystyle mathbb {C} setminus mathbb {R} ^{leq 0}}{displaystyle mathbb {C} setminus mathbb {R} ^{leq 0}} holomorph.

    Wegen {displaystyle left|z^{alpha } ight|=Theta left(|z|^{{ ext{Re}}\,alpha } ight)}left|z^alpha ight|=Thetaleft(|z|^{ ext{Re}\,alpha} ight) ist {displaystyle |f(z)|={frac {Theta left(|z|^{{ ext{Re}}(alpha )-1} ight)}{Theta left(|z|^{{ ext{Re}}(alpha +eta )} ight)}}=Theta left({frac {1}{|z|^{1+{ ext{Re}}(eta )}}} ight)={ ext{o}}left({frac {1}{|z|}} ight)}{displaystyle |f(z)|={frac {Theta left(|z|^{{ ext{Re}}(alpha )-1} ight)}{Theta left(|z|^{{ ext{Re}}(alpha +eta )} ight)}}=Theta left({frac {1}{|z|^{1+{ ext{Re}}(eta )}}} ight)={ ext{o}}left({frac {1}{|z|}} ight)} für {displaystyle |z| o infty \,}|z| oinfty\,.

    Als komplexe Zahl mit positivem Realteil besitzt {displaystyle w\,}{displaystyle w\,} eine Darstellung {displaystyle r\,e^{ivarphi }}{displaystyle r\,e^{ivarphi }} mit {displaystyle -{frac {pi }{2}}<varphi <{frac {pi }{2}}}{displaystyle -{frac {pi }{2}}<varphi <{frac {pi }{2}}}.

    Der Kehrwehrt {displaystyle {frac {1}{w}}={frac {1}{r}}\,e^{-ivarphi }}{displaystyle {frac {1}{w}}={frac {1}{r}}\,e^{-ivarphi }} besitzt dann auch einen positiven Realteil.

    Wegen {displaystyle |f(z)|={ ext{o}}left({frac {1}{|z|}} ight)}{displaystyle |f(z)|={ ext{o}}left({frac {1}{|z|}} ight)} ist nun {displaystyle int _{0}^{infty }f(x)dx=int _{0}^{infty }fleft({frac {x}{w}} ight){frac {dx}{w}}=int _{0}^{infty }{frac {left({frac {x}{w}} ight)^{alpha -1}}{left(1+{frac {x}{w}} ight)^{alpha +eta }}}\,dx}{displaystyle int _{0}^{infty }f(x)dx=int _{0}^{infty }fleft({frac {x}{w}} ight){frac {dx}{w}}=int _{0}^{infty }{frac {left({frac {x}{w}} ight)^{alpha -1}}{left(1+{frac {x}{w}} ight)^{alpha +eta }}}\,dx}

    {displaystyle =w^{eta }int _{0}^{infty }{frac {x^{alpha -1}}{left(w+x ight)^{alpha +eta }}}\,dx}{displaystyle =w^{eta }int _{0}^{infty }{frac {x^{alpha -1}}{left(w+x ight)^{alpha +eta }}}\,dx}. Also ist {displaystyle int _{0}^{infty }{frac {x^{alpha -1}}{left(w+x ight)^{alpha +eta }}}\,dx={frac {1}{w^{eta }}}int _{0}^{infty }f(x)dx}{displaystyle int _{0}^{infty }{frac {x^{alpha -1}}{left(w+x ight)^{alpha +eta }}}\,dx={frac {1}{w^{eta }}}int _{0}^{infty }f(x)dx}.

    Und das ist {displaystyle {frac {B(alpha ,eta )}{w^{eta }}}}{displaystyle {frac {B(alpha ,eta )}{w^{eta }}}}. Ersetzt man {displaystyle eta \,}{displaystyle eta \,} durch {displaystyle eta -alpha \,}{displaystyle eta -alpha \,}, so ist {displaystyle int _{0}^{infty }{frac {x^{alpha -1}}{(w+x)^{eta }}}\,dx={frac {B(alpha ,eta -alpha )}{w^{eta -alpha }}}}{displaystyle int _{0}^{infty }{frac {x^{alpha -1}}{(w+x)^{eta }}}\,dx={frac {B(alpha ,eta -alpha )}{w^{eta -alpha }}}}.

     
    3.2Bearbeiten
    {displaystyle int _{0}^{infty }{frac {x^{alpha -1}}{(1+x^{eta })^{gamma }}}\,dx={frac {1}{eta }}\,Bleft({frac {alpha }{eta }},gamma -{frac {alpha }{eta }} ight)}{displaystyle int _{0}^{infty }{frac {x^{alpha -1}}{(1+x^{eta })^{gamma }}}\,dx={frac {1}{eta }}\,Bleft({frac {alpha }{eta }},gamma -{frac {alpha }{eta }} ight)}
    ohne Beweis
     
    3.3Bearbeiten
    {displaystyle int _{0}^{infty }prod _{1leq ell leq 3}{frac {1}{(x^{2}+2xcos alpha _{ell }+1)}}\,dx={frac {1}{4}}sum _{(i,j,k)in A_{3}}{frac {alpha _{i}\,csc alpha _{i}\,cos 2alpha _{i}}{(cos alpha _{i}-cos alpha _{j})(cos alpha _{i}-cos alpha _{k})}}}{displaystyle int _{0}^{infty }prod _{1leq ell leq 3}{frac {1}{(x^{2}+2xcos alpha _{ell }+1)}}\,dx={frac {1}{4}}sum _{(i,j,k)in A_{3}}{frac {alpha _{i}\,csc alpha _{i}\,cos 2alpha _{i}}{(cos alpha _{i}-cos alpha _{j})(cos alpha _{i}-cos alpha _{k})}}}
    ohne Beweis
     
    4.1Bearbeiten
    {displaystyle int _{-infty }^{infty }{frac {dx}{(u+ix)^{alpha }\,(v-ix)^{eta }}}={frac {2pi }{(u+v)^{alpha +eta -1}}}\,{frac {Gamma (alpha +eta -1)}{Gamma (alpha )\,Gamma (eta )}}qquad { ext{Re}}(u),{ ext{Re}}(v)>0;,;{ ext{Re}}(alpha +eta )>1}{displaystyle int _{-infty }^{infty }{frac {dx}{(u+ix)^{alpha }\,(v-ix)^{eta }}}={frac {2pi }{(u+v)^{alpha +eta -1}}}\,{frac {Gamma (alpha +eta -1)}{Gamma (alpha )\,Gamma (eta )}}qquad { ext{Re}}(u),{ ext{Re}}(v)>0;,;{ ext{Re}}(alpha +eta )>1}
    Beweis

    Setze {displaystyle f(z)={frac {1}{(u+iz)^{alpha }\,(v-iz)^{eta }}}}{displaystyle f(z)={frac {1}{(u+iz)^{alpha }\,(v-iz)^{eta }}}}.

    Wegen {displaystyle { ext{Re}}(u),{ ext{Re}}(v)>0\,}{displaystyle { ext{Re}}(u),{ ext{Re}}(v)>0\,} ist {displaystyle { ext{Im}}(iu)>0,{ ext{Im}}(-iv)<0\,}{displaystyle { ext{Im}}(iu)>0,{ ext{Im}}(-iv)<0\,} und {displaystyle f\,}f\, ist auf {displaystyle mathbb {C} setminus left(iu+imathbb {R} ^{geq 0}cup -iv-imathbb {R} ^{geq 0} ight)}{displaystyle mathbb {C} setminus left(iu+imathbb {R} ^{geq 0}cup -iv-imathbb {R} ^{geq 0} ight)} holomorph.

    Integrationsweg6.PNG

    Wegen {displaystyle left|z^{alpha } ight|=Theta left(|z|^{{ ext{Re}}\,alpha } ight)}left|z^alpha ight|=Thetaleft(|z|^{ ext{Re}\,alpha} ight) ist {displaystyle |f(z)|={frac {1}{Theta left(|z|^{{ ext{Re}}(alpha )} ight)\,Theta left(|z|^{{ ext{Re}}(eta )} ight)}}=Theta left({frac {1}{|z|^{{ ext{Re}}(alpha +eta )}}} ight)={ ext{o}}left({frac {1}{|z|}} ight)}{displaystyle |f(z)|={frac {1}{Theta left(|z|^{{ ext{Re}}(alpha )} ight)\,Theta left(|z|^{{ ext{Re}}(eta )} ight)}}=Theta left({frac {1}{|z|^{{ ext{Re}}(alpha +eta )}}} ight)={ ext{o}}left({frac {1}{|z|}} ight)} für {displaystyle |z| o infty \,}|z| oinfty\,.

    Die Integrale über den Kreisbögen {displaystyle K_{1},K_{2}\,}{displaystyle K_{1},K_{2}\,} verschwinden daher wenn ihr Radius gegen unendlich geht.

    Vorübergehend mache man die zusätzliche Einschränkung {displaystyle { ext{Re}}(alpha )<0\,}{displaystyle { ext{Re}}(alpha )<0\,}.

    Dann ist {displaystyle lim _{z o iu}f(z)=0\,}{displaystyle lim _{z o iu}f(z)=0\,}, und somit verschwindet das Integral über dem Halbkreis {displaystyle kappa \,}{displaystyle kappa \,} mit Radius {displaystyle varepsilon \,}{displaystyle varepsilon \,}, wenn {displaystyle varepsilon o 0+\,}{displaystyle varepsilon o 0+\,} geht.

    Damit ist {displaystyle int _{-infty }^{infty }f(x)dx=lim _{varepsilon o 0+}left(int _{0}^{infty }f(iu+varepsilon +it)idt-int _{0}^{infty }f(iu-varepsilon +it)idt ight)}{displaystyle int _{-infty }^{infty }f(x)dx=lim _{varepsilon o 0+}left(int _{0}^{infty }f(iu+varepsilon +it)idt-int _{0}^{infty }f(iu-varepsilon +it)idt ight)},

    wobei {displaystyle f(iupm varepsilon +it)={frac {1}{(-tpm ivarepsilon )^{alpha }}}\,{frac {1}{(u+v+tmp ivarepsilon )^{eta }}}={frac {e^{mp ipi alpha }}{(tmp ivarepsilon )^{alpha }}}\,{frac {1}{(u+v+tmp ivarepsilon )^{eta }}}}{displaystyle f(iupm varepsilon +it)={frac {1}{(-tpm ivarepsilon )^{alpha }}}\,{frac {1}{(u+v+tmp ivarepsilon )^{eta }}}={frac {e^{mp ipi alpha }}{(tmp ivarepsilon )^{alpha }}}\,{frac {1}{(u+v+tmp ivarepsilon )^{eta }}}} ist.

    Also ist {displaystyle int _{-infty }^{infty }f(x)dx=left(e^{-ipi alpha }-e^{ipi alpha } ight)int _{0}^{infty }{frac {1}{t^{alpha }\,(u+v+t)^{eta }}}idt=2sin alpha pi int _{0}^{infty }{frac {t^{(1-alpha )-1}}{(u+v+t)^{eta }}}\,dt}{displaystyle int _{-infty }^{infty }f(x)dx=left(e^{-ipi alpha }-e^{ipi alpha } ight)int _{0}^{infty }{frac {1}{t^{alpha }\,(u+v+t)^{eta }}}idt=2sin alpha pi int _{0}^{infty }{frac {t^{(1-alpha )-1}}{(u+v+t)^{eta }}}\,dt}

    {displaystyle =2sin alpha \,{frac {B(1-alpha ,eta -(1-alpha ))}{(u+v)^{eta -(1-alpha )}}}}{displaystyle =2sin alpha \,{frac {B(1-alpha ,eta -(1-alpha ))}{(u+v)^{eta -(1-alpha )}}}}. Und das ist {displaystyle {frac {2pi }{Gamma (alpha )\,Gamma (1-alpha )}}\,{frac {1}{(u+v)^{alpha +eta -1}}}\,{frac {Gamma (1-alpha )\,Gamma (alpha +eta -1)}{Gamma (eta )}}}{displaystyle {frac {2pi }{Gamma (alpha )\,Gamma (1-alpha )}}\,{frac {1}{(u+v)^{alpha +eta -1}}}\,{frac {Gamma (1-alpha )\,Gamma (alpha +eta -1)}{Gamma (eta )}}}

    {displaystyle ={frac {2pi }{(u+v)^{alpha +eta -1}}}\,{frac {Gamma (alpha +eta -1)}{Gamma (alpha )\,Gamma (eta )}}}{displaystyle ={frac {2pi }{(u+v)^{alpha +eta -1}}}\,{frac {Gamma (alpha +eta -1)}{Gamma (alpha )\,Gamma (eta )}}}.


    Dass die Formel auch ohne die Einschränkung {displaystyle { ext{Re}}(alpha )<0\,}{displaystyle { ext{Re}}(alpha )<0\,} gilt, sieht man, wenn man sie wiederholt partiell integriert:

    {displaystyle int _{-infty }^{infty }{frac {1}{(u+ix)^{alpha }\,(v-ix)^{eta }}}\,dx=underbrace {left[{frac {1}{(u+ix)^{alpha }}}\,{frac {1}{i(eta -1)}}\,{frac {1}{(v-ix)^{eta -1}}} ight]_{-infty }^{infty }} _{=0}}{displaystyle int _{-infty }^{infty }{frac {1}{(u+ix)^{alpha }\,(v-ix)^{eta }}}\,dx=underbrace {left[{frac {1}{(u+ix)^{alpha }}}\,{frac {1}{i(eta -1)}}\,{frac {1}{(v-ix)^{eta -1}}} ight]_{-infty }^{infty }} _{=0}}

    {displaystyle -int _{-infty }^{infty }{frac {-ialpha }{(u+ix)^{alpha +1}}}\,{frac {1}{i(eta -1)}}\,{frac {1}{(v-ix)^{eta -1}}}\,dx={frac {alpha }{eta -1}}int _{-infty }^{infty }{frac {1}{(u+ix)^{alpha +1}}}\,{frac {1}{(v-ix)^{eta -1}}}\,dx}{displaystyle -int _{-infty }^{infty }{frac {-ialpha }{(u+ix)^{alpha +1}}}\,{frac {1}{i(eta -1)}}\,{frac {1}{(v-ix)^{eta -1}}}\,dx={frac {alpha }{eta -1}}int _{-infty }^{infty }{frac {1}{(u+ix)^{alpha +1}}}\,{frac {1}{(v-ix)^{eta -1}}}\,dx}.

    Und damit ist {displaystyle int _{-infty }^{infty }{frac {1}{(u+ix)^{alpha +1}}}\,{frac {1}{(v-ix)^{eta -1}}}\,dx={frac {2pi }{(u+v)^{alpha +eta -1}}}\,{frac {Gamma (alpha +eta -1)}{Gamma (alpha )\,Gamma (eta )}}\,{frac {eta -1}{alpha }}}{displaystyle int _{-infty }^{infty }{frac {1}{(u+ix)^{alpha +1}}}\,{frac {1}{(v-ix)^{eta -1}}}\,dx={frac {2pi }{(u+v)^{alpha +eta -1}}}\,{frac {Gamma (alpha +eta -1)}{Gamma (alpha )\,Gamma (eta )}}\,{frac {eta -1}{alpha }}}

    {displaystyle ={frac {2pi }{(u+v)^{alpha +eta -1}}}\,{frac {Gamma (alpha +eta -1)}{Gamma (alpha +1)\,Gamma (eta -1)}}}{displaystyle ={frac {2pi }{(u+v)^{alpha +eta -1}}}\,{frac {Gamma (alpha +eta -1)}{Gamma (alpha +1)\,Gamma (eta -1)}}}.
  • 相关阅读:
    如何将DataTable转换成List<T>呢？
    mySql中SUBSTRING_INDEX函数用法
    常用 Git 命令清单
    git学习笔记
    MySql 获取表的字段名
    mysql从身份证号中提取生日、性别
    年月日转大写汉字
    ExtJs服务器端代理（Ajax）
    ExtJS客户端代理
    ExtJS 数据模型
  • 原文地址：https://www.cnblogs.com/Eufisky/p/14730789.html
Copyright © 2011-2022 走看看