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

0.1Bearbeiten

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

ohne Beweis (Abels Integral)

 

1.1Bearbeiten

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

ohne Beweis

 

1.2Bearbeiten

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

Beweis

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

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

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

 

1.3Bearbeiten

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

Beweis

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

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

 

1.4Bearbeiten

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

Beweis

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

 

1.5Bearbeiten

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

Beweis

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

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

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


Ansatz (Variation der Konstante):

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

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

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

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

{displaystyle {egin{pmatrix}cos x&-sin x\sin x&cos xend{pmatrix}}{egin{pmatrix}c'(x)\d'(x)end{pmatrix}}={egin{pmatrix} an {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} an {frac {x}{2}}\0end{pmatrix}}={egin{pmatrix}cos x\,\, an {frac {x}{2}}\-sin x\,\, an {frac {x}{2}}end{pmatrix}}}

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

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

 

2.1Bearbeiten

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

ohne Beweis