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

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{displaystyle int _{0}^{frac {pi }{2}}{frac {x}{sin x}}\,dx=2G}

Beweis

{displaystyle int _{0}^{frac {pi }{2}}{frac {x}{sin x}}\,dx} ist nach Substitution {displaystyle xmapsto 2arctan x} gleich {displaystyle 2int _{0}^{1}{frac {arctan x}{x}}\,dx=2G}.

 

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{displaystyle int _{0}^{frac {pi }{2}}{frac {x^{2}}{sin x}}\,dx=2pi G-{frac {7}{2}}\,zeta (3)}

Beweis

{displaystyle F(x)=2log left(2sin {frac {x}{2}} ight)-log(2sin x)} ist eine Stammfunktion von {displaystyle {frac {1}{sin x}}}.

{displaystyle int _{0}^{frac {pi }{2}}{frac {x^{2}}{sin x}}\,dx} ist damit nach partieller Integration

{displaystyle underbrace {left[x^{2}\,F(x) ight]_{0}^{frac {pi }{2}}} _{=0}-int _{0}^{frac {pi }{2}}2xF(x)\,dx=underbrace {int _{0}^{frac {pi }{2}}4xleft[-log left(2sin {frac {x}{2}} ight) ight]\,dx} _{=A}-underbrace {int _{0}^{frac {pi }{2}}2xleft[-log(2sin x) ight]\,dx} _{=B}}

Verwende nun die Fourierreihenentwicklung {displaystyle -log left(2sin {frac {x}{2}} ight)=sum _{k=1}^{infty }{frac {cos kx}{k}}},

dann ist {displaystyle A=int _{0}^{frac {pi }{2}}4xsum _{k=1}^{infty }{frac {cos kx}{k}}\,dx=sum _{k=1}^{infty }int _{0}^{frac {pi }{2}}{frac {4xcos kx}{k}}\,dx}

{displaystyle =sum _{k=1}^{infty }left[{frac {4cos kx+4kxsin kx}{k^{3}}} ight]_{0}^{frac {pi }{2}}=sum _{k=1}^{infty }left({frac {4cos {frac {kpi }{2}}}{k^{3}}}-{frac {4}{k^{3}}}+2pi \,{frac {sin {frac {kpi }{2}}}{k^{2}}} ight)=-{frac {3}{8}}zeta (3)-4zeta (3)+2pi G}

und {displaystyle B=int _{0}^{frac {pi }{2}}2xsum _{k=1}^{infty }{frac {cos 2kx}{k}}\,dx=sum _{k=1}^{infty }int _{0}^{frac {pi }{2}}{frac {2xcos 2kx}{k}}\,dx}

{displaystyle =sum _{k=1}^{infty }left[{frac {cos 2kx+2kxsin 2kx}{2k^{3}}} ight]_{0}^{frac {pi }{2}}=sum _{k=1}^{infty }left({frac {cos kpi }{2k^{3}}}-{frac {1}{2k^{3}}}+{frac {pi }{2}}\,{frac {sin kpi }{k^{2}}} ight)=-{frac {3}{8}}zeta (3)-{frac {1}{2}}zeta (3)}.

Also ist {displaystyle A-B=2pi G-{frac {7}{2}}zeta (3)}.

 

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{displaystyle int _{0}^{frac {pi }{4}}{frac {x^{2}}{sin ^{2}x}}\,dx=G+{frac {pi }{4}}log 2-{frac {pi ^{2}}{16}}}

Beweis

{displaystyle int _{0}^{frac {pi }{4}}{frac {x^{2}}{sin ^{2}x}}\,dx=left[x^{2}\,(-cot x) ight]_{0}^{frac {pi }{4}}+int _{0}^{frac {pi }{4}}2xcot x\,dx}

{displaystyle =-{frac {pi ^{2}}{16}}+{Big [}2xlog sin x{Big ]}_{0}^{frac {pi }{4}}-int _{0}^{frac {pi }{4}}2log sin x\,dx=-{frac {pi ^{2}}{16}}-{frac {pi }{4}}log 2+{frac {pi }{2}}log 2+G}

 

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{displaystyle int _{0}^{frac {pi }{4}}{frac {x^{3}}{sin ^{2}x}}\,dx={frac {3}{4}}pi G-{frac {pi ^{3}}{64}}+{frac {3}{32}}pi ^{2}log 2-{frac {105}{64}}zeta (3)}

Beweis

{displaystyle I:=int _{0}^{frac {pi }{4}}{frac {x^{3}}{sin ^{2}x}}\,dx={Big [}x^{3}\,(-cot x){Big ]}_{0}^{frac {pi }{4}}+int _{0}^{frac {pi }{4}}3x^{2}\,cot x\,dx}

{displaystyle =-{frac {pi ^{3}}{64}}+underbrace {{Big [}3x^{2}\,log sin x{Big ]}_{0}^{frac {pi }{4}}} _{=-{frac {3}{32}}pi ^{2}log 2}-int _{0}^{frac {pi }{4}}6xlog sin x\,dx}
Nach der Fourierreihenentwicklung {displaystyle -log sin x=log 2+sum _{k=1}^{infty }{frac {cos 2kx}{k}}} ist

{displaystyle -int _{0}^{frac {pi }{4}}6xlog sin x\,dx=int _{0}^{frac {pi }{4}}6x\,dxcdot log 2+6sum _{k=1}^{infty }{frac {1}{k}}int _{0}^{frac {pi }{4}}xcos 2kx\,dx}

{displaystyle =2cdot {frac {3}{32}}pi ^{2}log 2+6sum _{k=1}^{infty }{frac {1}{k}}left({frac {pi }{8k}}sin left({frac {kpi }{2}} ight)+{frac {1}{(2k)^{2}}}cos left({frac {kpi }{2}} ight)-{frac {1}{(2k)^{2}}} ight)}.

Also ist {displaystyle I=-{frac {pi ^{3}}{64}}+{frac {3}{32}}pi ^{2}log 2+{frac {3}{4}}pi underbrace {sum _{k=1}^{infty }{frac {sin {frac {kpi }{2}}}{k^{2}}}} _{=G}+{frac {3}{2}}underbrace {sum _{k=1}^{infty }{frac {cos {frac {kpi }{2}}}{k^{3}}}} _{=-{frac {3}{32}}zeta (3)}-{frac {3}{2}}zeta (3)}.

 

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{displaystyle int _{0}^{1}sin(pi x)\,x^{x}\,(1-x)^{1-x}\,dx={frac {pi e}{24}}}

Beweis (Formel nach Ramanujan)

Es sei {displaystyle S:=int _{0}^{1}e^{ipi x}\,x^{x}\,{frac {1-x}{(1-x)^{x}}}\,dx=int _{0}^{1}e^{ipi x}\,e^{(log x)\,x}\,{frac {1-x}{e^{log(1-x)\,x}}}\,dx=int _{0}^{1}(1-x)\,e^{(ipi +log x-log(1-x))x}\,dx}.

Substituiert man {displaystyle t=log x-log(1-x)\,left(x={frac {e^{t}}{e^{t}+1}} ight)}, so ist

{displaystyle S=int _{-infty }^{infty }{frac {1}{e^{t}+1}}\,e^{(ipi +t)\,{frac {e^{t}}{e^{t}+1}}}\,{frac {e^{t}}{(e^{t}+1)^{2}}}\,dt=int _{-infty +ipi }^{infty +ipi }e^{t\,{frac {e^{t}}{e^{t}-1}}}\,{frac {e^{t}}{(e^{t}-1)^{3}}}\,dt}.

Setzt man {displaystyle f(z)=e^{z\,{frac {e^{z}}{e^{z}-1}}}\,{frac {e^{z}}{(e^{z}-1)^{3}}}}, so ist {displaystyle f\,} auf {displaystyle D:=left{zin mathbb {C} \,|\,-pi leq { ext{Im}}(z)leq pi ight}} meromorph. 
Die einzige Polstelle liegt bei {displaystyle z=0\,} und dort ist {displaystyle { ext{res}}(f,0)=-{frac {e}{24}}}.

Setzt man {displaystyle kappa _{R}=gamma _{R}+delta _{R}+sigma _{R}+ au _{R}\,}, so ist {displaystyle oint _{kappa _{R}}f\,dz=-2pi icdot { ext{res}}(f,0)=2i\,{frac {pi e}{24}}}.

Für jede Folge {displaystyle (z_{n})subset D} mit {displaystyle |z_{n}| o infty \,} geht {displaystyle f(z_{n})\,} gegen null.

Daher verschwinden {displaystyle int _{delta _{R}}f\,dz} und {displaystyle int _{ au _{R}}f\,dz} für {displaystyle R o infty \,}.

Und nachdem {displaystyle f\,} ungerade ist, ist {displaystyle int _{gamma _{R}}f\,dz=int _{sigma _{R}}f\,dz}.

{displaystyle 2S=2lim _{R o infty }int _{gamma _{R}}f\,dz} ist demnach {displaystyle lim _{R o infty }oint _{kappa _{R}}f\,dz=2i\,{frac {pi e}{24}}}.

Daraus ergibt sich das gesuchte Integral:

{displaystyle int _{0}^{1}sin(pi x)\,x^{x}\,(1-x)^{1-x}\,dx={ ext{Im}}(S)={frac {pi e}{24}}}

 

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{displaystyle int _{0}^{1}{frac {sin(pi x)}{x^{x}\,(1-x)^{1-x}}}\,dx={frac {pi }{e}}}

ohne Beweis

 

1.1Bearbeiten

{displaystyle int _{0}^{frac {pi }{2}}sin ^{n+1}x\,dx={frac {n}{n+1}}int _{0}^{frac {pi }{2}}sin ^{n-1}x\,dx}

ohne Beweis

 

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{displaystyle int _{0}^{frac {pi }{2}}sin ^{2n}x\,dx={frac {1}{2^{2n}}}{2n choose n}{frac {pi }{2}}}

ohne Beweis

 

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{displaystyle int _{0}^{frac {pi }{2}}sin ^{2n+1}x\,dx={frac {1}{2n+1}}left[{frac {1}{2^{2n}}}{2n choose n} ight]^{-1}}

ohne Beweis

 

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{displaystyle int _{-infty }^{infty }{frac {sin alpha x}{x}}\,dx=pi qquad alpha >0}

1. Beweis

Betrachte die Formel {displaystyle int _{0}^{infty }e^{-ax}\,{frac {sin bx}{x}}\,dx=arctan left({frac {b}{a}} ight)} für {displaystyle a,b>0\,}.

Lässt man {displaystyle a o 0+\,} gehen, so erhält man {displaystyle int _{0}^{infty }{frac {sin bx}{x}}\,dx={frac {pi }{2}}}.

Also ist {displaystyle int _{-infty }^{infty }{frac {sin bx}{x}}\,dx=pi }.

2. Beweis

Nach der Formel von Lobatschewski ist {displaystyle int _{-infty }^{infty }1cdot {frac {sin x}{x}}\,dx=int _{0}^{pi }1\,dx=pi }.

Substituiert man {displaystyle x o alpha x\,}, so erhält man die behauptete Formel.

 

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{displaystyle int _{0}^{infty }{frac {alpha \,sin x}{alpha ^{2}+x^{2}}}\,dx={ ext{Shi}}(alpha )cosh(alpha )-{ ext{Chi}}(alpha )sinh(alpha )qquad { ext{Re}}(alpha )>0}

ohne Beweis

 

1.6Bearbeiten

{displaystyle int _{-infty }^{infty }sin left(x^{2}-{frac {alpha ^{2}}{x^{2}}} ight)dx={sqrt {frac {pi }{2}}}\,e^{-2alpha }}

Beweis

Siehe Berechnung von {displaystyle int _{-infty }^{infty }cos left(x^{2}-{frac {alpha ^{2}}{x^{2}}} ight)dx={sqrt {frac {pi }{2}}}\,e^{-2alpha }}

 

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

Beweis

{displaystyle int _{-infty }^{infty }sin left(x^{2}+{frac {alpha ^{2}}{x^{2}}} ight)dx=int _{-infty }^{infty }sin left(left(x-{frac {alpha }{x}} ight)^{2}+2alpha ight)dx}

ist nach der Formel {displaystyle int _{-infty }^{infty }fleft(x-{frac {b}{x}} ight)dx=int _{-infty }^{infty }f(x)dx}, gleich

{displaystyle int _{-infty }^{infty }sin(x^{2}+2alpha )dx=int _{-infty }^{infty }left(sin x^{2}\,cos 2alpha +cos x^{2}\,sin 2alpha ight)dx={sqrt {frac {pi }{2}}}\,cos 2alpha +{sqrt {frac {pi }{2}}}\,sin 2alpha }.

 

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{displaystyle int _{-infty }^{infty }{frac {x\,sin alpha x}{1+x^{2}}}\,dx=pi \,e^{-alpha }qquad alpha >0}

ohne Beweis

 

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{displaystyle int _{-infty }^{infty }{frac {x\,sin alpha x}{1-x^{2}}}\,dx=-pi \,cos alpha qquad alpha >0}

ohne Beweis

 

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{displaystyle int _{-infty }^{infty }{frac {|sin alpha x|}{1+x^{2}}}\,dx=4sinh alpha \,;{ ext{artanh}}\,e^{-alpha }qquad alpha >0}

Beweis

Aus der Fourierreihe {displaystyle |sin alpha x|={frac {2}{pi }}+{frac {2}{pi }}sum _{n=1}^{infty }left({frac {1}{2n+1}}-{frac {1}{2n-1}} ight)cos 2nalpha x} ergibt sich

{displaystyle int _{-infty }^{infty }{frac {|sin alpha x|}{1+x^{2}}}\,dx=2+2sum _{n=1}^{infty }left({frac {1}{2n+1}}-{frac {1}{2n-1}} ight)\,underbrace {{frac {1}{pi }}int _{-infty }^{infty }{frac {cos 2nalpha x}{1+x^{2}}}\,dx} _{e^{-2nalpha }}}

{displaystyle =2+2sum _{n=1}^{infty }{frac {(e^{-alpha })^{2n}}{2n+1}}-2sum _{n=1}^{infty }{frac {(e^{-alpha })^{2n}}{2n-1}}=2sum _{n=0}^{infty }{frac {(e^{-alpha })^{2n}}{2n+1}}-2sum _{n=0}^{infty }{frac {(e^{-alpha })^{2n+2}}{2n+1}}}

{displaystyle =2\,(e^{alpha }-e^{-alpha })sum _{n=0}^{infty }{frac {(e^{-alpha })^{2n+1}}{2n+1}}=4sinh alpha \,\,{ ext{artanh}}\,e^{-alpha }}.

 

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{displaystyle int _{0}^{frac {pi }{2}}{frac {k\,sin \,x}{sqrt {1-k^{2}sin ^{2}x}}}\,dx={ ext{artanh}}\,k}

ohne Beweis

 

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{displaystyle int _{-infty }^{infty }|sin x|^{alpha -1}\,{frac {sin x}{x}}\,dx=2^{alpha -1}\,{frac {Gamma ^{2}!left({frac {alpha }{2}} ight)}{Gamma (alpha )}}qquad { ext{Re}}(alpha )>0}

Beweis

Die Funktion {displaystyle f(x)=|sin x|^{alpha -1}\,} ist {displaystyle pi \,}-periodisch. Daher gilt nach der Formel von Lobatschewski

{displaystyle int _{-infty }^{infty }|sin x|^{alpha -1}\,{frac {sin x}{x}}\,dx=int _{0}^{pi }|sin x|^{alpha -1}\,dx=2int _{0}^{frac {pi }{2}}sin ^{alpha -1}x\,dx=Bleft({frac {alpha }{2}},{frac {1}{2}} ight)={frac {Gamma left({frac {alpha }{2}} ight)\,{sqrt {pi }}}{Gamma left({frac {alpha }{2}}+{frac {1}{2}} ight)}}}.

Und das ist unter Verwendung der Legendreschen Verdopplungsformel gleich {displaystyle 2^{alpha -1}\,{frac {Gamma ^{2}left({frac {alpha }{2}} ight)}{Gamma (alpha )}}}.

 

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{displaystyle int _{-infty }^{infty }{ ext{sinc}}(x)cdot { ext{sinc}}(u-x)\,dx=pi cdot { ext{sinc}}(u)}    oder für {displaystyle f(x)={ ext{sinc}}(pi x)\,} gilt {displaystyle f*f=f\,}.

1. Beweis (Selbst-Faltung der sinc-Funktion)

{displaystyle { ext{sinc}}(x)cdot { ext{sinc}}(u-x)={frac {sin x}{x}}cdot {frac {sin(u-x)}{u-x}}}

{displaystyle {frac {1}{u}}cdot left({frac {1}{u-x}}+{frac {1}{x}} ight)cdot sin xcdot sin(u-x)={frac {1}{u}}cdot left(sin xcdot {frac {sin(u-x)}{u-x}}+{frac {sin x}{x}}cdot sin(u-x) ight)}

{displaystyle Rightarrow \,int _{-infty }^{infty }{ ext{sinc}}(x)cdot { ext{sinc}}(u-x)\,dx={frac {1}{u}}\,int _{-infty }^{infty }{frac {sin(x-u)}{x-u}}\,sin x\,dx\,+\,{frac {1}{u}}\,int _{-infty }^{infty }{frac {sin x}{x}}\,sin(u-x)\,dx}

{displaystyle ={frac {1}{u}}int _{-infty }^{infty }{frac {sin x}{x}}{Big (}sin(u+x)+sin(u-x){Big )}\,dx={frac {1}{u}}int _{-infty }^{infty }{frac {sin x}{x}}cdot 2cdot sin ucdot cos x\,dx={frac {sin u}{u}}int _{-infty }^{infty }{frac {sin 2x}{x}}\,dx=pi cdot { ext{sinc}}(u)}

2. Beweis

Die Fouriertransformierte von {displaystyle f(t)={ ext{sinc}}\,pi t} ist die Rechtecksfunktion.

{displaystyle {mathcal {F}}[f](s)=int _{-infty }^{infty }f(t)\,e^{-2pi ist}\,dt=int _{-infty }^{infty }{frac {sin pi tcdot cos 2pi st}{pi t}}dt}

{displaystyle =int _{-infty }^{infty }{frac {sin pi (2s+1)t-sin pi (2s-1)t}{2pi t}} dt={frac {{ ext{sgn}}left(s+{frac {1}{2}} ight)}{2}}-{frac {{ ext{sgn}}left(s-{frac {1}{2}} ight)}{2}}=sqcap (s):={egin{cases}0&|x|>{frac {1}{2}}\{frac {1}{2}}&|x|={frac {1}{2}}\1&|x|<{frac {1}{2}}end{cases}}}

In der Faltungsformel {displaystyle {mathcal {F}}[f*g](s)={mathcal {F}}[f](s)cdot {mathcal {F}}[g](s)} setze {displaystyle f=g={ ext{sinc}}\,}:

{displaystyle {mathcal {F}}[f*f](s)={mathcal {F}}[f](s)cdot {mathcal {F}}[f](s)=sqcap (s)cdot sqcap (s)}.

Für ein {displaystyle s eq pm {frac {1}{2}}} stimmt dies mit {displaystyle sqcap (s)={mathcal {F}}[f](s)} überein.

Also ist {displaystyle f*f=f\,} oder {displaystyle int _{-infty }^{infty }{ ext{sinc}}\,pi tcdot { ext{sinc}}\,pi (u-t)\,dt={ ext{sinc}}\,pi u}.

 

2.1Bearbeiten

{displaystyle int _{0}^{pi }sin nx\,sin mx\,dx=delta _{mn}{frac {pi }{2}}qquad n,min mathbb {Z} ^{geq 1}}

Beweis

Aus der Formel {displaystyle 2\,sin nx\,sin mx=cos(n-m)x-cos(n+m)x} folgt

{displaystyle 2int _{0}^{pi }sin nx\,sin mx\,dx=underbrace {int _{0}^{pi }cos(n-m)x\,dx} _{=delta _{nm}\,pi }-underbrace {int _{0}^{pi }cos(n+m)x\,dx} _{=0}}.

 

2.2Bearbeiten

{displaystyle {frac {1}{pi }}int _{0}^{pi }x^{2}\,sin nx\,sin mx\,dx=left{{egin{matrix}{frac {pi ^{2}}{6}}pm {frac {1}{4nm}}&,&n=m\\{frac {(-1)^{n-m}}{(n-m)^{2}}}pm {frac {(-1)^{n+m}}{(n+m)^{2}}}&,&n eq mend{matrix}} ight.qquad n,min mathbb {Z} ^{>0}}

ohne Beweis

 

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{displaystyle int _{0}^{infty }{frac {sin ^{2n}x}{x^{2m}}}\,dx={frac {pi cdot (-1)^{n-m}}{2^{2n}\,(2m-1)!}}\,sum _{k=0}^{n}(-1)^{k}\,{2n choose k}\,(2n-2k)^{2m-1}qquad n,min mathbb {Z} ^{>0}qquad ngeq m}

Beweis

{displaystyle {frac {1}{x^{2m}}}={frac {1}{Gamma (2m)}}int _{0}^{infty }t^{2m-1}\,e^{-xt}\,dt}

{displaystyle Rightarrow \,I=int _{0}^{infty }{frac {sin ^{2n}x}{x^{2m}}}\,dx={frac {1}{(2m-1)!}}int _{0}^{infty }int _{0}^{infty }sin ^{2n}x\,\,e^{-xt}\,dx\,\,t^{2m-1}\,dt}

{displaystyle ={frac {1}{(2m-1)!}}int _{0}^{infty }{frac {(2n)!}{tcdot (t^{2}+2^{2})cdots (t^{2}+(2n)^{2})}}\,\,t^{2m-1}\,dt}

{displaystyle ={frac {(2n)!}{(2m-1)!}}int _{0}^{infty }{frac {t^{2m-2}}{(t^{2}+2^{2})cdots (t^{2}+(2n)^{2})}}\,dt={frac {(2n)!\,2^{2m-1}}{(2m-1)!\,2^{2n}}}int _{0}^{infty }{frac {t^{2m-2}}{(t^{2}+1)cdots (t^{2}+n^{2})}}\,dt}

Differenziert man die Formel

{displaystyle int _{0}^{infty }{frac {cos 2alpha t}{(t^{2}+1)cdots (t^{2}+n^{2})}}\,dt={frac {(-1)^{n-1}\,pi }{2cdot (2n)!}}sum _{k=0}^{n}(-1)^{k}\,{2n choose k}\,(2n-2k)\,e^{-alpha (2n-2k)}}

{displaystyle (2m-2)}-mal nach {displaystyle alpha \,} und setzt anschließend {displaystyle alpha =0\,}, so ist

{displaystyle (-1)^{m-1}\,2^{2m-2}\,int _{0}^{infty }{frac {t^{2m-2}}{(t^{2}+1)cdots (t^{2}+n^{2})}}\,dt={frac {(-1)^{n-1}cdot pi }{2cdot (2n)!}}sum _{k=0}^{n}(-1)^{k}\,{2n choose k}\,(2n-2k)^{2m-1}}.

Also ist {displaystyle I={frac {(-1)^{n-m}cdot pi }{(2m-1)!\,\,2^{2n}}}\,sum _{k=0}^{n}(-1)^{k}\,{2n choose k}\,(2n-2k)^{2m-1}}.

 

2.4Bearbeiten

{displaystyle int _{0}^{infty }{frac {sin ^{2n+1}x}{x^{2m+1}}}\,dx={frac {pi cdot (-1)^{n-m}}{2^{2n+1}\,(2m)!}}\,sum _{k=0}^{n}(-1)^{k}\,{2n+1 choose k}\,(2n+1-2k)^{2m}qquad n,min mathbb {Z} ^{>0}qquad ngeq m}

Beweis

{displaystyle {frac {1}{x^{2m+1}}}={frac {1}{Gamma (2m+1)}}int _{0}^{infty }t^{2m}\,e^{-xt}\,dt}

{displaystyle Rightarrow \,I:=int _{0}^{infty }{frac {sin ^{2n+1}x}{x^{2m+1}}}\,dx={frac {1}{(2m)!}}int _{0}^{infty }int _{0}^{infty }sin ^{2n+1}xcdot e^{-xt}\,dxcdot t^{2m}\,dt}

{displaystyle ={frac {1}{(2m)!}}int _{0}^{infty }{frac {(2n+1)!}{(t^{2}+1)(t^{2}+3^{2})cdots (t^{2}+(2n+1)^{2})}}cdot t^{2m}\,dt}

{displaystyle ={frac {(2n+1)!}{(2m)!}}int _{0}^{infty }{frac {2^{2m}\,t^{2m}}{(4t^{2}+1)(4t^{2}+3^{2})cdots (4t^{2}+(2n+1)^{2})}}cdot 2cdot dt}

{displaystyle ={frac {(2n+1)!\,2^{2m}}{(2m)!\,2^{2n+1}}}\,int _{0}^{infty }{frac {t^{2m}}{left[t^{2}+left({frac {1}{2}} ight)^{2} ight]cdot left[t^{2}+left({frac {3}{2}} ight)^{2} ight]cdots left[t^{2}+left({frac {2n+1}{2}} ight)^{2} ight]}}\,dt}

Differenziert man die Formel

{displaystyle int _{0}^{infty }{frac {cos 2alpha t}{left[t^{2}+left({frac {1}{2}} ight)^{2} ight]cdot left[t^{2}+left({frac {3}{2}} ight)^{2} ight]cdots left[t^{2}+left({frac {2n+1}{2}} ight)^{2} ight]}}\,dt={frac {(-1)^{n}\,pi }{(2n+1)!}}sum _{k=0}^{n}(-1)^{k}\,{2n+1 choose k}\,e^{-alpha (2n+1-2k)}}

{displaystyle 2m}-mal nach {displaystyle alpha \,} und setzt anschließend {displaystyle alpha =0}, so ist

{displaystyle (-1)^{m}\,2^{2m}int _{0}^{infty }{frac {t^{2m}}{left[t^{2}+left({frac {1}{2}} ight)^{2} ight]cdot left[t^{2}+left({frac {3}{2}} ight)^{2} ight]cdots left[t^{2}+left({frac {2n+1}{2}} ight)^{2} ight]}}\,dt={frac {(-1)^{n}cdot pi }{(2n+1)!}}sum _{k=0}^{n}(-1)^{k}\,{2n+1 choose k}\,(2n+1-2k)^{2m}}.

Also ist {displaystyle I={frac {pi \,(-1)^{n-m}}{(2m)!\,2^{2n+1}}}sum _{k=0}^{n}(-1)^{k}\,{2n+1 choose k}\,(2n+1-2k)^{2m}}.

 

2.5Bearbeiten

{displaystyle int _{0}^{infty }{frac {sin ^{2n+1}x}{x^{2m}}}\,dx={frac {(-1)^{n-m}}{2^{2n}\,(2m-1)!}}\,sum _{k=0}^{n-1}(-1)^{k}\,{2n+1 choose k}\,(2n+1-2k)^{2m-1}\,log(2n+1-2k)qquad n,min mathbb {Z} ^{>0}qquad ngeq m}

ohne Beweis

 

2.6Bearbeiten

{displaystyle int _{0}^{infty }{frac {sin ^{2n}x}{x^{2m+1}}}\,dx={frac {(-1)^{n-m-1}}{2^{2n-1}\,(2m)!}}\,sum _{k=0}^{n-1}(-1)^{k}\,{2n choose k}\,(2n-2k)^{2m}\,log(2n-2k)qquad n,min mathbb {Z} ^{geq 0}qquad n>m}

Beweis

{displaystyle {frac {1}{x^{2m+1}}}={frac {1}{Gamma (2m+1)}}int _{0}^{infty }t^{2m}\,e^{-xt}\,dt}

{displaystyle I:=int _{0}^{infty }{frac {sin ^{2n}x}{x^{2m+1}}}\,dx={frac {1}{(2m)!}}int _{0}^{infty }int _{0}^{infty }sin ^{2n}x\,e^{-xt}\,dxcdot t^{2m}\,dt}

{displaystyle ={frac {(2n)!}{(2m)!}}int _{0}^{infty }{frac {t^{2m-1}}{(t^{2}+2^{2})(t^{2}+4^{2})cdots (t^{2}+(2n)^{2})}}\,dt}

{displaystyle ={frac {(2n)!cdot 2^{2m}}{(2m)!cdot 2^{2n}}}int _{0}^{infty }{frac {t^{2m-1}}{(t^{2}+1)(t^{2}+2^{2})cdots (t^{2}+n^{2})}}\,dt}

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

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

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

 

2.7Bearbeiten

{displaystyle int _{0}^{infty }{frac {|sin alpha x|-|sin eta x|}{x}}\,dx={frac {2}{pi }}log {frac {alpha }{eta }}qquad alpha ,eta >0}

Beweis

Aus der Fourierreihe {displaystyle |sin alpha x|={frac {2}{pi }}+{frac {2}{pi }}sum _{n=1}^{infty }left({frac {1}{2n+1}}-{frac {1}{2n-1}} ight)cos 2nalpha x} ergibt sich

{displaystyle |sin alpha x|-|sin eta x|={frac {2}{pi }}sum _{n=1}^{infty }left({frac {1}{2n+1}}-{frac {1}{2n-1}} ight){Big (}cos 2nalpha x-cos 2neta x{Big )}}.

Also ist {displaystyle int _{0}^{infty }{frac {|sin alpha x|-|sin eta x|}{x}}\,dx={frac {2}{pi }}sum _{n=1}^{infty }left({frac {1}{2n+1}}-{frac {1}{2n-1}} ight)int _{0}^{infty }{frac {cos 2nalpha x-cos 2neta x}{x}}\,dx},

wobei das Frullanische Integral {displaystyle int _{0}^{infty }{frac {cos 2nalpha x-cos 2neta x}{x}}\,dx=log {frac {eta }{alpha }}} nicht von {displaystyle n\,} abhängt.

Und die Reihe {displaystyle sum _{n=1}^{infty }left({frac {1}{2n+1}}-{frac {1}{2n-1}} ight)} konvergiert gegen {displaystyle -1\,}.

 

2.8Bearbeiten

{displaystyle J_{ u }(z)={frac {1}{2pi }}int _{C}e^{-izsin t+i u t}\,dtqquad { ext{Re}}(z)>0\,,\, u in mathbb {C} }

{displaystyle C\,} ist hierbei die Kurve, die gradlinig von {displaystyle -pi +iinfty \,} über {displaystyle -pi ,pi \,} nach {displaystyle pi +iinfty \,} läuft.

Beweis (Formel nach Sommerfeld)

{displaystyle f(t):={frac {1}{2pi }}\,e^{-izsin t+i u t}\,} ist auf ganz {displaystyle mathbb {C} } holomorph.

{displaystyle int _{C}f(t)\,dt=int _{infty }^{0}f(-pi +it)\,i\,dt+int _{-pi }^{pi }f(t)\,dt+int _{0}^{infty }f(pi +it)\,i\,dt}

{displaystyle =int _{-pi }^{pi }f(t)\,dt+iint _{0}^{infty }left(f(pi +it)-f(-pi +it) ight)dt}.

Das erste Integral ist {displaystyle {frac {1}{pi }}int _{0}^{pi }cos(zsin t- u t)\,dt}

und das zweite Integral ist wegen {displaystyle f(pm pi +it)={frac {1}{2pi }}\,e^{-zsinh t- u t}\,e^{pm ipi u }}

gleich {displaystyle -{frac {sin pi u }{pi }}int _{0}^{infty }e^{-zsinh t- u t}\,dt}.

Also ist {displaystyle int _{C}f(t)\,dt={frac {1}{pi }}int _{0}^{pi }cos(zsin t- u t)\,dt-{frac {sin pi u }{pi }}int _{0}^{infty }e^{-zsinh t- u t}\,dt},

was nach der Schläfli Formel gerade eine Darstellung der Besselfunktion {displaystyle J_{ u }(z)\,} ist.

 

3.1Bearbeiten

{displaystyle int _{0}^{infty }sin left(alpha \,t^{frac {1}{z}}+eta ight)\,dt={frac {Gamma (z+1)}{alpha ^{z}}}\,sin left({frac {pi z}{2}}+eta ight)qquad 0<z<1\,,\,alpha >0\,,\,eta in mathbb {C} }

ohne Beweis

 

n.1Bearbeiten

{displaystyle int _{0}^{infty }prod _{k=0}^{n}{frac {sin(a_{k}\,x)}{x}}\,dx={frac {pi }{2}}cdot prod _{k=1}^{n}a_{k}qquad qquad a_{k}>0}     und     {displaystyle a_{0}>sum _{k=1}^{n}a_{k}}

ohne Beweis