0.1Bearbeiten

{\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 x\mapsto 2\arctan x} gleich {\displaystyle 2\int _{0}^{1}{\frac {\arctan x}{x}}\,dx=2G}.

 

0.2Bearbeiten

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

Beweis

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

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

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

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

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

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

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

 

0.3Bearbeiten

{\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)\right]_{0}^{\frac {\pi }{4}}+\int _{0}^{\frac {\pi }{4}}2x\cot x\,dx}

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

 

0.4Bearbeiten

{\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}}6x\log \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}}6x\log \sin x\,dx=\int _{0}^{\frac {\pi }{4}}6x\,dx\cdot \log 2+6\sum _{k=1}^{\infty }{\frac {1}{k}}\int _{0}^{\frac {\pi }{4}}x\cos 2kx\,dx}

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

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 {k\pi }{2}}}{k^{2}}}} _{=G}+{\frac {3}{2}}\underbrace {\sum _{k=1}^{\infty }{\frac {\cos {\frac {k\pi }{2}}}{k^{3}}}} _{=-{\frac {3}{32}}\zeta (3)}-{\frac {3}{2}}\zeta (3)}.

 

0.5Bearbeiten

{\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^{i\pi x}\,x^{x}\,{\frac {1-x}{(1-x)^{x}}}\,dx=\int _{0}^{1}e^{i\pi x}\,e^{(\log x)\,x}\,{\frac {1-x}{e^{\log(1-x)\,x}}}\,dx=\int _{0}^{1}(1-x)\,e^{(i\pi +\log x-\log(1-x))x}\,dx}.

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

{\displaystyle S=\int _{-\infty }^{\infty }{\frac {1}{e^{t}+1}}\,e^{(i\pi +t)\,{\frac {e^{t}}{e^{t}+1}}}\,{\frac {e^{t}}{(e^{t}+1)^{2}}}\,dt=\int _{-\infty +i\pi }^{\infty +i\pi }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\{z\in \mathbb {C} \,|\,-\pi \leq {\text{Im}}(z)\leq \pi \right\}} meromorph. 
Die einzige Polstelle liegt bei {\displaystyle z=0\,} und dort ist {\displaystyle {\text{res}}(f,0)=-{\frac {e}{24}}}.

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

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

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

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

{\displaystyle 2S=2\lim _{R\to \infty }\int _{\gamma _{R}}f\,dz} ist demnach {\displaystyle \lim _{R\to \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={\text{Im}}(S)={\frac {\pi e}{24}}}

 

0.6Bearbeiten

{\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

 

 

1.2Bearbeiten

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

ohne Beweis

 

 

1.3Bearbeiten

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

ohne Beweis

 

 

1.4Bearbeiten

{\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}}\right)} für {\displaystyle a,b>0\,}.

Lässt man {\displaystyle a\to 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 }1\cdot {\frac {\sin x}{x}}\,dx=\int _{0}^{\pi }1\,dx=\pi }.

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

 

1.5Bearbeiten

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

ohne Beweis

 

 

1.6Bearbeiten

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

Beweis

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

 

1.7Bearbeiten

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

Beweis

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

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

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

 

1.8Bearbeiten

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

ohne Beweis

 

 

1.9Bearbeiten

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

ohne Beweis

 

 

1.10Bearbeiten

{\displaystyle \int _{-\infty }^{\infty }{\frac {|\sin \alpha x|}{1+x^{2}}}\,dx=4\sinh \alpha \,\;{\text{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}}\right)\cos 2n\alpha x} ergibt sich

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

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

 

1.11Bearbeiten

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

ohne Beweis

 

 

1.12Bearbeiten

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

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

 

1.13Bearbeiten

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

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

{\displaystyle {\text{sinc}}(x)\cdot {\text{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}}\right)\cdot \sin x\cdot \sin(u-x)={\frac {1}{u}}\cdot \left(\sin x\cdot {\frac {\sin(u-x)}{u-x}}+{\frac {\sin x}{x}}\cdot \sin(u-x)\right)}

{\displaystyle \Rightarrow \,\int _{-\infty }^{\infty }{\text{sinc}}(x)\cdot {\text{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 2\cdot \sin u\cdot \cos x\,dx={\frac {\sin u}{u}}\int _{-\infty }^{\infty }{\frac {\sin 2x}{x}}\,dx=\pi \cdot {\text{sinc}}(u)}

2. Beweis

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

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

{\displaystyle =\int _{-\infty }^{\infty }{\frac {\sin \pi (2s+1)t-\sin \pi (2s-1)t}{2\pi t}}\ dt={\frac {{\text{sgn}}\left(s+{\frac {1}{2}}\right)}{2}}-{\frac {{\text{sgn}}\left(s-{\frac {1}{2}}\right)}{2}}=\sqcap (s):={\begin{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={\text{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\neq \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 }{\text{sinc}}\,\pi t\cdot {\text{sinc}}\,\pi (u-t)\,dt={\text{sinc}}\,\pi u}.

 

2.1Bearbeiten

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

Beweis

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

{\displaystyle 2\int _{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\{{\begin{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\neq m\end{matrix}}\right.\qquad n,m\in \mathbb {Z} ^{>0}}

ohne Beweis

 

 

2.3Bearbeiten

{\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,m\in \mathbb {Z} ^{>0}\qquad n\geq 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)!}{t\cdot (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 2\alpha t}{(t^{2}+1)\cdots (t^{2}+n^{2})}}\,dt={\frac {(-1)^{n-1}\,\pi }{2\cdot (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 }{2\cdot (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,m\in \mathbb {Z} ^{>0}\qquad n\geq 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}x\cdot e^{-xt}\,dx\cdot 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 2\cdot dt}

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

Differenziert man die Formel

{\displaystyle \int _{0}^{\infty }{\frac {\cos 2\alpha t}{\left[t^{2}+\left({\frac {1}{2}}\right)^{2}\right]\cdot \left[t^{2}+\left({\frac {3}{2}}\right)^{2}\right]\cdots \left[t^{2}+\left({\frac {2n+1}{2}}\right)^{2}\right]}}\,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}}\right)^{2}\right]\cdot \left[t^{2}+\left({\frac {3}{2}}\right)^{2}\right]\cdots \left[t^{2}+\left({\frac {2n+1}{2}}\right)^{2}\right]}}\,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,m\in \mathbb {Z} ^{>0}\qquad n\geq 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,m\in \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}\,dx\cdot 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 \beta x|}{x}}\,dx={\frac {2}{\pi }}\log {\frac {\alpha }{\beta }}\qquad \alpha ,\beta >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}}\right)\cos 2n\alpha x} ergibt sich

{\displaystyle |\sin \alpha x|-|\sin \beta x|={\frac {2}{\pi }}\sum _{n=1}^{\infty }\left({\frac {1}{2n+1}}-{\frac {1}{2n-1}}\right){\Big (}\cos 2n\alpha x-\cos 2n\beta x{\Big )}}.

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

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

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

 

2.8Bearbeiten

{\displaystyle J_{\nu }(z)={\frac {1}{2\pi }}\int _{C}e^{-iz\sin t+i\nu t}\,dt\qquad {\text{Re}}(z)>0\,,\,\nu \in \mathbb {C} }

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

Beweis (Formel nach Sommerfeld)

{\displaystyle f(t):={\frac {1}{2\pi }}\,e^{-iz\sin t+i\nu 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+i\int _{0}^{\infty }\left(f(\pi +it)-f(-\pi +it)\right)dt}.

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

und das zweite Integral ist wegen {\displaystyle f(\pm \pi +it)={\frac {1}{2\pi }}\,e^{-z\sinh t-\nu t}\,e^{\pm i\pi \nu }}

gleich {\displaystyle -{\frac {\sin \pi \nu }{\pi }}\int _{0}^{\infty }e^{-z\sinh t-\nu t}\,dt}.

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

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

 

3.1Bearbeiten

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