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

Zurück zu Bestimmte Integrale

##### 0.1
${\displaystyle \int _{0}^{\frac {\pi }{2}}{\frac {x}{\sin x}}\,dx=2G}$
Beweis

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

##### 0.2
${\displaystyle \int _{0}^{\frac {\pi }{2}}{\frac {x^{2}}{\sin x}}\,dx=2\pi G-{\frac {7}{2}}\,\zeta (3)}$
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 ${\displaystyle A-B=2\pi G-{\frac {7}{2}}\zeta (3)}$.

##### 0.3
${\displaystyle \int _{0}^{\frac {\pi }{4}}{\frac {x^{2}}{\sin ^{2}x}}\,dx=G+{\frac {\pi }{4}}\log 2-{\frac {\pi ^{2}}{16}}}$
Beweis

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

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

##### 0.4
${\displaystyle \int _{0}^{\frac {\pi }{4}}{\frac {x^{3}}{\sin ^{2}x}}\,dx={\frac {3}{4}}\pi G-{\frac {\pi ^{3}}{64}}+{\frac {3}{32}}\pi ^{2}\log 2-{\frac {105}{64}}\zeta (3)}$
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 ${\displaystyle I=-{\frac {\pi ^{3}}{64}}+{\frac {3}{32}}\pi ^{2}\log 2+{\frac {3}{4}}\pi \underbrace {\sum _{k=1}^{\infty }{\frac {\sin {\frac {k\pi }{2}}}{k^{2}}}} _{=G}+{\frac {3}{2}}\underbrace {\sum _{k=1}^{\infty }{\frac {\cos {\frac {k\pi }{2}}}{k^{3}}}} _{=-{\frac {3}{32}}\zeta (3)}-{\frac {3}{2}}\zeta (3)}$.

##### 0.5
${\displaystyle \int _{0}^{1}\sin(\pi x)\,x^{x}\,(1-x)^{1-x}\,dx={\frac {\pi e}{24}}}$
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:

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

##### 0.6
${\displaystyle \int _{0}^{1}{\frac {\sin(\pi x)}{x^{x}\,(1-x)^{1-x}}}\,dx={\frac {\pi }{e}}}$
ohne Beweis

##### 1.1
${\displaystyle \int _{0}^{\frac {\pi }{2}}\sin ^{n+1}x\,dx={\frac {n}{n+1}}\int _{0}^{\frac {\pi }{2}}\sin ^{n-1}x\,dx}$
ohne Beweis

##### 1.2
${\displaystyle \int _{0}^{\frac {\pi }{2}}\sin ^{2n}x\,dx={\frac {1}{2^{2n}}}{2n \choose n}{\frac {\pi }{2}}}$
ohne Beweis

##### 1.3
${\displaystyle \int _{0}^{\frac {\pi }{2}}\sin ^{2n+1}x\,dx={\frac {1}{2n+1}}\left[{\frac {1}{2^{2n}}}{2n \choose n}\right]^{-1}}$
ohne Beweis

##### 1.4
${\displaystyle \int _{-\infty }^{\infty }{\frac {\sin \alpha x}{x}}\,dx=\pi \qquad \alpha >0}$
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.

##### 1.5
${\displaystyle \int _{0}^{\infty }{\frac {\alpha \,\sin x}{\alpha ^{2}+x^{2}}}\,dx={\text{Shi}}(\alpha )\cosh(\alpha )-{\text{Chi}}(\alpha )\sinh(\alpha )\qquad {\text{Re}}(\alpha )>0}$
ohne Beweis

##### 1.6
${\displaystyle \int _{-\infty }^{\infty }\sin \left(x^{2}-{\frac {\alpha ^{2}}{x^{2}}}\right)dx={\sqrt {\frac {\pi }{2}}}\,e^{-2\alpha }}$
Beweis

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

##### 1.7
${\displaystyle \int _{-\infty }^{\infty }\sin \left(x^{2}+{\frac {\alpha ^{2}}{x^{2}}}\right)dx={\sqrt {\frac {\pi }{2}}}\,(\cos 2\alpha +\sin 2\alpha )}$
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

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

##### 1.8
${\displaystyle \int _{-\infty }^{\infty }{\frac {x\,\sin \alpha x}{1+x^{2}}}\,dx=\pi \,e^{-\alpha }\qquad \alpha >0}$
ohne Beweis

##### 1.9
${\displaystyle \int _{-\infty }^{\infty }{\frac {x\,\sin \alpha x}{1-x^{2}}}\,dx=-\pi \,\cos \alpha \qquad \alpha >0}$
ohne Beweis

##### 1.10
${\displaystyle \int _{-\infty }^{\infty }{\frac {|\sin \alpha x|}{1+x^{2}}}\,dx=4\sinh \alpha \,\;{\text{artanh}}\,e^{-\alpha }\qquad \alpha >0}$
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

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

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

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

##### 1.11
${\displaystyle \int _{0}^{\frac {\pi }{2}}{\frac {k\,\sin \,x}{\sqrt {1-k^{2}\sin ^{2}x}}}\,dx={\text{artanh}}\,k}$
ohne Beweis

##### 1.12
${\displaystyle \int _{-\infty }^{\infty }|\sin x|^{\alpha -1}\,{\frac {\sin x}{x}}\,dx=2^{\alpha -1}\,{\frac {\Gamma ^{2}\!\left({\frac {\alpha }{2}}\right)}{\Gamma (\alpha )}}\qquad {\text{Re}}(\alpha )>0}$
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 ${\displaystyle 2^{\alpha -1}\,{\frac {\Gamma ^{2}\left({\frac {\alpha }{2}}\right)}{\Gamma (\alpha )}}}$.

##### 1.13
${\displaystyle \int _{-\infty }^{\infty }{\text{sinc}}(x)\cdot {\text{sinc}}(u-x)\,dx=\pi \cdot {\text{sinc}}(u)}$    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 ${\displaystyle \int _{-\infty }^{\infty }{\text{sinc}}\,\pi t\cdot {\text{sinc}}\,\pi (u-t)\,dt={\text{sinc}}\,\pi u}$.

##### 2.1
${\displaystyle \int _{0}^{\pi }\sin nx\,\sin mx\,dx=\delta _{mn}{\frac {\pi }{2}}\qquad n,m\in \mathbb {Z} ^{\geq 1}}$
Beweis

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

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

##### 2.2
${\displaystyle {\frac {1}{\pi }}\int _{0}^{\pi }x^{2}\,\sin nx\,\sin mx\,dx=\left\{{\begin{matrix}{\frac {\pi ^{2}}{6}}\pm {\frac {1}{4nm}}&,&n=m\\\\{\frac {(-1)^{n-m}}{(n-m)^{2}}}\pm {\frac {(-1)^{n+m}}{(n+m)^{2}}}&,&n\neq m\end{matrix}}\right.\qquad n,m\in \mathbb {Z} ^{>0}}$
ohne Beweis

##### 2.3
${\displaystyle \int _{0}^{\infty }{\frac {\sin ^{2n}x}{x^{2m}}}\,dx={\frac {\pi \cdot (-1)^{n-m}}{2^{2n}\,(2m-1)!}}\,\sum _{k=0}^{n}(-1)^{k}\,{2n \choose k}\,(2n-2k)^{2m-1}\qquad n,m\in \mathbb {Z} ^{>0}\qquad n\geq m}$
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 ${\displaystyle I={\frac {(-1)^{n-m}\cdot \pi }{(2m-1)!\,\,2^{2n}}}\,\sum _{k=0}^{n}(-1)^{k}\,{2n \choose k}\,(2n-2k)^{2m-1}}$.

##### 2.4
${\displaystyle \int _{0}^{\infty }{\frac {\sin ^{2n+1}x}{x^{2m+1}}}\,dx={\frac {\pi \cdot (-1)^{n-m}}{2^{2n+1}\,(2m)!}}\,\sum _{k=0}^{n}(-1)^{k}\,{2n+1 \choose k}\,(2n+1-2k)^{2m}\qquad n,m\in \mathbb {Z} ^{>0}\qquad n\geq m}$
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 ${\displaystyle I={\frac {\pi \,(-1)^{n-m}}{(2m)!\,2^{2n+1}}}\sum _{k=0}^{n}(-1)^{k}\,{2n+1 \choose k}\,(2n+1-2k)^{2m}}$.

##### 2.5
${\displaystyle \int _{0}^{\infty }{\frac {\sin ^{2n+1}x}{x^{2m}}}\,dx={\frac {(-1)^{n-m}}{2^{2n}\,(2m-1)!}}\,\sum _{k=0}^{n-1}(-1)^{k}\,{2n+1 \choose k}\,(2n+1-2k)^{2m-1}\,\log(2n+1-2k)\qquad n,m\in \mathbb {Z} ^{>0}\qquad n\geq m}$
ohne Beweis

##### 2.6
${\displaystyle \int _{0}^{\infty }{\frac {\sin ^{2n}x}{x^{2m+1}}}\,dx={\frac {(-1)^{n-m-1}}{2^{2n-1}\,(2m)!}}\,\sum _{k=0}^{n-1}(-1)^{k}\,{2n \choose k}\,(2n-2k)^{2m}\,\log(2n-2k)\qquad n,m\in \mathbb {Z} ^{\geq 0}\qquad n>m}$
Beweis

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

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

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

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

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

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

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

##### 2.7
${\displaystyle \int _{0}^{\infty }{\frac {|\sin \alpha x|-|\sin \beta x|}{x}}\,dx={\frac {2}{\pi }}\log {\frac {\alpha }{\beta }}\qquad \alpha ,\beta >0}$
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 ${\displaystyle -1\,}$.

##### 2.8
${\displaystyle J_{\nu }(z)={\frac {1}{2\pi }}\int _{C}e^{-iz\sin t+i\nu t}\,dt\qquad {\text{Re}}(z)>0\,,\,\nu \in \mathbb {C} }$
${\displaystyle C\,}$ 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.

##### 3.1
${\displaystyle \int _{0}^{\infty }\sin \left(\alpha \,t^{\frac {1}{z}}+\beta \right)\,dt={\frac {\Gamma (z+1)}{\alpha ^{z}}}\,\sin \left({\frac {\pi z}{2}}+\beta \right)\qquad 00\,,\,\beta \in \mathbb {C} }$
ohne Beweis

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