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

Zurück zu Bestimmte Integrale

0.1

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

Beweis

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

0.2

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

Beweis

$F(x)=2\log \left(2\sin {\frac {x}{2}}\right)-\log(2\sin x)$ ist eine Stammfunktion von ${\frac {1}{\sin x}}$.

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

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

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

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

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

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

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

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

0.3

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

Beweis

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

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

0.4

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

Beweis

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

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

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

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

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

0.5

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

Beweis (Formel nach Ramanujan)

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

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

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

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

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

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

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

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

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

Daraus ergibt sich das gesuchte Integral:

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

0.6

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

ohne Beweis

1.1

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

ohne Beweis

1.2

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

ohne Beweis

1.3

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

ohne Beweis

1.4

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

1. Beweis

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

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

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

2. Beweis

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

Substituiert man $x\to \alpha x\,$, so erhält man die behauptete Formel.

1.5

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

ohne Beweis

1.6

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

Beweis

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

1.7

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

Beweis

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

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

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

1.8

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

ohne Beweis

1.9

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

ohne Beweis

1.10

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

Beweis

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

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

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

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

1.11

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

ohne Beweis

1.12

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

Beweis

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

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

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

1.13

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

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

${\text{sinc}}(x)\cdot {\text{sinc}}(u-x)={\frac {\sin x}{x}}\cdot {\frac {\sin(u-x)}{u-x}}$

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

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

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

2. Beweis

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

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

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

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

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

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

Also ist $f*f=f\,$ oder $\int _{-\infty }^{\infty }{\text{sinc}}\,\pi t\cdot {\text{sinc}}\,\pi (u-t)\,dt={\text{sinc}}\,\pi u$.

2.1

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

Beweis

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

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

2.2

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

ohne Beweis

2.3

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

Beweis

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

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

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

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

Differenziert man die Formel

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

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

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

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

2.4

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

Beweis

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

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

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

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

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

Differenziert man die Formel

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

$2m$-mal nach $\alpha \,$ und setzt anschließend $\alpha =0$, so ist

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

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

2.5

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

ohne Beweis

2.6

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

Beweis

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

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

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

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

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

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

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

2.7

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

Beweis

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

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

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

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

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

2.8

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

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

Beweis (Formel nach Sommerfeld)

$f(t):={\frac {1}{2\pi }}\,e^{-iz\sin t+i\nu t}\,$ ist auf ganz $\mathbb {C}$ holomorph.

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

$=\int _{-\pi }^{\pi }f(t)\,dt+i\int _{0}^{\infty }\left(f(\pi +it)-f(-\pi +it)\right)dt$.

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

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

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

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

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

3.1

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

ohne Beweis

n.1

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

ohne Beweis