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

Zurück zu Bestimmte Integrale

1.1

${\frac {1}{2\pi i}}\int _{n-i\infty }^{n+i\infty }{\frac {\pi }{x\,\cos \pi x}}\,dx=2\sum _{k=n}^{\infty }{\frac {(-1)^{k}}{2k+1}}\qquad n\in \mathbb {Z} ^{>0}$

ohne Beweis

1.2

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

1. Beweis

Es sei $R>1,\alpha \geq 0,f(z)={\frac {e^{i\alpha z}}{1+z^{2}}},$

$K_{R}\,$ der Halbkreis von $R\,$ nach $-R\,$

und $\gamma _{R}=[-R,R]+K_{R}\,$

der geschlossene halbmondförmige Integrationsweg.

Für alle $z=x+iy\in \gamma _{R}$ ist der Imaginärteil $y\geq 0$

und somit $\left|e^{i\alpha z}\right|\leq e^{-\alpha y}\leq 1$.

Nun gilt $\left|\int _{K_{R}}f\,dz\right|\leq |K_{R}|\cdot \max _{z\in K_{R}}|f(z)|\leq R\pi \,{\frac {1}{R^{2}-1}}\to 0$ für $R\to \infty \,$.

Also ist $\int _{-\infty }^{\infty }f\,dz=\lim _{R\to \infty }\oint _{\gamma _{R}}f\,dz=2\pi i\,{\text{res}}(f,i)=\pi \,e^{-\alpha }$ und somit $\int _{-\infty }^{\infty }{\frac {\cos \alpha x}{1+x^{2}}}\,dx=\pi \,e^{-\alpha }$.

2. Beweis

Aus ${\frac {x}{1+x^{2}}}=\int _{0}^{\infty }\sin tx\,e^{-t}\,dt$ folgt $2\,{\frac {\cos \alpha x}{1+x^{2}}}=\int _{0}^{\infty }{\frac {2\,\sin tx\,\cos \alpha x}{x}}\,e^{-t}\,dt$.

Und das ist $\int _{0}^{\infty }{\frac {\sin(t+\alpha )x+\sin(t-\alpha )x}{x}}\,e^{-t}\,dt$.

Also ist $2\int _{0}^{\infty }{\frac {\cos \alpha x}{1+x^{2}}}\,dx=\pi \int _{0}^{\infty }\left[{\text{sgn}}(t+\alpha )+{\text{sgn}}(t-\alpha )\right]\,e^{-t}\,dt=\pi \int _{\alpha }^{\infty }e^{-t}\,dt=\pi \,e^{-\alpha }$.

1.3

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

ohne Beweis

1.4

$\int _{-\infty }^{\infty }{\frac {|\cos \alpha x|}{1+x^{2}}}\,dx=4\cosh \alpha \,\;\arctan e^{-\alpha }\qquad \alpha >0$

Beweis

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

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

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

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

1.5

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

Beweis

Die Funktion $f(z)=\exp \left(-z^{2}-{\frac {\alpha ^{2}}{z^{2}}}\right)$ ist auf dem Kreissektor $S_{R}:=\left\{re^{i\varphi }\,{\Big |}\,0<\varphi <{\frac {\pi }{4}}\;,\;0<r<R\right\}$ holomorph.
Auf dem Kreisbogen $K_{R}\,$ fällt $f\,$ für $R\to \infty \,$ exponentiell gegen null ab.

Daher ist $\lim _{R\to \infty }\int _{K_{R}}fdz=0$ und somit $J:=\int _{0}^{\infty }f(x)dx=\int _{0}^{\infty }f\left(e^{i{\frac {\pi }{4}}}x\right)\,e^{i{\frac {\pi }{4}}}\,dx$.

Also ist $J:=\int _{0}^{\infty }e^{-i\left(x^{2}-{\frac {\alpha ^{2}}{x^{2}}}\right)}\,{\frac {1+i}{\sqrt {2}}}\,dx=\int _{0}^{\infty }\left[\cos \left(x^{2}-{\frac {\alpha ^{2}}{x^{2}}}\right)-i\sin \left(x^{2}-{\frac {\alpha ^{2}}{x^{2}}}\right)\right]\,{\frac {1+i}{\sqrt {2}}}\,dx$

$={\frac {1}{\sqrt {2}}}\int _{0}^{\infty }\left[\cos \left(x^{2}-{\frac {\alpha ^{2}}{x^{2}}}\right)+\sin \left(x^{2}-{\frac {\alpha ^{2}}{x^{2}}}\right)\right]dx+{\frac {i}{\sqrt {2}}}\int _{0}^{\infty }\left[\cos \left(x^{2}-{\frac {\alpha ^{2}}{x^{2}}}\right)-\sin \left(x^{2}-{\frac {\alpha ^{2}}{x^{2}}}\right)\right]dx$.

Nun ist ${\sqrt {2}}\int _{0}^{\infty }\cos \left(x^{2}-{\frac {\alpha ^{2}}{x^{2}}}\right)dx={\text{Re}}(J)+{\text{Im}}(J)={\frac {\sqrt {\pi }}{2}}\,e^{-2\alpha }+0$

und ${\sqrt {2}}\int _{0}^{\infty }\sin \left(x^{2}-{\frac {\alpha ^{2}}{x^{2}}}\right)dx={\text{Re}}(J)-{\text{Im}}(J)={\frac {\sqrt {\pi }}{2}}\,e^{-2\alpha }-0$.

Also sind $\int _{-\infty }^{\infty }\cos \left(x^{2}-{\frac {\alpha ^{2}}{x^{2}}}\right)dx$ und $\int _{-\infty }^{\infty }\sin \left(x^{2}-{\frac {\alpha ^{2}}{x^{2}}}\right)dx$ jeweils ${\sqrt {\frac {\pi }{2}}}\,e^{-2\alpha }$.

1.6

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

Beweis

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

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

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

1.7

$J_{\nu }(z)={\frac {2}{\Gamma \left({\frac {1}{2}}\right)\Gamma \left(\nu +{\frac {1}{2}}\right)}}\left({\frac {z}{2}}\right)^{\nu }\int _{0}^{1}\cos(zx)\,(1-x^{2})^{\nu -{\frac {1}{2}}}\,dx\qquad {\text{Re}}(\nu )>-{\frac {1}{2}}$

Beweis (Poissonsche Darstellung der Besselfunktion)

Die Formel ist äquivalent zur Formel

$J_{\nu }(z)={\frac {1}{\Gamma \left({\frac {1}{2}}\right)\Gamma \left(\nu +{\frac {1}{2}}\right)}}\left({\frac {z}{2}}\right)^{\nu }\int _{-1}^{1}e^{izx}\,(1-x^{2})^{\nu -{\frac {1}{2}}}\,dx\qquad {\text{Re}}(\nu )>-{\frac {1}{2}}$

1.8

$\int _{0}^{\frac {\pi }{2}}\cos ^{2n}(x)\,dx={\frac {1}{2^{2n}}}\,{2n \choose n}\,{\frac {\pi }{2}}\qquad n\in \mathbb {N}$

Beweis

Für $m>n$ ist $\sum _{k=0}^{m-1}\left(2\cos {\frac {k\pi }{m}}\right)^{2n}=\sum _{k=0}^{m-1}\left[\exp \left(i\pi \cdot {\frac {k}{m}}\right)+\exp \left(-i\pi \cdot {\frac {k}{m}}\right)\right]^{2n}$

$=\sum _{k=0}^{m-1}\sum _{\ell =0}^{2n}{2n \choose \ell }\,\exp \left(i\pi \cdot {\frac {k}{m}}\cdot \ell \right)\exp \left(-i\pi \cdot {\frac {k}{m}}\cdot (2n-\ell )\right)=\sum _{\ell =0}^{2n}{2n \choose \ell }\sum _{k=0}^{m-1}\exp \left(2\pi i\cdot {\frac {\ell -n}{m}}\cdot k\right)$.

Nun ist $\sum _{k=0}^{m-1}\exp \left(2\pi i\cdot {\frac {\ell -n}{m}}\cdot k\right)=\left\{{\begin{matrix}0&{\text{für}}&\ell \neq n\\m&{\text{für}}&\ell =n\end{matrix}}\right.$.

Also ist $\sum _{k=0}^{m-1}\left(2\cos {\frac {k\pi }{m}}\right)^{2n}={2n \choose n}\cdot m$ und somit ist ${\frac {1}{m}}\sum _{k=0}^{m-1}\cos ^{2n}\left({\frac {k\pi }{m}}\right)={\frac {1}{2^{2n}}}{2n \choose n}$.

Durch den Grenzübergang $m\to \infty$ erhält man $\int _{0}^{1}\cos ^{2n}(\pi x)\,dx={\frac {1}{2^{2n}}}{2n \choose n}$.

Nach Substitution $x\mapsto {\frac {x}{\pi }}$ ist $\int _{0}^{\pi }\cos ^{2n}(x)\,dx={\frac {1}{2^{2n}}}{2n \choose n}\cdot \pi$.

Nachdem $\int _{0}^{\frac {\pi }{2}}\cos ^{2n}(x)\,dx=\int _{\frac {\pi }{2}}^{\pi }\cos ^{2n}(x)\,dx$ ist, folgt daraus die Behauptung.

1.9

$\int _{0}^{\frac {\pi }{2}}\cos ^{2n+1}(x)\,dx={\frac {1}{2n+1}}\left[{\frac {1}{2^{2n}}}\,{2n \choose n}\right]^{-1}\qquad n\in \mathbb {N}$

ohne Beweis

2.1

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

Beweis

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

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

2.2

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

ohne Beweis

2.3

$\int _{0}^{\pi }{\frac {\cos nx-\cos na}{\cos x-\cos a}}\,dx=\pi \,{\frac {\sin na}{\sin a}}\qquad n\in \mathbb {N} \,,\,a\in \mathbb {C}$

Beweis

Die Funktion $f_{n}(x)={\frac {\cos nx-\cos na}{\cos x-\cos a}}$

erfüllt die Rekursion $f_{n+1}(x)-2\cos a\cdot f_{n}(x)+f_{n-1}(x)=2\cos nx$.

Begründung:

$\cos(n+1)x-\cos(n+1)a\,+\,\cos(n-1)x-\cos(n-1)a$

$=\cos(n+1)x+\cos(n-1)x\,-\,\cos(n+1)a-\cos(n-1)a$

$=2\cos nx\cdot \cos x-2\cos na\cdot \cos a=2\cos nx\cdot (\cos x-\cos a)+2\cos a\cdot (\cos nx-\cos na)$

$\Rightarrow {\frac {\cos(n+1)x-\cos(n+1)a}{\cos x-\cos a}}+{\frac {\cos(n-1)x-\cos(n-1)a}{\cos x-\cos a}}=2\cos nx+2\cos a\cdot {\frac {\cos nx-\cos na}{\cos x-\cos a}}$



Also ist $s_{n+1}-2\cos a\cdot s_{n}+s_{n-1}=0$, wobei $s_{n}=\int _{0}^{\pi }f_{n}(x)\,dx$ sein soll.

Das Polynom $x^{2}-2\cos a\cdot x+1=0$ besitzt die Wurzeln $x_{1/2}=e^{\pm ia}\,$.

Daher hat die Folge $s_{n}\,$ die Form $s_{n}=C_{1}\,e^{ian}+C_{2}\,e^{-ian}=D_{1}\cos na+D_{2}\sin na$.

Aus $s_{0}=0\,\Rightarrow \,D_{1}=0$ und $s_{1}=\pi \,\Rightarrow \,D_{2}={\frac {\pi }{\sin a}}$ folgt schließlich $s_{n}=\pi \,{\frac {\sin na}{\sin a}}$.

2.4

$\int _{-\pi }^{\pi }{\frac {\cos nx}{1-2r\cos x+r^{2}}}\,dx={\frac {2\pi \,r^{n}}{1-r^{2}}}\qquad |r|<1\,,\,n\in \mathbb {Z} ^{\geq 0}$

Beweis

Betrachte die Poissonsche Integralformel

${\frac {1}{2\pi }}\int _{-\pi }^{\pi }P_{R}(r,\phi -\varphi )\,f(Re^{i\varphi })\,d\varphi =f(re^{i\phi })$, wobei der Kern $P_{R}(r,\phi )={\frac {R^{2}-r^{2}}{R^{2}-2Rr\cos \phi +r^{2}}}$ ist.

Setzt man $R=1\,,\,\phi =0$ und $f(z)=z^{n}\,$, so ist $P_{R}(r,\phi -x)={\frac {1-r^{2}}{1-2r\cos x+r^{2}}}$ und $f(Re^{ix})=e^{inx}\,$.

Also ist $\int _{-\pi }^{\pi }{\frac {\cos nx+i\sin nx}{1-2r\cos x+r^{2}}}\,dx={\frac {2\pi \,r^{n}}{1-r^{2}}}$. Der ungerade Anteil $\int _{-\pi }^{\pi }{\frac {\sin nx}{1-2r\cos x+r^{2}}}\,dx$ verschwindet dabei aus symmetriegründen.

2.5

$\int _{0}^{\infty }{\frac {|\cos \alpha x|-|\cos \beta x|}{x}}\,dx=\left(1-{\frac {2}{\pi }}\right)\log {\frac {\beta }{\alpha }}\qquad \alpha ,\beta >0$

Beweis

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

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

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

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

2.6

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

ohne Beweis

2.7

$\int _{-\infty }^{\infty }{\frac {\cos(\alpha x)}{1+2\cos \theta \,x+x^{2}}}\,dx={\frac {\pi }{\sin \theta }}\,{\frac {\cos(\alpha \cos \theta )}{e^{\alpha \sin \theta }}}\qquad \alpha \geq 0\,,\,\theta \in \mathbb {C} \setminus \pi \mathbb {Z}$

ohne Beweis

2.8

$\int _{-\infty }^{\infty }{\frac {\cos(\alpha x)}{x^{4}+\beta ^{4}}}\,dx={\frac {\pi }{\beta ^{3}\,{\sqrt {2}}}}\,\left(\cos {\frac {\alpha \beta }{\sqrt {2}}}+\sin {\frac {\alpha \beta }{\sqrt {2}}}\right)\,e^{-{\frac {\alpha \beta }{\sqrt {2}}}}\qquad \alpha ,\beta >0$

ohne Beweis

2.9

$\int _{-\infty }^{\infty }{\frac {\cos \alpha x}{\prod \limits _{k=0}^{\infty }\left(1+{\frac {x^{2}}{(\beta +k)^{2}}}\right)}}\,dx={\sqrt {\pi }}\,{\frac {\Gamma \left(\beta +{\frac {1}{2}}\right)}{\Gamma (\beta )}}\,{\text{sech}}^{2\beta }\left({\frac {\alpha }{2}}\right)\qquad \alpha ,\beta >0$

Beweis

Multipliziert man die Formel

$\int _{-\infty }^{\infty }{\frac {\cos \alpha x}{(\beta ^{2}+x^{2})((\beta +1)^{2}+x^{2})\cdots ((\beta +n)^{2}+x^{2})}}\,dx=2\pi \sum _{k=0}^{n}(-1)^{k}\,{\frac {(2\beta -1+k)!}{(2\beta +n+k)!}}\,{\frac {1}{k!\,(n-k)!}}\,e^{-\alpha (\beta +k)}$

mit $\beta ^{2}\,(\beta +1)^{2}\cdots (\beta +n)^{2}={\frac {(\beta +n)!^{2}}{\Gamma ^{2}(\beta )}}$ durch, so ist

$I_{n}:=\int _{-\infty }^{\infty }{\frac {\cos \alpha x}{\prod \limits _{k=0}^{n}\left(1+{\frac {x^{2}}{(\beta +k)^{2}}}\right)}}\,dx=2\pi \,{\frac {\Gamma (2\beta )}{\Gamma ^{2}(\beta )}}\,\sum _{k=0}^{n}(-1)^{k}\,{\frac {(2\beta -1+k)!}{\Gamma (2\beta )\,k!}}\,{\frac {(\beta +n)!^{2}}{(2\beta +n+k)!\,(n-k)!}}\,e^{-\alpha (\beta +k)}$,

wobei $(-1)^{k}\,{\frac {(2\beta -1+k)!}{\Gamma (2\beta )\,k!}}={-2\beta \choose k}$ ist.

Setzt man $A_{n,k}:={\frac {(\beta +n)!^{2}}{(2\beta +n+k)!\,(n-k)!}}$ so ist $I_{n}=2\pi \,{\frac {\Gamma (2\beta )}{\Gamma ^{2}(\beta )}}\,\sum _{k=0}^{n}{-2\beta \choose k}\,A_{n,k}\,e^{-\alpha k}\,e^{-\alpha \beta }$.

Mit einem $0<M<n\,$ lässt sich letzte Summe folgendermaßen aufspalten:

$2\pi \,{\frac {\Gamma (2\beta )}{\Gamma ^{2}(\beta )}}\left(\sum _{k=0}^{M-1}{-2\beta \choose k}\,A_{n,k}\,e^{-\alpha k}+\sum _{k=M}^{n}{-2\beta \choose k}\,A_{n,k}\,e^{-\alpha k}\right)\,e^{-\alpha \beta }$

Die Folge $(A_{n,0}\,,\,A_{n,1}\,,\,...\,,\,A_{n,n})\subset [0,1]$ fällt monoton und für alle $0\leq k<M$ gilt $\lim _{n\to \infty }A_{n,k}=1\,$.

Also ist $I:=\lim _{n\to \infty }I_{n}=2\pi {\frac {\Gamma (2\beta )}{\Gamma ^{2}(\beta )}}\,\left(\sum _{k=0}^{M-1}{-2\beta \choose k}\,e^{-\alpha k}+\varepsilon _{M}\right)\,e^{-\alpha \beta }$

Für $M\to \infty \,$ geht $\varepsilon _{M}\,$ gegen null und $\sum _{k=0}^{M-1}{-2\beta \choose k}\,e^{-\alpha k}$ konvergiert gegen die Binomialreihenentwicklung von $(1+e^{-\alpha })^{-2\beta }\,$.

Also ist $I=2\pi \,{\frac {\Gamma (2\beta )}{\Gamma ^{2}(\beta )}}\,\left(1+e^{-\alpha }\right)^{-2\beta }\,\left(e^{\frac {\alpha }{2}}\right)^{-2\beta }$.

Unter Verwendung der Legendreschen Verdopplungsformel $2\pi \Gamma (2\beta )={\sqrt {\pi }}\,\Gamma (\beta )\,\Gamma \left(\beta +{\frac {1}{2}}\right)\,2^{2\beta }$

ist das ${\sqrt {\pi }}\,{\frac {\Gamma \left(\beta +{\frac {1}{2}}\right)}{\Gamma (\beta )}}\,2^{2\beta }\,\left(e^{\frac {\alpha }{2}}+e^{-{\frac {\alpha }{2}}}\right)^{-2\beta }={\sqrt {\pi }}\,{\frac {\Gamma \left(\beta +{\frac {1}{2}}\right)}{\Gamma (\beta )}}\,{\text{sech}}^{2\beta }\left({\frac {\alpha }{2}}\right)$.

2.10

$\int _{0}^{\frac {\pi }{2}}\cos ^{\alpha -1}x\,\cos \beta x\,dx={\frac {\pi }{2^{\alpha }}}\,{\frac {\Gamma (\alpha )}{\Gamma \left({\frac {\alpha +\beta +1}{2}}\right)\,\Gamma \left({\frac {\alpha -\beta +1}{2}}\right)}}\qquad {\text{Re}}(\alpha )>0\,,\,\beta \in \mathbb {C}$

1. Beweis (Cauchy Cosinus-Integralformel)

In der Formel $\int _{-\infty }^{\infty }{\frac {dx}{(u+ix)^{a}\,(v-ix)^{b}}}={\frac {2\pi }{(u+v)^{a+b-1}}}\,{\frac {\Gamma (a+b-1)}{\Gamma (a)\,\Gamma (b)}}$

für ${\text{Re}}(u),{\text{Re}}(v)>0\;,\;{\text{Re}}(a+b)>1$ setze $u=v=1\,$ und substituiere $x\mapsto \tan x$:

$I:=\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}{\frac {1}{\left(1+i\,{\frac {\sin x}{\cos x}}\right)^{a}\,\left(1-i\,{\frac {\sin x}{\cos x}}\right)^{b}}}\,{\frac {dx}{\cos ^{2}x}}={\frac {2\pi }{2^{a+b-1}}}\,{\frac {\Gamma (a+b-1)}{\Gamma (a)\,\Gamma (b)}}$

Nun ist $I=\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}{\frac {(\cos x)^{a+b-2}}{(\cos x+i\sin x)^{a}\,(\cos x-i\sin x)^{b}}}\,dx=\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}(\cos x)^{a+b-2}\,e^{-i(a-b)x}\,dx$

aus symmetriegründen gleich $\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}(\cos x)^{a+b-2}\,\cos(a-b)x\,dx$.

Also ist $\int _{0}^{\frac {\pi }{2}}(\cos x)^{a+b-2}\,\cos(a-b)x\,dx={\frac {\pi }{2^{a+b-1}}}\,{\frac {\Gamma (a+b-1)}{\Gamma (a)\,\Gamma (b)}}$.

Substituiert man $\alpha =a+b-1,\,\beta =a-b$,

so ist $\int _{0}^{\frac {\pi }{2}}(\cos x)^{\alpha -1}\,\cos \beta x\,dx={\frac {\pi }{2^{\alpha }}}\,{\frac {\Gamma (\alpha )}{\Gamma \left({\frac {\alpha +\beta +1}{2}}\right)\,\Gamma \left({\frac {\alpha -\beta +1}{2}}\right)}}$.

2. Beweis

Nach der Formel $\cos 2\beta x=\sum _{n=0}^{\infty }{\frac {\beta \,(\beta -1+n)!}{(\beta -n)!}}\,{\frac {(2\sin x)^{2n}}{(2n)!}}$ ist

$2\int _{0}^{\frac {\pi }{2}}\cos ^{2\alpha -1}x\,\cos 2\beta x\,dx=\sum _{n=0}^{\infty }(-1)^{n}\,{\frac {\beta \,(\beta -1+n)!\,2^{2n}}{(\beta -n)!\,(2n)!}}\,\;2\int _{0}^{\frac {\pi }{2}}\sin ^{2n}x\,\cos ^{2\alpha -1}\,dx$

$=\sum _{n=0}^{\infty }(-1)^{n}\,{\frac {\beta \,(\beta -1+n)!\,2^{2n}}{(\beta -n)!\,(2n)!}}\,{\frac {\left(n-{\frac {1}{2}}\right)!\,(\alpha -1)!}{\left(\alpha +n-{\frac {1}{2}}\right)!}}$.

Nach der Legendreschen Verdopplungsformel ${\frac {2^{2n}\,\left(n-{\frac {1}{2}}\right)!}{(2n)!}}={\frac {\sqrt {\pi }}{n!}}$

ist dies $\beta \,(\alpha -1)!\,{\sqrt {\pi }}\,\sum _{n=0}^{\infty }(-1)^{n}\,{\frac {(\beta -1+n)!}{n!\,(\beta -n)!\,\left(\alpha -{\frac {1}{2}}+n\right)!}}$.

Ersetzt man $(-1)^{n}\,(\beta -1+n)!\,$ durch ${\frac {(\beta -1)!\,(-\beta )!}{(-\beta -n)!}}$, so ist das

$\beta !\,(-\beta )!\,(\alpha -1)!\,{\sqrt {\pi }}\,\sum _{n=0}^{\infty }{\frac {1}{n!\,\left(\alpha -{\frac {1}{2}}+n\right)!\,(\beta -n)!\,(-\beta -n)!}}$

$={\frac {{\sqrt {\pi }}\,\left(\alpha -{\frac {1}{2}}\right)!\,(\alpha -1)!}{\left(\alpha +\beta -{\frac {1}{2}}\right)!\,\left(\alpha -\beta -{\frac {1}{2}}\right)!}}={\frac {\pi }{2^{2\alpha -1}}}\,{\frac {\Gamma (2\alpha )}{\Gamma \left(\alpha +\beta +{\frac {1}{2}}\right)\,\Gamma \left(\alpha -\beta +{\frac {1}{2}}\right)}}$.

Also ist $2\int _{0}^{\frac {\pi }{2}}\cos ^{\alpha -1}x\,\cos \beta x\,dx={\frac {\pi }{2^{\alpha }}}\,{\frac {\Gamma (\alpha )}{\Gamma \left({\frac {\alpha +\beta +1}{2}}\right)\,\Gamma \left({\frac {\alpha -\beta +1}{2}}\right)}}$.

Beweis für $0<\alpha <\beta +1$

Es sei $H=\{z\in \mathbb {C} \,|\,{\text{Im}}(z)>0\}$ die obere komplexe Halbebene.

Die Funktion $f(z)=(1+z)^{\alpha -1}\,z^{\beta -1}\,$, mit $\alpha ,\beta >0\,$, ist holomorph auf $H\,$ und stetig auf ${\bar {H}}\,$.



Also gilt $\oint _{K}f\,dz=0$, gleichbedeutend mit $\underbrace {\int _{-1}^{0}f(x)\,dx} _{S_{1}}+\underbrace {\int _{0}^{1}f(x)\,dx} _{S_{2}}+\underbrace {\int _{0}^{\frac {\pi }{2}}f(e^{2ix})\,e^{2ix}\,2i\,dx} _{S_{3}}=0$.

Das erste Integral $S_{1}\,$ ist nach Substitution $x\mapsto -x\,$ gleich $\int _{0}^{1}f(-x)\,dx$

$=\int _{0}^{1}(1-x)^{\alpha -1}\,(-x)^{\beta -1}\,dx=e^{i\pi (\beta -1)}\int _{0}^{1}(1-x)^{\alpha -1}\,x^{\beta -1}\,dx=-e^{i\pi \beta }\,B(\alpha ,\beta )$.

$\Rightarrow \,{\text{Im}}(S_{1})=-\sin \pi \beta \,\,B(\alpha ,\beta )=-{\frac {\pi }{\Gamma (\beta )\,\Gamma (1-\beta )}}\,{\frac {\Gamma (\alpha )\,\Gamma (\beta )}{\Gamma (\alpha +\beta )}}=-\pi \,{\frac {\Gamma (\alpha )}{\Gamma (1-\beta )\,\Gamma (\alpha +\beta )}}$.

Das zweite Integral $S_{2}\,$ ist reell, d.h. ${\text{Im}}(S_{2})=0\,$.

Und das dritte Integral $S_{3}\,$ ist $2i\,\int _{0}^{\frac {\pi }{2}}(1+e^{2ix})^{\alpha -1}\,e^{2ix(\beta -1)}\,e^{2ix}\,dx$

$=2i\,\int _{0}^{\frac {\pi }{2}}(e^{-ix}+e^{ix})^{\alpha -1}\,e^{i(\alpha +2\beta -1)x}\,dx\,\Rightarrow \,{\text{Im}}(S_{3})=2^{\alpha }\int _{0}^{\frac {\pi }{2}}\cos ^{\alpha -1}x\,\cos(\alpha +2\beta -1)x\,dx$.

Aus der Betrachtung der Imaginärteile folgt $2^{\alpha }\,\int _{0}^{\frac {\pi }{2}}\cos ^{\alpha -1}x\,\cos(\alpha +2\beta -1)x\,dx=\pi \,{\frac {\Gamma (\alpha )}{\Gamma (1-\beta )\,\Gamma (\alpha +\beta )}}$.

Ersetzt man $\alpha +2\beta -1\,$ durch $\gamma \,$, also $\beta ={\frac {\gamma -\alpha +1}{2}}\,$, so ist $\int _{0}^{\frac {\pi }{2}}\cos ^{\alpha -1}x\,\cos \gamma x\,dx={\frac {\pi }{2^{\alpha }}}\,{\frac {\Gamma (\alpha )}{\Gamma \left({\frac {\alpha -\gamma +1}{2}}\right)\,\Gamma \left({\frac {\alpha +\gamma +1}{2}}\right)}}$.

3.1

$\int _{-\infty }^{\infty }{\frac {\cos \alpha x}{(\beta ^{2}+x^{2})((\beta +1)^{2}+x^{2})\cdots ((\beta +n)^{2}+x^{2})}}\,dx=2\pi \sum _{k=0}^{n}(-1)^{k}\,{\frac {(2\beta -1+k)!}{(2\beta +n+k)!}}\,{\frac {1}{k!\,(n-k)!}}\,e^{-\alpha (\beta +k)}\qquad n\in \mathbb {N} \;\;,\;\;\alpha ,\beta >0$

Beweis

Es sei $g(z)=\prod _{\ell =0}^{n}\left((\beta +\ell )^{2}+z^{2}\right)\,,\,f(z)={\frac {e^{i\alpha z}}{g(z)}}$ und $\gamma _{R}\,$ sei der Halbmond in der oberen komplexen Halbebene.

$I:=\int _{-\infty }^{\infty }{\frac {\cos \alpha x}{(\beta ^{2}+x^{2})((\beta +1)^{2}+x^{2})\cdots ((\beta +n)^{2}+x^{2})}}\,dx$

$=\lim _{R\to \infty }\oint _{\gamma _{R}}f(z)\,dz=2\pi i\,\sum _{k=0}^{n}{\text{res}}\left(f,i(\beta +k)\right)=2\pi i\,\sum _{k=0}^{n}{\frac {e^{-\alpha (\beta +k)}}{g'\left(i(\beta +k)\right)}}$.

Nach der Produktregel ist $g'(z)=\sum _{k=0}^{n}\prod _{\ell =0}^{k-1}\left((\beta +\ell )^{2}+z^{2}\right)\cdot 2z\cdot \prod _{\ell =k+1}^{n}\left((\beta +\ell )^{2}+z^{2}\right)$.

Setzt man $z=i(\beta +k)\,$, so ist $g'\left(i(\beta +k)\right)=\prod _{\ell =0}^{k-1}\left((\beta +\ell )^{2}-(\beta +k)^{2}\right)\cdot 2i(\beta +k)\cdot \prod _{\ell =k+1}^{n}\left((\beta +\ell )^{2}-(\beta +k)^{2}\right)$,

wobei $(\beta +\ell )^{2}-(\beta +k)^{2}=(\ell -k)(2\beta +\ell +k)$ ist.

Also ist $g'\left(i(\beta +k)\right)=(-1)^{k}\,k!\,{\frac {(2\beta +2k-1)!}{(2\beta -1+k)!}}\cdot 2i(\beta +k)\cdot (n-k)!\,{\frac {(2\beta +n+k)!}{(2\beta +2k)!}}=i\,(-1)^{k}\,{\frac {(2\beta +n+k)!}{(2\beta -1+k)!}}\,k!\,(n-k)!$.

Und somit ist $I=2\pi \sum _{k=0}^{n}(-1)^{k}\,{\frac {(2\beta -1+k)!}{(2\beta +n+k)!}}\,{\frac {1}{k!\,(n-k)!}}\,e^{-\alpha (\beta +k)}$.

3.2

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

ohne Beweis