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

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

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

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

der geschlossene halbmondförmige Integrationsweg.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Die Formel ist äquivalent zur Formel

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

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

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

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

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

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

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

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

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

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

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

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

Begründung:

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

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

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

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



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

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

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

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

Betrachte die Poissonsche Integralformel

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

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

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

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

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

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

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

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

Multipliziert man die Formel

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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



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

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

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

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

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

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

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

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

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