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

 

1.1Bearbeiten

{\displaystyle {\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

 

 

1.2Bearbeiten

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

1. Beweis

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 }}.

2. Beweis

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 }}.

 

1.3Bearbeiten

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

ohne Beweis

 

 

1.4Bearbeiten

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

Beweis

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 }}.

 

1.5Bearbeiten

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

Beweis

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 }}.

 

1.6Bearbeiten

{\displaystyle \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

{\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 }.

 

1.7Bearbeiten

{\displaystyle 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

{\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}}}

 

1.8Bearbeiten

{\displaystyle \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 {\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.

 

1.9Bearbeiten

{\displaystyle \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

 

 

2.1Bearbeiten

{\displaystyle \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 {\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}}.

 

2.2Bearbeiten

{\displaystyle {\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

 

 

2.3Bearbeiten

{\displaystyle \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 {\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}}}.

 

2.4Bearbeiten

{\displaystyle \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

{\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.

 

2.5Bearbeiten

{\displaystyle \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 {\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}.

 

2.6Bearbeiten

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

ohne Beweis

 

 

2.7Bearbeiten

{\displaystyle \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

 

 

2.8Bearbeiten

{\displaystyle \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

 

 

2.9Bearbeiten

{\displaystyle \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

{\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)}.

 

2.10Bearbeiten

{\displaystyle \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 {\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)}}}.

2. Beweis

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)}}}.

Beweis für {\displaystyle 0<\alpha <\beta +1}

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)}}}.

 

3.1Bearbeiten

{\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)}\qquad n\in \mathbb {N} \;\;,\;\;\alpha ,\beta >0}

Beweis

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

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

{\displaystyle =\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 {\displaystyle 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 {\displaystyle z=i(\beta +k)\,}, so ist {\displaystyle 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 {\displaystyle (\beta +\ell )^{2}-(\beta +k)^{2}=(\ell -k)(2\beta +\ell +k)} ist.

Also ist {\displaystyle 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 {\displaystyle 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)}}.

 

3.2Bearbeiten

{\displaystyle \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