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

Zurück zu Bestimmte Integrale

##### 1.1
${\displaystyle \int _{0}^{1}{\frac {1-x^{z-1}}{1-x}}\,dx=\gamma +\psi (z)\qquad {\text{Re}}(z)>0}$
Beweis (Formel nach Gauß)

${\displaystyle {\frac {1-x^{z-1}}{1-x}}=\sum _{k=1}^{\infty }\left(x^{k-1}-x^{k+z-2}\right)}$

${\displaystyle \Rightarrow \int _{0}^{1}{\frac {1-x^{z-1}}{1-x}}\,dx=\sum _{k=1}^{\infty }\int _{0}^{1}\left(x^{k-1}-x^{k+z-2}\right)dx=-{\frac {1}{z}}+\sum _{k=1}^{\infty }\left({\frac {1}{k}}-{\frac {1}{k+z}}\right)=\gamma +\psi (z+1)-{\frac {1}{z}}=\gamma +\psi (z)}$

##### 1.2
${\displaystyle \int _{0}^{1}{\frac {1-x^{\alpha -1}}{{\sqrt {1-x^{2}}}^{\,3}}}dx=2^{\alpha -2}{\frac {\Gamma ^{2}\left({\frac {\alpha }{2}}\right)}{\Gamma (\alpha -1)}}}$
ohne Beweis

##### 1.3
${\displaystyle \int _{0}^{1}x\cdot {\sqrt {\frac {1-\alpha ^{2}x^{2}}{1-x^{2}}}}\,dx={\frac {1}{2}}+{\frac {1-\alpha ^{2}}{2\alpha }}\,{\text{artanh}}\,\alpha }$
Beweis

${\displaystyle I:=\int _{0}^{1}x\cdot {\sqrt {\frac {1-\alpha ^{2}x^{2}}{1-x^{2}}}}\,dx}$

Nach der Substitution ${\displaystyle x={\sqrt {\frac {1-u^{2}}{1-\alpha ^{2}u^{2}}}}}$ bzw. ${\displaystyle u={\sqrt {\frac {1-x^{2}}{1-\alpha ^{2}x^{2}}}}}$ ist

${\displaystyle {\frac {dx}{du}}={\frac {{\sqrt {1-\alpha ^{2}u^{2}}}\cdot {\frac {-u}{\sqrt {1-u^{2}}}}-{\sqrt {1-u^{2}}}\cdot {\frac {-\alpha ^{2}u}{\sqrt {1-\alpha ^{2}u^{2}}}}}{1-\alpha ^{2}u^{2}}}=-{\frac {(1-\alpha ^{2}u^{2})\cdot u-(1-u^{2})\cdot \alpha ^{2}u}{{\sqrt {1-u^{2}}}\cdot {\sqrt {1-\alpha ^{2}u^{2}}}^{3}}}=-{\frac {(1-\alpha ^{2})\cdot u}{{\sqrt {1-u^{2}}}\cdot {\sqrt {1-\alpha ^{2}u^{2}}}^{3}}}}$.

${\displaystyle I=\int _{0}^{1}{\sqrt {\frac {1-u^{2}}{1-\alpha ^{2}u^{2}}}}\cdot {\frac {1}{u}}\cdot {\frac {(1-\alpha ^{2})\cdot u}{{\sqrt {1-u^{2}}}\cdot {\sqrt {1-\alpha ^{2}u^{2}}}^{3}}}\cdot du=(1-\alpha ^{2})\cdot \int _{0}^{1}{\frac {du}{(1-\alpha ^{2}u^{2})^{2}}}}$

${\displaystyle (1-\alpha ^{2})\cdot {\frac {1}{2}}\cdot \left[{\frac {u}{1-\alpha ^{2}u^{2}}}+{\frac {{\text{artanh}}(\alpha u)}{\alpha }}\right]_{0}^{1}=(1-\alpha ^{2})\cdot {\frac {1}{2}}\cdot \left({\frac {1}{1-\alpha ^{2}}}+{\frac {{\text{artanh}}\,\alpha }{\alpha }}\right)={\frac {1}{2}}+{\frac {1-\alpha ^{2}}{2\alpha }}\,{\text{artanh}}\,\alpha }$

##### 1.4
${\displaystyle \int _{0}^{1}{\frac {1}{\sqrt {(1-x^{2})(1-\alpha ^{2}x^{2})}}}\,\mathrm {d} x=K(\alpha )}$
ohne Beweis (Definition des vollständigen elliptischen Integrals erster Art)

##### 1.5
${\displaystyle \int _{0}^{1}{\sqrt {\frac {1-\alpha ^{2}x^{2}}{1-x^{2}}}}\,\mathrm {d} x=E(\alpha )}$
ohne Beweis (Definition des vollständigen elliptischen Integrals zweiter Art)

##### 1.6
${\displaystyle \int _{0}^{\infty }\left\{{\frac {1}{x}}\right\}x^{s-1}\,dx=-{\frac {\zeta (s)}{s}}\qquad 0<{\text{Re}}(s)<1}$
Beweis

${\displaystyle I_{N}:=-s\int _{1/N}^{\infty }\left\{{\frac {1}{x}}\right\}x^{s-1}\,dx}$ ist nach Substitution ${\displaystyle x\mapsto {\frac {1}{x}}}$ gleich ${\displaystyle -s\int _{0}^{N}\{x\}\,{\frac {1}{x^{s-1}}}\,{\frac {dx}{x^{2}}}=\int _{0}^{N}\{x\}\,{\frac {d}{dx}}{\frac {1}{x^{s}}}\,dx}$.

Dies ist nach Eulerscher Summenformel ${\displaystyle \sum _{n=1}^{N}{\frac {1}{n^{s}}}-\int _{0}^{N}{\frac {1}{x^{s}}}\,dx=\sum _{n=1}^{N}{\frac {1}{n^{s}}}-{\frac {N^{1-s}}{1-s}}}$, woraus ${\displaystyle \lim _{N\to \infty }I_{N}=\zeta (s)\,}$ folgt.

##### 2.1
${\displaystyle \int _{0}^{\infty }{\frac {x^{m-1}}{1+x+...+x^{n-1}}}\,dx={\frac {\pi }{n}}\left[\cot \left({\frac {m\pi }{n}}\right)-\cot \left({\frac {(m+1)\pi }{n}}\right)\right]\qquad 0
Beweis

Integriere ${\displaystyle f(z)={\frac {z^{m-1}}{1+...+z^{n-1}}}}$ entlang dem Kreissektor, der durch den Ursprung, ${\displaystyle R\in \mathbb {R} _{+}}$ und ${\displaystyle R\,e^{i\pi /n}}$ als Eckpunkte beschränkt wird.

Das Integral über dem Kreisbogen geht gegen null für ${\displaystyle R\to \infty \,}$. Also ist ${\displaystyle I=\int _{\mathbb {R} _{+}}f\,dz=\int _{e^{i\pi /n}\,\mathbb {R} _{+}}f\,dz=\int _{0}^{\infty }f\left(e^{i\pi /n}z\right)\,e^{i\pi /n}\,dz}$.

Nachdem sich ${\displaystyle f(z)\,}$ auch als ${\displaystyle {\frac {1-z}{1-z^{n}}}\,z^{m-1}}$ schreiben lässt, ist ${\displaystyle I=\int _{0}^{\infty }{\frac {1-e^{i\pi /n}z}{1+z^{n}}}\,e^{i(m-1)\pi /n}\,z^{m-1}\,e^{i\pi /n}\,dz}$

${\displaystyle =e^{im\pi /n}\,\int _{0}^{\infty }{\frac {z^{m-1}}{1+z^{n}}}\,dz+e^{i(m+1)\pi /n}\,\int _{0}^{\infty }{\frac {z^{m}}{1+z^{n}}}\,dz}$

${\displaystyle =\left[\cos \left({\frac {m\pi }{n}}\right)+i\sin \left({\frac {m\pi }{n}}\right)\right]\,{\frac {\pi }{n}}\,{\frac {1}{\sin \left({\frac {m\pi }{n}}\right)}}-\left[\cos \left({\frac {(m+1)\pi }{n}}\right)+i\sin \left({\frac {(m+1)\pi }{n}}\right)\right]\,{\frac {\pi }{n}}\,{\frac {1}{\sin \left({\frac {(m+1)\pi }{n}}\right)}}}$.

Der Imaginärteil hebt sich auf und übrig bleibt ${\displaystyle I={\frac {\pi }{n}}\left[\cot \left({\frac {m\pi }{n}}\right)-\cot \left({\frac {(m+1)\pi }{n}}\right)\right]}$.

##### 2.2
${\displaystyle B(\alpha ,\beta )=\int _{0}^{1}x^{\alpha -1}(1-x)^{\beta -1}\,dx\qquad {\text{Re}}(\alpha )\,,\,{\text{Re}}(\beta )>0}$
ohne Beweis (Definition der Betafunktion)

##### 2.3
${\displaystyle \int _{0}^{\infty }{\frac {x^{\alpha -1}}{(1+x)^{\alpha +\beta }}}\,dx=B(\alpha ,\beta )\qquad {\text{Re}}(\alpha )\,,\,{\text{Re}}(\beta )>0}$
Beweis

${\displaystyle B(\alpha ,\beta )=\int _{0}^{1}x^{\alpha -1}(1-x)^{\beta -1}\,dx}$ ist nach der Substitution ${\displaystyle x\to {\frac {1}{1+x}}}$ gleich ${\displaystyle \int _{0}^{\infty }{\frac {x^{\beta -1}}{(1+x)^{\alpha +\beta }}}\,dx}$.

Und auf Grund der Symmetrie ${\displaystyle B(\alpha ,\beta )=B(\beta ,\alpha )\,}$ ist das das selbe wie ${\displaystyle \int _{0}^{\infty }{\frac {x^{\alpha -1}}{(1+x)^{\alpha +\beta }}}\,dx}$.

##### 2.4
${\displaystyle \int _{0}^{\infty }{\frac {x^{\alpha -1}}{1+x^{\beta }}}\,dx={\frac {\pi }{\beta }}\;\csc \left({\frac {\alpha \pi }{\beta }}\right)\qquad 0<\operatorname {Re} (\alpha )<\operatorname {Re} (\beta )}$
1. Beweis

Es sei ${\displaystyle {\text{Log}}:\mathbb {C} \setminus \{0\}\to \mathbb {C} }$ definiert durch ${\displaystyle re^{i\varphi }\mapsto \log r+i\varphi }$ für ${\displaystyle r>0\,}$ und ${\displaystyle 0\leq \varphi <2\pi }$.

Für ${\displaystyle 0 gilt die Partialbruchzerlegung ${\displaystyle {\frac {x^{m-1}}{1+x^{n}}}={\frac {1}{n}}\,\sum _{k=0}^{n-1}{\frac {\left(e^{\frac {im\pi }{n}}\right)^{2k+1}}{\left(e^{\frac {i\pi }{n}}\right)^{2k+1}-x}}}$.

Also ist ${\displaystyle \int _{0}^{R}{\frac {x^{m-1}}{1+x^{n}}}\,dx={\frac {1}{n}}\sum _{k=0}^{n-1}\left[-{\text{Log}}\left(e^{{\frac {i\pi }{n}}(2k+1)}-x\right)\right]_{0}^{R}\,\left(e^{\frac {im\pi }{n}}\right)^{2k+1}}$

${\displaystyle =\underbrace {{\frac {1}{n}}\sum _{k=0}^{n-1}-{\text{Log}}\left(e^{{\frac {i\pi }{n}}(2k+1)}-R\right)\,\left(e^{\frac {im\pi }{n}}\right)^{2k+1}} _{\text{1.Summe}}+\underbrace {{\frac {1}{n}}\sum _{k=0}^{n-1}{\frac {i\pi }{n}}(2k+1)\left(e^{\frac {im\pi }{n}}\right)^{2k+1}} _{\text{2.Summe}}}$.

Nun soll gezeigt werden, dass die 1.Summe für ${\displaystyle R\to \infty \,}$ gegen null konvergiert und die 2.Summe gleich ${\displaystyle {\frac {\pi }{n}}\,{\frac {1}{\sin {\frac {m\pi }{n}}}}}$ ist.

Für ${\displaystyle x\neq \pm 1\,}$ gilt ${\displaystyle \sum _{k=0}^{n-1}x^{2k+1}={\frac {x^{2n}-1}{x-{\frac {1}{x}}}}}$ und ${\displaystyle \sum _{k=0}^{n-1}(2k+1)\,x^{2k+1}={\frac {n\,x^{2n}}{{\frac {1}{2}}\left(x-{\frac {1}{x}}\right)}}-(1-x^{2n})\,{\frac {x+{\frac {1}{x}}}{\left(x-{\frac {1}{x}}\right)^{2}}}}$.

Setzt man ${\displaystyle x=e^{\frac {im\pi }{n}}}$, so ist ${\displaystyle \sum _{k=0}^{n-1}(2k+1)\left(e^{\frac {im\pi }{n}}\right)^{2k+1}={\frac {n}{i\,\sin {\frac {m\pi }{n}}}}}$. Also ist die 2.Summe gleich ${\displaystyle {\frac {\pi }{n}}\,{\frac {1}{\sin {\frac {m\pi }{n}}}}}$.

Und wegen ${\displaystyle \sum _{k=0}^{n-1}\left(e^{\frac {im\pi }{n}}\right)^{2k+1}=0}$ lässt sich die 1.Summe schreiben als

${\displaystyle {\frac {1}{n}}\sum _{k=0}^{n-1}\underbrace {\left[{\text{Log}}(-R)-{\text{Log}}\left(e^{{\frac {i\pi }{n}}(2k+1)}-R\right)\right]} _{\to 0\,{\text{wenn}}\,R\to \infty }\,\left(e^{\frac {im\pi }{n}}\right)^{2k+1}}$.

Damit ist ${\displaystyle \int _{0}^{\infty }{\frac {x^{m-1}}{1+x^{n}}}\,dx={\frac {\pi }{n}}\,{\frac {1}{\sin {\frac {m\pi }{n}}}}}$ gezeigt. Substituiert man ${\displaystyle x\,}$ durch ${\displaystyle x^{\frac {\beta }{n}}}$, so ist ${\displaystyle \int _{0}^{\infty }{\frac {x^{\beta {\frac {m}{n}}-1}}{1+x^{\beta }}}\,dx={\frac {\pi }{\beta }}\,{\frac {1}{\sin {\frac {m\pi }{n}}}}}$.

Für reelle ${\displaystyle \alpha ,\beta \,}$ folgt die Behauptung, wenn man eine Folge rationaler Zahlen ${\displaystyle {\frac {m}{n}}}$ konstruiert, die gegen ${\displaystyle {\frac {\alpha }{\beta }}}$ konvergiert.

2. Beweis

Spalte auf in ${\displaystyle \int _{0}^{1}{\frac {x^{\alpha -1}}{1+x^{\beta }}}\,dx+\int _{1}^{\infty }{\frac {x^{\alpha -1}}{1+x^{\beta }}}\,dx}$. Das erste Integral ist ${\displaystyle {\frac {1}{2\beta }}\left[\psi \left({\frac {1}{2}}+{\frac {\alpha }{2\beta }}\right)-\psi \left({\frac {\alpha }{2\beta }}\right)\right]}$.

Und das zweite Integral ist nach Substitution ${\displaystyle x\mapsto {\frac {1}{x}}}$ gleich ${\displaystyle \int _{0}^{1}{\frac {x^{\beta -\alpha -1}}{1+x^{\beta }}}\,dx}$, und somit gleich ${\displaystyle {\frac {1}{2\beta }}\left[\underbrace {\psi \left({\frac {1}{2}}+{\frac {\beta -\alpha }{2\beta }}\right)} _{\psi \left(1-{\frac {\alpha }{2\beta }}\right)}-\underbrace {\psi \left({\frac {\beta -\alpha }{2\beta }}\right)} _{\psi \left({\frac {1}{2}}-{\frac {\alpha }{2\beta }}\right)}\right]}$.

Also ist ${\displaystyle \int _{0}^{\infty }{\frac {x^{\alpha -1}}{1+x^{\beta }}}\,dx={\frac {1}{2\beta }}\left[\underbrace {\psi \left(1-{\frac {\alpha }{2\beta }}\right)-\psi \left({\frac {\alpha }{2\beta }}\right)} _{\pi \,\cot {\frac {\alpha \pi }{2\beta }}}+\underbrace {\psi \left({\frac {1}{2}}+{\frac {\alpha }{2\beta }}\right)-\psi \left({\frac {1}{2}}-{\frac {\alpha }{2\beta }}\right)} _{_{\pi \,\tan {\frac {\alpha \pi }{2\beta }}}}\right]={\frac {\pi }{\beta }}\,{\frac {1}{\sin {\frac {\alpha \pi }{\beta }}}}}$.

Anders formuliert kann das erste Integral ${\displaystyle \int _{0}^{1}{\frac {x^{\alpha -1}}{1+x^{\beta }}}\,dx}$ geschrieben werden als ${\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}}{\alpha +\beta k}}}$

und das zweite Integral ${\displaystyle \int _{0}^{1}{\frac {x^{\beta -\alpha -1}}{1+x^{\beta }}}\,dx}$ geschrieben werden als ${\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}}{\beta -\alpha +\beta k}}=\sum _{k=-\infty }^{-1}{\frac {(-1)^{k}}{\alpha +\beta k}}}$.

Also ist ${\displaystyle \int _{0}^{\infty }{\frac {x^{\alpha -1}}{1+x^{\beta }}}\,dx=\sum _{k\in \mathbb {Z} }{\frac {(-1)^{k}}{\alpha +\beta \,k}}}$, was gerade die Partialbruchentwicklung von ${\displaystyle {\frac {\pi }{\beta }}\,\csc \left({\frac {\alpha \pi }{\beta }}\right)}$ ist.

3. Beweis

Für ${\displaystyle 0<{\text{Re}}(\alpha )<\beta \in \mathbb {R} }$ und ein ${\displaystyle n\in \mathbb {Z} ^{>2}}$ definiere ${\displaystyle f(z)={\frac {z^{n{\frac {\alpha }{\beta }}-1}}{1+z^{n}}}}$.

Integriere ${\displaystyle f\,}$ entlang dem Kreissektor ${\displaystyle \gamma _{R}\,}$, der durch den Ursprung, ${\displaystyle R>0\,}$ und ${\displaystyle R\,e^{\frac {2\pi i}{n}}}$ als Eckpunkte beschränkt wird.

Das Integral über dem Kreisbogen geht gegen null für ${\displaystyle R\to \infty \,}$.

Also ist ${\displaystyle I:=\lim _{R\to \infty }\oint _{\gamma _{R}}f\,dz=\int _{\mathbb {R} _{+}}f\,dz-\int _{e^{\frac {2\pi i}{n}}\mathbb {R} _{+}}f\,dz}$, wobei letztes Integral

${\displaystyle \int _{\mathbb {R} _{+}}f\left(e^{\frac {2\pi i}{n}}z\right)\,e^{\frac {2\pi i}{n}}\,dz=e^{2\pi i{\frac {\alpha }{\beta }}}\,\int _{\mathbb {R} _{+}}f\,dz}$ ist. Und somit ist ${\displaystyle I=\left(1-e^{2\pi i{\frac {\alpha }{\beta }}}\right)\int _{\mathbb {R} _{+}}f\,dz}$.

Nach dem Residuensatz ist ${\displaystyle I=2\pi i\,{\text{res}}\left(f,z=e^{\frac {i\pi }{n}}\right)=-{\frac {2\pi i}{n}}e^{i\pi {\frac {\alpha }{\beta }}}}$.

Daher muss gelten ${\displaystyle \int _{\mathbb {R} _{+}}f\,dz={\frac {2\pi i}{n}}\,{\frac {e^{i\pi {\frac {\alpha }{\beta }}}}{e^{2\pi i{\frac {\alpha }{\beta }}}-1}}={\frac {\pi }{n}}\,{\frac {1}{\sin {\frac {\alpha \pi }{\beta }}}}}$.

Nach Substitution ${\displaystyle z\mapsto z^{\frac {\beta }{n}}}$ ergibt sich die Behauptung zumindest für reelles ${\displaystyle \beta \,}$.

4. Beweis

${\displaystyle I:=\int _{0}^{\infty }{\frac {x^{\alpha -1}}{1+x^{\beta }}}\,dx=\int _{0}^{\infty }x^{\alpha -1}\int _{0}^{\infty }e^{-(1+x^{\beta })\cdot t}\,dt\,dx=\int _{0}^{\infty }\int _{0}^{\infty }x^{\alpha -1}\,e^{-t}\,e^{-t\cdot x^{\beta }}\,dx\,dt}$

Nach Substitution ${\displaystyle u=t\cdot x^{\beta }\quad \left[x=\left({\frac {u}{t}}\right)^{\frac {1}{\beta }}\right]}$ ist

${\displaystyle I=\int _{0}^{\infty }\int _{0}^{\infty }\left({\frac {u}{t}}\right)^{{\frac {\alpha }{\beta }}-{\frac {1}{\beta }}}\,e^{-t}\,e^{-u}\,{\frac {1}{\beta }}\,\left({\frac {u}{t}}\right)^{{\frac {1}{\beta }}-1}\,{\frac {1}{t}}\,du\,dt={\frac {1}{\beta }}\int _{0}^{\infty }\int _{0}^{\infty }\left({\frac {u}{t}}\right)^{{\frac {\alpha }{\beta }}-1}\,e^{-u}\,e^{-t}\,{\frac {1}{t}}\,du\,dt}$

${\displaystyle ={\frac {1}{\beta }}\int _{0}^{\infty }u^{{\frac {\alpha }{\beta }}-1}\,e^{-u}\,du\,\,\int _{0}^{\infty }t^{-{\frac {\alpha }{\beta }}}\,e^{-t}\,dt={\frac {1}{\beta }}\,\Gamma \left({\frac {\alpha }{\beta }}\right)\,\Gamma \left(1-{\frac {\alpha }{\beta }}\right)={\frac {1}{\beta }}\,{\frac {\pi }{\sin \left({\frac {\alpha \pi }{\beta }}\right)}}}$.

##### 2.5
${\displaystyle \int _{0}^{\infty }{\frac {x^{\alpha -1}}{1-x^{\beta }}}\,dx={\frac {\pi }{\beta }}\;\cot \left({\frac {\alpha \pi }{\beta }}\right)\qquad 0<{\text{Re}}(\alpha )<{\text{Re}}(\beta )}$
ohne Beweis

##### 2.6
${\displaystyle \int _{0}^{1}{\frac {x^{\alpha -1}}{1+x^{\beta }}}\,dx={\frac {1}{2\beta }}\left[\psi \left({\frac {1}{2}}+{\frac {\alpha }{2\beta }}\right)-\psi \left({\frac {\alpha }{2\beta }}\right)\right]}$
Beweis

${\displaystyle \int _{0}^{1}{\frac {x^{\alpha -1}}{1+x^{\beta }}}\,dx=\int _{0}^{1}x^{\alpha -1}\,\sum _{k=0}^{\infty }(-x^{\beta })^{k}\,dx=\sum _{k=0}^{\infty }(-1)^{k}\,\int _{0}^{1}x^{\alpha -1+\beta \,k}\,dx=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{\alpha +\beta \,k}}}$

Und diese Reihe konvergiert gegen ${\displaystyle {\frac {1}{2\beta }}\left[\psi \left({\frac {1}{2}}+{\frac {\alpha }{2\beta }}\right)-\psi \left({\frac {\alpha }{2\beta }}\right)\right]}$.

##### 2.7
${\displaystyle \int _{0}^{\infty }{\frac {x^{\alpha }}{x^{2}+2\cos \theta \cdot x+1}}\,dx={\frac {\pi }{\sin \alpha \pi }}\,{\frac {\sin \alpha \theta }{\sin \theta }}\qquad -1<{\text{Re}}(\alpha )<1\,,\,-\pi <{\text{Re}}(\theta )<\pi }$
Beweis

Verwende die Formel ${\displaystyle \int _{0}^{\infty }f(x)\,x^{\alpha }\,dx={\frac {\pi }{\sin \alpha \pi }}\,\sum {\text{res}}{\Big (}f(-z)\,z^{\alpha }{\Big )}}$.

${\displaystyle f(z)={\frac {1}{z^{2}+2\cos \theta \cdot z+1}}={\frac {1}{2i\sin \theta }}\left({\frac {1}{z+e^{-i\theta }}}-{\frac {1}{z+e^{i\theta }}}\right)}$

${\displaystyle \Rightarrow \,f(-z)\,z^{\alpha }={\frac {1}{2i\sin \theta }}\left({\frac {z^{\alpha }}{z-e^{i\theta }}}-{\frac {z^{\alpha }}{z-e^{-i\theta }}}\right)}$

${\displaystyle \Rightarrow \,\sum {\text{res}}{\Big (}f(-z)\,z^{\alpha }{\Big )}={\frac {1}{2i\sin \theta }}\,{\Big (}\left(e^{i\theta }\right)^{\alpha }-\left(e^{-i\theta }\right)^{\alpha }{\Big )}={\frac {\sin \alpha \theta }{\sin \theta }}}$

##### 2.8
${\displaystyle \int _{0}^{\infty }{\frac {1\;\;\;{\text{oder}}\;\;\;x^{2}}{(x^{2}+2x\cos \alpha +1)(x^{2}+2x\cos \beta +1)}}\,dx={\frac {1}{2}}\,{\frac {\alpha \cot \alpha -\beta \cot \beta }{\cos \alpha -\cos \beta }}}$
ohne Beweis

##### 2.9
${\displaystyle \int _{0}^{\infty }{\frac {x}{(x^{2}+2x\cos \alpha +1)(x^{2}+2x\cos \beta +1)}}\,dx=-{\frac {1}{2}}\,{\frac {\alpha \csc \alpha -\beta \csc \beta }{\cos \alpha -\cos \beta }}}$
Beweis

${\displaystyle {\frac {x}{(x^{2}+2x\cos \alpha +1)(x^{2}+2x\cos \beta +1)}}}$

${\displaystyle ={\frac {1}{2\,(\cos \beta -\cos \alpha )}}\,\left({\frac {1}{x^{2}+2x\cos \alpha +1}}-{\frac {1}{x^{2}+2x\cos \beta +1}}\right)}$

${\displaystyle ={\frac {1}{2\,(\cos \beta -\cos \alpha )}}\,\left({\frac {1}{(x+\cos \alpha )^{2}+\sin ^{2}\alpha }}-{\frac {1}{(x+\cos \beta )^{2}+\sin ^{2}\beta }}\right)}$

Also ist

${\displaystyle \int _{0}^{\infty }{\frac {x}{(x^{2}+2x\cos \alpha +1)(x^{2}+2x\cos \beta +1)}}\,dx}$

${\displaystyle ={\frac {1}{2\,(\cos \beta -\cos \alpha )}}\,\left(\int _{0}^{\infty }{\frac {dx}{(x+\cos \alpha )^{2}+\sin ^{2}\alpha }}-\int _{0}^{\infty }{\frac {dx}{(x+\cos \beta )^{2}+\sin ^{2}\beta }}\right)}$

${\displaystyle ={\frac {1}{2\,(\cos \beta -\cos \alpha )}}\,\left(\left[{\frac {1}{\sin \alpha }}\arctan \left({\frac {x+\cos \alpha }{\sin \alpha }}\right)\right]_{0}^{\infty }-\left[{\frac {1}{\sin \beta }}\arctan \left({\frac {x+\cos \beta }{\sin \beta }}\right)\right]_{0}^{\infty }\right)}$

${\displaystyle ={\frac {1}{2\,(\cos \beta -\cos \alpha )}}\,\left({\frac {\alpha }{\sin \alpha }}-{\frac {\beta }{\sin \beta }}\right)}$.

##### 2.10
${\displaystyle \int _{-\infty }^{\infty }{\frac {1}{\sqrt {[x^{2}+2x\sin(\alpha )+1][x^{2}+2x\sin(\beta )+1]}}}\,dx=2\sec {\bigl (}{\frac {\alpha }{2}}+{\frac {\beta }{2}}{\bigr )}K{\bigl [}\sin {\bigl (}{\frac {\alpha }{2}}-{\frac {\beta }{2}}{\bigr )}\sec {\bigl (}{\frac {\alpha }{2}}+{\frac {\beta }{2}}{\bigr )}{\bigr ]}\qquad -\pi <\alpha +\beta <\pi }$
Beweis

Diese Formel ist gültig:

${\displaystyle {\biggl [}{\frac {1}{2}}{\sqrt {(1-v^{2})(1-w^{2})}}-{\frac {1}{2}}\,v\,w+{\frac {1}{2}}{\biggr ]}^{1/2}\,{\frac {1}{\sqrt {(x^{2}+2vx+1)(x^{2}+2wx+1)}}}=}$

${\displaystyle ={\frac {\mathrm {d} }{\mathrm {d} x}}\,F{\biggl \{}\arcsin {\biggl [}{\frac {{\sqrt {1-w^{2}}}(x+v)+{\sqrt {1-v^{2}}}(x+w)}{{\sqrt {1-w^{2}}}{\sqrt {x^{2}+2vx+1}}+{\sqrt {1-v^{2}}}{\sqrt {x^{2}+2wx+1}}}}{\biggr ]};{\frac {v-w}{{\sqrt {(1-v^{2})(1-w^{2})}}-v\,w+1}}{\biggr \}}}$

Durch Einsetzen von ${\displaystyle v=\sin(\alpha )}$ und ${\displaystyle w=\sin(\beta )}$ erhält man die dargestellte Form.

Und durch Einfügen der Untergrenze und Obergrenze wird das unvollständige elliptische Integral F zum vollständigen elliptischen Intetgral K umgewandelt.

##### 2.11
${\displaystyle \int _{-\infty }^{\infty }{\frac {1}{1+{\frac {x^{2}}{a^{2}}}}}\,\prod _{k=1}^{\infty }{\frac {1+{\frac {x^{2}}{(b+k)^{2}}}}{1+{\frac {x^{2}}{(a+k)^{2}}}}}\,dx={\sqrt {\pi }}\,\,{\frac {\Gamma \left(a+{\frac {1}{2}}\right)\cdot \Gamma (b+1)\cdot \Gamma \left(b-a+{\frac {1}{2}}\right)}{\Gamma (a)\cdot \Gamma \left(b+{\frac {1}{2}}\right)\cdot \Gamma \left(b-a+1\right)}}\qquad 0
Beweis (Formel nach Ramanujan)

Es sei ${\displaystyle f(z)=\prod _{\ell =1}^{n}\left((b+\ell )^{2}+z^{2}\right)\,,\,g(z)=\prod _{\ell =0}^{n}\left((a+\ell )^{2}+z^{2}\right)}$ und ${\displaystyle \gamma _{R}\,}$ der Halbmond in der oberen komplexen Halbebene.

${\displaystyle f{\Big (}i(a+k){\Big )}=\prod _{\ell =1}^{n}\left((b+\ell )^{2}-(a+k)^{2}\right)=\prod _{\ell =1}^{n}{\Big (}(b+a+k+\ell )(b-a-k+\ell ){\Big )}={\frac {(b+a+k+n)!}{(b+a+k)!}}\,{\frac {(b-a-k+n)!}{(b-a-k)!}}}$

${\displaystyle g'{\Big (}i(a+k){\Big )}=\prod _{\ell =0}^{k-1}\underbrace {\left((a+\ell )^{2}-(a+k)^{2}\right)} _{(\ell -k)(2a+k+\ell )}\cdot 2i(a+k)\cdot \prod _{\ell =k+1}^{n}\underbrace {\left((a+\ell )^{2}-(a+k)^{2}\right)} _{(\ell -k)(2a+k+\ell )}}$

${\displaystyle =(-1)^{k}\,k!\,{\frac {(2a+2k-1)!}{(2a+k-1)!}}\cdot 2i(a+k)\cdot (n-k)!\,{\frac {(2a+k+n)!}{(2a+2k)!}}=i\,(-1)^{k}\,{\frac {(2a+k+n)!}{(2a+k-1)!}}\,k!\,(n-k)!}$

${\displaystyle \int _{-\infty }^{\infty }{\frac {1}{a^{2}+x^{2}}}\,{\frac {(b+1)^{2}+x^{2}}{(a+1)^{2}+x^{2}}}\cdots {\frac {(b+n)^{2}+x^{2}}{(a+n)^{2}+x^{2}}}\,dx=\lim _{R\to \infty }\oint _{\gamma _{R}}{\frac {f(z)}{g(z)}}\,dz=2\pi i\sum _{k=0}^{n}{\text{res}}\left({\frac {f}{g}},i(a+k)\right)}$

${\displaystyle =2\pi i\,\sum _{k=0}^{n}{\frac {f{\Big (}i(a+k){\Big )}}{g'{\Big (}i(a+k){\Big )}}}=2\pi \sum _{k=0}^{n}(-1)^{k}\,{\frac {(2a+k-1)!}{(2a+k+n)!}}\,{\frac {1}{k!\,(n-k)!}}\,{\frac {(b+a+k+n)!\,(b-a-k+n)!}{(b+a+k)!\,(b-a-k)!}}}$

Wenn man beide Seiten mit ${\displaystyle a^{2}\,(a+1)^{2}\cdots (a+n)^{2}={\frac {(a+n)!^{2}}{(a-1)!^{2}}}}$ durchmultipliziert

und mit ${\displaystyle (b+1)^{2}\,(b+2)^{2}\cdots (b+n)^{2}={\frac {(b+n)!^{2}}{b!^{2}}}}$ durchdividiert, so ist ${\displaystyle I_{n}:=\int _{-\infty }^{\infty }{\frac {1}{1+{\frac {x^{2}}{a^{2}}}}}\cdot {\frac {1+{\frac {x^{2}}{(b+1)^{2}}}}{1+{\frac {x^{2}}{(a+1)^{2}}}}}\cdots {\frac {1+{\frac {x^{2}}{(b+n)^{2}}}}{1+{\frac {x^{2}}{(a+n)^{2}}}}}\,dx}$

${\displaystyle =2\pi \,{\frac {b!^{2}}{(a-1)!^{2}}}\sum _{k=0}^{n}\underbrace {(-1)^{k}\,{\frac {(2a+k-1)!}{k!}}} _{={\frac {(-2a)!\,(2a-1)!}{k!\,(-2a-k)!}}}\,{\frac {1}{(b+a+k)!\,(b-a-k)!}}\,\underbrace {{\frac {(b+a+k+n)!\,(b-a-k+n)!}{(2a+k+n)!\,(n-k)!}}\,{\frac {(a+n)!^{2}}{(b+n)!^{2}}}} _{A_{n,k}}}$.

Mit einem ${\displaystyle 0\leq k lässt sich letzte Summe folgendermaßen aufspalten:

${\displaystyle 2\pi \,{\frac {b!^{2}\,(-2a)!\,(2a-1)!}{(a-1)!^{2}}}\,\left(\sum _{k=0}^{M-1}{\frac {A_{n,k}}{k!\,(b+a+k)!\,(-2a-k)!\,(b-a-k)!}}+\sum _{k=M}^{n}{\frac {A_{n,k}}{k!\,(b+a+k)!\,(-2a-k)!\,(b-a-k)!}}\right)}$

Für alle ${\displaystyle 0\leq k gilt ${\displaystyle \lim _{n\to \infty }A_{n,k}=1\,}$.

Also ist ${\displaystyle I:=\lim _{n\to \infty }I_{n}=2\pi \,{\frac {b!^{2}\,(-2a)!\,(2a-1)!}{(a-1)!^{2}}}\,\left(\sum _{k=0}^{M-1}{\frac {1}{k!\,(b+a+k)!\,(-2a-k)!\,(b-a-k)!}}+\varepsilon _{M}\right)}$.

Für ${\displaystyle M\to \infty \,}$ geht ${\displaystyle \varepsilon _{M}\,}$ gegen null und die hypergeometrische Reihe ${\displaystyle \sum _{k=0}^{\infty }{\frac {1}{k!\,(b+a+k)!\,(-2a-k)!\,(b-a-k)!}}}$ lässt sich

nach der Formel ${\displaystyle \sum _{k=0}^{\infty }{\frac {1}{k!\,(\alpha +k)!\,(\beta -k)!\,(\gamma -k)!}}={\frac {(\alpha +\beta +\gamma )!}{\beta !\,\gamma !\,(\alpha +\beta )!\,(\alpha +\gamma )!}}}$ für ${\displaystyle \alpha +\beta +\gamma >-1\,}$,

schreiben als ${\displaystyle {\frac {(2b-2a)!}{(-2a)!\,(b-a)!\,(b-a)!\,(2b)!}}}$ wenn ${\displaystyle 2b-2a>-1\,}$ bzw. ${\displaystyle a ist. Also ist ${\displaystyle I=2\pi \,{\frac {b!^{2}\,(2a-1)!\,(2b-2a)!}{(a-1)!^{2}\,(b-a)!^{2}\,(2b)!}}}$.

Und das lässt sich unter Verwendung der Legendreschen Verdopplungsformel schreiben als ${\displaystyle {\sqrt {\pi }}\,{\frac {\Gamma \left(a+{\frac {1}{2}}\right)\cdot \Gamma (b+1)\cdot \Gamma \left(b-a+{\frac {1}{2}}\right)}{\Gamma (a)\cdot \Gamma \left(b+{\frac {1}{2}}\right)\cdot \Gamma \left(b-a+1\right)}}}$.

##### 2.12
${\displaystyle \int _{0}^{\infty }{\frac {x^{2m-1}}{\prod \limits _{k=1}^{n}(x^{2}+k^{2})}}\,dx=2\cdot (-1)^{m-1}\sum _{k=1}^{n}{\frac {(-1)^{k}\,k^{2m}\,\log k}{(n+k)!\,(n-k)!}}\qquad m
Beweis

Aus der Partialbruchzerlegung ${\displaystyle {\frac {x^{2m-1}}{\prod \limits _{k=1}^{n}(x^{2}+k^{2})}}=\sum _{k=1}^{n}{\frac {(-1)^{m+k}\,k^{2m}}{(n+k)!\,(n-k)!}}\cdot {\frac {2x}{x^{2}+k^{2}}}}$

folgt ${\displaystyle \sum _{k=1}^{n}{\frac {(-1)^{m+k}\,k^{2m}\cdot 2}{(n+k)!\,(n-k)!}}=\lim _{x\to \infty }\sum _{k=1}^{n}{\frac {(-1)^{m+k}\,k^{2m}}{(n+k)!\,(n-k)!}}\cdot {\frac {2x^{2}}{x^{2}+k^{2}}}=\lim _{x\to \infty }{\frac {x^{2m}}{\prod \limits _{k=1}^{n}(x^{2}+k^{2})}}=0}$.

Also verschwindet ${\displaystyle \sum _{k=1}^{n}{\frac {(-1)^{m+k}\,k^{2m}}{(n+k)!\,(n-k)!}}\cdot \log(R^{2})}$ für alle ${\displaystyle R>0}$.

Nun ist ${\displaystyle \int _{0}^{\infty }{\frac {x^{2m-1}}{\prod \limits _{k=1}^{n}(x^{2}+k^{2})}}\,dx=\lim _{R\to \infty }\sum _{k=1}^{n}{\frac {(-1)^{m+k}\,k^{2m}}{(n+k)!\,(n-k)!}}\cdot \int _{0}^{R}{\frac {2x}{x^{2}+k^{2}}}\,dx}$

${\displaystyle =\lim _{R\to \infty }\sum _{k=1}^{n}{\frac {(-1)^{m+k}\,k^{2m}}{(n+k)!\,(n-k)!}}\cdot \left[\log(R^{2}+k^{2})-\log(k^{2})\right]}$

${\displaystyle =\lim _{R\to \infty }\sum _{k=1}^{n}{\frac {(-1)^{m+k}\,k^{2m}}{(n+k)!\,(n-k)!}}\cdot \left[\log \left(1+{\frac {k^{2}}{R^{2}}}\right)-\log(k^{2})\right]=\sum _{k=1}^{n}{\frac {(-1)^{m+k}\,k^{2m}}{(n+k)!\,(n-k)!}}\,(-2\cdot \log k)}$.

##### 2.13
${\displaystyle \int _{0}^{\infty }{\frac {x^{2m-1}}{\prod \limits _{k=0}^{n}{\Big (}x^{2}+(2k+1)^{2}{\Big )}}}\,dx=\sum _{k=0}^{n}{\frac {(-1)^{m+k}\,(2k+1)^{2m-1}\,\log(2k+1)}{2^{2n}\,(n+k+1)!\,(n-k)!}}\qquad m\leq n}$
Beweis

Aus der Partialbruchzerlegung ${\displaystyle {\frac {x^{2m-1}}{\prod \limits _{k=0}^{n}{\Big (}x^{2}+(2k+1)^{2}{\Big )}}}=\sum _{k=0}^{n}{\frac {(-1)^{m-1+k}\,(2k+1)^{2m-1}}{2^{2n}\,(n+k+1)!\,(n-k)!}}\cdot {\frac {x}{x^{2}+(2k+1)^{2}}}}$

folgt ${\displaystyle \sum _{k=0}^{n}{\frac {(-1)^{m-1+k}\,(2k+1)^{2m-1}}{2^{2n}\,(n+k+1)!\,(n-k)!}}=\lim _{x\to \infty }\sum _{k=0}^{n}{\frac {(-1)^{m-1+k}\,(2k+1)^{2m-1}}{2^{2n}\,(n+k+1)!\,(n-k)!}}\cdot {\frac {x^{2}}{x^{2}+(2k+1)^{2}}}=\lim _{x\to \infty }{\frac {x^{2m}}{\prod \limits _{k=0}^{n}{\Big (}x^{2}+(2k+1)^{2}{\Big )}}}=0}$.

Also verschwindet ${\displaystyle \sum _{k=0}^{n}{\frac {(-1)^{m-1+k}\,(2k+1)^{2m-1}}{2^{2n}\,(n+k+1)!\,(n-k)!}}\,\log(R^{2})}$ für alle ${\displaystyle R>0}$.

Nun ist ${\displaystyle \int _{0}^{\infty }{\frac {x^{2m-1}}{\prod \limits _{k=0}^{n}{\Big (}x^{2}+(2k+1)^{2}{\Big )}}}\,dx=\lim _{R\to \infty }\sum _{k=0}^{n}{\frac {(-1)^{m-1+k}\,(2k+1)^{2m-1}}{2^{2n+1}\,(n+k+1)!\,(n-k)!}}\,\int _{0}^{R}{\frac {2x}{x^{2}+(2k+1)^{2}}}\,dx}$

${\displaystyle =\lim _{R\to \infty }\sum _{k=0}^{n}{\frac {(-1)^{m-1+k}\,(2k+1)^{2m-1}}{2^{2n+1}\,(n+k+1)!\,(n-k)!}}\left[\log {\Big (}R^{2}+(2k+1)^{2}{\Big )}-\log {\Big (}(2k+1)^{2}{\Big )}\right]}$

${\displaystyle =\lim _{R\to \infty }\sum _{k=0}^{n}{\frac {(-1)^{m-1+k}\,(2k+1)^{2m-1}}{2^{2n+1}\,(n+k+1)!\,(n-k)!}}\left[\log \left(1+{\frac {(2k+1)^{2}}{R^{2}}}\right)-\log {\Big (}(2k+1)^{2}{\Big )}\right]}$

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

##### 2.14
${\displaystyle \int _{0}^{\infty }{\frac {x^{2m}}{\prod \limits _{k=1}^{n}(x^{2}+k^{2})}}\,dx=\pi \cdot \sum _{k=1}^{n}{\frac {(-1)^{m-1+k}\,k^{2m+1}}{(n+k)!\,(n-k)!}}\qquad m
Beweis

Aus der Partialbruchzerlegung ${\displaystyle {\frac {x^{2m}}{\prod \limits _{k=1}^{n}(x^{2}+k^{2})}}=\sum _{k=1}^{n}{\frac {(-1)^{m-1+k}\,k^{2m+2}}{(n+k)!\,(n-k)!}}\cdot {\frac {2}{x^{2}+k^{2}}}}$

folgt unmittelbar ${\displaystyle \int _{0}^{\infty }{\frac {x^{2m}}{\prod \limits _{k=1}^{n}(x^{2}+k^{2})}}\,dx=\sum _{k=1}^{n}{\frac {(-1)^{m-1+k}\,k^{2m+2}}{(n+k)!\,(n-k)!}}\cdot \underbrace {\int _{0}^{\infty }{\frac {2}{x^{2}+k^{2}}}\,dx} _{={\frac {\pi }{k}}}}$.

##### 2.15
${\displaystyle \int _{0}^{\infty }{\frac {x^{2m}}{\prod \limits _{k=0}^{n}{\Big (}x^{2}+(2k+1)^{2}{\Big )}}}\,dx=\pi \cdot \sum _{k=0}^{n}{\frac {(-1)^{m+k}\,(2k+1)^{2m}}{2^{2n+1}\,(n+k+1)!\,(n-k)!}}\qquad m\leq n}$
Beweis

Aus der Partialbruchzerlegung ${\displaystyle {\frac {x^{2m}}{\prod \limits _{k=0}^{n}{\Big (}x^{2}+(2k+1)^{2}{\Big )}}}=\sum _{k=0}^{n}{\frac {(-1)^{m+k}\,(2k+1)^{2m+1}}{2^{2n}\,(n+k+1)!\,(n-k)!}}\cdot {\frac {1}{x^{2}+(2k+1)^{2}}}}$

folgt unmittelbar ${\displaystyle \int _{0}^{\infty }{\frac {x^{2m}}{\prod \limits _{k=0}^{n}{\Big (}x^{2}+(2k+1)^{2}{\Big )}}}\,dx=\sum _{k=0}^{n}{\frac {(-1)^{m+k}\,(2k+1)^{2m+1}}{2^{2n}\,(n+k+1)!\,(n-k)!}}\cdot \underbrace {\int _{0}^{\infty }{\frac {dx}{x^{2}+(2k+1)^{2}}}} _{={\frac {1}{2}}\cdot {\frac {\pi }{2k+1}}}}$.

##### 2.16
${\displaystyle \int _{0}^{\infty }\left({\frac {2}{1+{\sqrt {1+4x}}}}\right)^{p}\cdot x^{s-1}\,dx={\frac {p\cdot \Gamma (s)\cdot \Gamma (p-2s)}{\Gamma (p-s+1)}}\qquad 0<{\text{Re}}(s)<{\frac {1}{2}}\cdot {\text{Re}}(p)}$
Beweis

${\displaystyle I:=\int _{0}^{\infty }\left({\frac {2}{1+{\sqrt {1+4x}}}}\right)^{p}\cdot x^{s-1}\,dx}$

Nach Substitution ${\displaystyle x=y+y^{2}}$ ist

${\displaystyle I=\int _{0}^{\infty }\left({\frac {2}{1+(1+2y)}}\right)^{p}\cdot y^{s-1}\,(1+y)^{s-1}\cdot (1+2y)\,dy=\int _{0}^{\infty }y^{s-1}\cdot (1+y)^{s-p-1}\cdot (1+2y)\,dy}$.

Nach Substitution ${\displaystyle y={\frac {z}{1-z}}}$ ist

${\displaystyle I=\int _{0}^{1}{\frac {z^{s-1}}{(1-z)^{s-1}}}\cdot {\frac {1}{(1-z)^{s-p-1}}}\cdot {\frac {1+z}{1-z}}\cdot {\frac {dz}{(1-z)^{2}}}=\int _{0}^{1}z^{s-1}\cdot (1-z)^{p-2s-1}\,(1+z)\,dz}$

${\displaystyle ={\frac {\Gamma (s)\cdot \Gamma (p-2s)}{\Gamma (p-s)}}+{\frac {\Gamma (s+1)\cdot \Gamma (p-2s)}{\Gamma (p-s+1)}}={\frac {(p-s)\cdot \Gamma (s)\cdot \Gamma (p-2s)}{(p-s)\cdot \Gamma (p-s)}}+{\frac {s\cdot \Gamma (s)\cdot \Gamma (p-2s)}{\Gamma (p-s+1)}}={\frac {p\cdot \Gamma (s)\cdot \Gamma (p-2s)}{\Gamma (p-s+1)}}}$.

##### 3.1
${\displaystyle \int _{0}^{\infty }{\frac {x^{\alpha -1}}{(w+x)^{\beta }}}\,dx={\frac {B(\alpha ,\beta -\alpha )}{w^{\beta -\alpha }}}\qquad 0<{\text{Re}}(\alpha )<{\text{Re}}(\beta )\,,\;{\text{Re}}(w)>0}$
Beweis

Setzt man ${\displaystyle f(z)={\frac {z^{\alpha -1}}{(1+z)^{\alpha +\beta }}}}$, so ist ${\displaystyle f\,}$ auf ${\displaystyle \mathbb {C} \setminus \mathbb {R} ^{\leq 0}}$ holomorph.

Wegen ${\displaystyle \left|z^{\alpha }\right|=\Theta \left(|z|^{{\text{Re}}\,\alpha }\right)}$ ist ${\displaystyle |f(z)|={\frac {\Theta \left(|z|^{{\text{Re}}(\alpha )-1}\right)}{\Theta \left(|z|^{{\text{Re}}(\alpha +\beta )}\right)}}=\Theta \left({\frac {1}{|z|^{1+{\text{Re}}(\beta )}}}\right)={\text{o}}\left({\frac {1}{|z|}}\right)}$ für ${\displaystyle |z|\to \infty \,}$.

Als komplexe Zahl mit positivem Realteil besitzt ${\displaystyle w\,}$ eine Darstellung ${\displaystyle r\,e^{i\varphi }}$ mit ${\displaystyle -{\frac {\pi }{2}}<\varphi <{\frac {\pi }{2}}}$.

Der Kehrwehrt ${\displaystyle {\frac {1}{w}}={\frac {1}{r}}\,e^{-i\varphi }}$ besitzt dann auch einen positiven Realteil.

Wegen ${\displaystyle |f(z)|={\text{o}}\left({\frac {1}{|z|}}\right)}$ ist nun ${\displaystyle \int _{0}^{\infty }f(x)dx=\int _{0}^{\infty }f\left({\frac {x}{w}}\right){\frac {dx}{w}}=\int _{0}^{\infty }{\frac {\left({\frac {x}{w}}\right)^{\alpha -1}}{\left(1+{\frac {x}{w}}\right)^{\alpha +\beta }}}\,dx}$

${\displaystyle =w^{\beta }\int _{0}^{\infty }{\frac {x^{\alpha -1}}{\left(w+x\right)^{\alpha +\beta }}}\,dx}$. Also ist ${\displaystyle \int _{0}^{\infty }{\frac {x^{\alpha -1}}{\left(w+x\right)^{\alpha +\beta }}}\,dx={\frac {1}{w^{\beta }}}\int _{0}^{\infty }f(x)dx}$.

Und das ist ${\displaystyle {\frac {B(\alpha ,\beta )}{w^{\beta }}}}$. Ersetzt man ${\displaystyle \beta \,}$ durch ${\displaystyle \beta -\alpha \,}$, so ist ${\displaystyle \int _{0}^{\infty }{\frac {x^{\alpha -1}}{(w+x)^{\beta }}}\,dx={\frac {B(\alpha ,\beta -\alpha )}{w^{\beta -\alpha }}}}$.

##### 3.2
${\displaystyle \int _{0}^{\infty }{\frac {x^{\alpha -1}}{(1+x^{\beta })^{\gamma }}}\,dx={\frac {1}{\beta }}\,B\left({\frac {\alpha }{\beta }},\gamma -{\frac {\alpha }{\beta }}\right)}$
ohne Beweis

##### 3.3
${\displaystyle \int _{0}^{\infty }\prod _{1\leq \ell \leq 3}{\frac {1}{(x^{2}+2x\cos \alpha _{\ell }+1)}}\,dx={\frac {1}{4}}\sum _{(i,j,k)\in A_{3}}{\frac {\alpha _{i}\,\csc \alpha _{i}\,\cos 2\alpha _{i}}{(\cos \alpha _{i}-\cos \alpha _{j})(\cos \alpha _{i}-\cos \alpha _{k})}}}$
ohne Beweis

##### 4.1
${\displaystyle \int _{-\infty }^{\infty }{\frac {dx}{(u+ix)^{\alpha }\,(v-ix)^{\beta }}}={\frac {2\pi }{(u+v)^{\alpha +\beta -1}}}\,{\frac {\Gamma (\alpha +\beta -1)}{\Gamma (\alpha )\,\Gamma (\beta )}}\qquad {\text{Re}}(u),{\text{Re}}(v)>0\;,\;{\text{Re}}(\alpha +\beta )>1}$
Beweis

Setze ${\displaystyle f(z)={\frac {1}{(u+iz)^{\alpha }\,(v-iz)^{\beta }}}}$.

Wegen ${\displaystyle {\text{Re}}(u),{\text{Re}}(v)>0\,}$ ist ${\displaystyle {\text{Im}}(iu)>0,{\text{Im}}(-iv)<0\,}$ und ${\displaystyle f\,}$ ist auf ${\displaystyle \mathbb {C} \setminus \left(iu+i\mathbb {R} ^{\geq 0}\cup -iv-i\mathbb {R} ^{\geq 0}\right)}$ holomorph.

Wegen ${\displaystyle \left|z^{\alpha }\right|=\Theta \left(|z|^{{\text{Re}}\,\alpha }\right)}$ ist ${\displaystyle |f(z)|={\frac {1}{\Theta \left(|z|^{{\text{Re}}(\alpha )}\right)\,\Theta \left(|z|^{{\text{Re}}(\beta )}\right)}}=\Theta \left({\frac {1}{|z|^{{\text{Re}}(\alpha +\beta )}}}\right)={\text{o}}\left({\frac {1}{|z|}}\right)}$ für ${\displaystyle |z|\to \infty \,}$.

Die Integrale über den Kreisbögen ${\displaystyle K_{1},K_{2}\,}$ verschwinden daher wenn ihr Radius gegen unendlich geht.

Vorübergehend mache man die zusätzliche Einschränkung ${\displaystyle {\text{Re}}(\alpha )<0\,}$.

Dann ist ${\displaystyle \lim _{z\to iu}f(z)=0\,}$, und somit verschwindet das Integral über dem Halbkreis ${\displaystyle \kappa \,}$ mit Radius ${\displaystyle \varepsilon \,}$, wenn ${\displaystyle \varepsilon \to 0+\,}$ geht.

Damit ist ${\displaystyle \int _{-\infty }^{\infty }f(x)dx=\lim _{\varepsilon \to 0+}\left(\int _{0}^{\infty }f(iu+\varepsilon +it)idt-\int _{0}^{\infty }f(iu-\varepsilon +it)idt\right)}$,

wobei ${\displaystyle f(iu\pm \varepsilon +it)={\frac {1}{(-t\pm i\varepsilon )^{\alpha }}}\,{\frac {1}{(u+v+t\mp i\varepsilon )^{\beta }}}={\frac {e^{\mp i\pi \alpha }}{(t\mp i\varepsilon )^{\alpha }}}\,{\frac {1}{(u+v+t\mp i\varepsilon )^{\beta }}}}$ ist.

Also ist ${\displaystyle \int _{-\infty }^{\infty }f(x)dx=\left(e^{-i\pi \alpha }-e^{i\pi \alpha }\right)\int _{0}^{\infty }{\frac {1}{t^{\alpha }\,(u+v+t)^{\beta }}}idt=2\sin \alpha \pi \int _{0}^{\infty }{\frac {t^{(1-\alpha )-1}}{(u+v+t)^{\beta }}}\,dt}$

${\displaystyle =2\sin \alpha \,{\frac {B(1-\alpha ,\beta -(1-\alpha ))}{(u+v)^{\beta -(1-\alpha )}}}}$. Und das ist ${\displaystyle {\frac {2\pi }{\Gamma (\alpha )\,\Gamma (1-\alpha )}}\,{\frac {1}{(u+v)^{\alpha +\beta -1}}}\,{\frac {\Gamma (1-\alpha )\,\Gamma (\alpha +\beta -1)}{\Gamma (\beta )}}}$

${\displaystyle ={\frac {2\pi }{(u+v)^{\alpha +\beta -1}}}\,{\frac {\Gamma (\alpha +\beta -1)}{\Gamma (\alpha )\,\Gamma (\beta )}}}$.

Dass die Formel auch ohne die Einschränkung ${\displaystyle {\text{Re}}(\alpha )<0\,}$ gilt, sieht man, wenn man sie wiederholt partiell integriert:

${\displaystyle \int _{-\infty }^{\infty }{\frac {1}{(u+ix)^{\alpha }\,(v-ix)^{\beta }}}\,dx=\underbrace {\left[{\frac {1}{(u+ix)^{\alpha }}}\,{\frac {1}{i(\beta -1)}}\,{\frac {1}{(v-ix)^{\beta -1}}}\right]_{-\infty }^{\infty }} _{=0}}$

${\displaystyle -\int _{-\infty }^{\infty }{\frac {-i\alpha }{(u+ix)^{\alpha +1}}}\,{\frac {1}{i(\beta -1)}}\,{\frac {1}{(v-ix)^{\beta -1}}}\,dx={\frac {\alpha }{\beta -1}}\int _{-\infty }^{\infty }{\frac {1}{(u+ix)^{\alpha +1}}}\,{\frac {1}{(v-ix)^{\beta -1}}}\,dx}$.

Und damit ist ${\displaystyle \int _{-\infty }^{\infty }{\frac {1}{(u+ix)^{\alpha +1}}}\,{\frac {1}{(v-ix)^{\beta -1}}}\,dx={\frac {2\pi }{(u+v)^{\alpha +\beta -1}}}\,{\frac {\Gamma (\alpha +\beta -1)}{\Gamma (\alpha )\,\Gamma (\beta )}}\,{\frac {\beta -1}{\alpha }}}$

${\displaystyle ={\frac {2\pi }{(u+v)^{\alpha +\beta -1}}}\,{\frac {\Gamma (\alpha +\beta -1)}{\Gamma (\alpha +1)\,\Gamma (\beta -1)}}}$.