0.1Bearbeiten

{\displaystyle \int _{-\infty }^{\infty }e^{-x^{2}}\,dx={\sqrt {\pi }}}

1. Beweis

{\displaystyle I^{2}=\left(\int _{-\infty }^{\infty }e^{-x^{2}}\,dx\right)\,\left(\int _{-\infty }^{\infty }e^{-y^{2}}\,dy\right)=\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }e^{-(x^{2}+y^{2})}\,dx\,dy}

lässt sich in Polarkoordinaten schreiben als {\displaystyle \int _{0}^{2\pi }\int _{0}^{\infty }e^{-r^{2}}\,r\,dr\,d\varphi }.

Und das ist {\displaystyle \int _{0}^{2\pi }\left[-{\frac {e^{-r^{2}}}{2}}\right]_{0}^{\infty }\,d\varphi =\int _{0}^{2\pi }{\frac {1}{2}}\,d\varphi =\pi \Rightarrow I={\sqrt {\pi }}}.

2. Beweis

Die Fläche, die entsteht wenn {\displaystyle f(x)=e^{-x^{2}}} um die z-Achse rotiert, schließt mit der xy-Ebene das gleiche Volumen ein

wie die Fläche, die entsteht, wenn {\displaystyle f^{-1}(x)={\sqrt {-\log x}}} um die x-Achse rotiert, mit der yz-Ebene.

Also {\displaystyle I^{2}=\pi \int _{0}^{1}{\sqrt {-\log x}}^{\,2}\,dx=\pi \Rightarrow I={\sqrt {\pi }}}.

3. Beweis

Definiert man {\displaystyle F(x)=\left(\int _{0}^{x}e^{-t^{2}}\,dt\right)^{2}} und {\displaystyle G(x)=\int _{0}^{1}{\frac {e^{-x^{2}\,(1+t^{2})}}{1+t^{2}}}\,dt}, so ist {\displaystyle F'(x)=2\,\int _{0}^{x}e^{-t^{2}}\,dt\;e^{-x^{2}}}

und {\displaystyle G'(x)=\int _{0}^{1}e^{-x^{2}\,(1+t^{2})}\,(-2x)\,dt=-2\,\int _{0}^{1}e^{-x^{2}\,t^{2}}\,x\,dt\;e^{-x^{2}}=-2\int _{0}^{x}e^{-t^{2}}\,dt\;e^{-x^{2}}}.

Es ist also {\displaystyle F'(x)+G'(x)=0\,}. Folglich muss {\displaystyle F(x)+G(x)\,} konstant sein.

{\displaystyle F(\infty )+G(\infty )=F(0)+G(0)\Rightarrow \left({\frac {I}{2}}\right)^{2}+0=0+{\frac {\pi }{4}}\Rightarrow I={\sqrt {\pi }}}

4. Beweis

Es sei {\displaystyle a={\sqrt {i\pi }}=(1+i){\sqrt {\frac {\pi }{2}}}} und {\displaystyle f(z)={\frac {e^{-z^{2}}}{1+e^{-2az}}}}. 

Wegen {\displaystyle {\text{Re}}(a)>0\,} gilt {\displaystyle e^{-2ax}\to \left\{{\begin{matrix}0&,&x\to \infty \\\infty &,&x\to -\infty \end{matrix}}\right.}.

Ist {\displaystyle 0\leq y\leq {\sqrt {\frac {\pi }{2}}}}, so geht für {\displaystyle x\to \pm \infty \,} der Nenner von {\displaystyle f(x+iy)={\frac {e^{-x^{2}}\,e^{y^{2}-2ixy}}{1+e^{-2ax}\,e^{-2iay}}}} gegen {\displaystyle 1\,} oder {\displaystyle \infty \,}

und der Zähler geht gegen Null. Also verschwinden die beiden Integrale {\displaystyle \int _{\pm R}^{\pm R+a}\,f\,dz} für {\displaystyle R\to \infty \,}.

Wegen {\displaystyle f(z)-f(z+a)=e^{-z^{2}}} gilt nun {\displaystyle \int _{-\infty }^{\infty }e^{-z^{2}}\,dz=\lim _{R\to \infty }\oint f\,dz=2\pi i\,{\text{res}}\left(f,{\frac {a}{2}}\right)={\sqrt {\pi }}}.

 

0.2Bearbeiten

{\displaystyle \int _{0}^{\infty }\left({\frac {1}{e^{x}-1}}-{\frac {1}{x\,e^{x}}}\right)\,dx=\gamma }

Beweis

Für {\displaystyle {\text{Re}}(z)>1\,} gilt {\displaystyle \int _{0}^{\infty }\left({\frac {x^{z-1}}{e^{x}-1}}-{\frac {x^{z-1}}{x\,e^{x}}}\right)\,dx=\Gamma (z)\,\left(\zeta (z)-{\frac {1}{z-1}}\right)\to \gamma } für {\displaystyle z\to 1\,}.

 

1.1Bearbeiten

{\displaystyle \int _{0}^{\infty }x^{z-1}e^{-x}\,dx=\Gamma (z)\qquad {\text{Re}}(z)>0}

ohne Beweis

 

 

1.2Bearbeiten

{\displaystyle \int _{0}^{\infty }\exp \left(-x^{2}-{\frac {\alpha ^{2}}{x^{2}}}\right)\,dx={\frac {\sqrt {\pi }}{2}}\,e^{-2\alpha }\qquad \alpha >0}

1. Beweis

{\displaystyle 2I=\int _{0}^{\infty }\exp \left(-x^{2}-{\frac {\alpha ^{2}}{x^{2}}}\right)\,dx=\int _{-\infty }^{\infty }\exp \left(-\left(x-{\frac {\alpha }{x}}\right)^{2}-2\alpha \right)\,dx=\int _{-\infty }^{\infty }e^{-\left(x-{\frac {\alpha }{x}}\right)^{2}}dx\cdot e^{-2\alpha }}.

Nach der Formel {\displaystyle \int _{-\infty }^{\infty }f\left(x-{\frac {b}{x}}\right)dx=\int _{-\infty }^{\infty }f(x)dx}, gilt im Fall {\displaystyle f(x)=e^{-x^{2}}},

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

2. Beweis

{\displaystyle I'(\alpha )=\int _{0}^{\infty }\exp \left(-x^{2}-{\frac {\alpha ^{2}}{x^{2}}}\right)\,{\frac {-2\alpha }{x^{2}}}\,dx} ist nach Substitution {\displaystyle x\mapsto {\frac {\alpha }{x}}} gleich {\displaystyle -2\int _{0}^{\infty }\exp \left(-x^{2}-{\frac {\alpha ^{2}}{x^{2}}}\right)\,dx}.

Die Differenzialgleichung {\displaystyle I'(\alpha )=-2\,I(\alpha )} wird gelöst durch {\displaystyle I(\alpha )=C\,e^{-2\alpha }}, wobei {\displaystyle C=I(0)=\int _{0}^{\infty }e^{-x^{2}}\,dx={\frac {\sqrt {\pi }}{2}}} ist.

 

1.3Bearbeiten

{\displaystyle \int _{0}^{\infty }\left({\frac {1}{x\,e^{x}}}-{\frac {e^{-x(z-1)}}{e^{x}-1}}\right)dx=\psi (z)\qquad {\text{Re}}(z)>0}

Beweis

In der Formel {\displaystyle \int _{0}^{1}{\frac {1-x^{z-1}}{1-x}}\,dx=\psi (z)+\gamma } wird das Integral

nach Substitution {\displaystyle x\mapsto e^{-x}} zu {\displaystyle \int _{0}^{\infty }\left({\frac {1}{e^{x}-1}}-{\frac {e^{-x(z-1)}}{e^{x}-1}}\right)\,dx},

und {\displaystyle \gamma \,} lässt sich schreiben als {\displaystyle \int _{0}^{\infty }\left({\frac {1}{e^{x}-1}}-{\frac {1}{x\,e^{x}}}\right)\,dx}.

 

1.4Bearbeiten

{\displaystyle \int _{0}^{\infty }\left({\frac {z-1}{x\,e^{x}}}-{\frac {1-e^{-x(z-1)}}{x\,(e^{x}-1)}}\right)\,dx=\log \Gamma (z)\qquad {\text{Re}}(z)>0}

Beweis (Formel nach Malmstén)

Integriere die Formel {\displaystyle \int _{0}^{\infty }\left({\frac {1}{x\,e^{x}}}-{\frac {e^{-x(z'-1)}}{e^{x}-1}}\right)dx=\psi (z')} nach z' von 1 bis z.

 

1.5Bearbeiten

{\displaystyle \int _{0}^{\infty }\left({\frac {1}{e^{x}}}-{\frac {1}{(1+x)^{z}}}\right)\,{\frac {dx}{x}}=\psi (z)\qquad {\text{Re}}(z)>0}

Beweis (Formel nach Cauchy)

{\displaystyle \forall \varepsilon >0} ist {\displaystyle \int _{0}^{\infty }\left({\frac {x^{\varepsilon -1}}{e^{x}}}-{\frac {x^{\varepsilon -1}}{(1+x)^{z}}}\right)dx=\Gamma (\varepsilon )-{\frac {\Gamma (z-\varepsilon )\,\Gamma (\varepsilon )}{\Gamma (z)}}}

{\displaystyle ={\frac {\Gamma (z)\,\Gamma (\varepsilon )-\Gamma (z-\varepsilon )\,\Gamma (\varepsilon )}{\Gamma (z)}}={\frac {\Gamma (1+\varepsilon )}{\Gamma (z)}}\cdot {\frac {\Gamma (z)-\Gamma (z-\varepsilon )}{\varepsilon }}\,{\xrightarrow {\,\,\varepsilon \to 0\,\,}}\,{\frac {1}{\Gamma (z)}}\cdot \Gamma '(z)=\psi (z)}.

 

1.6Bearbeiten

{\displaystyle \int _{0}^{\infty }{\frac {x^{z-1}}{e^{x}-1}}\,dx=\Gamma (z)\,\zeta (z)\qquad {\text{Re}}(z)>1}

Beweis

Wegen {\displaystyle {\frac {1}{e^{x}-1}}={\frac {e^{-x}}{1-e^{-x}}}=\sum _{k=1}^{\infty }e^{-kz}} ist {\displaystyle {\frac {x^{z-1}}{e^{x}-1}}=\sum _{k=1}^{\infty }x^{z-1}\,e^{-kx}}

und somit {\displaystyle \int _{0}^{\infty }{\frac {x^{z-1}}{e^{x}-1}}\,dx=\sum _{k=1}^{\infty }\int _{0}^{\infty }x^{z-1}\,e^{-kx}\,dx=\sum _{k=1}^{\infty }{\frac {\Gamma (z)}{k^{z}}}=\Gamma (z)\,\zeta (z)}.

 

1.7Bearbeiten

{\displaystyle \int _{0}^{\infty }{\frac {x^{z-1}}{e^{x}+1}}\,dx=\Gamma (z)\,\eta (z)\qquad {\text{Re}}(z)>0}

Beweis

Wegen {\displaystyle {\frac {1}{e^{x}+1}}={\frac {e^{-x}}{1+e^{-x}}}={\frac {-(-e^{-x})}{1-(-e^{-x})}}=-\sum _{k=1}^{\infty }(-e^{-x})^{k}=\sum _{k=1}^{\infty }(-1)^{k-1}\,e^{-kx}} ist {\displaystyle {\frac {x^{z-1}}{e^{x}+1}}=\sum _{k=1}^{\infty }(-1)^{k-1}\,x^{z-1}\,e^{-kx}}

und somit {\displaystyle \int _{0}^{\infty }{\frac {x^{z-1}}{e^{x}+1}}\,dx=\sum _{k=1}^{\infty }(-1)^{k-1}\,\int _{0}^{\infty }x^{z-1}\,e^{-kx}\,dx=\sum _{k=1}^{\infty }(-1)^{k-1}\,{\frac {\Gamma (z)}{k^{z}}}=\Gamma (z)\,\eta (z)}.

 

1.8Bearbeiten

{\displaystyle {\frac {1}{\Gamma (z)}}={\frac {1}{2\pi i}}\int _{C}t^{-z}e^{t}\,dt\qquad C\,} ist eine Kurve in {\displaystyle \mathbb {C} \setminus \mathbb {R} ^{\geq 0}}, die von {\displaystyle -\infty -i\varepsilon \,} nach {\displaystyle -\infty +i\varepsilon \,} läuft.

 

Für {\displaystyle {\text{Re}}(z)>0\,} kann man als Integrationsweg {\displaystyle C\,} auch die Gerade {\displaystyle a+i\mathbb {R} }, mit {\displaystyle a>0\,}, hernehmen.

Beweis für Re(z)>0 (Hankelsche Integraldarstellung für die reziproke Gammafunktion)

Die Funktion {\displaystyle f(t)=t^{-z}\,e^{t}} mit {\displaystyle {\text{Re}}(z)>0\,} ist auf {\displaystyle \mathbb {C} \setminus \mathbb {R} ^{\geq 0}} holomorph.

Für {\displaystyle \varepsilon \leq t\leq R\,} ist {\displaystyle |f(-R+it)|\,}

{\displaystyle =\left|(-R+it)^{-z}\right|\cdot \left|e^{-R+it}\right|=\Theta \left(R^{-{\text{Re}}(z)}\right)\cdot e^{-R}}.

Daher verschwinden die Integrale über den Abschnitten {\displaystyle \sigma _{1},\sigma _{2}\,} für {\displaystyle R\to \infty \,}.


Und es ist

{\displaystyle \left|\int _{-R}^{a}f(t\pm iR)\,dt\right|\leq \int _{-R}^{a}\left|(t\pm iR)^{-z}\right|\cdot \left|e^{t\pm iR}\right|\,dt}

{\displaystyle \leq \max _{-R\leq t\leq a}\left|(t\pm iR)^{-z}\right|\cdot \int _{-R}^{a}e^{t}\,dt=\Theta \left(R^{-{\text{Re}}(z)}\right)\cdot e^{a}}.

Daher verschwinden auch die Integrale über den Abschnitten {\displaystyle \tau _{1},\tau _{2}\,} für {\displaystyle R\to \infty \,}.


Also ist

{\displaystyle {\frac {1}{2\pi i}}\int _{C}t^{-z}e^{t}\,dt={\frac {1}{2\pi i}}\int _{a-i\infty }^{a+i\infty }s^{-z}\,e^{s}\,ds}

{\displaystyle ={\mathcal {L}}^{-1}\left[s^{-z}\right](1)=\left.{\frac {t^{z-1}}{\Gamma (z)}}\right|_{t=1}={\frac {1}{\Gamma (z)}}}.

Beweis für Re(z)<1

Die Funktion {\displaystyle f(t)=t^{z-1}e^{t}\,} mit {\displaystyle {\text{Re}}(z)>0\,} ist auf {\displaystyle \mathbb {C} \setminus \mathbb {R} ^{\geq 0}} holomorph.



Das Integral über dem Kreisbogen {\displaystyle K_{\varepsilon }\,} verschwindet für {\displaystyle \varepsilon \to 0+\,}, weil wegen {\displaystyle \left|t^{z-1}\right|=\Theta \left(|t|^{{\text{Re}}(z)-1}\right)=o\left({\frac {1}{|t|}}\right)} für {\displaystyle |t|\to 0\,}

ist {\displaystyle \max _{t\in K_{\varepsilon }}\left|t^{z-1}e^{-t}\right|=o\left({\frac {1}{\varepsilon }}\right)}, und daher gilt {\displaystyle \left|\int _{K_{\varepsilon }}f(t)\,dt\right|\leq \pi \varepsilon \cdot o\left({\frac {1}{\varepsilon }}\right)=o(1)}.

Für die horizontalen Integrationswege gilt:

{\displaystyle \int _{-\infty }^{0}f(t\pm i\varepsilon )\,dt=\int _{-\infty }^{0}(t\pm i\varepsilon )^{z-1}\,e^{t\pm i\varepsilon }\,dt=e^{\pm i\pi (z-1)}\int _{-\infty }^{0}(-t\mp i\varepsilon )^{z-1}\,e^{t\pm i\varepsilon }\,dt}

{\displaystyle =-e^{\pm i\pi z}\int _{0}^{\infty }(t\mp i\varepsilon )^{z-1}\,e^{-t\pm i\varepsilon }\,dt\to -e^{\pm i\pi z}\,\Gamma (z)} für {\displaystyle \varepsilon \to 0+\,}.

Daher ist {\displaystyle \int _{C}t^{z-1}\,e^{t}\,dt=\left(e^{i\pi z}-e^{-i\pi z}\right)\Gamma (z)=2i\sin \pi z\,\Gamma (z)={\frac {2\pi i}{\Gamma (z)\,\Gamma (1-z)}}\,\Gamma (z)={\frac {2\pi i}{\Gamma (1-z)}}}.

Ersetzt man {\displaystyle z\,} durch {\displaystyle 1-z\,}, so ist {\displaystyle {\frac {1}{\Gamma (z)}}={\frac {1}{2\pi i}}\int _{C}t^{-z}e^{t}\,dt} für {\displaystyle {\text{Re}}(z)<1\,}.

 

1.9Bearbeiten

{\displaystyle J_{\nu }(x)={\frac {1}{2\pi i}}\int _{C}e^{{\frac {x}{2}}\left(t-{\frac {1}{t}}\right)}\,{\frac {dt}{t^{\nu +1}}}\qquad {\text{Re}}(x)>0\qquad C\,} ist eine Kurve in {\displaystyle \mathbb {C} \setminus \mathbb {R} ^{\geq 0}}, die von {\displaystyle -\infty -i\varepsilon \,} nach {\displaystyle -\infty +i\varepsilon \,} läuft.

Beweis für x>0 (Hankelsche Integraldarstellung für die Besselfunktion)

{\displaystyle J_{\nu }(x)=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{n!\,\Gamma (n+\nu +1)}}\left({\frac {x}{2}}\right)^{\nu +2n}}

Ersetze {\displaystyle {\frac {1}{\Gamma (n+\nu +1)}}} durch die Hankelsche Integraldarstellung {\displaystyle {\frac {1}{2\pi i}}\int _{C}t^{-n-\nu -1}e^{t}\,dt}.

{\displaystyle J_{\nu }(x)={\frac {1}{2\pi i}}\int _{C}\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{n!}}\,{\frac {\left({\frac {x}{2}}\right)^{2n}}{t^{n}}}\,\left({\frac {x}{2}}\right)^{\nu }\,{\frac {e^{t}}{t^{\nu +1}}}\,dt={\frac {1}{2\pi i}}\int _{C}e^{t-{\frac {x^{2}}{4t}}}\,\left({\frac {x}{2}}\right)^{\nu }{\frac {dt}{t^{\nu +1}}}}

Nach Substitution {\displaystyle t\mapsto {\frac {x}{2}}\cdot t} ändert sich am Integrationsweg {\displaystyle C\,} nichts, und es ist {\displaystyle J_{\nu }(x)={\frac {1}{2\pi i}}\int _{C}e^{{\frac {x}{2}}\left(t-{\frac {1}{t}}\right)}\,{\frac {dt}{t^{\nu +1}}}}.

 

1.10Bearbeiten

{\displaystyle \int _{0}^{\infty }\left({\frac {1}{x}}-{\frac {1}{e^{x}-1}}\right)e^{-zx}\,dx=\psi (z+1)-\log z\qquad {\text{Re}}(z)>0}

Beweis

Dies folgt unmittelbar aus den Formeln {\displaystyle \psi (z+1)=\int _{0}^{\infty }\left({\frac {e^{-x}}{x}}-{\frac {e^{-zx}}{e^{x}-1}}\right)dx} und {\displaystyle \log z=\int _{0}^{\infty }\left({\frac {e^{-x}}{x}}-{\frac {e^{-zx}}{x}}\right)dx}.

 

1.11Bearbeiten

{\displaystyle \int _{0}^{\infty }\left({\frac {1}{2}}-{\frac {1}{x}}+{\frac {1}{e^{x}-1}}\right)\,{\frac {e^{-zx}}{x}}\,dx=\log \left({\frac {z!\,e^{z}}{z^{z}\,{\sqrt {2\pi z}}}}\right)\qquad {\text{Re}}(z)>0}

Beweis (Erste Binetsche Formel)

In der Formel {\displaystyle \psi (z)+{\frac {1}{z}}=\log z+\int _{0}^{\infty }e^{-zx}\left({\frac {1}{x}}-{\frac {1}{e^{x}-1}}\right)dx} ersetze {\displaystyle {\frac {1}{z}}} durch {\displaystyle \int _{0}^{\infty }e^{-zx}\,dx}:

{\displaystyle \psi (z)=\log z-{\frac {1}{2z}}+\int _{0}^{\infty }e^{-zx}\left(-{\frac {1}{2}}+{\frac {1}{x}}-{\frac {1}{e^{x}-1}}\right)dx}, wobei {\displaystyle \lim _{x\to 0}{\frac {{\frac {1}{2}}-{\frac {1}{x}}+{\frac {1}{e^{x}-1}}}{x}}={\frac {1}{12}}} ist.

Integriert man beide Seiten unbestimmt nach {\displaystyle z\,}, so ist

{\displaystyle \log \Gamma (z)=\left(z-{\frac {1}{2}}\right)\log z-z+\int _{0}^{\infty }e^{-zx}\left({\frac {1}{2}}-{\frac {1}{x}}+{\frac {1}{e^{x}-1}}\right){\frac {dx}{x}}+C}.

Daraus folgt {\displaystyle \int _{0}^{\infty }e^{-zx}\left({\frac {1}{2}}-{\frac {1}{x}}+{\frac {1}{e^{x}-1}}\right){\frac {dx}{x}}+C=\log \left({\frac {(z-1)!\,e^{z}}{z^{z-{\frac {1}{2}}}}}\right)}.

Nachdem für {\displaystyle z\to \infty \,} das Integral verschwindet, ist {\displaystyle C=\lim _{z\to \infty }\log \left({\frac {z!\,e^{z}}{z^{z}\,{\sqrt {z}}}}\right)=\log {\sqrt {2\pi }}}.

 

1.12Bearbeiten

{\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}}}

Beweis (Poissonsche Darstellung der Besselfunktion)

Setze {\displaystyle I:=\int _{-1}^{1}e^{izx}\,(1-x^{2})^{\nu -{\frac {1}{2}}}\,dx}.

Verwende die Reihenentwicklung {\displaystyle e^{izx}=\sum _{n=0}^{\infty }{\frac {(iz)^{n}}{n!}}\,x^{n}}:

{\displaystyle I=\sum _{n=0}^{\infty }{\frac {(iz)^{n}}{n!}}\int _{-1}^{1}x^{n}(1-x^{2})^{\nu -{\frac {1}{2}}}\,dx}

Letzter Integrand ist für gerade {\displaystyle n\,} gerade und für ungerade {\displaystyle n\,} ungerade.

{\displaystyle I=\sum _{n=0}^{\infty }{\frac {(iz)^{2n}}{(2n)!}}\cdot 2\int _{0}^{1}x^{2n}\,(1-x^{2})^{\nu -{\frac {1}{2}}}\,dx}

Nach Substitution {\displaystyle x={\sqrt {t}}} ist {\displaystyle I=\sum _{n=0}^{\infty }{\frac {(-1)^{n}\,z^{2n}}{(2n)!}}\,\int _{0}^{1}t^{n-{\frac {1}{2}}}\,(1-t)^{\nu -{\frac {1}{2}}}\,dt}.

Dabei ist {\displaystyle \int _{0}^{1}t^{n-{\frac {1}{2}}}\,(1-t)^{\nu -{\frac {1}{2}}}\,dt=B\left(n+{\frac {1}{2}},\nu +{\frac {1}{2}}\right)}   {\displaystyle ={\frac {\Gamma \left(n+{\frac {1}{2}}\right)\Gamma \left(\nu +{\frac {1}{2}}\right)}{\Gamma (\nu +n+1)}}},

wobei nach Legendrescher Verdopplungsformel {\displaystyle \Gamma \left(n+{\frac {1}{2}}\right)={\frac {\Gamma \left({\frac {1}{2}}\right)}{2^{2n}}}\,{\frac {(2n)!}{n!}}} ist.

Also ist {\displaystyle I=\sum _{n=0}^{\infty }(-1)^{n}\,\left({\frac {z}{2}}\right)^{2n}\,{\frac {\Gamma \left({\frac {1}{2}}\right)\,\Gamma \left(\nu +{\frac {1}{2}}\right)}{n!\,\Gamma (\nu +n+1)}}},

und damit ist {\displaystyle {\frac {1}{\Gamma \left({\frac {1}{2}}\right)\Gamma \left(\nu +{\frac {1}{2}}\right)}}\left({\frac {z}{2}}\right)^{\nu }I=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{n!\,\Gamma (\nu +n+1)}}\left({\frac {z}{2}}\right)^{\nu +2n}=J_{\nu }(z)}.

 

2.1Bearbeiten

{\displaystyle \int _{0}^{\infty }e^{-sx}\,{\frac {a}{a^{2}+x^{2}}}\,dx=\sin(as)\,{\text{Ci}}(as)-\cos(as)\,\left({\text{Si}}(as)-{\frac {\pi }{2}}\right)}

Beweis

{\displaystyle \int _{0}^{\infty }e^{-sx}\,{\frac {a}{a^{2}+x^{2}}}\,dx=\int _{0}^{\infty }e^{-sx}\int _{0}^{\infty }\sin(at)\,e^{-xt}\,dt\,dx=\int _{0}^{\infty }\int _{0}^{\infty }\sin(at)\,e^{-(s+t)x}\,dx\,dt}

{\displaystyle =\int _{0}^{\infty }{\frac {\sin(at)}{s+t}}\,dt=\int _{s}^{\infty }{\frac {\sin \left(a(t-s)\right)}{t}}\,dt=\int _{s}^{\infty }{\frac {\sin(at)\,\cos(as)-\cos(at)\,\sin(as)}{t}}\,dt}

{\displaystyle =\cos(as)\int _{s}^{\infty }{\frac {\sin(at)}{t}}\,dt-\sin(as)\int _{s}^{\infty }{\frac {\cos(at)}{t}}\,dt=\cos(as)\,\left({\frac {\pi }{2}}-{\text{Si}}(as)\right)+\sin(as)\,{\text{Ci}}(as)}

 

2.2Bearbeiten

{\displaystyle \int _{0}^{\infty }e^{-sx}\,{\frac {x}{a^{2}+x^{2}}}\,dx=\sin(as)\,\left({\frac {\pi }{2}}-{\text{Si}}(as)\right)-\cos(as)\,{\text{Ci}}(as)}

Beweis

{\displaystyle \int _{0}^{\infty }e^{-sx}\,{\frac {x}{a^{2}+x^{2}}}\,dx=\int _{0}^{\infty }e^{-sx}\int _{0}^{\infty }\cos(at)\,e^{-xt}\,dt\,dx=\int _{0}^{\infty }\int _{0}^{\infty }\cos(at)\,e^{-(s+t)x}\,dx\,dt}

{\displaystyle =\int _{0}^{\infty }{\frac {\cos(at)}{s+t}}\,dt=\int _{s}^{\infty }{\frac {\cos \left(a(t-s)\right)}{t}}\,dt=\int _{s}^{\infty }{\frac {\cos(at)\,\cos(as)+\sin(at)\,\sin(as)}{t}}\,dt}

{\displaystyle =\cos(as)\int _{s}^{\infty }{\frac {\cos(at)}{t}}\,dt+\sin(as)\int _{s}^{\infty }{\frac {\sin(at)}{t}}\,dt=-\cos(as)\,{\text{Ci}}(as)+\sin(as)\,\left({\frac {\pi }{2}}-{\text{Si}}(as)\right)}

 

2.3Bearbeiten

{\displaystyle \int _{0}^{\infty }x^{z-1}e^{-\mu x}\,dx={\frac {\Gamma (z)}{\mu ^{z}}}\qquad {\text{Re}}(z),{\text{Re}}(\mu )>0}

ohne Beweis

 

 

2.4Bearbeiten

{\displaystyle \int _{0}^{\infty }x^{z-1}\,e^{-i\mu x}\,dx={\frac {\Gamma (z)}{(i\mu )^{z}}}\qquad 0<\operatorname {Re} (z)<1\,,\,\mu >0}

ohne Beweis

 

 

2.5Bearbeiten

{\displaystyle \int _{-\infty }^{\infty }{\frac {a^{2}}{(e^{x}-ax-b)^{2}+(a\pi )^{2}}}\,dx={\frac {1}{1+W\left({\frac {1}{a}}\,e^{-b/a}\right)}}\qquad a>0\,,\,b\in \mathbb {R} }

Beweis

Betrachte die Funktion {\displaystyle f(z)={\frac {1}{a\log(-z)+b-z}}\cdot {\frac {1}{z}}} auf dem Gebiet {\displaystyle D:=\mathbb {C} \setminus \mathbb {R} ^{\geq 0}}.

{\displaystyle \forall z\in D} gibt es genau ein {\displaystyle r>0\,} und genau ein {\displaystyle -{\frac {\pi }{2}}<\varphi <{\frac {\pi }{2}}}, so dass {\displaystyle -z=re^{i\varphi }\,} ist.

Beim Nenner {\displaystyle a\log(-z)+b-z=a\log \left(re^{i\varphi }\right)+b+re^{i\varphi }}

{\displaystyle =a\log r+ia\varphi +b+r\cos \varphi +ir\sin \varphi =(a\log r+r\cos \varphi +b)+i(a\varphi +r\sin \varphi )\,}

hat der Imaginärteil das selbe Vorzeichen wie {\displaystyle \varphi \,} und der Realteil steigt streng monoton in {\displaystyle r\,}.

Daher ist {\displaystyle z=-a\cdot W\left({\frac {1}{a}}\,e^{-b/a}\right)} die einzige Polstelle von {\displaystyle f\,}.

Diese erhält man, wenn man {\displaystyle \varphi =0\,} und {\displaystyle r=a\cdot W\left({\frac {1}{a}}\,e^{-b/a}\right)} setzt.

Nun ist {\displaystyle {\text{res}}\left(f,-a\cdot W\left({\frac {1}{a}}\,e^{-b/a}\right)\right)={\frac {1}{a}}\cdot {\frac {1}{1+W\left({\frac {1}{a}}\,e^{-b/a}\right)}}}.

Also gilt nach dem Residuensatz {\displaystyle \int _{\gamma _{R,\varepsilon }}f\,dz+\int _{C_{R}}f\,dz+\int _{\delta _{R,\varepsilon }}f\,dz+\int _{c_{\varepsilon }}f\,dz={\frac {1}{a}}\cdot {\frac {2\pi i}{1+W\left({\frac {1}{a}}\,e^{-b/a}\right)}}}.

Aus {\displaystyle L(C_{R})\sim 2\pi R\,} und {\displaystyle M(C_{R})=\max _{z\in C_{R}}|f(z)|\sim {\frac {1}{R^{2}}}} folgt {\displaystyle \left|\int _{C_{R}}f\,dz\right|\leq L(C_{R})\,M(C_{R})\sim {\frac {2\pi }{R}}}.

Daher geht {\displaystyle \int _{C_{R}}f\,dz} gegen null für {\displaystyle R\to \infty \,}.

Und aus {\displaystyle L(c_{\varepsilon })=\pi \varepsilon \,} und {\displaystyle M(c_{\varepsilon })=\max _{z\in c_{\varepsilon }}|f(z)|\sim {\frac {-1}{a\,\log \varepsilon }}\cdot {\frac {1}{\varepsilon }}} folgt {\displaystyle \left|\int _{c_{\varepsilon }}f\,dz\right|\leq L(c_{\varepsilon })\,M(c_{\varepsilon })\sim {\frac {-\pi }{a\,\log \varepsilon }}}.

Daher geht {\displaystyle \int _{c_{\varepsilon }}f\,dz} auch gegen null für {\displaystyle \varepsilon \to 0+\,}.

Im Grenzübergang {\displaystyle R\to \infty \,,\,\varepsilon \to 0+} ergibt sich

{\displaystyle \int _{0}^{\infty }\left({\frac {1}{a\log(-x-i0^{+})+b-x}}-{\frac {1}{a\log(-x+i0^{+})+b-x}}\right){\frac {dx}{x}}={\frac {1}{a}}\cdot {\frac {2\pi i}{1+W\left({\frac {1}{a}}\,e^{-b/a}\right)}}}.

Dabei ist {\displaystyle {\frac {1}{a(\log x-i\pi )+b-x}}-{\frac {1}{a(\log x+i\pi )+b-x}}={\frac {2\pi i\cdot a}{(a\log x+b-x)^{2}-(i\pi a)^{2}}}},

und somit gilt {\displaystyle \int _{0}^{\infty }{\frac {a^{2}}{(a\log x+b-x)^{2}+(a\pi )^{2}}}\cdot {\frac {dx}{x}}={\frac {1}{1+W\left({\frac {1}{a}}\,e^{-b/a}\right)}}}.

Substituiert man {\displaystyle x\mapsto e^{x}}, so ist {\displaystyle \int _{-\infty }^{\infty }{\frac {a^{2}}{(ax+b-e^{x})^{2}+(a\pi )^{2}}}\,dx={\frac {1}{1+W\left({\frac {1}{a}}\,e^{-b/a}\right)}}}.

 

4.1Bearbeiten

{\displaystyle \int _{0}^{\infty }{\frac {e^{kx}-e^{\lambda x}}{e^{\mu x}-e^{\nu x}}}\,dx={\frac {1}{\mu -\nu }}\left(\psi \left({\frac {\mu -\lambda }{\mu -\nu }}\right)-\psi \left({\frac {\mu -k}{\mu -\nu }}\right)\right)}

Beweis

Aus der Gaußschen Formel {\displaystyle \psi (z)+\gamma =\int _{0}^{1}{\frac {1-u^{z-1}}{1-u}}\,du}

folgt {\displaystyle \psi (1-\beta )-\psi (1-\alpha )=\int _{0}^{1}{\frac {u^{-\alpha }-u^{-\beta }}{1-u}}\,du=\int _{0}^{\infty }{\frac {e^{\alpha x}-e^{\beta x}}{e^{x}-1}}\,dx\quad \left({\text{nach Substitution}}\,u=e^{-x}\right)}.

Nun ist {\displaystyle \int _{0}^{\infty }{\frac {e^{kx}-e^{\lambda x}}{e^{\mu x}-e^{\nu x}}}\,dx=\int _{0}^{\infty }{\frac {e^{(k-\nu )x}-e^{(\lambda -\nu )x}}{e^{(\mu -\nu )x}-1}}\,dx={\frac {1}{\mu -\nu }}\int _{0}^{\infty }{\frac {e^{{\frac {k-\nu }{\mu -\nu }}\,x}-e^{{\frac {\lambda -\nu }{\mu -\nu }}\,x}}{e^{x}-1}}\,dx}

{\displaystyle ={\frac {1}{\mu -\nu }}\left(\psi \left(1-{\frac {\lambda -\nu }{\mu -\nu }}\right)-\psi \left(1-{\frac {k-\nu }{\mu -\nu }}\right)\right)={\frac {1}{\mu -\nu }}\left(\psi \left({\frac {\mu -\lambda }{\mu -\nu }}\right)-\psi \left({\frac {\mu -k}{\mu -\nu }}\right)\right)}