2.1Bearbeiten
- {\displaystyle \int _{0}^{\infty }{\text{Ci}}(ax)\,{\text{Ci}}(bx)\,dx={\frac {1}{\max\{a,b\}}}\cdot {\frac {\pi }{2}}\qquad a,b>0}
In der Formel
{\displaystyle \int {\text{Ci}}(ax)\,{\text{Ci}}(bx)\,dx=x\,{\text{Ci}}(ax)\,{\text{Ci}}(bx)-{\frac {\sin ax}{a}}\,{\text{Ci}}(bx)-{\frac {\sin bx}{b}}\,{\text{Ci}}(ax)+{\frac {1}{2a}}{\Big (}{\text{Si}}(ax+bx)+{\text{Si}}(ax-bx){\Big )}+{\frac {1}{2b}}{\Big (}{\text{Si}}(ax+bx)-{\text{Si}}(ax-bx){\Big )}}
setze {\displaystyle 0\,} und {\displaystyle \infty }
als Integrationsgrenzen ein.
Asymptotisch verhalten sich {\displaystyle {\text{Ci}}(ax)} und {\displaystyle {\text{Ci}}(bx)}
für {\displaystyle x\to 0+}
wie {\displaystyle \log x}
und für {\displaystyle x\to \infty \,}
wie {\displaystyle {\frac {\cos x}{x}}}
.
Also sind {\displaystyle {\Big [}x\,{\text{Ci}}(ax)\,{\text{Ci}}(bx){\Big ]}_{0}^{\infty }\,\,,\,\,{\Big [}{\frac {\sin ax}{a}}\,{\text{Ci}}(bx){\Big ]}_{0}^{\infty }\,\,,\,\,{\Big [}{\frac {\sin bx}{b}}\,{\text{Ci}}(ax){\Big ]}_{0}^{\infty }} jeweils gleich {\displaystyle 0-0=0}
.
Der übrige Term {\displaystyle \left[{\frac {1}{2a}}{\Big (}{\text{Si}}(ax+bx)+{\text{Si}}(ax-bx){\Big )}+{\frac {1}{2b}}{\Big (}{\text{Si}}(ax+bx)-{\text{Si}}(ax-bx){\Big )}\right]_{0}^{\infty }} verschwindet für {\displaystyle x=0}
.
Für {\displaystyle x\to \infty } geht der Term gegen
{\displaystyle \bullet \quad {\frac {1}{2a}}\left({\frac {\pi }{2}}+{\frac {\pi }{2}}\right)+{\frac {1}{2b}}\left({\frac {\pi }{2}}-{\frac {\pi }{2}}\right)={\frac {1}{a}}\cdot {\frac {\pi }{2}}} falls {\displaystyle a>b}
.
{\displaystyle \bullet \quad {\frac {1}{2a}}\left({\frac {\pi }{2}}+0\right)+{\frac {1}{2b}}\left({\frac {\pi }{2}}+0\right)={\frac {1}{a}}\cdot {\frac {\pi }{2}}={\frac {1}{b}}\cdot {\frac {\pi }{2}}} falls {\displaystyle a=b}
.
{\displaystyle \bullet \quad {\frac {1}{2a}}\left({\frac {\pi }{2}}-{\frac {\pi }{2}}\right)+{\frac {1}{2b}}\left({\frac {\pi }{2}}+{\frac {\pi }{2}}\right)={\frac {1}{b}}\cdot {\frac {\pi }{2}}} falls {\displaystyle a<b}
.