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

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##### 2.1
${\displaystyle \int _{0}^{\infty }{\text{Ci}}(ax)\,{\text{Ci}}(bx)\,dx={\frac {1}{\max\{a,b\}}}\cdot {\frac {\pi }{2}}\qquad a,b>0}$
Beweis

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.