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

2021-05-05 03:03:17 作者：互联网

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}

\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 )}} $\int {\text{Ci}}(ax)\,{\text{Ci}}(bx)\,dx=x\,{\text{Ci}}(ax)\,{\text{Ci}}(bx)-{\frac {\sin ax}{a}}\,{\text{Ci}}(bx)-{\frac {\sin bx}{b}}\,{\text{Ci}}(ax)+{\frac {1}{2a}}{\Big (}{\text{Si}}(ax+bx)+{\text{Si}}(ax-bx){\Big )}+{\frac {1}{2b}}{\Big (}{\text{Si}}(ax+bx)-{\text{Si}}(ax-bx){\Big )}$

setze {\displaystyle 0\,} $0\,$ und {\displaystyle \infty } $\infty$ als Integrationsgrenzen ein.

Asymptotisch verhalten sich {\displaystyle {\text{Ci}}(ax)} ${\text{Ci}}(ax)$ und {\displaystyle {\text{Ci}}(bx)} ${\text{Ci}}(bx)$ für {\displaystyle x\to 0+} $x\to 0+$ wie {\displaystyle \log x} $\log x$ und für {\displaystyle x\to \infty \,} $x\to \infty \,$ wie {\displaystyle {\frac {\cos x}{x}}} ${\frac {\cos x}{x}}$ .

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 }} ${\Big [}x\,{\text{Ci}}(ax)\,{\text{Ci}}(bx){\Big ]}_{0}^{\infty }\,\,,\,\,{\Big [}{\frac {\sin ax}{a}}\,{\text{Ci}}(bx){\Big ]}_{0}^{\infty }\,\,,\,\,{\Big [}{\frac {\sin bx}{b}}\,{\text{Ci}}(ax){\Big ]}_{0}^{\infty }$ jeweils gleich {\displaystyle 0-0=0} $0-0=0$ .

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 }} $\left[{\frac {1}{2a}}{\Big (}{\text{Si}}(ax+bx)+{\text{Si}}(ax-bx){\Big )}+{\frac {1}{2b}}{\Big (}{\text{Si}}(ax+bx)-{\text{Si}}(ax-bx){\Big )}\right]_{0}^{\infty }$ verschwindet für {\displaystyle x=0} $x=0$ .

Für {\displaystyle x\to \infty } $x\to \infty$ 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}}} $\bullet \quad {\frac {1}{2a}}\left({\frac {\pi }{2}}+{\frac {\pi }{2}}\right)+{\frac {1}{2b}}\left({\frac {\pi }{2}}-{\frac {\pi }{2}}\right)={\frac {1}{a}}\cdot {\frac {\pi }{2}}$ falls {\displaystyle a>b} $a>b$ .

{\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}}} $\bullet \quad {\frac {1}{2a}}\left({\frac {\pi }{2}}+0\right)+{\frac {1}{2b}}\left({\frac {\pi }{2}}+0\right)={\frac {1}{a}}\cdot {\frac {\pi }{2}}={\frac {1}{b}}\cdot {\frac {\pi }{2}}$ falls {\displaystyle a=b} $a=b$ .

{\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}}} $\bullet \quad {\frac {1}{2a}}\left({\frac {\pi }{2}}-{\frac {\pi }{2}}\right)+{\frac {1}{2b}}\left({\frac {\pi }{2}}+{\frac {\pi }{2}}\right)={\frac {1}{b}}\cdot {\frac {\pi }{2}}$ falls {\displaystyle a<b} $a<b$ .

标签：Integrale,Ci,frac,Form,text,ax,pi,bx,displaystyle
来源： https://www.cnblogs.com/Eufisky/p/14730806.html