Formelsammlung Mathematik: Bestimmte Integrale: Form R(x,exp,sin)

 

1.1Bearbeiten
{\displaystyle \int _{0}^{\infty }\left({\frac {\sin x}{x}}\right)^{2}\,e^{-2ax}\,dx=a\,\log \left({\frac {a}{\sqrt {1+a^{2}}}}\right)+\operatorname {arccot} a}{\displaystyle \int _{0}^{\infty }\left({\frac {\sin x}{x}}\right)^{2}\,e^{-2ax}\,dx=a\,\log \left({\frac {a}{\sqrt {1+a^{2}}}}\right)+\operatorname {arccot} a}
ohne Beweis

 

 
2.1Bearbeiten
{\displaystyle \int _{0}^{\infty }e^{-\alpha x}\,\sin ^{2n}x\,dx={\frac {(2n)!}{\alpha \,(\alpha ^{2}+2^{2})(\alpha ^{2}+4^{2})\cdots (\alpha ^{2}+(2n)^{2})}}\qquad n\in \mathbb {N} \,\,,\,\,{\text{Re}}(\alpha )>0}{\displaystyle \int _{0}^{\infty }e^{-\alpha x}\,\sin ^{2n}x\,dx={\frac {(2n)!}{\alpha \,(\alpha ^{2}+2^{2})(\alpha ^{2}+4^{2})\cdots (\alpha ^{2}+(2n)^{2})}}\qquad n\in \mathbb {N} \,\,,\,\,{\text{Re}}(\alpha )>0}
Beweis

Es sei {\displaystyle I_{n}=\int _{0}^{\infty }e^{-\alpha x}\,\sin ^{n}x\,dx}{\displaystyle I_{n}=\int _{0}^{\infty }e^{-\alpha x}\,\sin ^{n}x\,dx}.

Durch zweimalige partielle Integration erhält man die Rekursion {\displaystyle I_{n}=I_{n-2}\,{\frac {(n-1)\,n}{\alpha ^{2}+n^{2}}}}{\displaystyle I_{n}=I_{n-2}\,{\frac {(n-1)\,n}{\alpha ^{2}+n^{2}}}}.

Also ist

{\displaystyle I_{2n}=I_{0}\cdot {\frac {1\cdot 2}{\alpha ^{2}+2^{2}}}\cdot {\frac {3\cdot 4}{\alpha ^{2}+4^{2}}}\cdots {\frac {(2n-1)\,2n}{\alpha ^{2}+(2n)^{2}}}={\frac {(2n)!}{\alpha \,(\alpha ^{2}+2^{2})(\alpha ^{2}+4^{2})\cdots (\alpha ^{2}+(2n)^{2})}}}{\displaystyle I_{2n}=I_{0}\cdot {\frac {1\cdot 2}{\alpha ^{2}+2^{2}}}\cdot {\frac {3\cdot 4}{\alpha ^{2}+4^{2}}}\cdots {\frac {(2n-1)\,2n}{\alpha ^{2}+(2n)^{2}}}={\frac {(2n)!}{\alpha \,(\alpha ^{2}+2^{2})(\alpha ^{2}+4^{2})\cdots (\alpha ^{2}+(2n)^{2})}}}.

 
2.2Bearbeiten
{\displaystyle \int _{0}^{\infty }e^{-\alpha x}\,\sin ^{2n+1}x\,dx={\frac {(2n+1)!}{(\alpha ^{2}+1)(\alpha ^{2}+3^{2})\cdots (\alpha ^{2}+(2n+1)^{2})}}\qquad n\in \mathbb {N} \,\,,\,\,{\text{Re}}(\alpha )>0}{\displaystyle \int _{0}^{\infty }e^{-\alpha x}\,\sin ^{2n+1}x\,dx={\frac {(2n+1)!}{(\alpha ^{2}+1)(\alpha ^{2}+3^{2})\cdots (\alpha ^{2}+(2n+1)^{2})}}\qquad n\in \mathbb {N} \,\,,\,\,{\text{Re}}(\alpha )>0}
Beweis

Es sei {\displaystyle I_{n}=\int _{0}^{\infty }e^{-\alpha x}\,\sin ^{n}x\,dx}{\displaystyle I_{n}=\int _{0}^{\infty }e^{-\alpha x}\,\sin ^{n}x\,dx}.

Durch zweimalige partielle Integration erhält man die Rekursion {\displaystyle I_{n}=I_{n-2}\,{\frac {(n-1)\,n}{\alpha ^{2}+n^{2}}}}{\displaystyle I_{n}=I_{n-2}\,{\frac {(n-1)\,n}{\alpha ^{2}+n^{2}}}}.

Also ist

{\displaystyle I_{2n+1}=I_{1}\cdot {\frac {2\cdot 3}{\alpha ^{2}+3^{2}}}\cdot {\frac {4\cdot 5}{\alpha ^{2}+5^{2}}}\cdots {\frac {n\,(2n+1)}{\alpha ^{2}+(2n+1)^{2}}}={\frac {(2n+1)!}{(\alpha ^{2}+1)(\alpha ^{2}+3^{2})\cdots (\alpha ^{2}+(2n+1)^{2})}}}{\displaystyle I_{2n+1}=I_{1}\cdot {\frac {2\cdot 3}{\alpha ^{2}+3^{2}}}\cdot {\frac {4\cdot 5}{\alpha ^{2}+5^{2}}}\cdots {\frac {n\,(2n+1)}{\alpha ^{2}+(2n+1)^{2}}}={\frac {(2n+1)!}{(\alpha ^{2}+1)(\alpha ^{2}+3^{2})\cdots (\alpha ^{2}+(2n+1)^{2})}}}.

 
2.3Bearbeiten
{\displaystyle \int _{0}^{\infty }{\frac {\sin \alpha x}{1-e^{\beta x}}}\,dx={\frac {1}{2\alpha }}-{\frac {\pi }{2\beta }}\;{\text{coth}}\left({\frac {\alpha \pi }{\beta }}\right)}{\displaystyle \int _{0}^{\infty }{\frac {\sin \alpha x}{1-e^{\beta x}}}\,dx={\frac {1}{2\alpha }}-{\frac {\pi }{2\beta }}\;{\text{coth}}\left({\frac {\alpha \pi }{\beta }}\right)}
Beweis

Aus {\displaystyle {\frac {\sin \alpha x}{1-e^{\beta x}}}=-\sum _{k=1}^{\infty }e^{-k\beta x}\,\sin \alpha x}{\displaystyle {\frac {\sin \alpha x}{1-e^{\beta x}}}=-\sum _{k=1}^{\infty }e^{-k\beta x}\,\sin \alpha x} folgt {\displaystyle \int _{0}^{\infty }{\frac {\sin \alpha x}{1-e^{\beta x}}}\,dx=-\sum _{k=1}^{\infty }\int _{0}^{\infty }e^{-k\beta x}\,\sin \alpha x\,dx}{\displaystyle \int _{0}^{\infty }{\frac {\sin \alpha x}{1-e^{\beta x}}}\,dx=-\sum _{k=1}^{\infty }\int _{0}^{\infty }e^{-k\beta x}\,\sin \alpha x\,dx}.

Und das ist {\displaystyle -\sum _{k=1}^{\infty }{\frac {\alpha }{k^{2}\beta ^{2}+\alpha ^{2}}}={\frac {1}{2\alpha }}-{\frac {1}{2}}\sum _{k=-\infty }^{\infty }{\frac {\alpha }{k^{2}\beta ^{2}+\alpha ^{2}}}={\frac {1}{2\alpha }}-{\frac {\pi }{2\beta }}\;{\text{coth}}\left({\frac {\alpha \pi }{\beta }}\right)}{\displaystyle -\sum _{k=1}^{\infty }{\frac {\alpha }{k^{2}\beta ^{2}+\alpha ^{2}}}={\frac {1}{2\alpha }}-{\frac {1}{2}}\sum _{k=-\infty }^{\infty }{\frac {\alpha }{k^{2}\beta ^{2}+\alpha ^{2}}}={\frac {1}{2\alpha }}-{\frac {\pi }{2\beta }}\;{\text{coth}}\left({\frac {\alpha \pi }{\beta }}\right)}.

 
2.4Bearbeiten
{\displaystyle \int _{0}^{\infty }{\frac {\sin \alpha x}{1+e^{\beta x}}}\,dx={\frac {1}{2\alpha }}-{\frac {\pi }{2\beta }}\;{\text{csch}}\left({\frac {\alpha \pi }{\beta }}\right)}{\displaystyle \int _{0}^{\infty }{\frac {\sin \alpha x}{1+e^{\beta x}}}\,dx={\frac {1}{2\alpha }}-{\frac {\pi }{2\beta }}\;{\text{csch}}\left({\frac {\alpha \pi }{\beta }}\right)}
Beweis

Aus {\displaystyle {\frac {\sin \alpha x}{1+e^{\beta x}}}=-\sum _{k=1}^{\infty }(-1)^{k}\,e^{-k\beta x}\,\sin \alpha x}{\displaystyle {\frac {\sin \alpha x}{1+e^{\beta x}}}=-\sum _{k=1}^{\infty }(-1)^{k}\,e^{-k\beta x}\,\sin \alpha x} folgt {\displaystyle \int _{0}^{\infty }{\frac {\sin \alpha x}{1+e^{\beta x}}}\,dx=-\sum _{k=1}^{\infty }(-1)^{k}\int _{0}^{\infty }e^{-k\beta x}\,\sin \alpha x\,dx}{\displaystyle \int _{0}^{\infty }{\frac {\sin \alpha x}{1+e^{\beta x}}}\,dx=-\sum _{k=1}^{\infty }(-1)^{k}\int _{0}^{\infty }e^{-k\beta x}\,\sin \alpha x\,dx}.

Und das ist {\displaystyle -\sum _{k=1}^{\infty }(-1)^{k}\,{\frac {\alpha }{k^{2}\beta ^{2}+\alpha ^{2}}}={\frac {1}{2\alpha }}-{\frac {1}{2}}\sum _{k=-\infty }^{\infty }(-1)^{k}\,{\frac {\alpha }{k^{2}\beta ^{2}+\alpha ^{2}}}={\frac {1}{2\alpha }}-{\frac {\pi }{2\beta }}\;{\text{csch}}\left({\frac {\alpha \pi }{\beta }}\right)}{\displaystyle -\sum _{k=1}^{\infty }(-1)^{k}\,{\frac {\alpha }{k^{2}\beta ^{2}+\alpha ^{2}}}={\frac {1}{2\alpha }}-{\frac {1}{2}}\sum _{k=-\infty }^{\infty }(-1)^{k}\,{\frac {\alpha }{k^{2}\beta ^{2}+\alpha ^{2}}}={\frac {1}{2\alpha }}-{\frac {\pi }{2\beta }}\;{\text{csch}}\left({\frac {\alpha \pi }{\beta }}\right)}.

 
2.5Bearbeiten
{\displaystyle \int _{0}^{\infty }e^{-ax}\,\sin bx\,dx={\frac {b}{a^{2}+b^{2}}}}{\displaystyle \int _{0}^{\infty }e^{-ax}\,\sin bx\,dx={\frac {b}{a^{2}+b^{2}}}}
ohne Beweis

 

 
2.6Bearbeiten
{\displaystyle \int _{0}^{\infty }e^{-ax}\,{\frac {\sin bx}{x}}\,dx=\arctan \left({\frac {b}{a}}\right)\qquad {\text{Re}}(a)\geq |{\text{Im}}(b)|\quad ,\quad {\frac {b}{a}}\neq \pm i}{\displaystyle \int _{0}^{\infty }e^{-ax}\,{\frac {\sin bx}{x}}\,dx=\arctan \left({\frac {b}{a}}\right)\qquad {\text{Re}}(a)\geq |{\text{Im}}(b)|\quad ,\quad {\frac {b}{a}}\neq \pm i}
Beweis für a>b>0

Aus der Reihenentwicklung {\displaystyle \sin bx=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{(2k+1)!}}\,(bx)^{2k+1}}{\displaystyle \sin bx=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{(2k+1)!}}\,(bx)^{2k+1}}

folgt {\displaystyle e^{-ax}\,{\frac {\sin bx}{x}}=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{(2k+1)!}}\,b^{2k+1}\,x^{2k}\,e^{-ax}}{\displaystyle e^{-ax}\,{\frac {\sin bx}{x}}=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{(2k+1)!}}\,b^{2k+1}\,x^{2k}\,e^{-ax}}.

Also ist {\displaystyle \int _{0}^{\infty }e^{-ax}\,{\frac {\sin bx}{x}}\,dx=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{(2k+1)!}}\,b^{2k+1}\,\int _{0}^{\infty }x^{2k}\,e^{-ax}\,dx}{\displaystyle \int _{0}^{\infty }e^{-ax}\,{\frac {\sin bx}{x}}\,dx=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{(2k+1)!}}\,b^{2k+1}\,\int _{0}^{\infty }x^{2k}\,e^{-ax}\,dx}

{\displaystyle =\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{(2k+1)!}}\,b^{2k+1}\,{\frac {(2k)!}{a^{2k+1}}}=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{2k+1}}\,\left({\frac {b}{a}}\right)^{2k+1}=\arctan \left({\frac {b}{a}}\right)}{\displaystyle =\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{(2k+1)!}}\,b^{2k+1}\,{\frac {(2k)!}{a^{2k+1}}}=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{2k+1}}\,\left({\frac {b}{a}}\right)^{2k+1}=\arctan \left({\frac {b}{a}}\right)}.

 
3.1Bearbeiten
{\displaystyle \int _{0}^{\infty }e^{-ax}\,\sin(bx)\,x^{s-1}\,dx={\frac {\Gamma (s)}{{\sqrt {a^{2}+b^{2}}}^{s}}}\,\sin \left(s\,\arctan {\frac {b}{a}}\right)\qquad a>0\,,\,b\in \mathbb {R} \,,\,{\text{Re}}(s)>0}{\displaystyle \int _{0}^{\infty }e^{-ax}\,\sin(bx)\,x^{s-1}\,dx={\frac {\Gamma (s)}{{\sqrt {a^{2}+b^{2}}}^{s}}}\,\sin \left(s\,\arctan {\frac {b}{a}}\right)\qquad a>0\,,\,b\in \mathbb {R} \,,\,{\text{Re}}(s)>0}
ohne Beweis

posted on 2021-05-05 02:35  Eufisky  阅读(30)  评论(0)  编辑  收藏  举报

导航