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

2021-05-05 03:02:52 作者：互联网

{\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}

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

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

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

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}

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

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

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

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{\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)}

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

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

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{\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)}

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

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

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

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

ohne Beweis

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{\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}

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

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

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

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

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}

\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

标签：Integrale,infty,frac,Form,Mathematik,sin,beta,alpha,displaystyle
来源： https://www.cnblogs.com/Eufisky/p/14730808.html

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

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