Zurück zu Bestimmte Integrale
Es sei I n = ∫ 0 ∞ e − α x sin n x d x {\displaystyle I_{n}=\int _{0}^{\infty }e^{-\alpha x}\,\sin ^{n}x\,dx} . Durch zweimalige partielle Integration erhält man die Rekursion I n = I n − 2 ( n − 1 ) n α 2 + n 2 {\displaystyle I_{n}=I_{n-2}\,{\frac {(n-1)\,n}{\alpha ^{2}+n^{2}}}} . Also ist I 2 n = I 0 ⋅ 1 ⋅ 2 α 2 + 2 2 ⋅ 3 ⋅ 4 α 2 + 4 2 ⋯ ( 2 n − 1 ) 2 n α 2 + ( 2 n ) 2 = ( 2 n ) ! α ( α 2 + 2 2 ) ( α 2 + 4 2 ) ⋯ ( α 2 + ( 2 n ) 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})}}} .
Es sei I n = ∫ 0 ∞ e − α x sin n x d x {\displaystyle I_{n}=\int _{0}^{\infty }e^{-\alpha x}\,\sin ^{n}x\,dx} . Durch zweimalige partielle Integration erhält man die Rekursion I n = I n − 2 ( n − 1 ) n α 2 + n 2 {\displaystyle I_{n}=I_{n-2}\,{\frac {(n-1)\,n}{\alpha ^{2}+n^{2}}}} . Also ist I 2 n + 1 = I 1 ⋅ 2 ⋅ 3 α 2 + 3 2 ⋅ 4 ⋅ 5 α 2 + 5 2 ⋯ n ( 2 n + 1 ) α 2 + ( 2 n + 1 ) 2 = ( 2 n + 1 ) ! ( α 2 + 1 ) ( α 2 + 3 2 ) ⋯ ( α 2 + ( 2 n + 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})}}} .
Aus sin α x 1 − e β x = − ∑ k = 1 ∞ e − k β x sin α x {\displaystyle {\frac {\sin \alpha x}{1-e^{\beta x}}}=-\sum _{k=1}^{\infty }e^{-k\beta x}\,\sin \alpha x} folgt ∫ 0 ∞ sin α x 1 − e β x d x = − ∑ k = 1 ∞ ∫ 0 ∞ e − k β x sin α x d x {\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 − ∑ k = 1 ∞ α k 2 β 2 + α 2 = 1 2 α − 1 2 ∑ k = − ∞ ∞ α k 2 β 2 + α 2 = 1 2 α − π 2 β coth ( α π β ) {\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)} .
Aus sin α x 1 + e β x = − ∑ k = 1 ∞ ( − 1 ) k e − k β x sin α x {\displaystyle {\frac {\sin \alpha x}{1+e^{\beta x}}}=-\sum _{k=1}^{\infty }(-1)^{k}\,e^{-k\beta x}\,\sin \alpha x} folgt ∫ 0 ∞ sin α x 1 + e β x d x = − ∑ k = 1 ∞ ( − 1 ) k ∫ 0 ∞ e − k β x sin α x d x {\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 − ∑ k = 1 ∞ ( − 1 ) k α k 2 β 2 + α 2 = 1 2 α − 1 2 ∑ k = − ∞ ∞ ( − 1 ) k α k 2 β 2 + α 2 = 1 2 α − π 2 β csch ( α π β ) {\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)} .
Aus der Reihenentwicklung sin b x = ∑ k = 0 ∞ ( − 1 ) k ( 2 k + 1 ) ! ( b x ) 2 k + 1 {\displaystyle \sin bx=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{(2k+1)!}}\,(bx)^{2k+1}} folgt e − a x sin b x x = ∑ k = 0 ∞ ( − 1 ) k ( 2 k + 1 ) ! b 2 k + 1 x 2 k e − a x {\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 ∫ 0 ∞ e − a x sin b x x d x = ∑ k = 0 ∞ ( − 1 ) k ( 2 k + 1 ) ! b 2 k + 1 ∫ 0 ∞ x 2 k e − a x d x {\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} = ∑ k = 0 ∞ ( − 1 ) k ( 2 k + 1 ) ! b 2 k + 1 ( 2 k ) ! a 2 k + 1 = ∑ k = 0 ∞ ( − 1 ) k 2 k + 1 ( b a ) 2 k + 1 = arctan ( b a ) {\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)} .
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