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Question Number 194868 by Erico last updated on 17/Jul/23

$$\mathrm{Prove}\mathrm{that}\mathrm{\forall }n\in \mathrm{IN}$$$${\underset{0}{\int }}^{1}t{sin}^{2n}\left(lnt\right)dt=\frac{1}{1-e^{-2\pi }}{\underset{0}{\int }}^{\pi }{e}^{-2t}{sin}^{2n}\left(t\right)dt$$

Answered by witcher3 last updated on 17/Jul/23

$$\ln \left(\mathrm{t}\right)=-\mathrm{x}$$$$\Rightarrow {\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-2\mathrm{x}}{\sin }^{2\mathrm{n}}\left(\mathrm{x}\right)\mathrm{dx}$$$$=\sum _{\mathrm{k}⩾0}{\int }_{\mathrm{k}\pi }^{(\mathrm{k}+1)\pi }{\mathrm{e}}^{-2\mathrm{x}}{\sin }^{2\mathrm{n}}\left(\mathrm{x}\right)\mathrm{dx}$$$$\mathrm{x}\to \mathrm{k}\pi +\mathrm{t}$$$$=\sum _{\mathrm{k}⩾0}{\int }_0^{\pi }{\mathrm{e}}^{-2\mathrm{k}\pi -2\mathrm{t}}{\sin }^{2\mathrm{n}}(\mathrm{k}\pi +\mathrm{t})\mathrm{dt}$$$$=\sum _{\mathrm{k}⩾0}{\mathrm{e}}^{-2\mathrm{k}\pi }{\int }_0^{\pi }{\mathrm{e}}^{-2\mathrm{t}}{\sin }^{2\mathrm{n}}\left(\mathrm{t}\right)\mathrm{dt}$$$$={\int }_0^{\pi }{\mathrm{e}}^{-2\mathrm{t}}{\sin }^{2\mathrm{n}}\left(\mathrm{t}\right)\mathrm{dt}.\sum _{\mathrm{k}⩾0}{\mathrm{e}}^{-2\mathrm{k}\pi }$$$$=\frac{1}{1-{\mathrm{e}}^{-2\pi }}{\int }_0^{\pi }{\mathrm{e}}^{-2\mathrm{t}}{\sin }^{2\mathrm{n}}\left(\mathrm{t}\right)\mathrm{dt}$$

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