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

Prove that ∀n∈IN ∫^( 1) _( 0) t sin^(2n) (lnt)dt= (1/(1−e^(−2π) )) ∫^( π) _( 0) e^(−2t) sin^(2n) (t)dt

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

Answered by witcher3 last updated on 17/Jul/23

ln(t)=−x ⇒∫_0 ^∞ e^(−2x) sin^(2n) (x)dx =Σ_(k≥0) ∫_(kπ) ^((k+1)π) e^(−2x) sin^(2n) (x)dx x→kπ+t =Σ_(k≥0) ∫_0 ^π e^(−2kπ−2t) sin^(2n) (kπ+t)dt =Σ_(k≥0) e^(−2kπ) ∫_0 ^π e^(−2t) sin^(2n) (t)dt =∫_0 ^π e^(−2t) sin^(2n) (t)dt.Σ_(k≥0) e^(−2kπ) =(1/(1−e^(−2π) ))∫_0 ^π e^(−2t) sin^(2n) (t)dt

$$\mathrm{ln}\left(\mathrm{t}\right)=−\mathrm{x} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{2x}} \mathrm{sin}^{\mathrm{2n}} \left(\mathrm{x}\right)\mathrm{dx} \\ $$$$=\underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{k}\pi} ^{\left(\mathrm{k}+\mathrm{1}\right)\pi} \mathrm{e}^{−\mathrm{2x}} \mathrm{sin}^{\mathrm{2n}} \left(\mathrm{x}\right)\mathrm{dx} \\ $$$$\mathrm{x}\rightarrow\mathrm{k}\pi+\mathrm{t} \\ $$$$=\underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{\pi} \mathrm{e}^{−\mathrm{2k}\pi−\mathrm{2t}} \mathrm{sin}^{\mathrm{2n}} \left(\mathrm{k}\pi+\mathrm{t}\right)\mathrm{dt} \\ $$$$=\underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\mathrm{e}^{−\mathrm{2k}\pi} \int_{\mathrm{0}} ^{\pi} \mathrm{e}^{−\mathrm{2t}} \mathrm{sin}^{\mathrm{2n}} \left(\mathrm{t}\right)\mathrm{dt} \\ $$$$=\int_{\mathrm{0}} ^{\pi} \mathrm{e}^{−\mathrm{2t}} \mathrm{sin}^{\mathrm{2n}} \left(\mathrm{t}\right)\mathrm{dt}.\underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\mathrm{e}^{−\mathrm{2k}\pi} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{e}^{−\mathrm{2}\pi} }\int_{\mathrm{0}} ^{\pi} \mathrm{e}^{−\mathrm{2t}} \mathrm{sin}^{\mathrm{2n}} \left(\mathrm{t}\right)\mathrm{dt} \\ $$

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