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Question Number 32042 by abdo imad last updated on 18/Mar/18

let u_n =∫_0 ^1 x^n sin(πx)dx 1) prove that Σ u_n converges 2) prove that Σ u_n = ∫_0 ^π ((sint)/t)dt .

$${let}\:{u}_{{n}} \:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{x}^{{n}} \:{sin}\left(\pi{x}\right){dx} \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:\Sigma\:{u}_{{n}} \:{converges} \\ $$$$\left.\mathrm{2}\right)\:{prove}\:{that}\:\Sigma\:{u}_{{n}} =\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{sint}}{{t}}{dt}\:. \\ $$

Commented by abdo imad last updated on 20/Mar/18

let put S_n =Σ_(k=0) ^n ∫_0 ^1 x^k sin(πx)dx S_n = ∫_0 ^1 (Σ_(k=0) ^n x^k )sin(πx)dx =∫_0 ^1 ((1−x^(n+1) )/(1−x)) sin(πx)dx ⇒ S_(n ) − ∫_0 ^1 ((sin(πx))/(1−x))dx = −∫_0 ^1 (x^(n+1) /(1−x)) sin(πx)dx and ∃m>0/ ∣ S_n − ∫_0 ^1 ((sin(πx))/(1−x))dx∣ ≤ m ∫_0 ^1 x^(n+1) dx =(m/(n+2)) →_(n→∞) 0 ⇒ S_n converged and lim S_n = ∫_0 ^1 ((sin(πx))/(1−x)) dx 2) ch .1−x=t give ∫_0 ^1 ((sin(πx))/(1−x)) dx =∫_0 ^1 ((sin(π(1−t)))/t)= dt = ∫_0 ^1 ((sin(πt))/t) dt after ch. πt =u give lim _(n→∞) S_n = ∫_0 ^π ((sin(u))/(u/π)) (du/π) =∫_0 ^π ((sinu)/u) du . so Σ_(n=0) ^∞ u_n = ∫_0 ^π ((sinu)/u) du .

$${let}\:{put}\:{S}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{k}} \:{sin}\left(\pi{x}\right){dx} \\ $$$${S}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\sum_{{k}=\mathrm{0}} ^{{n}} \:{x}^{{k}} \right){sin}\left(\pi{x}\right){dx}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}−{x}^{{n}+\mathrm{1}} }{\mathrm{1}−{x}}\:{sin}\left(\pi{x}\right){dx} \\ $$$$\Rightarrow\:{S}_{{n}\:} \:−\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{sin}\left(\pi{x}\right)}{\mathrm{1}−{x}}{dx}\:=\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{x}^{{n}+\mathrm{1}} }{\mathrm{1}−{x}}\:{sin}\left(\pi{x}\right){dx}\:{and}\:\exists{m}>\mathrm{0}/ \\ $$$$\mid\:{S}_{{n}} \:−\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{sin}\left(\pi{x}\right)}{\mathrm{1}−{x}}{dx}\mid\:\:\leqslant\:{m}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}+\mathrm{1}} {dx}\:=\frac{{m}}{{n}+\mathrm{2}}\:\rightarrow_{{n}\rightarrow\infty} \mathrm{0} \\ $$$$\Rightarrow\:{S}_{{n}} {converged}\:{and}\:{lim}\:{S}_{{n}} \:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{sin}\left(\pi{x}\right)}{\mathrm{1}−{x}}\:{dx} \\ $$$$\left.\mathrm{2}\right)\:{ch}\:.\mathrm{1}−{x}={t}\:{give}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{sin}\left(\pi{x}\right)}{\mathrm{1}−{x}}\:{dx}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{sin}\left(\pi\left(\mathrm{1}−{t}\right)\right)}{{t}}=\:{dt} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{sin}\left(\pi{t}\right)}{{t}}\:{dt}\:{after}\:{ch}.\:\pi{t}\:={u}\:{give} \\ $$$${lim}\:_{{n}\rightarrow\infty} \:{S}_{{n}} =\:\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{sin}\left({u}\right)}{\frac{{u}}{\pi}}\:\frac{{du}}{\pi}\:=\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{sinu}}{{u}}\:{du}\:.\:{so} \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:{u}_{{n}} =\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{sinu}}{{u}}\:{du}\:\:. \\ $$

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