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

$$let{u}_n={\int }_0^1{x}^{n}sin\left(\pi x\right)dx$$$$1)provethat\mathrm{\Sigma }{u}_{n}converges$$$$2)provethat\mathrm{\Sigma }{u}_n={\int }_0^{\pi }\frac{sint}{t}dt.$$

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

$$letput{S}_n=\sum _{k=0}^n{\int }_0^1{x}^{k}sin\left(\pi x\right)dx$$$$S_n={\int }_0^1\left(\sum _{k=0}^n{x}^k\right)sin\left(\pi x\right)dx={\int }_0^1\frac{1-x^{n+1}}{1-x}sin\left(\pi x\right)dx$$$$\Rightarrow S_n-{\int }_0^1\frac{sin\left(\pi x\right)}{1-x}dx=-{\int }_0^1\frac{x^{n+1}}{1-x}sin\left(\pi x\right)dxand\mathrm{\exists }m>0/$$$$\mid S_n-{\int }_0^1\frac{sin\left(\pi x\right)}{1-x}dx\mid ⩽m{\int }_0^1{x}^{n+1}dx=\frac{m}{n+2}{\to }_{n\to \mathrm{\infty }}0$$$$\Rightarrow S_{n}convergedandlim{S}_n={\int }_0^1\frac{sin\left(\pi x\right)}{1-x}dx$$$$2)ch.1-x=tgive{\int }_0^1\frac{sin\left(\pi x\right)}{1-x}dx={\int }_0^1\frac{sin\left(\pi (1-t)\right)}{t}=dt$$$$={\int }_0^1\frac{sin\left(\pi t\right)}{t}dtafterch.\pi t=ugive$$$$lim{}_{n\to \mathrm{\infty }}{S}_n={\int }_0^{\pi }\frac{sin\left(u\right)}{\frac{u}{\pi }}\frac{du}{\pi }={\int }_0^{\pi }\frac{sinu}{u}du.so$$$$\sum _{n=0}^{\mathrm{\infty }}{u}_n={\int }_0^{\pi }\frac{sinu}{u}du.$$

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