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Question Number 47651 by prof Abdo imad last updated on 12/Nov/18

$$calculate{A}_n={\int }_0^{1}sin\left(\left[nx\right]\right)e^{-2x}dxwithn$$$$integrnatural.$$

Commented by maxmathsup by imad last updated on 13/Nov/18

$$wehave{A}_n=\sum _{k=0}^{n-1}{\int }_{\frac{k}{n}}^{\frac{k+1}{n}}sin\left(\left[nx\right]\right)e^{-2x}dx$$$$=\sum _{k=0}^{n-1}{\int }_{\frac{k}{n}}^{\frac{k+1}{n}}sin\left(k\right)e^{-2x}dx=\sum _{k=0}^{n}sin\left(k\right){\int }_{\frac{k}{n}}^{\frac{k+1}{n}}{e}^{-2x}dx$$$$=-\frac{1}{2}\sum _{k=0}^{n-1}sin\left(k\right)(e^{-2\frac{k+1}{n}}-e^{\frac{-2k}{n}})$$$$=\frac{1}{2}\sum _{k=0}^{n-1}{e}^{\frac{-2k}{n}}-\frac{1}{2}\sum _{k=0}^{n-1}{e}^{-2\frac{k+1}{n}}sin\left(k\right)$$$$=\frac{1-e^{-\frac{2}{n}}}{2}\sum _{k=0}^{n-1}{e}^{\frac{-2k}{n}}sin\left(k\right)but\sum _{k=0}^{n-}{e}^{\frac{-2k}{n}}sink=Im\left(\sum _{k=0}^{n-1}{e}^{ik-\frac{2}{n}k}\right)and$$$$\sum _{k=0}^{n-1}{e}^{ik-\frac{2}{n}k}=\sum _{k=0}^{n-1}{e}^{(i-\frac{2}{n})k}=\frac{1-{\left(e^{(i-\frac{2}{n})}\right)}^n}{1-e^{i-\frac{2}{n}}}=\frac{1-e^{-2}{e}^{in}}{1-e^{-\frac{2}{n}}{e}^i}$$$$=\frac{1-e^{-2}cosn-{ie}^{-2}sin\left(n\right)}{1-e^{-\frac{2}{n}}cos\left(1\right)-i{e}^{-\frac{2}{n}}sin\left(1\right)}$$$$=\frac{(1-e^{-2}cos\left(n\right)-i{e}^{-2}sinn)(1-e^{-\frac{2}{n}}cos\left(1\right)+i{e}^{-\frac{2}{n}}sin\left(1\right))}{{(1-e^{-2}cos\left(1\right))}^2+e^{-\frac{4}{n}}{sin}^2\left(1\right)}$$$$=\frac{(1-e^{-2}cos\left(n\right))(1-e^{-\frac{2}{n}}cos\left(1\right))+i(1-e^{-2}cos\left(n\right))e^{-\frac{2}{n}}sin\left(1\right)-{ie}^{-2}sin\left(n\right)(1-e^{-\frac{2}{n}}cos\left(1\right))+e^{\frac{4}{n}}sin\left(n\right)sin\left(1\right)}{{(1-e^{-2}cos\left(1\right))}^2+e^{-\frac{4}{n}}{sin}^2\left(1\right)}$$$$sothevalueof{A}_{n}iscleardeterminedafterextractingIm(\mathrm{\Sigma }....).$$

Answered by tanmay.chaudhury50@gmail.com last updated on 13/Nov/18

$${\int }_0^{\frac{1}{n}}sin\left(\left[nx\right]\right)e^{-2x}dx+{\int }_{\frac{1}{n}}^{1}sin\left(\left[nx\right]\right)e^{-2x}dx$$$${\int }_0^{\frac{1}{n}}sin\left(0\right)e^{-2x}dx+sin\left(1\right){\int }_{\frac{1}{n}}^1{e}^{-2x}dx$$$$sin\left(1\right)\times \mid \frac{e^{-2x}}{-2}{\mid }_{\frac{1}{n}}^1$$$$sin\left(1\right)\times \left(\frac{e^{-2}-e^{\frac{-2}{n}}}{-2}\right)$$

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