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Question Number 47651 by prof Abdo imad last updated on 12/Nov/18
calculate  A_n =∫_0 ^1   sin([nx])e^(−2x) dx with n  integr natural .
$${calculate}\:\:{A}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{sin}\left(\left[{nx}\right]\right){e}^{−\mathrm{2}{x}} {dx}\:{with}\:{n} \\ $$$${integr}\:{natural}\:. \\ $$
Commented by maxmathsup by imad last updated on 13/Nov/18
we have A_n =Σ_(k=0) ^(n−1)  ∫_(k/n) ^((k+1)/n)  sin([nx])e^(−2x) dx  =Σ_(k=0) ^(n−1)  ∫_(k/n) ^((k+1)/n)  sin(k)e^(−2x) dx =Σ_(k=0) ^n  sin(k)∫_(k/n) ^((k+1)/n)  e^(−2x) dx  =−(1/2)Σ_(k=0) ^(n−1)  sin(k)( e^(−2((k+1)/n)) −e^((−2k)/n) )  =(1/2) Σ_(k=0) ^(n−1)  e^((−2k)/n)  −(1/2)Σ_(k=0) ^(n−1)  e^(−2((k+1)/n))  sin(k)  =((1−e^(−(2/n)) )/2) Σ_(k=0) ^(n−1)  e^((−2k)/n)  sin(k) but Σ_(k=0) ^(n−)  e^((−2k)/n)  sink =Im(Σ_(k=0) ^(n−1)  e^(ik−(2/n)k) ) and  Σ_(k=0) ^(n−1)  e^(ik−(2/n)k)  =Σ_(k=0) ^(n−1)  e^((i−(2/n))k)  =((1−(e^((i−(2/n))) )^n )/(1−e^(i−(2/n)) )) =((1−e^(−2)  e^(in) )/(1−e^(−(2/n))  e^i ))  =((1−e^(−2) cosn −ie^(−2) sin(n))/(1−e^(−(2/n)) cos(1)−i e^(−(2/n)) sin(1)))  =(((1−e^(−2) cos(n)−i e^(−2) sinn)(1−e^(−(2/n)) cos(1)+i e^(−(2/n)) sin(1)))/((1−e^(−2) cos(1))^2  +e^(−(4/n)) sin^2 (1)))  =(((1−e^(−2) cos(n))(1−e^(−(2/n)) cos(1))+i(1−e^(−2) cos(n))e^(−(2/n)) sin(1)−ie^(−2) sin(n)(1−e^(−(2/n)) cos(1))+e^(4/n) sin(n)sin(1))/((1−e^(−2) cos(1))^2  +e^(−(4/n))  sin^2 (1)))  so the value of  A_n  is clear determined after extracting Im(Σ....).
$${we}\:{have}\:{A}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\int_{\frac{{k}}{{n}}} ^{\frac{{k}+\mathrm{1}}{{n}}} \:{sin}\left(\left[{nx}\right]\right){e}^{−\mathrm{2}{x}} {dx} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\int_{\frac{{k}}{{n}}} ^{\frac{{k}+\mathrm{1}}{{n}}} \:{sin}\left({k}\right){e}^{−\mathrm{2}{x}} {dx}\:=\sum_{{k}=\mathrm{0}} ^{{n}} \:{sin}\left({k}\right)\int_{\frac{{k}}{{n}}} ^{\frac{{k}+\mathrm{1}}{{n}}} \:{e}^{−\mathrm{2}{x}} {dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{sin}\left({k}\right)\left(\:{e}^{−\mathrm{2}\frac{{k}+\mathrm{1}}{{n}}} −{e}^{\frac{−\mathrm{2}{k}}{{n}}} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{e}^{\frac{−\mathrm{2}{k}}{{n}}} \:−\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{e}^{−\mathrm{2}\frac{{k}+\mathrm{1}}{{n}}} \:{sin}\left({k}\right) \\ $$$$=\frac{\mathrm{1}−{e}^{−\frac{\mathrm{2}}{{n}}} }{\mathrm{2}}\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{e}^{\frac{−\mathrm{2}{k}}{{n}}} \:{sin}\left({k}\right)\:{but}\:\sum_{{k}=\mathrm{0}} ^{{n}−} \:{e}^{\frac{−\mathrm{2}{k}}{{n}}} \:{sink}\:={Im}\left(\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{e}^{{ik}−\frac{\mathrm{2}}{{n}}{k}} \right)\:{and} \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{e}^{{ik}−\frac{\mathrm{2}}{{n}}{k}} \:=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{e}^{\left({i}−\frac{\mathrm{2}}{{n}}\right){k}} \:=\frac{\mathrm{1}−\left({e}^{\left({i}−\frac{\mathrm{2}}{{n}}\right)} \right)^{{n}} }{\mathrm{1}−{e}^{{i}−\frac{\mathrm{2}}{{n}}} }\:=\frac{\mathrm{1}−{e}^{−\mathrm{2}} \:{e}^{{in}} }{\mathrm{1}−{e}^{−\frac{\mathrm{2}}{{n}}} \:{e}^{{i}} } \\ $$$$=\frac{\mathrm{1}−{e}^{−\mathrm{2}} {cosn}\:−{ie}^{−\mathrm{2}} {sin}\left({n}\right)}{\mathrm{1}−{e}^{−\frac{\mathrm{2}}{{n}}} {cos}\left(\mathrm{1}\right)−{i}\:{e}^{−\frac{\mathrm{2}}{{n}}} {sin}\left(\mathrm{1}\right)} \\ $$$$=\frac{\left(\mathrm{1}−{e}^{−\mathrm{2}} {cos}\left({n}\right)−{i}\:{e}^{−\mathrm{2}} {sinn}\right)\left(\mathrm{1}−{e}^{−\frac{\mathrm{2}}{{n}}} {cos}\left(\mathrm{1}\right)+{i}\:{e}^{−\frac{\mathrm{2}}{{n}}} {sin}\left(\mathrm{1}\right)\right)}{\left(\mathrm{1}−{e}^{−\mathrm{2}} {cos}\left(\mathrm{1}\right)\right)^{\mathrm{2}} \:+{e}^{−\frac{\mathrm{4}}{{n}}} {sin}^{\mathrm{2}} \left(\mathrm{1}\right)} \\ $$$$=\frac{\left(\mathrm{1}−{e}^{−\mathrm{2}} {cos}\left({n}\right)\right)\left(\mathrm{1}−{e}^{−\frac{\mathrm{2}}{{n}}} {cos}\left(\mathrm{1}\right)\right)+{i}\left(\mathrm{1}−{e}^{−\mathrm{2}} {cos}\left({n}\right)\right){e}^{−\frac{\mathrm{2}}{{n}}} {sin}\left(\mathrm{1}\right)−{ie}^{−\mathrm{2}} {sin}\left({n}\right)\left(\mathrm{1}−{e}^{−\frac{\mathrm{2}}{{n}}} {cos}\left(\mathrm{1}\right)\right)+{e}^{\frac{\mathrm{4}}{{n}}} {sin}\left({n}\right){sin}\left(\mathrm{1}\right)}{\left(\mathrm{1}−{e}^{−\mathrm{2}} {cos}\left(\mathrm{1}\right)\right)^{\mathrm{2}} \:+{e}^{−\frac{\mathrm{4}}{{n}}} \:{sin}^{\mathrm{2}} \left(\mathrm{1}\right)} \\ $$$${so}\:{the}\:{value}\:{of}\:\:{A}_{{n}} \:{is}\:{clear}\:{determined}\:{after}\:{extracting}\:{Im}\left(\Sigma….\right). \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 13/Nov/18
∫_0 ^(1/n) sin([nx])e^(−2x) dx+∫_(1/n) ^1 sin([nx])e^(−2x) dx  ∫_0 ^(1/n) sin(0)e^(−2x) dx+sin(1)∫_(1/n) ^1 e^(−2x) dx  sin(1)×∣(e^(−2x) /(−2))∣_(1/n) ^1   sin(1)×(((e^(−2) −e^((−2)/n) )/(−2)))
$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{{n}}} {sin}\left(\left[{nx}\right]\right){e}^{−\mathrm{2}{x}} {dx}+\int_{\frac{\mathrm{1}}{{n}}} ^{\mathrm{1}} {sin}\left(\left[{nx}\right]\right){e}^{−\mathrm{2}{x}} {dx} \\ $$$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{{n}}} {sin}\left(\mathrm{0}\right){e}^{−\mathrm{2}{x}} {dx}+{sin}\left(\mathrm{1}\right)\int_{\frac{\mathrm{1}}{{n}}} ^{\mathrm{1}} {e}^{−\mathrm{2}{x}} {dx} \\ $$$${sin}\left(\mathrm{1}\right)×\mid\frac{{e}^{−\mathrm{2}{x}} }{−\mathrm{2}}\mid_{\frac{\mathrm{1}}{{n}}} ^{\mathrm{1}} \\ $$$${sin}\left(\mathrm{1}\right)×\left(\frac{{e}^{−\mathrm{2}} −{e}^{\frac{−\mathrm{2}}{{n}}} }{−\mathrm{2}}\right) \\ $$

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