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