Question Number 121498 by Lordose last updated on 08/Nov/20
$$ \\ $$$$\mathrm{Evaluate} \\ $$$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{sin}\left(\mathrm{n}\right)}{\mathrm{n}} \\ $$
Answered by Bird last updated on 09/Nov/20
$${let}\:{find}\:{S}\left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty\:} \:\frac{{sin}\left({nx}\right)}{{n}}{dx} \\ $$$${S}\left({x}\right)={Im}\left(\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{e}^{{inx}} }{{n}}\right)\:{and} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty\:} \:\frac{{e}^{{inx}} }{{n}}\:=\sum_{{n}=\mathrm{1}} ^{\infty\:} \:\frac{\left({e}^{{ix}} \right)^{{n}} }{{n}} \\ $$$$=−{ln}\left(\mathrm{1}−{e}^{{ix}} \right) \\ $$$$=−{ln}\left(\mathrm{1}−{cosx}−{isinx}\:\right) \\ $$$$=−{ln}\left(\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)−\mathrm{2}{isin}\left(\frac{{x}}{\mathrm{2}}\right){cos}\left(\frac{{x}}{\mathrm{2}}\right)\right) \\ $$$$=−{ln}\left(−\mathrm{2}{isin}\left(\frac{{x}}{\mathrm{2}}\right){e}^{\frac{{ix}}{\mathrm{2}}} \right) \\ $$$$=−{ln}\left(−\mathrm{2}\right)−{ln}\left({i}\right)−{ln}\left({sin}\left(\frac{{x}}{\mathrm{2}}\right)\right)−\frac{{ix}}{\mathrm{2}} \\ $$$$=−{ln}\left(\mathrm{2}\right)−{i}\pi−\frac{{i}\pi}{\mathrm{2}}−\frac{{ix}}{\mathrm{2}}−{ln}\left({sin}\left(\frac{{x}}{\mathrm{2}}\right)\right) \\ $$$$=−{ln}\mathrm{2}−{ln}\left({sin}\left(\frac{{x}}{\mathrm{2}}\right)\right)−\frac{{ix}}{\mathrm{2}}−\frac{\mathrm{3}{i}\pi}{\mathrm{2}} \\ $$$$\Rightarrow\Sigma\:\frac{{sin}\left({nx}\right)}{{n}}\:=−\frac{{x}+\mathrm{3}\pi}{\mathrm{2}} \\ $$$${x}=\mathrm{1}\:\Rightarrow\:\Sigma\:\:\frac{{sin}\left({n}\right)}{{n}}\:=−\frac{\mathrm{3}\pi+\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$