Question Number 45960 by maxmathsup by imad last updated on 19/Oct/18
$${find}\:{f}\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{x}^{{n}} {sin}\left({nx}\right)}{{n}} \\ $$
Answered by Smail last updated on 22/Oct/18
$${f}\left({x}\right)={Im}\left(\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} {e}^{{inx}} }{{n}}\right) \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} {e}^{{inx}} }{{n}}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left({xe}^{{ix}} \right)^{{n}} }{{n}} \\ $$$${ln}\left(\mathrm{1}−{x}\right)=−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}}\:\:{with}\:\mid{x}\mid<\mathrm{1} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left({xe}^{{ix}} \right)^{{n}} }{{n}}=−{ln}\left(\mathrm{1}−{xe}^{{ix}} \right) \\ $$$$=−{ln}\left(\mathrm{1}−{xcos}\left({x}\right)−{xisin}\left({x}\right)\right)={a}+{ib} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−{xcosx}−{ixsinx}}={e}^{{a}+{ib}} \\ $$$${e}^{{a}} \left({cosb}+{isinb}\right)=\frac{\mathrm{1}−{xcosx}+{ixsinx}}{\mathrm{1}+{x}^{\mathrm{2}} −\mathrm{2}{xcosx}} \\ $$$${e}^{{a}} {cos}\left({b}\right)=\frac{\mathrm{1}−{xcosx}}{\mathrm{1}+{x}^{\mathrm{2}} −\mathrm{2}{xcosx}} \\ $$$${e}^{{a}} {sin}\left({b}\right)=\frac{{xsinx}}{\mathrm{1}+{x}^{\mathrm{2}} −\mathrm{2}{xcosx}} \\ $$$${tan}\left({b}\right)=\frac{{xsinx}}{\mathrm{1}−{xcosx}} \\ $$$${b}={tan}^{−\mathrm{1}} \left(\frac{{xsinx}}{\mathrm{1}−{xcosx}}\right) \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} {sin}\left({nx}\right)}{{n}}={tan}^{−\mathrm{1}} \left(\frac{{xsinx}}{\mathrm{1}−{xcosx}}\right) \\ $$
Commented by maxmathsup by imad last updated on 23/Oct/18
$${correct}\:{answer}\:{thanks}\:{a}\:{lots}.. \\ $$