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Question Number 33127 by prof Abdo imad last updated on 10/Apr/18
$$\:{find}\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{sin}\left({na}\right)}{\left({sina}\right)^{{n}} }\:\frac{{x}^{{n}} }{{n}!}\:\:{with}\:\mathrm{0}<{a}<\pi\:. \\ $$
Commented byprof Abdo imad last updated on 12/Apr/18
$${S}\left({x}\right)=\:{Im}\left(\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{e}^{{ina}} \:{x}^{{n}} }{{n}!\left({sina}\right)^{{n}} }\right)\:\:{but} \\ $$ $${w}\left({x}\right)\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{{e}^{{ina}} \:{x}^{{n}} }{{n}!\left({sina}\right)^{{n}} }\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{{n}!}\:\:\left(\:\frac{{e}^{{ia}} \:{x}}{{sina}}\right)^{{n}} \\ $$ $${but}\:{we}\:{have}\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{u}^{{n}} }{{n}!}\:\:={e}^{{u}} \:\:\:\forall\:{u}\:\in\:{C}\:\Rightarrow \\ $$ $${w}\left({x}\right)\:={e}^{\frac{{e}^{{ia}} }{{sina}}{x}} \:=\:{e}^{\frac{{x}}{{sina}}\left(\:{coa}\:+{isina}\right)} \\ $$ $$=\:{e}^{\frac{{x}\:{cosa}}{{sins}}} \:\:{e}^{{ix}} \:=\:{e}^{\frac{{xcosa}}{{sina}}} \left(\:{cosx}\:+{isinx}\right)\:\Rightarrow \\ $$ $${S}\left({x}\right)\:=\:{sinx}\:{e}^{\frac{{xcosa}}{{sina}}} \:\:. \\ $$