Question Number 139479 by mnjuly1970 last updated on 27/Apr/21
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….{advanced}\:….\bigstar\bigstar\bigstar…..{calculus}….. \\ $$$$\:\:\:\:\:\:\:\:\:\:\: :=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{{sin}\left({n}\right)}{{n}}\right)^{\mathrm{3}} =?\: \\ $$$$\:\:\:\:\:\:\:\:\:\: \\ $$
Answered by Dwaipayan Shikari last updated on 27/Apr/21
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{sin}^{\mathrm{3}} \left({n}\right)}{{n}^{\mathrm{3}} }=\frac{\mathrm{3}}{\mathrm{4}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{sin}\left({n}\right)}{{n}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{sin}\left(\mathrm{3}{n}\right)}{{n}^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{12}}+\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\frac{\pi}{\mathrm{4}}\right)−\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{3}}{\mathrm{12}}+\frac{\pi^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{9}\pi}{\mathrm{4}}\right) \\ $$$$=\frac{\mathrm{9}\pi}{\mathrm{16}}−\frac{\mathrm{3}\pi}{\mathrm{16}}=\frac{\mathrm{3}\pi}{\mathrm{8}}\:\:\:\:\:\:\left(\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{sin}\left({n}\theta\right)}{{n}^{\mathrm{3}} }=\frac{\theta^{\mathrm{3}} }{\mathrm{12}}+\frac{\pi^{\mathrm{2}} \theta}{\mathrm{6}}−\frac{\pi\theta^{\mathrm{2}} }{\mathrm{4}}\right) \\ $$
Commented by mnjuly1970 last updated on 27/Apr/21
$$\:{mercey}\:{mr}\:{Dwaipayan}.. \\ $$