Question Number 133432 by Dwaipayan Shikari last updated on 22/Feb/21
$$\frac{{sin}\sqrt{\pi}}{\mathrm{1}^{\mathrm{3}} }+\frac{{sin}\sqrt{\mathrm{4}\pi}}{\mathrm{2}^{\mathrm{3}} }+\frac{{sin}\sqrt{\mathrm{9}\pi}}{\mathrm{3}^{\mathrm{3}} }+\frac{{sin}\sqrt{\mathrm{16}\pi}}{\mathrm{4}^{\mathrm{3}} }+….=\frac{\pi\sqrt{\pi}}{\mathrm{12}}\left(\mathrm{1}−\mathrm{3}\sqrt{\pi}+\mathrm{2}\pi\right) \\ $$
Answered by mindispower last updated on 24/Feb/21
$$\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{{sim}\left({nx}\right)}{{n}}=\frac{\pi−{x}}{\mathrm{2}_{} } \\ $$$$\Rightarrow\Sigma\frac{{cos}\left({nx}\right)}{{n}^{\mathrm{2}} }=−\frac{\pi}{\mathrm{2}}{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{4}}+\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\Rightarrow\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{sin}\left({nx}\right)}{{n}^{\mathrm{3}} }=−\frac{\pi{x}^{\mathrm{2}} }{\mathrm{4}}+\frac{{x}^{\mathrm{3}} }{\mathrm{12}}+\frac{\pi^{\mathrm{2}} }{\mathrm{6}}{x}={f}\left({x}\right) \\ $$$$\frac{{sin}\left(\sqrt{\pi}\right)}{\mathrm{1}^{\mathrm{3}} }+\frac{{sin}\left(\sqrt{\mathrm{4}\pi}\right)}{\mathrm{2}^{\mathrm{2}} }+…..+\frac{{sin}\left(\sqrt{{n}^{\mathrm{2}} \pi}\right)}{{n}^{\mathrm{2}} }+… \\ $$$$=\Sigma\frac{{sin}\left({n}\sqrt{\pi}\right)}{{n}^{\mathrm{3}} }={f}\left(\sqrt{\pi}\right)=−\frac{\pi^{\mathrm{2}} }{\mathrm{4}}+\frac{\pi\sqrt{\pi}}{\mathrm{12}}+\frac{\pi^{\mathrm{2}} \sqrt{\pi}}{\mathrm{6}} \\ $$$$=\frac{\pi\sqrt{\pi}}{\mathrm{12}}\left(\mathrm{1}−\mathrm{3}\sqrt{\pi}+\mathrm{2}\pi\right) \\ $$