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Question Number 133432 by Dwaipayan Shikari last updated on 22/Feb/21

$$\frac{sin\sqrt{\pi }}{1^3}+\frac{sin\sqrt{4\pi }}{2^3}+\frac{sin\sqrt{9\pi }}{3^3}+\frac{sin\sqrt{16\pi }}{4^3}+....=\frac{\pi \sqrt{\pi }}{12}(1-3\sqrt{\pi }+2\pi )$$

Answered by mindispower last updated on 24/Feb/21

$$\sum _{n⩾0}\frac{sim\left(nx\right)}{n}=\frac{\pi -x}{2_{}}$$$$\Rightarrow \mathrm{\Sigma }\frac{cos\left(nx\right)}{n^2}=-\frac{\pi }{2}x+\frac{x^2}{4}+\frac{{\pi }^2}{6}$$$$\Rightarrow \sum _{n⩾1}\frac{sin\left(nx\right)}{n^3}=-\frac{\pi x^2}{4}+\frac{x^3}{12}+\frac{{\pi }^2}{6}x=f\left(x\right)$$$$\frac{sin\left(\sqrt{\pi }\right)}{1^3}+\frac{sin\left(\sqrt{4\pi }\right)}{2^2}+.....+\frac{sin\left(\sqrt{n^2\pi }\right)}{n^2}+...$$$$=\mathrm{\Sigma }\frac{sin\left(n\sqrt{\pi }\right)}{n^3}=f\left(\sqrt{\pi }\right)=-\frac{{\pi }^2}{4}+\frac{\pi \sqrt{\pi }}{12}+\frac{{\pi }^2\sqrt{\pi }}{6}$$$$=\frac{\pi \sqrt{\pi }}{12}(1-3\sqrt{\pi }+2\pi )$$

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