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Question Number 151954 by mnjuly1970 last updated on 24/Aug/21

$$$$$$nice...calculus$$$$$$$$\mathit{\varphi }:={\int }_0^{\frac{\pi }{2}}{x}^3.cot\left(x\right)dx=\frac{a}{16}$$$$a:=?$$$$m.n...$$

Answered by Kamel last updated on 24/Aug/21

$$$$$$\mathrm{\Omega }={\int }_0^{\frac{\pi }{2}}{x}^{3}cot\left(x\right)dx\stackrel{IBP}{=}-3{\int }_0^{\frac{\pi }{2}}{x}^{2}Ln\left(sin\left(x\right)\right)dx$$$$Ln\left(sin\left(x\right)\right)=-Ln\left(2\right)-\underset{n=1}{\sum ^{+\mathrm{\infty }}}\frac{cos\left(2nx\right)}{n},0<x⩽\frac{\pi }{2}.$$$$\therefore \mathrm{\Omega }=\frac{{\pi }^3}{8}Ln\left(2\right)+3\underset{n=1}{\sum ^{+\mathrm{\infty }}}\frac{1}{n}{\int }_0^{\frac{\pi }{2}}{x}^{2}cos\left(2nx\right)dx$$$$=\frac{{\pi }^3}{8}Ln\left(2\right)+\frac{3\pi }{2}\underset{n=1}{\sum ^{+\mathrm{\infty }}}\frac{1}{n}\left(\frac{{(-1)}^n}{2n^2}\right)$$$$=\frac{{\pi }^3}{8}Ln\left(2\right)+\frac{3\pi }{4}(\frac{1}{8}\zeta \left(3\right)-(\zeta \left(3\right)-\frac{1}{8}\zeta \left(3\right)))$$$$=\frac{{\pi }^3}{8}Ln\left(2\right)-\frac{9\pi }{16}=\frac{{\pi }^{3}Ln\left(4\right)-9\pi }{16}$$

Commented by mnjuly1970 last updated on 25/Aug/21

$$excellent...$$

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