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Question Number 151641 by Tawa11 last updated on 22/Aug/21

Answered by Kamel last updated on 22/Aug/21

$$S_n=\underset{k=0}{\sum ^{n-1}}{\int }_0^1{x}^{3k}dx={\int }_0^1\frac{1-x^{3n}}{1-x^3}dx$$$$\stackrel{t=x^3}{=}\frac{1}{3}{\int }_0^1\frac{t^{-\frac{2}{3}}-t^{n-\frac{2}{3}}}{1-t}dt=\frac{1}{3}(\mathrm{\Psi }(n+\frac{1}{3})+\gamma -{\int }_0^1\frac{1-t^{-\frac{2}{3}}}{1-t}dt)$$$$=\frac{\mathrm{\Psi }(n+\frac{1}{3})}{3}+\frac{\gamma }{3}-{\int }_0^1\frac{x^2-1}{1-x^3}dx=\frac{\mathrm{\Psi }(n+\frac{1}{3})}{3}+\frac{\gamma }{3}+\frac{1}{2}{\int }_0^1\frac{1+2x+1}{1+x+x^2}dx$$$$=\frac{\mathrm{\Psi }(n+\frac{1}{3})}{3}+\frac{\gamma }{3}+\frac{Ln\left(3\right)}{2}+\frac{1}{2}{\int }_0^1\frac{1}{{(x+\frac{1}{2})}^2+\frac{3}{4}}dx$$$$=\frac{\mathrm{\Psi }(n+\frac{1}{3})}{3}+\frac{\gamma }{3}+\frac{Ln\left(3\right)}{2}+\frac{1}{\sqrt{3}}(Arctan(\frac{2}{\sqrt{3}}+\frac{1}{\sqrt{3}})-Arctan\left(\frac{1}{\sqrt{3}}\right))$$$$=\frac{\mathrm{\Psi }(n+\frac{1}{3})}{3}+\frac{\gamma }{3}+\frac{Ln\left(3\right)}{2}+\frac{1}{\sqrt{3}}\left(Arctan\left(\frac{1}{\sqrt{3}}\right)\right)$$$$=\frac{\mathrm{\Psi }(n+\frac{1}{3})}{3}+\frac{\gamma }{3}+\frac{Ln\left(3\right)}{2}+\frac{\pi \sqrt{3}}{18}$$$$$$

Commented by Tawa11 last updated on 22/Aug/21

$$\mathrm{Wow}.\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}.$$

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