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Question Number 158405 by Eric002 last updated on 03/Nov/21

$$prove:$$$${\int }_0^{\mathrm{\infty }}\frac{1}{x^5+x^4+x^3+x^2+x+1}dx=\frac{\pi }{3\sqrt{3}}$$

Answered by MJS_new last updated on 03/Nov/21

$$x^5+x^4+x^3+x^2+x+1=\frac{x^6-1}{x-1}$$$$\Rightarrow$$$$\int \frac{dx}{x^5+x^4+x^3+x^2+x+1}=$$$$=\int \frac{dx}{(x+1)(x^2+x+1)(x^2+x+1)}=$$$$=\frac{1}{3}\int \frac{dx}{x+1}+\frac{1}{2}\int \frac{dx}{x^2+x+1}-\frac{1}{6}\int \frac{2x-1}{x^2-x+1}=$$$$=\frac{1}{3}\ln \mid x+1\mid +\frac{\sqrt{3}}{3}\arctan \frac{2x+1}{\sqrt{3}}-\frac{1}{6}\ln (x^2-x+1)+C$$$$\Rightarrow$$$$\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{dx}{x^5+x^4+x^3+x^2+x+1}=\frac{\pi }{3\sqrt{3}}$$

Commented by Eric002 last updated on 03/Nov/21

$$welldone$$

Commented by Tawa11 last updated on 04/Nov/21

$$\mathrm{Great}\mathrm{sir}$$

Answered by Ar Brandon last updated on 03/Nov/21

$$I={\int }_0^{\mathrm{\infty }}\frac{dx}{x^5+x^4+x^3+x^2+x+1}={\int }_0^{\mathrm{\infty }}\frac{1-x}{1-x^6}dx$$$$={\int }_0^1\frac{1-x}{1-x^6}dx+{\int }_1^{\mathrm{\infty }}\frac{1-x}{1-x^6}dx={\int }_0^1\frac{1-x}{1-x^6}dx+{\int }_0^1\frac{x^3-x^4}{1-x^6}dx$$$$={\int }_0^1\frac{1-x+x^3-x^4}{1-x^6}dx=\frac{1}{6}{\int }_0^1\frac{{\mathrm{v}}^{-\frac{5}{6}}-{\mathrm{v}}^{-\frac{4}{6}}+{\mathrm{v}}^{-\frac{2}{6}}-{\mathrm{v}}^{-\frac{1}{6}}}{1-\mathrm{v}}d\mathrm{v}$$$$=\frac{1}{6}[-\psi \left(\frac{1}{6}\right)+\psi \left(\frac{2}{6}\right)-\psi \left(\frac{4}{6}\right)+\psi \left(\frac{5}{6}\right)]$$$$=\frac{1}{6}[(\psi \left(\frac{5}{6}\right)-\psi \left(\frac{1}{6}\right))+(\psi \left(\frac{1}{3}\right)-\psi \left(\frac{2}{3}\right))]$$$$=\frac{1}{6}[-\pi \cot \left(\frac{5}{6}\pi \right)-\pi \cot \left(\frac{\pi }{3}\right)]=\frac{1}{6}[\pi \sqrt{3}-\frac{\pi }{\sqrt{3}}]=\frac{2\sqrt{3}\pi }{18}=\frac{\sqrt{3}}{9}\pi$$

Commented by Ar Brandon last updated on 03/Nov/21

$$\mathrm{\Phi }={\int }_1^{\mathrm{\infty }}\frac{1-x}{1-x^6}dx,u=\frac{1}{x}\Rightarrow x=\frac{1}{u}\Rightarrow dx=-\frac{1}{u^2}du$$$$\Rightarrow \mathrm{\Phi }={\int }_0^1\frac{1-\frac{1}{u}}{1-\frac{1}{u^6}}\cdot \frac{du}{u^6}={\int }_0^1\frac{u^4-u^3}{u^6-1}du={\int }_0^1\frac{x^3-x^4}{1-x^6}dx$$

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