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Question Number 183977 by cortano1 last updated on 01/Jan/23

$${\int }_0^{\mathrm{\infty }}\frac{dx}{x^8+x^4+1}=?$$

Answered by ARUNG_Brandon_MBU last updated on 01/Jan/23

$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{dx}{x^8+x^4+1}={\int }_0^{\mathrm{\infty }}\frac{1-x^4}{1-x^{12}}dx={\int }_0^1\frac{1-x^4}{1-x^{12}}dx+{\int }_1^{\mathrm{\infty }}\frac{1-x^4}{1-x^{12}}dx$$$$={\int }_0^1\frac{1-x^4}{1-x^{12}}dx+{\int }_0^1\frac{1-\frac{1}{x^4}}{1-\frac{1}{x^{12}}}\cdot \frac{1}{x^2}dx={\int }_0^1\frac{1-x^4}{1-x^{12}}dx+{\int }_0^1\frac{x^{10}-x^6}{x^{12}-1}dx$$$$={\int }_0^1\frac{1-x^4+x^6-x^{10}}{1-x^{12}}dx=\frac{1}{12}{\int }_0^1\frac{x^{-\frac{11}{12}}-x^{-\frac{7}{12}}+x^{-\frac{5}{12}}-x^{-\frac{1}{12}}}{1-x}dx$$$$=\frac{1}{12}(\psi \left(\frac{11}{12}\right)-\psi \left(\frac{1}{12}\right)+\psi \left(\frac{5}{12}\right)-\psi \left(\frac{7}{12}\right))=\frac{1}{12}(\pi \cot \left(\frac{\pi }{12}\right)-\pi \cot \left(\frac{5\pi }{12}\right))$$$$=\frac{\pi }{12}(\frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}})=\frac{\pi }{12}\left(\frac{(8+4\sqrt{3})-(8-4\sqrt{3})}{4}\right)=\frac{\pi \sqrt{3}}{6}$$

Commented by MJS_new last updated on 01/Jan/23

$$\mathrm{great}!$$$$\mathrm{happy}\mathrm{new}\mathrm{year}!$$

Commented by Ar Brandon last updated on 01/Jan/23

Thank you, Sir �� Happy New Year ����

Answered by Ar Brandon last updated on 01/Jan/23

$$x^8+x^4+1={(x^4+1)}^2-x^4=(x^4-x^2+1)(x^4+x^2+1)$$$$I=\int \frac{dx}{x^8+x^4+1}=\int \frac{dx}{(x^4-x^2+1)(x^4+x^2+1)}$$$$=\frac{1}{2}\int \frac{(x^6+1)-(x^6-1)}{(x^4-x^2+1)(x^4+x^2+1)}dx$$$$=\frac{1}{2}\int \frac{x^2+1}{x^4+x^2+1}dx-\frac{1}{2}\int \frac{x^2-1}{x^4-x^2+1}dx$$$$=\frac{1}{2}\int \frac{1+\frac{1}{x^2}}{x^2+1+\frac{1}{x^2}}dx-\frac{1}{2}\int \frac{1-\frac{1}{x^2}}{x^2-1+\frac{1}{x^2}}dx$$$$=\frac{1}{2}\int \frac{1+\frac{1}{x^2}}{{(x-\frac{1}{x})}^2+3}dx-\frac{1}{2}\int \frac{1-\frac{1}{x^2}}{{(x+\frac{1}{x})}^2-3}dx$$$$=\frac{1}{2\sqrt{3}}\arctan \left(\frac{x^2-1}{\sqrt{3}x}\right)+\frac{1}{2\sqrt{3}}\mathrm{argcoth}\left(\frac{x^2+1}{\sqrt{3}x}\right)+C$$$$=\frac{1}{2\sqrt{3}}\arctan \left(\frac{x^2-1}{\sqrt{3}x}\right)+\frac{1}{4\sqrt{3}}\ln \mid \frac{x^2+\sqrt{3}x+1}{x^2-\sqrt{3}x+1}\mid +C$$$${\int }_0^{\mathrm{\infty }}\frac{dx}{x^8+x^4+1}={[\frac{1}{2\sqrt{3}}\arctan \left(\frac{x^2-1}{\sqrt{3}x}\right)+\frac{1}{4\sqrt{3}}\ln \mid \frac{x^2+\sqrt{3}x+1}{x^2-\sqrt{3}x+1}\mid ]}_0^{\mathrm{\infty }}$$$$\Rightarrow \overline{)\begin{array}{c}{\int }_0^{\mathrm{\infty }}\frac{dx}{x^8+x^4+1}=\frac{\pi }{2\sqrt{3}}\end{array}}$$

Answered by universe last updated on 01/Jan/23

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