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Question Number 160560 by HongKing last updated on 01/Dec/21

$$\mathrm{Find}:$$$$\mathbf{\Omega }=\underset{1}{\stackrel{\mathrm{\infty }}{\int }}\frac{\ln \left(\mathrm{x}\right)}{{\mathrm{x}}^4+{\mathrm{x}}^2+1}\mathrm{dx}=?$$

Answered by Ar Brandon last updated on 02/Dec/21

$$\mathrm{\Omega }={\int }_1^{\mathrm{\infty }}\frac{\ln x}{x^4+x^2+1}dx={\int }_1^{\mathrm{\infty }}\frac{1-x^2}{1-x^6}\ln xdx,x=u^{-1}\Rightarrow dx=-u^{-2}du$$$$=-{\int }_0^1\frac{u^4-u^2}{u^6-1}\ln udu=\frac{1}{36}{\int }_0^1\frac{t^{-\frac{1}{6}}-t^{-\frac{3}{6}}}{1-t}\ln tdt$$$$=\frac{1}{36}({\psi }^{\prime }\left(\frac{1}{2}\right)-{\psi }^{\prime }\left(\frac{5}{6}\right))=\frac{1}{36}(\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(n+\frac{1}{2})}^2}-\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(n+\frac{5}{6})}^2})$$

Commented by HongKing last updated on 02/Dec/21

$$\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}\mathrm{my}\mathrm{dear}\mathbf{Sir}$$$$\mathrm{answer}:\frac{{\pi }^2}{72}-\frac{1}{36}{\psi }^{\left(1\right)}\left(\frac{5}{6}\right)$$

Commented by Ar Brandon last updated on 02/Dec/21

$$\frac{1}{36}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(n+\frac{1}{2})}^2}=\frac{1}{9}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(2n+1)}^2}$$$$=\frac{1}{9}\times \frac{3}{4}\zeta \left(2\right)=\frac{1}{9}\times \frac{3}{4}\times \frac{{\pi }^2}{6}=\frac{{\pi }^2}{72}$$

Commented by HongKing last updated on 02/Dec/21

$$\mathrm{Thank}\mathrm{you}\mathrm{my}\mathrm{dear}\mathbf{Sir}$$

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