Find-1-ln-x-x-4-x-2-1-dx-

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

Find :

Ω = \underset{1}{\int^{\infty}} \frac{\ln (x)}{x^{4} + x^{2} + 1} dx = ?

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

Ω = \int_{1}^{\infty} \frac{\ln x}{x^{4} + x^{2} + 1} d x = \int_{1}^{\infty} \frac{1 - x^{2}}{1 - x^{6}} \ln x d x, x = u^{- 1} \Rightarrow d x = - u^{- 2} d u

= - \int_{0}^{1} \frac{u^{4} - u^{2}}{u^{6} - 1} \ln u d u = \frac{1}{36} \int_{0}^{1} \frac{t^{- \frac{1}{6}} - t^{- \frac{3}{6}}}{1 - t} \ln t d t

= \frac{1}{36} (ψ^{'} (\frac{1}{2}) - ψ^{'} (\frac{5}{6})) = \frac{1}{36} (\underset{n = 0}{\sum^{\infty}} \frac{1}{{(n + \frac{1}{2})}^{2}} - \underset{n = 0}{\sum^{\infty}} \frac{1}{{(n + \frac{5}{6})}^{2}})

Commented by HongKing last updated on 02/Dec/21

thank you so much my dear Sir

answer : \frac{π^{2}}{72} - \frac{1}{36} ψ^{(1)} (\frac{5}{6})

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

\frac{1}{36} \underset{n = 0}{\sum^{\infty}} \frac{1}{{(n + \frac{1}{2})}^{2}} = \frac{1}{9} \underset{n = 0}{\sum^{\infty}} \frac{1}{{(2 n + 1)}^{2}}

= \frac{1}{9} \times \frac{3}{4} ζ (2) = \frac{1}{9} \times \frac{3}{4} \times \frac{π^{2}}{6} = \frac{π^{2}}{72}

Commented by HongKing last updated on 02/Dec/21

Thank you my dear Sir