solve-0-1-ln-2-x-tanh-1-x-x-dx-0-1-tanh-1-x-2-1-x-

Question Number 164653 by mnjuly1970 last updated on 20/Jan/22

s o l v e

ϕ = \int_{0}^{1} \frac{\ln^{2} (x) . {t a n h}^{- 1} (x)}{x} d x = ?

Ω = \int_{0}^{1} \frac{{({t a n h}^{- 1} (x))}^{2}}{1 + x} = ?

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Answered by Lordose last updated on 20/Jan/22

Ω = \int_{0}^{1} \frac{{(\tanh^{- 1} (x))}^{2}}{1 + x} dx

Ω = \frac{1}{4} \int_{0}^{1} \frac{\ln^{2} (\frac{1 - x}{1 + x})}{1 + x} dx

Ω \overset{x = \frac{1 - x}{1 + x}}{=} \frac{1}{4} \int_{0}^{1} \frac{\ln^{2} (x)}{1 + x} = \frac{1}{4} \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n - 1} \int_{0}^{1} x^{n - 1} \ln^{2} (x) dx

Ω \overset{IBP \times 2}{=} \frac{1}{2} \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{n^{3}} = η (3) = \frac{3}{8} ζ (3)

Ω = \frac{3}{8} ζ (3)

Answered by Ar Brandon last updated on 20/Jan/22

ϕ = \int_{0}^{1} \frac{\ln^{2} x \tanh^{- 1} (x)}{x} d x

= {[\frac{\ln^{3} x}{3} \tanh^{- 1} (x)]}_{0}^{1} - \frac{1}{3} \int_{0}^{1} \frac{\ln^{3} x}{1 - x^{2}} d x

= - \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{1}{8} \int_{0}^{1} \frac{u^{- \frac{1}{2}} \ln^{3} u}{1 - u} d u = \frac{1}{48} ψ^{(3)} (\frac{1}{2})

= 2 (ζ (4) - \frac{1}{16} ζ (4)) = \frac{15}{8} ζ (4) = \frac{15}{720} π^{4}

Commented by mnjuly1970 last updated on 20/Jan/22

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