calculate-0-1-ln-1-x-ln-1-x-1-x-dx-

Question Number 157469 by mnjuly1970 last updated on 23/Oct/21

c a l c u l a t e :

Ω := \int_{0}^{1} \frac{l n (1 + x) . l n (1 - x)}{1 + x} d x = ?

Answered by qaz last updated on 23/Oct/21

\int_{0}^{1} \frac{\ln (1 + x) \ln (1 - x)}{1 + x} dx

= \int_{1}^{2} \frac{lnxln (2 - x)}{x} dx

= \int_{1 / 2}^{1} \frac{(\ln 2 + lnx) (\ln 2 + \ln (1 - x))}{x} dx

= \int_{1 / 2}^{1} \frac{\ln^{2} 2 + \ln 2 (lnx + \ln (1 - x)) + lnxln (1 - x)}{x} dx

= \ln^{3} 2 + \ln 2 \cdot (- \frac{1}{2} \ln^{2} 2 + \int_{1 / 2}^{1} \frac{\ln (1 - x)}{x} dx) + \int_{1 / 2}^{1} \frac{lnxln (1 - x)}{x} dx

= \frac{1}{2} \ln^{3} 2 + \ln 2 ({Li}_{2} (\frac{1}{2}) - {Li}_{2} (1)) - lnx \cdot {Li}_{2} (x) ∣_{1 / 2}^{1} + {Li}_{3} (1) - {Li}_{3} (\frac{1}{2})

= \frac{1}{2} \ln^{3} 2 - \frac{π^{2}}{6} \ln 2 - \ln 2 \cdot {Li}_{2} (\frac{1}{2}) - \ln 2 \cdot {Li}_{2} (\frac{1}{2}) + ζ (3) - {Li}_{3} (\frac{1}{2})

= - \frac{π^{2}}{4} \ln 2 + \frac{4}{3} \ln^{3} 2 + \frac{1}{8} ζ (3)

- - - - - - - - - - - - - - - - -

{Li}_{2} (\frac{1}{2}) = - \frac{1}{2} \ln^{2} 2 + \frac{π^{2}}{12}

{Li}_{3} (\frac{1}{2}) = - \frac{π^{2}}{12} \ln 2 + \frac{1}{6} \ln^{3} 2 + \frac{7}{8} ζ (3)

Commented by mnjuly1970 last updated on 23/Oct/21

t h a n k s a l o t \dots

Commented by mnjuly1970 last updated on 23/Oct/21

p l e a s e r e c h e c k t h e a n s w e r